supriyag

Linear Equations In Two Variables Introduction

First of all we shall understand the meaning of terms used in the name of the chapter “Linear equations in two variables”.

Consider the following equations:

⇒ \(3 x-8 y=21,2 x+5 y=14, \frac{7}{4} x+\frac{9}{5} y=\frac{5}{6}\)

Each of these equations contains two variables over R (real numbers). An equation of the form an ax + by + c = 0, where a, b, c are real numbers (a ≠ 0, b ≠ 0) is called a linear equation in two variables x and b, where a is the coefficient of x, b is the coefficient of y and c is a constant term.

Solution Of The Simultaneous Linear Equations

Any pair of values of x and y which satisfies the equation is called the solution of the equation. It means a solution of two equations is the point (x, y) of the intersection of two straight lines or any two curves. This point will lie on both the equations (curves).

If the equation contains only one variable, the graph of such type of equation will be a straight line parallel to the x-axis or y-axis according to a form of equation as ay = b or ax = b, where a ≠ 0. The graph of a linear equation in two variables is also a straight line.

The general form of a pair of linear equations in two variables x and y is

a₁x +b₁y +c₁ = 0

a₂x + b₂y + c₂ = 0

where a₁,b₁,C₁,a₂,b₂ and C₂ are all real numbers.

A system of linear equations will have either a unique solution or infinitely many solutions or no solution. If a system of simultaneous linear equations has a solution (either unique or infinitely many), then the system is said to be consistent otherwise it is said to be an inconsistent system.

Consistent System: A system of simultaneous linear equations is said to be consistent if it has at least one solution.

In-consistent System: A system of simultaneous linear equations is said to be inconsistent if it has no solution.

Graphical Method To Solve The Simultaneous Linear Equations

Let the given system of linear equations be

a₁x +b₁y +c₁ = 0 ……(1)

a₂x + b₂y + c₂ = 0 ……(2)

A pair of linear equations in two variables will be represented by two straight lines, both to be considered together. The following three possibilities can happen:

1. The two lines intersect at exactly one point. Let the coordinates of a point of intersection be (a, b).
Solution:

Given

The two lines intersect at exactly one point. Let the coordinates of a point of intersection be (a, b)

Then, x = a and y = b are the unique solutions of the given system of equations.

2. The two lines are parallel to each other.
Solution:

Then there is no common solution to the given system of equations. Thus, in this case, the given system is inconsistent.

3. One line overlaps the other i.e., the two lines are coincident.
Solution:

Then the given system of equations has infinitely many solutions.

Simultaneous Linear Equations Solved Examples

Question 1. Solve graphically the system of linear equations x +y =10 and x-y = 4.
Solution:

Given

x +y =10 and x-y = 4

For x +y= 10, it can be written as y = 10- x

For equation x-y = 4, it can be written as y = x- 4

Now we plot the points A(10, 0), B(0, 10) and P(4, 0), Q(0, -4)

We observe that the two lines representing the two equations intersect at points (7, and 3). Hence, \(\left.\begin{array}{l}
x=7 \\
y=3
\end{array}\right\}\) is a solution of given equation.

Question 2. Show graphically that the system of equations 2x + 4y = 10 and 3x + 6y = 12 has no solution.
Solution:

The given equation is

2x + 4y = 1 0

4y = 1 0 – 2x

⇒ \(y=\frac{5-x}{2}\)

Now for equation 3x + 6y – 12

6y = 12-3x

⇒ \(y=\frac{4-x}{2}\)

Now plot the points 2l(1, 2), B(3, 1), C(5, 0) and P(2, 1), Q(0, 2), R(4, 0) on the same graph paper.

We observe from the graph that the two lines are parallel to each other i.e., they do not intersect each other at any point. Hence, the given system of equations has no solution.

Question 3. Show graphically that the system of equations 3x -y=2, 9x-3y = 6 has an infinite = number of solutions. When this system has infinitely many solutions, then find any two solutions. Also, find the y-intercept made by this line.
Solution:

For equation 3x -y = 2

y = 3x-2 ……..(1)

and for equation 9x- 3y = 6

3y = 9x- 6

y = \(\frac{9 x-6}{3}\)

Now, plot the points A(-1, -5), B(0, -2), C(2, 4) and P(-2, -8), Q(1, 1), R(2, 4) on the graph paper. We observe from the graph that the two lines coincide. Hence, a given system of equations has an infinite number of solutions.

For any two solutions, Put x = 0 in equation (1), we get = -2 If we put x = 1 in the equation (1), we get = 1

So, two solutions are \(\left.\left.\begin{array}{l}
x=0 \\
y=-2
\end{array}\right\} \text { and } \begin{array}{l}
x=1 \\
y=1
\end{array}\right\}\)

This line (both lines are the same) cuts the y-axis at 2 units below the origin. So, its y-intercept (part) is -2

Algebraic Methods Of Solving Simultaneous Linear Equations In Two Variables

Elimination by “Substitution” Method (or Simply “Substitution Method”)

Step 1: From any of the given two equations, find the value of one variable in terms of other.

Step 2: Substitute the value of the variable obtained in step 1 in the other equation. In this way, have eliminated one variable with the help of substitution. So the name fell as “elimination by substitution”. Now, solve it to get the value of one variable.]

Step 3: Substituting the value of the variable obtained in step 2 in the result of step 1 get the value of the remaining unknown variable.

Algebraic Methods Solved Examples

Question 1. Solve x +y = 7 and 3x- 2y = 11.
Solution:

Given x +y = 7 …(1)

and 3x-2y= 11 …(2)

From equation (1) we get y = 7 -x …(3)

Substituting the value of y from equation (3) in (2), we get

3x-2 (7-x) = 11

3x- 14 + 2r= 11

5x- 14= 11

5x = 25 or x = 5

Substituting the value of x in equation (3), we get

y = 7 – 5 ⇒ y = 2

∴ Solution is \(\left.\begin{array}{l}
x=5 \\
y=2
\end{array}\right\}\)

Question 2. Solve 15x- 8y = 29 and 17x + 12y = 75.
Solution:

Given equations are 15x- 8y = 29 ….(1)

and 17x + 12y = 75 …….(2)

From equation (1), we get

⇒ \(x=\frac{29+8 y}{15}\) …..(3)

Substituting the value of x from equation (3) in (2), we get

⇒ \(17\left(\frac{29+8 y}{15}\right)+12 y=75\)

⇒ 493 + 136y + 180y = 1125

⇒ 316y= 1125-493

⇒ 316y = 632

⇒ y = 2

Substituting the value of y in equation (3) we get

⇒ \(x=\frac{29+8 \times 2}{15}\)

⇒ \(x=\frac{45}{15}\)

⇒ x = 3

Hence, the solution is \(\left.\begin{array}{l}
x=3 \\
y=2
\end{array}\right\}\)

Elimination by Equating the Coefficients (Elimination Method)

Step 1: Multiply both the equations by a suitable number, so that the coefficients of x or the coefficients of y in both the equations become equal.

Step 2: If these coefficients are of the same sign then subtract the new equations otherwise add them.

Step 3: Solve the resulting equation to find the value of unknown. In this process, you will see that we have eliminated one variable. So, the name fell as “elimination by equating the coefficients”.

Step 4: Substitute this value in any of the two equations given and find the value of the other unknown.

Example 3. Solve 3x- 4y = 20 and x + 2y = 5.
Solution:

The given equations are 3x- 4y = 20 ….(1)

and x+2y=5 …..(2)

Multiplying equation (1) by 1 and (2) by 2, we get

3x- 4y = 20 ……(3)

2x+4y = 10 ……(4)

Adding these equations, we get

5x = 30

x=6

Substituting x = 6 in equation (1), we get

3 × 6- 4y = 20

18 -4y = 20

-4y = 2

y = \(-\frac{1}{2}\)

Hence, the solution is \(\left.\begin{array}{l}
x=6 \\
y=-\frac{1}{2}
\end{array}\right\}\)

Question 4. Solve 3x – 17y = 23 and \(\frac{x}{3}+\frac{y}{4}=4 .\)
Solution:

The given equations are 3x-y = 23 ….(1)

and \(\frac{x}{3}+\frac{y}{4}=4\)

4x + 3y = 48 ….(2)

Multiplying equation (1) by 3 and (2) by 1, we get

9x-3y = 69 …..(3)

4x + 3y = 48 …..(4)

Adding equations (3) and (4), we get

13x = 117

⇒ x = 9

Substituting x = 9 in equation (1), we get

3 x 9-y = 23

⇒ 27 -y = 23

⇒ -y =-4

⇒ y = 4

Hence, the solution is \(\left.\begin{array}{l}
x=9 \\
y=4
\end{array}\right\}\)

Question 5. Solve \(4 x+\frac{6}{y}=15 \text { and } 6 x-\frac{8}{y}=14 \text {. }\)
Solution:

The given equation are \(4 x+\frac{6}{y}=15\)…(1)

and \(6 x-\frac{8}{y}=14\)

Multiplying equation (1) by 4 and (2) by 3, we get

⇒ \(16 x+\frac{24}{y}=60\) …..(3)

⇒ \(18 x-\frac{24}{y}=42\) …..(4)

Adding equations (3) and (4), we get

34x = 102 ⇒ x = 3

Substituting x = 3 in equation (1), we get

⇒ \(4 \times 3+\frac{6}{y}=15\)

⇒ \(\frac{6}{y}=15-12\)

⇒ \(\frac{6}{y}=3\)

⇒ 3y = 6

or y = 2

Hence, the solution is \(\left.\begin{array}{l}
x=3 \\
y=2
\end{array}\right\}\)

Question 6. Solve for x and y, \(\frac{1}{2 x}-\frac{1}{y}=-1 \text { and } \frac{1}{x}+\frac{1}{2 y}=8 \text {. }\)
Solution:

The given equations are \(\frac{1}{2 x}-\frac{1}{y}=-1\) ….(1)

and \(\frac{1}{x}+\frac{1}{2 y}=8\) …..(2)

Let \(\frac{1}{x}=u \text { and } \frac{1}{y}=v\) then equations (1) and (2) can be written as

⇒ \(\frac{n}{2}-v=-1\)

⇒ u – 2v = -2 …….(3)

and \(u+\frac{v}{2}=8\)

⇒ 2u + v = 16 …….(4)

Multiplying equation (3) by 1 and (4) by 2, we get

⇒ u – 2v = -2 ….(5)

⇒ 4u + 2v = 32 ….(6)

Adding equations (5) and (6), we get

⇒ 5u = 30

⇒ u = 6

Substituting u = 6 in equation (3), we get

⇒ 6-2v = -2

⇒ -2v = -8

⇒ v = 4

Now, \(\frac{1}{x}=u\)

⇒ \(\frac{1}{x}=6\)

⇒ \(x=\frac{1}{6}\)

and \(\frac{1}{y}=v\)

⇒\(\frac{1}{y}=4\)

⇒\(x=\frac{1}{4}\)

Hence, the solution is \(\left.\begin{array}{l}
x=\frac{1}{6} \\
y=\frac{1}{6}
\end{array}\right\}\)

Question 7. Solve \(\frac{20}{x+y}+\frac{3}{x-y}=7 \text { and } \frac{8}{x-y}-\frac{15}{x+y}=5\)
Solution:

Given equations are

⇒ \(\frac{20}{x+y}+\frac{3}{x-y}=7\) ……..(1)

and \(\frac{8}{x-y}-\frac{15}{x+y}=5\) ……(2)

Let \(\) then equation (1) and (2) can be written as

20a + 3b = 7 ……(3)

and -15a + 8b = 5 …….(4)

Multiplying equation (3) by 8 and (4) by 3, we get

160a + 24b = 56 ……(5)

-45a + 24b = 15 ……(6)

Subtracting equation (6) from equation (5), we get

205a = 41

⇒ \(a=\frac{1}{5}\)

Substituting \(a=\frac{1}{5}\) in equation (3), we get

⇒ \(20 \times \frac{1}{5}+3 b=7\)

⇒ 32b = 3

⇒ b = 1

Now, \(\frac{1}{x+y}=a \quad \Rightarrow \quad \frac{1}{x+y}=\frac{1}{5} \quad \Rightarrow x+y=5\) ……(7)

and \(\frac{1}{x-y}=b \quad \Rightarrow \quad \frac{1}{x-y}=1 \quad \Rightarrow x-y=1\) …..(8)

Adding equations (7) and (8), we get

2x= 6

x = 3

Substituting x = 3 in equation (7), we get

3 +y = 5

⇒ y = 2

Hence, the solution is \(\left.\begin{array}{l}
x=3 \\
y=2
\end{array}\right\}\)

Question 8.

1. Solve 8x- 3y = 5xy and 6x-5y = -2xy. How many solutions, does this system of equations have?
2. Solve for x and y, by reducing the following equations in a pair of linear equations: 2x + 3y = 5xy and 3x-2y=xy.

Solution:

Given equations are 8x-3y = 5xy

and 6x- 5y- -2xy

The given equations are not linear in the variables x and y. These can be reduced into linear equations.

If we put x = 0 in either of the equations, we get y = 0. Hence, x = 0 and y = 0 is a solution of these equations.

To find another solution, divide each of the given equations by xy

So, \(\frac{8}{y}-\frac{3}{x}=5\) ……(1)

and \(\frac{6}{y}-\frac{5}{x}=-2\) …..(2)

Let \(\frac{1}{y}=a \text { and } \frac{1}{x}=b\) then from equations (1) and (2), we get

⇒ \(\left.\begin{array}{l}
8 a-3 b=5 \\
6 a-5 b=-2
\end{array}\right\} \text { linear equations }\) …….(3)and (4)

Multiplying equation (3) by 5 and (4) by 3, we get

40a -15b = 25 ……(5)

18a – 15b = -6 ……(6)

Subtracting equation (6) from (5), we get

22a = 31

⇒ \(a=\frac{31}{22}\)

Putting a = \(\frac{31}{22}\) in equation (3), we get

⇒ \(8 \times \frac{31}{22}-3 b=5\)

⇒ \(\frac{124}{11}-3 b=5\Rightarrow-3 b=5-\frac{124}{11}\)

⇒ \(-3 b=-\frac{69}{11}\Rightarrow=\frac{23}{11}\)

Now, \(\frac{1}{y}=a \quad \Rightarrow \quad \frac{1}{y}=\frac{31}{22} \quad \Rightarrow \quad y=\frac{22}{31}\)

and \(\frac{1}{x}=b \quad \Rightarrow \quad \frac{1}{x}=\frac{23}{11} \quad \Rightarrow \quad x=\frac{11}{23}\)

Hence, the solution of the given two equations are \(\left.\left.\begin{array}{l}
x=0 \\
y=0
\end{array}\right\} \text { and } \quad \begin{array}{l}
x=\frac{11}{23} \\
y=\frac{22}{31}
\end{array}\right\}\)

So, the given system of equations has two solutions.

Note:

In the above example 10,

1. If it is given that x ≠ 0, y ≠ 0 or reduce the given equations in a pair of linear equations, then we have a right to divide both the equations by xy. In this way, we get only one solution i.e., \(x=\frac{11}{23}, y=\frac{22}{31}\) only. When no condition is given to the equations then we get the two solutions of this system of equations i.e.,

⇒ \(\left.\left.\begin{array}{l}
x=0 \\
y=0
\end{array}\right\} \text { and } \begin{array}{r}
x=\frac{11}{23} \\
y=\frac{22}{31}
\end{array}\right\}\) (for more details, see “Gold coins” at the end of this chapter).

2. We have,

2x + 3y = 5xy ……(1)

3x-2y=xy …..(2)

Dividing equations (1) and (2) by xy, we get

⇒ \(\frac{2 x}{x y}+\frac{3 y}{x y}=\frac{5 x y}{x y}\Rightarrow \frac{2}{y}+\frac{3}{x}=5\) …..(3)

and \(\frac{3 x}{x y}-\frac{2 y}{x y}=\frac{x y}{x y}\Rightarrow \frac{3}{y}-\frac{2}{x}=1\) …….(4)

Let \(\frac{1}{x}=u \text { and } \frac{1}{y}=v\)

∴ Equations (3) and (4) become,

2v + 3u = 5 ….(5)

3v- 2u = 1 ….(6)

Now this is a pair of linear equations, multiplying equation (5) by 2 and equation (6) by 3, we get

4v + 6u = 10 …(7)

9v – 6u = 3 …..(8)

On adding equations (7) and (8), we get

13v = 13

⇒ v = 1

Putting v = 1 in equation (5), we get

2(1) + 3u = 5

3u = 3

u=1

Now, u = 1 ⇒ \(\frac{1}{x}=1\) => x =1

and v = 1 ⇒ \(\frac{1}{y}=1\) => y =l

Hence, the required solution is \(\left.\begin{array}{l}
x=1 \\
y=1
\end{array}\right\}\)

Question 9. Solve \(\frac{1}{2(2 x+3 y)}+\frac{12}{7(3 x-2 y)}=\frac{1}{2} \text { and } \frac{7}{2 x+3 y}+\frac{4}{3 x-2 y}=2,\) by reducing them into a pair of linear equations.
Solution:

The given equations

⇒ \(\frac{1}{2(2 x+3 y)}+\frac{12}{7(3 x-2 y)}=\frac{1}{2}\) ….(1)

and \(\frac{7}{2 x+3 y}+\frac{4}{3 x-2 y}=2\) ….(2)

are not linear equations. So, first of all, we shall make these as a pair of linear equations, by putting

⇒ \(\frac{1}{2 x+3 y}=a \text { and } \frac{1}{3 x-2 y}=b\)

Therefore, equations (1) and (2) become,

⇒ \(\frac{1}{2} a+\frac{12}{7} b=\frac{1}{2}\) ….(3)

and 7a + 4b = 2 …….(4)

Multiplying equation (3) by 14, we get

7a + 24b = 7 ……(5)

Subtracting equation (5) from (4), we get

b = \(\frac{1}{4}\)

Putting b = \(\frac{1}{4}\) in equation (4), we get

⇒ \(7 a+4\left(\frac{1}{4}\right)=2 \quad \Rightarrow 7 a=2-1 \quad \Rightarrow a=\frac{1}{7}\)

since, \(a=\frac{1}{7} \quad \Rightarrow \frac{1}{2 x+3 y}=\frac{1}{7} \quad \Rightarrow 2 x+3 y=7\) ……..(6)

and \(b=\frac{1}{4} \quad \Rightarrow \frac{1}{3 x-2 y}=\frac{1}{4} \quad \Rightarrow 3 x-2 y=4\) ……….(7)

Multiplying equation (6) by 2 and equation (7) by 3, we get

⇒ x = 2

On adding, we get

Putting x = 2 in equation (8), we get,

4(2) + 6y= 14

6y = 14- 8

y = 1

Hence, the solution is \(\left.\begin{array}{r}
x=2 \\
y=1
\end{array}\right\}\)

Question 10. Solve 41x + 53y = 135 and 53x + 41y = 147
Solution:

Given equations are 41x + 53y = 135 ……..(1)

53x + 41y = 147 ……..(2)

On adding equations (1) and (2) we get

94x + 94y = 282

⇒ x+y = 3 ………(3)

On subtracting equation (2) from (1), we get

-12x+ 12y = -12

⇒ x-y- 1 …………(4)

Adding equations (3) and (4), we get

2x= 4  ⇒ x = 2

Putting x = 2 in equation (3)

2 + y = 3 ⇒ y = 1

Hence, the solution is \(\left.\begin{array}{r}
x=2 \\
y=1
\end{array}\right\}\)

Question 11. Solve (a – b)x + (a+b)y = a²– 2ab – b² and (a + b)(x +y) = a² + b².
Solution:

Given equations are (a- b)x + (a + b]y = a² – 2ab – b² ….(1)

and (a + b)(x+y) – a² +b²

or (a + b)x + (a +b)y = a² + b² …….(2)

On subtracting equation (2) from (1), we get

(a – b)k- [a + b)x = a² – 2ab- b² – a² – b²

⇒  x(a -b -a -b) = -2b² – 2ab

⇒ -2bx = -2b(b + a)

⇒ x = a +b

Putting x= (a + b) in equation (1), we get

(a – b)(a + b) + (a + b)y = a²– 2ab- b²

a² -b² + (a +b)y = a² – 2ab – b²

(a +b)y = -2ab

⇒ \(y=\frac{-2 a b}{a+b}\)

Hence, the solution is \(\left.\begin{array}{l}
x=a+b \\
y=\frac{-2 a b}{a+b}
\end{array}\right\}\)

Method of Cross Multiplication

Theorem : If a₁x+b₁y +c₁ =0 and a₂x +b₂y + c₂ = 0 be a system of simultaneous linear equations in two variables x and y such that \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \text { i.e., } a_1 b_2-a_2 b_1 \neq 0\).

Then the system has a unique solution given by

⇒ \(x=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1} \text { and } y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}\)

Proof: The given equations are

a₁x +b₁y + c₁ = 0

a₂x +b₂y + c₂ = 0

Multiplying equation (1) by b₂ and (2) by b₁ and subtracting we get

⇒ \(\left(a_1 b_2-a_2 b_1\right) x=\left(b_1 c_2-b_2 c_1\right)\)

⇒ \(x=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}\)

Again multiplying equation (1) by a2, (2) by A J and subtracting we get

⇒ \(y\left(a_2 b_1-a_1 b_2\right)=c_2 a_1-c_1 a_2\)

⇒ \(y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}\)

Hence, x = \(\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1} \text { and } y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1} \text {. }\)

⇒ \(y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-c_2 b_1}\)

Solved Examples

Question 1. Solve the following system of equations by using the method of cross multiplication 2x+3y = 7 and 6x + 5y = 11.
Solution:

Given

2x+3y = 7 and 6x + 5y = 11

The given equations can be written as 2x + 3y- 7 = 0

⇒ \(\frac{x}{3 \times(-11)-5 \times(-7)}=\frac{y}{-7 \times 6-(-11) \times 2}=\frac{1}{2 \times 5-6 \times 3}\)

⇒ \(\frac{x}{-33+35}=\frac{y}{-42+22}=\frac{1}{10-18}\)

⇒ \(\frac{x}{2}=\frac{y}{-20}=\frac{1}{-8}\)

when \(\frac{x}{2}=-\frac{1}{8} \quad \text { or } \quad x=-\frac{1}{4}\)

and \(\frac{y}{-20}=-\frac{1}{8} \quad \text { or } \quad y=\frac{20}{8}=\frac{5}{2}\)

Hence, \(\left.\begin{array}{l}
x=-\frac{1}{4} \\
y=\frac{5}{2}
\end{array}\right\}\) is the required solution.

Question 2. Solve the following system of equations bx + ay = 2ab and ax -by- a² – b².

Solution:

Given

bx + ay = 2ab and ax -by- a² – b²

Given equations can be written as

bx + ay- 2ab = 0 and ax- by – a² + b² = 0

By cross multiplication method, we have

⇒ \(\frac{x}{a \times\left[-\left(a^2-b^2\right)\right]-(-b) \times(-2 a b)}=\frac{y}{(-2 a b) \times a-\left[-\left(a^2-b^2\right)\right] \times b}=\frac{1}{b \times(-b)-a \times a}\)

⇒ \(\frac{x}{a\left(-a^2+b^2-2 b^2\right)}=\frac{y}{-2 a^2 b+a^2 b-b^3}=\frac{1}{-b^2-a^2}\)

⇒ \(\frac{x}{a\left(-a^2-b^2\right)}=\frac{y}{-a^2 b-b^3}=\frac{1}{-b^2-a^2}\)

⇒ \(\frac{x}{a\left(-a^2-b^2\right)}=\frac{1}{-b^2-a^2}\)

⇒ \(x=\frac{-a\left(a^2+b^2\right)}{-1\left(b^2+a^2\right)}\) or x = a

⇒ \(\frac{y}{-a^2 b-b^3}=\frac{1}{-b^2-a^2}\)

⇒ \(y=\frac{-b\left(a^2+b^2\right)}{-1\left(b^2+a^2\right)}\) or y = b

Hence, \(\left.\begin{array}{l}
x=a \\
y=b
\end{array}\right\}\) is the required solution.

Question 3. Solve the following system of equations for x and y, a(x +y) + b(x -y) = a² – ab + b² and a(x +y) -b(x -y) = a² + ab + b².
Solution:

Given

a(x +y) + b(x -y) = a² – ab + b² and a(x +y) -b(x -y) = a² + ab + b²

Given equations can be written as

⇒ ax + ay +bx -by = a² – ab + b²

⇒ (a + b)x + (a -b)y = a² – ab + b²

and ax+ ay -bx + by =a² +ab +b² ….(1)

⇒ (a – b)x + (a + b)y = a² + ab + b² …..(2)

For equations (1) and (2) by cross multiplication method, we have

⇒ \(\frac{y}{\left[-\left(a^2-a b+b^2\right)\right] \times(a-b)-\left[-\left(a^2+a b+b^2\right) \times(a+b)\right]}=\frac{1}{(a+b)(a+b)-(a-b)(a-b)}\)

⇒ \(\frac{x}{-\left(a^3-b^3\right)+\left(a^3+b^3\right)}\)

⇒ \(\frac{y}{-\left(a^3-a^2 b+a b^2-a^2 b+a b^2-b^3\right)+\left(a^3+a^2 b+a b^2+a^2 b+a b^2+b^3\right)}=\frac{1}{(a+b)^2-(a-b)^2}\)

⇒ \(\frac{x}{2 b^3}=\frac{y}{2 a^2 b-2 a b^2+b^3+2 a^2 b+2 a b^2+b^3}=\frac{1}{4 a b}\)

⇒ \(\frac{x}{2 b^3}=\frac{y}{4 a^2 b+2 b^3}=\frac{1}{4 a b}\)

⇒ \(\frac{x}{2 b^3}=\frac{y}{2 b\left(2 a^2+b^2\right)}=\frac{1}{4 a b}\)

when \(\frac{x}{2 b^3}=\frac{1}{4 a b} \quad \Rightarrow \quad x=\frac{2 b^3}{4 a b} \quad \Rightarrow \quad x=\frac{b^2}{2 a}\)

and \(\frac{y}{2 b\left(2 a^2+b^2\right)}=\frac{1}{4 a b} \Rightarrow y=\frac{2 b\left(2 a^2+b^2\right)}{4 a b} \Rightarrow y=\frac{2 a^2+b^2}{2 a}\)

Hence, \(\left.\begin{array}{rl}
x=\frac{b^2}{2 a} \\
y=\frac{2 a^2+b^2}{2 a}
\end{array}\right\}\) is the required solution.

Condition For Solvability Of Linear Equations

Consistent and In-consistent System: When a system of equations a₁ x + b₁ y = c₁ and a₂x + b₂y = c₂ has a solution, the system is said to be consistent. A consistent system has one or more solutions. When a system has no solution, the system is said to be inconsistent.

Conditions for Solvability

1. If \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2},\) then the given system of equations has a unique solution and thus it is consistent.
Solution:

In this case, the two lines intersect at exactly one point.

2. If \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) then there is no solution and system is in-consistent.
Solution:

In this case, the two lines are parallel to each other. They will not meet at any point.

3. If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\) then the system has infinitely many solutions; s and thus the system is consistent and the given pair of linear equal.
Solution:

In this case, the two lines coincide each other, i.e., the two lines are the same. So, we say such lines as coincident lines.

Remark:

If c₁ and c₂ are both equal to zero, the solution can be found more easily as follows:

1. When \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2} ，\) then only solution is x = y = 0
2. When \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\) then there are infinite number of non-zero solution.

Linear Equations Solved Examples

Question 1. Show that the following system of equations has a unique solution:

1. 7x – 2y = 3
   22x – 3y = 16
2. 3x + y = 17
   8x + 11y = 37

and also solve the system of equations in each case.

Solution:

Given equations are 7x- 2y = 3 …….(1)

22x- 3y = 16 …….(2)

Here, a₁ = 7, b₁ = -2 and c₁ = 3

a₂ = 22, b₂ = -3 and c₂= 16

Now, \(\frac{a_1}{a_2}=\frac{7}{22} \text { and } \frac{b_1}{b_2}=\frac{-2}{-3}=\frac{2}{3}\)

Since, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\) Hence, the given system has a unique solution.

By cross multiplication method, we have

⇒ \(\frac{x}{(-2) \times(-16)-(-3) \times(-3)}=\frac{y}{(-3) \times 22-(-16) \times 7}=\frac{1}{7 \times(-3)-22 \times(-2)}\)

⇒ \(\frac{x}{32-9}=\frac{y}{-66+112}=\frac{1}{-21+44}\)

⇒ \(\frac{x}{23}=\frac{y}{46}=\frac{1}{23}\)

when \(\frac{x}{23}=\frac{1}{23}\)

x = 1 and \(\frac{y}{46}=\frac{1}{23}\) y = 2

Hence, \(\left.\begin{array}{l}
x=1 \\
y=2
\end{array}\right\}\) is the required solution.

The given system of equations is 3x + y = 17 ……..(1)

8x + 11y = 37 ……..(2)

Here a₁ = 3, b₁ =1 and c₁ = 17

a₂ = 8, b₂ = 11 and c₂ = 37

Now, \(\frac{a_1}{a_2}=\frac{3}{8}, \frac{b_1}{b_2}=\frac{1}{11}\)

Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\). Hence, the given system has a unique solution.

We can write the equations as

3x+y- 17 = 0 and 8r + 1ly- 37 = 0

By cross multiplication method, we have

⇒ \(\frac{x}{1 \times(-37)-11 \times(-17)}=\frac{y}{(-17) \times 8-(-37) \times 3}=\frac{1}{3 \times 11-8 \times 1}\)

⇒ \(\frac{x}{-37+187}=\frac{y}{-136+111}=\frac{1}{33-8}\)

⇒ \(\frac{x}{150}=\frac{y}{-25}=\frac{1}{25}\)

When \(\frac{x}{150}=\frac{1}{25}\) = x = 6 and \(\frac{y}{-25}=\frac{1}{25}\) = y = -1

Hence, \(\left.\begin{array}{l}
x=6 \\
y=-1
\end{array}\right\}\) is the required solution

Question 2. Find the value of k for which the system of equations 2x + ky = 1 and 3x- 5y = 7 has a unique solution.
Solution:

The given system of equations is

2x +ky = 1 ……..(1)

and 3x- 5y = 7 ……..(2)

Here, a₁ =2, b₁=k and c₁ = 1

a₂ = 3, b₂ = -5 and c₂ = 7

The system has unique solution if \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

⇒ \(\frac{2}{3} \neq \frac{k}{-5} \quad \text { or } \quad k \neq \frac{-10}{3}\)

So, k can take any real value except \(\frac{-10}{3}\)

Question 3. Find the value of k for which the system of equations x + 2y = 5, 3x + ky – 1 5 = 0 has no solution.
Solution:

The given system of equations is

x + 2y- 5 = 0 ………(1)

3x + ky- 15 = 0 ……..(2)

Here, a₁ = 1, b₁ = 2, c₁ =-5

a₂ = 3, b₂ = k, c₂ = -15

If the equations have no solution then

⇒ \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

⇒ \(\frac{1}{3}=\frac{2}{k} \neq \frac{-5}{-15} \Rightarrow \frac{1}{3}=\frac{2}{k} \quad \text { and } \quad \frac{2}{k} \neq \frac{-5}{-15}\)

k = 6 and k ≠ 6 which is impossible.

Hence, there is no such value of k for which the given system has no solution.

Question 4. For what value(s) of a will the system of linear equations αx + 3y = α – 3 and 12x + αy = a have a unique solution?
Solution:

The given equations are

αx + 3y = α – 3

12x + αy = α

Here a₁ = α, b₁ = 3 and c₁ = α- 3

a₂ = 12, b₂ = α and c₂ = α

The system has a unique solution if \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

⇒ \(\frac{\alpha}{12} \neq \frac{3}{\alpha}\)

⇒ a² ≠  36

⇒ a ≠ ± 6

i.e., a ≠ 6 and a ≠ -6

So, α can take any real value except -6 and 6

Question 5. Find the value(s) of k for which the system of equations 5x + 2y = k and 10x + 4y = 3 has infinitely many solutions.
Solution:

The given system of equations is

5x + 2y – k = 0 and 10k + 4y- 3 = 0

Here a₁ = 5, b₁ = 2 and c₁ = k

a₂= 10, b₂ = 4 and c₂ = -3

The system has infinitely many solutions if

⇒ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

⇒ \(\frac{5}{10}=\frac{2}{4}=\frac{-k}{-3}\)

Hence, for k = \(\frac{3}{2}\), the given system of equations will have infinitely many solutions.

Question 6. Find the values of p and q for which the following system of equations has an infinite number of solutions: 2x + 3y = 7 and (p+q)x+(2p-q)y = 21.
Solution:

The given system of equations is

2x + 3y = 7

(p + q)x + (2p- q)y = 21

Here a₁ = 2, b₁ = 3,

a₂ =P + q, b₂ = q and c₂ = 21

The given system will have an infinite number of solutions if

⇒ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

⇒ \(\frac{2}{p+q}=\frac{3}{2 p-q}=\frac{7}{21}\)

Taking the first two and last two expressions

⇒ \(\frac{2}{p+q}=\frac{1}{3} \quad \text { and } \quad \frac{3}{2 p-q}=\frac{1}{3}\)

p + q = 6 and 2p-q = 9

Solving p + q = 6 and 2p- q = 9, we get

p = 5 and q = 1

Hence, the given system of equations will have infinitely many solutions, if p = 5 and q = 1.

Question 7. For which value(s) of A, will the pair of equations represent coincident lines k x + 3y – (k + 3) = 0 and 12x + ky + k = 0?
Solution:

Comparing the given equations with

a₁x + b₁y + c₁ = 0

a₂x+ b₂y + c₂ = 0

We have

⇒ \(\frac{a_1}{a_2}=\frac{k}{12}, \frac{b_1}{b_2}=\frac{3}{k} \text { and } \frac{c_1}{c_2}=\frac{-(k+3)}{k}\)

Now, for coincident lines

⇒ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

⇒ \(\frac{k}{12}=\frac{3}{k}=-\frac{(k+3)}{k}\)

Taking the first two expressions, we get

⇒ \(\frac{k}{12}=\frac{3}{k}\)….(1)

k² = 36

k = +6

Taking the last two expressions, we get

⇒ \(\frac{3}{k}=-\frac{(k+3)}{k}\) …..(2)

3k = -k² – 3k

k² + 6k = 0

k(k + 6) = 0

∴ k = 0 or k = -6

But k = -6 is a value which satisfies both the equations (1) and (2).

∴ k = -6

Linear Equations Solved Examples

Question 1. The sum of the two numbers is 85. If the larger number exceeds four times the smaller one by 5. Find the numbers.
Solution:

Given

The sum of the two numbers is 85. If the larger number exceeds four times the smaller one by 5.

Let the two numbers x andy, and x > y.

According to question, x + y = 85 ……(1)

and x – 4y = 5 …..(2)

Subtracting equation (2) from (1), we get

5y = 80

⇒ y = 16

Putting, y = 16 in equation (1), we get

x+ 16 = 85

⇒ x = 69

Hence, the required numbers are 69 and 16.

Question 2. The sum of the two numbers is 18. The sum of their reciprocals is \(\frac{1}{4}\) Find the numbers.
Solution:

Given

The sum of the two numbers is 18. The sum of their reciprocals is \(\frac{1}{4}\)

Let the two numbers be x and y.

∴ According to the condition,

x + y = 18 …….(1)

According to the 2 condition, \(\frac{1}{x}+\frac{1}{y}=\frac{1}{4}\) ……(2)

⇒ \(\frac{y+x}{x y}=\frac{1}{4}\)

⇒ xy = 4(x+y)

⇒ xy = 4 × 18

⇒ xy = 72

Now, we know that

(x-y)² = (x+y)²-4xy

⇒ (x -y)² = (18)² – 4(72) = 324 – 288 = 36

∴ x-y = ± 6

If x-y = 6 ….(3)

Now, from equations (3) and (1),

∴ x= 12

∴ y = 18-12

= 6

∴ Two numbers are 12 and 6 in both cases.

If x – y = -6 …….(4)

Now, from equations (4) and (1),

∴ x= 6

∴ y = 18-6

= 12

Question 3. Five years hence father’s age will he three times the age of his son. Five years ago, the father was seven times as old as his son. Find their present ages.
Solution:

Given

Five years hence father’s age will he three times the age of his son. Five years ago, the father was seven times as old as his son.

Let the present age of the son be x years and the present age of the father by years.

Five years hence, the age of son = x + 5

and the age of father = y + 5

Using given information

3(x + 5)=y + 5

⇒ 3x-y = -10 …….(1)

Five years ago, age of son = (x- 5) years

age of father = (y- 5) years

∴ 7(x-5)=y-5

7x -y = 30 ………(2)

Subtracting equation (1) from (2), we get

4x = 40

⇒ x = 10

On substituting x = 10 in equation (1)

3 × 10 – y = -10

⇒ 30 -y =-10

⇒ -y = -40

⇒ y = 40

Hence, the present age of the son is 10 years and the present age of a father is 40 years.

Question 4. The sum of the digits of a two-digit number is 9. The number obtained by reversing the order of the digits of the given number exceeds the given number by 27. Find the given number.
Solution:

Given

The sum of the digits of a two-digit number is 9. The number obtained by reversing the order of the digits of the given number exceeds the given number by 27.

Let the digit at the unit’s place be x and the ten’s place be y.

∴ Number = 10y +x

On reversing the digits, the new number = 10x+y

According to the given conditions,

x+y = 9 …..(1)

10x+y = 21 + (10y +x)

9x-9y = 21

x-y = 3 …….(2)

Adding equations (1) and (2), we get

2x= 12

x = 6

Putting x= 6 in equation (1), we get

6+y = 9

y = 3

Hence, the required number = 10y +x= 10 x 3 + 6 = 36

Question 5. If 2 is added to the numerator of a fraction it reduces to \(\frac{1}{2}\) and if 1 is subtracted from the denominator, it reduces to \(\frac{1}{3}\) Find the fraction.
Solution:

Given

If 2 is added to the numerator of a fraction it reduces to \(\frac{1}{2}\) and if 1 is subtracted from the denominator, it reduces to \(\frac{1}{3}\)

Let the fraction be \(\frac{x}{y}\), where the numerator is x and the denominator is y.

According to 1 condition, \(\frac{x+2}{y}=\frac{1}{2}\)

2x+ 4 = y

⇒ 2x-y = -4 ……(1)

According to 2 condition, \(\frac{x}{y-1}=\frac{1}{3}\)

⇒ 3x=y- 1

⇒ 3x -y = -1 …..(2)

Subtracting equation (2) from (1), we get

-x = -3

⇒ x = 3

Substituting x = 3 in equation (1), we get

2 x 3 -y = -4

⇒ -y = -10

⇒ y = 10

Hence, the required fraction is \(\frac{x}{y}=\frac{3}{10}\)

Question 6. A man sold a chair and table together for ₹760 thereby making a profit of 25% on chairs and 10% on tables. By selling them together for ₹767.50, he would have made a profit of 1 0% on a chair and 25% on the table. Find the cost price of each.
Solution:

Given

A man sold a chair and table together for ₹760 thereby making a profit of 25% on chairs and 10% on tables. By selling them together for ₹767.50, he would have made a profit of 1 0% on a chair and 25% on the table.

Let the cost price of the chair = ₹x and the cost price of the table = ₹y

Now, two cases arise:

Case 1: 25% profit on chair and 10% profit on table

Selling price of chair = \(₹\left(x+\frac{25 x}{100}\right)=₹ \frac{125 x}{100}\)

Selling price of table = \(₹\left(y+\frac{10 y}{100}\right)=₹ \frac{110 y}{100}\)

But total S.P = ₹760

⇒ \(\frac{125 x}{100}+\frac{110 y}{100}=760 \quad \Rightarrow 125 x+110 y=76000\)

∴ 25x + 22y = 15200 ……(1)

Case 2: 10% profit on chair and 25% profit on table

Selling price of chair = \(₹\left(x+\frac{10 x}{100}\right)=₹ \frac{110 x}{100}\)

Selling price of table = \(₹\left(y+\frac{25 y}{100}\right)=₹ \frac{125 y}{100}\)

But total S.P. = ₹767.50

∴ \(\frac{110 x}{100}+\frac{125 y}{100}=767.50\)

⇒ 110x + 125y = 76750

⇒ 22x + 25y = 15350 ……(2)

To solve equations (1) and (2), adding equations (1) and (2), we get

47x + 47y = 30550

⇒ x + y = 650 …….(3)

Subtracting equation (2) from (1), we get

3x- 3y = -150

⇒ x-y = -50 ….(4)

On adding equations (3) and (4), we get

2x = 600

⇒ x = 300

Putting x = 300 in equation (3), we get

300 +y = 650

⇒ y = 650- 300 = 350

Cost of chair = ₹300 and cost of table = ₹350

Question 7. A lady has 25 paise and 50 paise coins in her purse. If in all she has 40 coins and the value of her money is ₹12.75, how many coins of each type does she have?
Solution:

Given

A lady has 25 paise and 50 paise coins in her purse. If in all she has 40 coins and the value of her money is ₹12.75,

Let a number of 25 paise coins be x and a number of 50 paise coins be y.

According to the given conditions, we have

x + y = 40 ….(1)

and 25x + 50y= 1275 …..(2)

Multiplying equation (1) by 50, we get

50x+50y = 2000

Subtracting equation (3) from (2), we get

-25x = -725

⇒ x = 29

Substituting x = 29 in equation (1), we get

29 + y = 40

⇒ y = 11

Hence, the number of 25 paise coins = 29 and the number of 50 paise coins = 11.

Question 8. 2 men and 7 boys can finish a work in 4 days, while 4 men and 4 boys can finish the same work in 3 days. How long would it take one man or one boy to do it?
Solution:

Given

2 men and 7 boys can finish a work in 4 days, while 4 men and 4 boys can finish the same work in 3 days.

Let 1 man can do a piece of work in* days while a boy can do a piece of work in y ways.

∴ One man’s one day work = \(\frac{1}{x}\)

and one boy’s one day work = \(\frac{1}{y}\)

According to the given conditions,

⇒ \(\frac{2}{x}+\frac{7}{y}=\frac{1}{4}\) ……(1)

and \(\frac{4}{x}+\frac{4}{y}=\frac{1}{3}\) …….(2)

Multiplying equation (1) by 2 and (2) by 1, we get

⇒ \(\frac{4}{x}+\frac{14}{y}=\frac{1}{2}\) ……..(3)

⇒ \(\frac{4}{x}+\frac{4}{y}=\frac{1}{3}\) ……..(4)

Subtracting equation (4) from (3), we get

⇒ \(\frac{14}{y}-\frac{4}{y}=\frac{1}{2}-\frac{1}{3} \Rightarrow \frac{10}{y}=\frac{1}{6}\)

⇒ y = 60

Substituting y = 60 in equation (1), we get

⇒ \(\frac{2}{x}+\frac{7}{60}=\frac{1}{4} \quad \Rightarrow \quad \frac{2}{x}=\frac{1}{4}-\frac{7}{60}\)

⇒ \(\frac{2}{x}=\frac{8}{60}\)

⇒ 8x= 120 or x = 15

Hence, one man would do the work in 15 days and one boy in 60 days.

Question 9. A boat goes 301km upstream and 44 km downstream in 10 hours. In 1 3 hours it can go 40 1cm upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.
Solution:

Given

A boat goes 301km upstream and 44 km downstream in 10 hours. In 1 3 hours it can go 40 1cm upstream and 55 km downstream.

Let the speed of the boat in still water is x km/hr and the speed of the stream is y km/hr.

∴ speed of boat upstream = (x-y) km/hr

and speed of boat downstream = (x +y) km/hr

We know that, speed = \(\frac{\text { distance }}{\text { time }}\)

According to the given conditions,

Time taken in going 30 km upstream = \(\frac{30}{x-y} \mathrm{hr}\)

Time taken in 44 lem downstream = \(\frac{40}{x+y} \mathrm{hr}\)

∴ \(\frac{30}{x-y}+\frac{44}{x+y}=10\)

Similarly in the second condition,

⇒ \(\frac{40}{x-y}+\frac{55}{x+y}=13\)

Let \(\frac{1}{x-y}=a \text { and } \frac{1}{x+y}=b,\), we get

30a + 44b = 10

and 40a + 55b =13

Multiplying equation (3) by 4 and (4) by 3, we get

120a+ 1766 = 40

120a+ 1656 = 39

Subtracting equation (6) from (5), we get

11b = 1

⇒ b = \(\frac{1}{11}\)

Putting b = \(\frac{1}{11}\) inequation (3), we get

⇒ \(30 a+44 \times \frac{1}{11}=10\)

30a = 6

⇒ \(a=\frac{1}{5}\)

⇒ \(\frac{1}{x-y}=\frac{1}{5}\text { or }x-y=5\) ……(7)

and \(\frac{1}{x+y}=11\text { or }x+y=11\) …….(8)

Solving equations (7) and (8), we get

x = 8 and y = 3

Hence, the speed of a boat in still water = 8 km/hr and the speed of the water stream = 3 km/hr.

Question 10. A boat which travels at the rate of 10.5 km/hr downstream takes 3 times as long time to go to a certain distance up a river as to go the same distance down. Find the rate at which the stream flows.
Solution:

Given

A boat which travels at the rate of 10.5 km/hr downstream takes 3 times as long time to go to a certain distance up a river as to go the same distance down.

Let the speed of the stream = x km/hr

and the speed of the boat in still water = y km/hr

∴ Speed of boat downstream = (x +y) km/hr

and speed of boat upstream = (y- x) km/hr

Let the distance be D km.

∴ According to the problem,

x+y = 10.5 ……(1)

Also, time taken by boat upstream = 3 x time taken by boat downstream

⇒ \(\frac{D}{y-x}=3 \times \frac{D}{10.5}\)

y – x = 3.5 ……(2)

Adding equations (1) and (2), we get

2y= 14 ⇒ y = 7

Putting y = 7 in equation (1), we get

x+7 = 10.5 ⇒ x= 3.5

Hence, speed of stream = 3.5 km/hr

Linear Equations In Two Variables Exercise 3.1

Question 1. Aftab tells his daughter, “Seven years ago, 1 was seven times as old as you were then. Also, three years from now, 1 shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.
Solution:

Given

Aftab tells his daughter, “Seven years ago, 1 was seven times as old as you were then. Also, three years from now, 1 shall be three times as old as you will be.”

Let the present age of Aftab be x years and the present age of his daughter be x years.

7 years ago,

age of Aftab = (x- 7) years

age of his daughter = (y- 7) years

According to the problem,

x-7 = 7(y -7)

⇒ x-7 = 7y- 49

⇒ x-7y = -42

3 years later,

age of Aftab = (x + 3) years

age of his daughter = (y + 3) years

According to the problem,

x + 3 = 3(y + 3)

⇒ x + 3 = 3y + 9

⇒ x- 3y = 6

∴ The algebraic form of the given problem is as follows:

x- 7y = -42 and x- 3y = 6

Now, x-7y = -42

x = 7y – 42

y = 5 then x=7 × 5-42=-7

y = 6 then x-7 × 6 – 42 = 0

y = 7 then A=7 × 7-42 = 7

and x- 3y = 6 ⇒ x = 3y + 6

y = -1 then x = 3 × -1 +6 = 3

y = 0 then x = 3 × 0 + 6 = 6

y = -2 then x = 3 × -2 + 6 = 0

Mark and join the given points in the two tables. It is the graphical form of the given problem.

Question 2. The coach of a cricket team buys 3 bats and 6 balls for ₹3900. Later, she buys another bat and 3 more balls of the same kind for ₹1300. Represent this situation algebraically and geometrically.
Solution:

Given

The coach of a cricket team buys 3 bats and 6 balls for ₹3900. Later, she buys another bat and 3 more balls of the same kind for ₹1300.

Let the cost of a bat = ₹x

and cost of a ball = ₹y

According to the problem,

3x + 6y = 3900

x + 2y= 1300 ……….(1)

x + 3y = 1300 ……….(2)

Equations (1 ) and (2) represent the algebraic form of the given problem.

From equation (1),

x = 1300 -2y

Mark the points A(700, 300), B(300, 500), C(100, 600) and join them.

From equation (2), x- 1300- 3y

Mark the points D = (100, 400), E = (400, 300), E(700, 200) and join them.

It is the geometric form of the given problem.

Question 3. The cost of 2 kg of apples and I kg of grapes in a day was found to be ‘ 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ‘ 300. Represent the situation algebraically and geometrically.
Solution:

Given

The cost of 2 kg of apples and I kg of grapes in a day was found to be ‘ 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ‘ 300.

Let the cost of 1 kg apple = ₹x

and cost of 1kg grapes = ₹y

According to the problem,

2x+y = 160 ……(1)

4x + 2y = 300

2x+y = 150 ……(2)

Equation (1) and equation (2) is the algebraic form of the given problem.

From equation (1),y = 160- 2x

Mark the points A(50, 60), B(60, 40) and C(70, 20) and join them.

From equation (2), y = 150-2x

Mark the points D(50, 50), E(60, 30) and F(70, 10) and join them.

The lines shown in the graph represent the geometric form of the given problem.

Linear Equations In Two Variables Exercise 3.2

Question 1. Form the pair of linear equations in the following problems and find their solutions graphically.

1. 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
2. 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.

Solution:

1. Let many boys = x and a number of girls =y.

According to the problem,

x +y = 10 ……(1)

y – x + 4 …….(2)

From equation (1), x +y = 10 ⇒ y = 10 – x

Mark the points A(5, 5), B(4, 6) and C(3, 7) and join them.

From equation (2), y = x + 4

Mark the points D(1, 5), E(2, 6) and C(3, 7) and join them.

Both lines intersect at points C(3, 7).

∴ x = 3, y = 7

⇒ Number of boys = 3 and number of girls = 7

2. Let the cost of 1 pencil = ₹x

and the cost of 1 pen = ₹y

According to the problem,

5x + 7y = 50

⇒ \(x=\frac{50-7 y}{5}\)

Mark the points A(10, 0), B(3, 5) and C(-4, 10) and join them

And 7x + 5y – 46 ⇒ \(x=\frac{46-5 y}{7}\)

Mark the points B(3, 5) and E(8, -3) and join them.

Two lines intersect at point (3,5).

∴ x = 3,y = 5

∴ Cost of 1 pencil= ₹3

and cost of 1 pen = ₹5.

Question 2. On comparing the ratios \(\frac{a_1}{a_2}, \frac{b_1}{b_2} \text { and } \frac{c_1}{c_2}\), find representing the following pairs of linear equations intersect at a point, are parallel or coincident :

1. 5x- 4y + 8 = 0
   7x + by- 9 = 0
2.  9x + 3y+12 = 0
   18r + 6y + 24 = 0
3. 6x-3y+ 10 = 0
   2x-y + 9 = 0

Solution:

1. Given equations, 5x- 4y + 8 = 0 and 7x + 6y – 9 = 0

Here, a₁ = 5, b₁ = -4, c₁ = 8

a₂ = 7, b₂ = 6, c₂ = -9

Now, \(\frac{a_1}{a_2}=\frac{5}{7}, \frac{b_1}{b_2}=\frac{-4}{6}=\frac{-2}{3}, \frac{c_1}{c_2}=\frac{8}{-9}\)

∵ \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

∴ Two lines intersect at a point.

2. Given equations, 9x + 3y+ 12 = 0 and 18x + 6y + 24 =0

Here,

a₁ = 9, b₁ = 3, c₁ = 12

a₂ = 18, b₂ = 6, c₂ = 24

Now \(\frac{a_1}{a_2}=\frac{9}{18}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}\)

and \(\frac{c_1}{c_2}=\frac{12}{24}=\frac{1}{2}\)

∵ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

∴ Two lines are coincident

3. Given equations, 6x- 3y + 10 = 0 and 2x-y + 9 =0

Here, a₁ = 6, b₁ = -3, c₁ = 10

a₂ = 2, b₂=-1, c₂ = 9

Now \(\frac{a_1}{a_2}=\frac{6}{2}=\frac{3}{1}, \frac{b_1}{b_2}=\frac{-3}{-1}=\frac{3}{1}\)

and \(\frac{c_1}{c_2}=\frac{10}{9}\)

∵ \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

∴ Two lines are parallel.

Question 3. On comparing the ratios, \(\frac{a_1}{a_2}, \frac{b_1}{b_2} \text { and } \frac{c_1}{c_2}\), find out whether the following pair of linear equations are consistent, or inconsistent.

1. 3x + 2y = 5; 2x- 3y = 7
2. 2x- 3y = 8; 4x- 6y = 9
3. \(\frac{3}{2} x+\frac{5}{3} y=7 ; 9 x-10 y=14\)
4. 5x-3y = 11;-10x+6y = -22
5. \(\frac{4}{3} x+2 y=8 ; 2 x+3 y=12\)

Solution:

1. Given equations,

3x + 2y = 5 ⇒ 3x + 2y – 5 = 0

and 2x- 3y = 7 ⇒ 2x- 3y- 7 = 0

Here, a₁ =3, b₁ = 2, c₁ = -5

a₂ = 2, b₂ =-3, c₂ = -7

Now, \(\frac{a_1}{a_2}=\frac{3}{2}, \frac{b_1}{b_2}=\frac{2}{-3} \text { and } \frac{c_1}{c_2}=\frac{-5}{-7}=\frac{5}{7}\)

∵ \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

∴ Given equations have a unique solution.

⇒ Given pair of equations is consistent.

2. Given equations,

2x-3y = 8 ⇒ 2x-3y-8 = 0

and 4x- 6y= 9 ⇒ 4r-6y-9 = 0

Here, a₁= 2, b₁ = -3, c₁ = -8

a₂ = 4, b₂ = -6, c₂ = -9

Now, \(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{-3}{-6}=\frac{1}{2}\)

and \(\frac{c_1}{c_2}=\frac{-8}{-9}=\frac{8}{9}\)

∵ \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

∴ Given equations have no solution.

⇒  Given pair of equations is inconsistent.

3. Given equations, \(\frac{3}{2} x+\frac{5}{3} y=7 \Rightarrow \frac{3}{2} x+\frac{5}{3} y-7=0\)

and 9x- 10y = 14 ⇒ 9x- 10y- 14 = 0

Here, \(a_1=\frac{3}{2}, b_1=\frac{5}{3}, c_1=-7\)

and \(a_2=9, b_2=-10, c_2=-14\)

Now, \(\frac{a_1}{a_2}=\frac{3 / 2}{9}=\frac{1}{6}, \frac{b_1}{b_2}=\frac{5 / 3}{-10}=-\frac{1}{6}\)

and \(\frac{c_1}{c_2}=\frac{-7}{-14}=\frac{1}{2}\)

∵ \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

∴ Given equations have a unique solution.

⇒ Given pair of equations is consistent.

4. Given equations,

5x-3y =11 ⇒ 5x- 3y – 11 = 0

and -10x + 6y = -22 ⇒ -10x + 6y + 22 = 0

Here, a₁ = 5, b₁ -3, c₁ = -11

a₂ = -10, b₂ = 6, c₂ = 22

Now, \(\frac{a_1}{a_2}=\frac{5}{-10}=-\frac{1}{2}, \frac{b_1}{b_2}=\frac{-3}{6}=-\frac{1}{2}\)

and \(\frac{c_1}{c_2}=\frac{-11}{22}=\frac{-1}{2}\)

∵ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

∴ Two lines are coincident i.e., they will have infinitely many solutions.

⇒ Given pair of equations is consistent.

5. Given equations, \(\frac{4}{3} x+2 y=8\)

⇒ \(\frac{4}{3} x+2 y-8=0\)

and 2x + 3y = 12

⇒ 2x + 3y – 12 = 0

Here, \(a_1=\frac{4}{3}, b_1=2, c_1=-8\)

a₂ = 2, b₂ = 3, c₂ =-12

Now, \(\frac{a_1}{a_2}=\frac{4 / 3}{2}=\frac{2}{3}, \frac{b_1}{b_2}=\frac{2}{3}, \frac{c_1}{c_2}=\frac{-8}{-12}=\frac{2}{3}\)

∵ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

∴ Two lines are coincident i.e., they will have infinitely many solutions.

⇒ Given pair of equations is consistent.

Question 4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:

1. x + y = 5, 2x + 2y = 10
2. x-y = 8, 3x- 3y = 16
3. 2x+y-6 = 0, 4x- 2y – 4 = 0
4. 2x-2y-2 = 0, 4x-4y-5 = 0

Solution:

1. Given equations,

x+y = 5 ⇒ x+y-5 = 0

and 2x + 2y = 10 ⇒ 2x + 2y – 10 = 0

Here, a₁ = 1, b₁ = 1, c₁ = -5

a₂ = 2, b₂ = 2, c₂ = -10

Now, \(\frac{a_1}{a_2}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{1}{2} \text { and } \frac{c_1}{c_2}=\frac{-5}{-10}=\frac{1}{2}\)

∵ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

∴ Given equations have infinitely many solutions.

⇒ Given pair of equations is consistent.

Now, x + y = 5 ⇒ y = 5-x

Mark the points (5, 0), (3, 2) and (0, 5) and join them.

It is the required solution of given equations.

2. Given equations,

x-y = 8 ⇒ x -y- 8 = 0

3x-3y = 16 ⇒ 3x- 3y – 16 = 0

a₁ = 1, b₁ =-1, c₁ =-8

a₂= 3, b₂ = -3, c₂= -16

Now, \(\frac{a_1}{a_2}=\frac{1}{3}, \frac{b_1}{b_2}=\frac{-1}{-3}=\frac{1}{3} \text { and } \frac{c_1}{c_2}=\frac{-8}{-16}=\frac{1}{2}\)

∵ \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

∴ Given system of equations has no solution

⇒ Given pair of equations is inconsistent.

3. Given equations,

2x+y – 6 =0 and 4x-2y -4 = 0

Here, a₁ = 2, b₁ = 1, c₁ = -6

a₂ = 4, b₂= -2, c₂ = -4

Now, \(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{1}{-2}=-\frac{1}{2}\)

∵ \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

∴ Given equations have a unique solution.

⇒ Given pair of equations is consistent.

From the first equation,

2x +y- 6 =0 ⇒ y = 6- 2x

Mark the points (3, 0), (2, 2) and (.0, 6) and join them.

From the second equation,

4x- 2y- 4 = 0 ⇒ 2x -y – 2 = 0

y = 2x-2

Mark the points (3, 4), (2, 2) and (1,0) and join them.

Two lines intersect at point (2, 2).

∴ Solution of given pair of equations: x = 2, y = 2

4. Given equations,

2x- 2y – 2 = 0 and 4x- 4y- 5 = 0

Here, a₁ = 2, b₁ = -2, c₁ = -2

a₂ = 4, b₂ = -4, c₂ = -5

Now \(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{-2}{-4}=\frac{1}{2}, \frac{c_1}{c_2}=\frac{-2}{-5}=\frac{2}{5}\)

∵ \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

∴ Given system of equations has no solution.

⇒ Given pair of equations is inconsistent.

Question5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Solution:

Given

Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m.

Let the length of the garden = x metre

and breadth =y metre.

∴ The perimeter of a rectangular garden

= 2(length + breadth)

= 2(x+y)

So, half the perimeter = x+y

but it is given that, half the perimeter = 36 m

∴ x+y = 36 ….(1)

and x=y + 4 i.e., x-y = 4 …(2)

To solve the equations (1) and (2) we get x=20, y = 16

Therefore, length = 20 m and breadth = 16 m.

Question 6. Given the linear equation 2x + 3y- 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

1. Intersecting lines
2. Parallel lines
3. Coincident lines

Solution: Given a linear equation,

2x+ 3y- 8 = 0

Here, a₁ = 2, b₁ = 3, c₁ =-8

1. We know that, intersecting lines

⇒ \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

∴ Taking a₂ = 3, b₂ = 4 and c₂ = 2

Required equation, 3x + 4y + 2 = 0

2. We know that, for parallel lines,

⇒ \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

∴ Taking = 4, = 6 and = 5

Required equation, 4x + 6y + 5 = 0

3. We know that, for coincident lines,

⇒ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

∴ Taking a₂ = 4, b₂ = 6, c₂ = -16

Required equation, 4x+ 6y- 16 = 0

Question 7. Draw the graphs of the equations x -y +1=0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Solution:

x-y =-1 ⇒ x=y – 1

3x + 2y =12 ⇒ \(x=\frac{12-2 y}{3}\)

Mark the points Q(0, 1),B(-1, 0), P(0, 6) and C(4, 0).

Obtain the graph of the given lines on the graph paper by joining BQ and PC.

Two lines intersect at point A(2, 3).

∴ x = 2, y = 3

Lines intersect x-axis at points B(-l, 0) and C(4, 0).

Draw the perpendicular AM from A to BC

∴ Area of ΔABC = \(\frac{1}{2} \times B C \times A M\)

⇒ \(\frac{1}{2} \times 5 \times 3\)

= 7.5 square units.

Linear Equations In Two Variables Exercise 3.3

Question 1. Solve the following pair of linear equations by the substitution method :

1. x+y= 14, x-y = 4
2. s-t = 3, \(\frac{s}{3}+\frac{t}{2}=6\)
3. 3x-y = 3, 9x-3y = 9
4. 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3
5. \(\sqrt{2} x+\sqrt{3} y=0\), \(\sqrt{3} x-\sqrt{8} y=0\)
6. \(\frac{3 x}{2}-\frac{5 y}{3}=-2\), \(\frac{1}{2 x}+\frac{1}{3 y}=2\)

Solution:

1. x+y = 14 …..(1)

x-y = 4 ……(2)

From equation (1),

x+y =14 ⇒ x= 14 -y

Put this value of x in equation (2),

14 -y -y = 4

⇒ -2y = 4-14 =-10

⇒ y = 5

Put the value of y in equation (3),

x = 14-5 = 9

∴ x = 9, y = 5

2. s-t = 3 …..(1)

⇒ \(\frac{s}{3}+\frac{t}{2}=6\) …….(2)

From equation (1)

s-t =3 ⇒ s = 3 + t ……(3)

Put the value of s from equation (3) to equation (2),

⇒ \(\frac{3+t}{3}+\frac{t}{2}=6 \Rightarrow \frac{6+2 t+3 t}{6}=6\)

⇒ 6+5t = 36

⇒ 5t = 36-6 = 30

⇒ t = 6

Put this value of t in equation (3),

s = 3 + t = 3 + 6 = 9

s =9, t = 6

3. 3x -y = 3 …….(1)

9x- 3y = 9 ……(2)

From equation (1),

3x-y=3 ⇒ y = 3x-3 ,..(3)

Put the value of y from equation (3) to equation (2),

9x- 3(3x- 3) = 9

⇒ 9x- 9x + 9 =9

⇒ 9 = 9

From here, we do not obtain the values of x and y.

∴ Given equations have infinitely many solutions.

4. 0.2x + 0.3y = 1.3 + 3y = 13 …….(1)

and 0.4r + 0.5y = 2.3 => 4x + 5y = 23 …..(2)

From equation (1),

2x + 3y = 13 => 2x = 13-3y-

⇒ \(x=\frac{13-3 y}{2}\)…….(3)

Put the value of x from equation (3) to equation (2)

⇒ \(\frac{4(13-3 y)}{2}+5 y=23\)

⇒ 2(13 -3y) + 5y = 23

⇒ 26 – 6y + 5y = 23

⇒ -y = 23 – 26 =-3

⇒ y = 3

Put the value of y in equation (3)

⇒ \(x=\frac{13-3 \times 3}{2}\)

⇒ \(x=\frac{13-9}{2}=2\)

∴ x = 2,y = 3

5. \(\sqrt{2} x+\sqrt{3} y=0\) …….(1)

⇒ \(\sqrt{3} x-\sqrt{8} y=0\) ……(2)

From equation (2),

⇒ \(\sqrt{3} x=\sqrt{8} y\)

⇒ \(x=\frac{\sqrt{8}}{\sqrt{3}} y\) ……..(3)

Put the value of x from equation (3) to equation (1),

⇒ \(\sqrt{2} \cdot \frac{\sqrt{8}}{\sqrt{3}} y+\sqrt{3} y=0 \Rightarrow \frac{4 y+3 y}{\sqrt{3}}=0\)

⇒ 7y = 0

⇒ y = 0

Put this value of y in equation (3),

⇒ \(x=\frac{\sqrt{8}}{\sqrt{3}} \times 0=0\)

∴ x = 0,y = 0

6. \(\frac{3 x}{2}-\frac{5 y}{3}=-2\) ……(1)

⇒ \(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\) …..(2)

From equation (1)

⇒ \(\frac{3 x}{2}=-2+\frac{5 y}{3}=\frac{-6+5 y}{3}\)

⇒ \(x=\frac{10 y-12}{9}\) ………(3)

Put the value of x from equation (3) to equation (2),

⇒ \(\frac{1}{3} \cdot\left(\frac{10 y-12}{9}\right)+\frac{y}{2}=\frac{13}{6}\)

⇒ \(\frac{20 y-24+27 y}{-54}=\frac{13}{6}\)

⇒ \(47 y-24=\frac{13}{6} \times 54=117\)

⇒ 47y = 117 + 24= 141

⇒ y = 3

Put this value of/ in equation (3),

⇒ x = \(\frac{10 \times 3-12}{9}=2\)

∴ x = 2, y = 3

Question 2. Solve 2x + 3y = 11 and 2x -4y = -24 and hence find the value of ‘m’ for which y = mx + 3.
Solution:

Here, 2x + 3y= 11 ……(1)

2x- 4y = -24 …….(2)

From equation (2), 4y = 2x+ 24

y = \(\frac{x+12}{2}\) …….(3)

Put the value of y from equation (3) to equation (1),

⇒ \(2 x+3\left(\frac{x+12}{2}\right)=11\)

⇒ 4x + 3(x + 12)= 11 x 2

⇒ 4x + 3x + 36 = 22

⇒ 4x = 22- 36

⇒ 7x = -14

⇒ x = -2

Put x = -2 in equation (3),

⇒ \(y=\frac{-2+12}{2} \Rightarrow y=\frac{10}{2}=5\)

Put x =-2 andy = 5 in the equation, y = mx + 3

⇒ 5 = m x (-2) + 3

⇒ 5 = -2m + 3

⇒ 2m = 3-5=-2

⇒ m = -1

The value of ‘m’ = -1

Question 3. Form the pair of linear equations for the following problems and find their solution by substitution method :

1. The difference between two numbers is 26 and one number is three times the other. Find them.
2. The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
3. The coach of a cricket team buys 7 bats and 6 balls for ₹3800. Later, she buys 3 bats and 5 balls for ₹1750. Find the cost of each bat and each ball.
4. The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ?₹105 and for a journey of 15 km, the charge paid is ₹155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
5. A fraction becomes \(\frac{9}{11}\) if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\) Find the fraction.
6. Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Solution:

1. Let two numbers be x and y. According to the problem,

x-y = 26

⇒ x = 26 +y ……(1)

and x = 3y …..(2)

⇒ 26 +y = 3y

Put the value of x from equation (1),

⇒ 3y -y= 26

⇒ 2y = 26

⇒ y= 13

From equation (2),

x = 3 x 13 = 39

Therefore numbers =39 and 13

2. Let two angles are x and y According to the problem,

x+y = 180°

⇒ x = 180° -y ……(1)

and x =y+ 18° …..(2)

⇒ 180° -y =y+ 18°

Put the value of x from equation (1)

⇒ -y – y = 18°- 180°

⇒ -2y = -162°

⇒ y = 81°

From equation (2),

x = 81°+ 18° = 99°

∴ Required angle= 99° and 81°

3. Let the cost of one bat = ₹x

and the cost of one ball = ₹y

According to the question,

⇒ 7x + 6y = 3800

⇒ 7x = 3800 – 6y

⇒ \(x=\frac{3800-6 y}{7}\) …….(1)

and 3x+5y = 1750 …….(2)

Put the value of x from equation (1),

⇒ \(\frac{11400-18 y+35 y}{7}=1750\)

11400+ 17y = 12250

⇒ 17y= 12250- 11400 = 850

⇒ y = 50

Put the value of y in equation (1),

⇒ \(x=\frac{3800-6 \times 50}{7}=\frac{3500}{7}=500\)

∴ Cost of 1 bat = ₹500

and Cost of 1 ball = ₹50

4. Let fixed charge = ₹x

and the fare for each lem = ₹y

According to the problem,

x+10y = 105

⇒ x = 105 – 10y …..(1)

and x+15y = 155

⇒ 105 – 10y + 15y = 155

Put the value of x from equation (1),

⇒ 5y = 155- 105 = 50

⇒ y = 10

Put the value of y in equation (1),

x = 105- 10 x 10 = 5

∴ Fixed charge = ₹5

and the fare of each km = ₹10

Fare for travelling 25 km

= x + 25y = 5 + 25 x 10

= 5 + 250 = ₹255

5. Let fraction \(\frac{x}{y}\)

According to the problem,

⇒ \(\frac{x+2}{y+2}=\frac{9}{11}\)

⇒ 11x + 22 = 9y + 18

⇒ 11x = 9y-4

⇒ x = \(\frac{9 y-4}{11}\) ……(1)

and \(\frac{x+3}{y+3}=\frac{5}{6}\)

6x + 18 = 5y + 15

6x = 5y- 3

⇒ \(\frac{6(9 y-4)}{11}=5 y-3\)

Put the value of* from equation (1),

⇒ 54y – 24 = 55y- 33

⇒ 54y- 55y = -33 + 24

⇒ -y = -9 ⇒ y = 9

Put the value of y in equation (1)

⇒ \(x=\frac{9 \times 9-4}{11}=7\)

∴ Fraction = \(\frac{x}{y}=\frac{7}{9}\)

6. Let the present age of Jacob = x years

and present age of son = y years

After 5 years,

Jacob’s age = (x + 5) years

son’s age = (y + 5) years

According to the problem,

x + 5 = 3(y + 5)

⇒ x = 3y+ 10 …….(1)

Five years ago,

Jacob’s age = (x- 5) years

son’s age = (y- 5) years

According to the problem, x- 5 = 7(y-5)

⇒ 3y+10-5 = 7y – 35

Put the value of x from equation (1),

⇒ 3y-7y = -35-5 ⇒ -4y =-40

⇒ y = 10

Put the value of y in equation (1),

⇒ x = 3 x 10+10 = 40

∴ The present age of Jacob = 40 years

Present age of son = 10 years

Linear Equations In Two Variables Exercise 3.4

Question 1. Solve the following pair of linear equations bv the elimination method and the substitution method:

1. x +y = 5 and 2x- 3y = 4
2. 3a + 4y = 10 and 2x- 2y = 2
3. 3x- 5y -4 = 0 and 9x = 2y + 7
4. \(\frac{x}{2}+\frac{2 y}{3}=-1 \text { and } x-\frac{y}{3}=3\)

Solution:

1. x +y = 5 ……(1)

2x-3y = 4 …….(2)

Elimination method

Multiply equation (1) by 3 and adding in equation (2),

⇒ \(x=\frac{19}{5}\)

Put this value of or in equation (1),

⇒ \(\frac{19}{5}+y=5\)

⇒ \(y=5-\frac{19}{5}=\frac{6}{5}\)

∴ \(x=\frac{19}{5}, y=\frac{6}{5}\)

Substitution method

From equation (1),

⇒ x + y = 5 ⇒ x = 5 -y ……(3)

Put the value of x from equation (3) to equation (2)

⇒ 2(5 -y) -3y = 4 ⇒ 10 -2y- 3y = 4

⇒ -5y = 4-10 = -6

⇒ \(y=\frac{6}{5}\)

Put this value of y in equation (3),

⇒ \(x=5-\frac{6}{5}=\frac{25-6}{5}=\frac{19}{5}\)

∴ \(x=\frac{19}{5}, y=\frac{6}{5}\)

Here both methods are appropriate.

2. 3x + 4y = 10 …….(1)

2x-2y = 2 ……(2)

Elimination method

Multiply equation (2) by 2 and adding in equation (1),

x = 2

Put the value of x in equation (1)

3×2+4y = 10

⇒ 4y = 10-6 = 4

⇒ y = 1

∴ x = 2, y = 1

Substitution method

From equation (2),

2x- 2y = 2 ⇒ x-y = 1

⇒ x = 1+y ……(3)

Put the value of x from equation (3) to equation (1),

3(1+y) + 4y = 10

⇒ 3 + 3y + 4y = 10

⇒ 7y = 10- 3 = 7

⇒ y = 1

Put this value of y in equation (1),

x = 1+y

⇒ x = 1+1=2

∴ x = 2, y = 1

Here both methods are appropriate.

3. 3x- 5y- 4 = 0

⇒ 3x- 5y = 4 …(1)

9a = 2y + 7

⇒ 9x-2y= 7 …(2)

Elimination method

Multiply equation (1) by 3 and subtracting from equation (2),

⇒ \(y=\frac{-5}{13}\)

Put this value of y in equation (1),

⇒ \(3 x-5\left(\frac{-5}{13}\right)=4\)

⇒ \(3 x=4-\frac{25}{13}=\frac{52-25}{13}=\frac{27}{13}\)

⇒ \(x=\frac{9}{13}\)

⇒ \(x=\frac{9}{13}\), \(x=\frac{-5}{13}\)

Substitution method

From equation (1),

3x- 5y = 4 ⇒ 3x = 4 + 5y

⇒ \(x=\frac{4+5 y}{3}\) ….(3)

Put the value of x from equation (3) to equation (2),

⇒ \(\frac{9(4+5 y)}{3}-2 y=7\)

3(4 + 5y) – 2y = 7k

⇒ 12 + 15y- 2y = 7

⇒ 13y = 7 – 12= -5 ⇒ \(y=\frac{-5}{13}\)

Put this value of y in equation (3),

⇒ \(x=\frac{4+5\left(\frac{-5}{13}\right)}{3}=\frac{52-25}{3 \times 13}=\frac{27}{3 \times 13}=\frac{9}{13}\)

⇒ \(x=\frac{9}{13}, y=\frac{-5}{13}\)

Here both methods are appropriate.

4. \(\frac{x}{2}+\frac{2 y}{3}=-1 \Rightarrow 3 x+4 y=-6\) …..(1)

⇒ \(x-\frac{y}{3}=3 \quad \Rightarrow 3 x-y=9\) => 3x-y = 9 …..(2)

Elimination method

Subtracting equation (2) from (1),

y = -3

Put the value of y in equation (2),

3x- (-3) = 9

⇒ 3x = 9-3 = 6 ⇒ x = 2

x = 2, y- -3

Substitution method

From equation (2),

3x-y= 9

y = 3x- 9 …..(3)

Put this value of y in equation (1)

⇒ 3x + 4(3x- 9) = -6

⇒ 3x + 12x- 36 = -6

⇒ 15x = -6 + 36 = 30

⇒ x = 2

Put this value of x in equation (3),

y = 3×2-9 = -3

∴ x = 2, y = -3

Here both methods are appropriate.

Question 2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :

1. If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \(\frac{1}{2}\) if we only add 1 to the denominator. What is the fraction?
2. Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
3. The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
4. Meena went to a bank to withdraw ₹2000. She asked the cashier to give her ₹50 and ₹100 notes only. Meena got 25 notes in all. Find how many notes of ₹50 and ₹100 she received.
5. A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹27 for a book kept for seven days, while Susy paid ₹21 for a book she kept for five days. Find the fixed charge and the charge for each extra day.

Solution:

1. Let fraction = \(\frac{x}{y}\)

According to the problem,

⇒ \(\frac{x+1}{y-1}=1\)

⇒ x + 1 = y- 1

⇒ x-y = -2 …..(1)

and \(\frac{x}{y+1}=\frac{1}{2}\)

⇒ 2x = y + 1

⇒ 2x-y = 1 ……(2)

Subtracting equation (1) from equation (2),

Put this value of x in equation (1),

3 -y = -2 ⇒ y = 3 + 2 = 5

∴ Fraction = \(\frac{x}{y}=\frac{3}{5}\)

2. Let present age of Noori = A years and present age of Sonu =y years 5 years ago,

the age of Noori = (x- 5) years

the age of Sonu = (y – 5) years

According to the problem,

(x-5) =3(y-5)

⇒ x- 5 = 3y – 15

⇒ x-3y = -10 ……(1)

After 10 years,

the age of Noori = (x + 10) years

the age of Sonu = (y + 10) years

According to the problem,

(x + 10) = 2(y + 10)

⇒ x+ 10 = 2y + 20

⇒ x- 2y = 10 …..(2)

Subtracting equation (2) from equation (1),

⇒ y = 20

Put the value of y in equation (2),

⇒ x- 2 × 20 = 10

⇒ x = 10 + 40 = 50

∴ The present age of Noori = 50 years

and present age of Sonu = 20 years

3. Let two digit number = 10x +y

According to the problem,

x+y = 9 ….(1)

and 9(10x +y) = 2 x number formed by reversing the digits.

= 2(10 +x)

⇒ 90x+ 9y = 20y + 2x

⇒ 88x- 11y = 0

⇒ 8x -y = 0 …..(2)

Adding equations (1) and (2),

⇒ x = 1

Put the value of A in equation (2),

8×1 -y – 0 ⇒ y = 8

∴ Number = 10A +y = 10 x 1 + 8 = 18

4. Let ₹50 note = x

and ₹100 note = y

According to the problem,

x+y = 25

and ₹100 note =y

According to the problem,

⇒ x+y = 25 …..(1)

and 50x + 100y = 2000

⇒ x + 2y= 40 …..(2)

Subtracting equation (1) from equation (2),

Put this value of y in equation (1),

x + 15 = 25

x = 10

∴ ₹50 note =10

and ₹100 note = 15 …(2)

5. Let the fixed charges for the first 3 days = ₹x

and the charge per day after it = ₹y

According to the problem,

charges for 7 days = ? 27

⇒ x+ (7 – 3)y = 27 ⇒ x + 4y = 27 …(1)

and charges for 5 days = ₹21

⇒ x+ (5 – 2)y = 21

⇒ x + 2y = 21 …..(2)

Subtracting equation (2) from equation (1)

⇒ y = 3

Put this value of y in equation (l),

⇒ x + 4×3 = 27 ⇒  x = 27 – 12 = 15

Fixed charges = ₹15

and charges per day after the first 3 days = ₹3

Linear Equations In Two Variables Exercise 3.5

Question 1. Which of the following pairs of linear equations has a unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method.

1. x- 3y – 3 = 0, 3x- 9y -2 = 0
2. 2x + y = 5, 3x + 2y = 8
3. 3x- 5y = 20, 6x- 10y = 40
4. x- 3y -7 = 0, 3x-3y- 15 = 0

Solution:

1. x-3y-3 = 0

3x- 9y- 2 = 0

Here, a₁ = 1, b₁ = -3, c₁ =-3

a₂ = 3, b₂ = -9, c₂ = -2

∴ \(\frac{a_1}{a_2}=\frac{1}{3}, \frac{b_1}{b_2}=\frac{-3}{-9}=\frac{1}{3} \text { and } \frac{c_1}{c_2}=\frac{-3}{-2}=\frac{3}{2}\)

∵ \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

∴ Given pair of equations has no solution.

2. 2x +y = 5 ⇒ 2x +y – 5 = 0

3x + 2y =8 ⇒ 3x + 2y – 8 = 0

Here, a₁ = 2, b₁ = 1, c₁ = -5

a₂ = 3, b₂ = 2, c₂ = -8

∴ \(\frac{a_1}{a_2}=\frac{2}{3}, \frac{b_1}{b_2}=\frac{1}{2}\)

∵ \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

∴ Given pairofequations has a unique solution.

From the cross-multiplication method,

⇒ \(\frac{x}{-8-(-10)}=\frac{y}{-15-(-16)}=\frac{1}{4-3}\)

⇒ \(\frac{x}{2}=\frac{y}{1}=\frac{1}{1} \quad \Rightarrow x=2, y=1\)

3. 3x- 5y = 20 ⇒ 3x-5y- 20 = 0

6x- 10y = 40 ⇒ 6x- 10y- 40 = 0

Here, a₁ =3, b₁, =-5, c₁, =-20

a₂ = 6, b₂ =-10 c₂ = -40

∴ \(\frac{a_1}{a_2}=\frac{3}{6}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{-5}{-10}=\frac{1}{2}, \frac{c_1}{c_2}=\frac{-20}{-40}=\frac{1}{2}\)

∵ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

∴ Given pair of equations has infinitely many solutions.

4. x-3y – 7 = 0

3x-3y- 15 = 0

Here, a₁ = 1, b₁ = -3, c₁ = -7

a₂ = 3, b₂ = -3, c₂ = -15

∴ \(\frac{a_1}{a_2}=\frac{1}{3}, \frac{b_1}{b_2}=\frac{-3}{-3}=1\)

∵ \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

∴ Given pair ofequations has aunique solution.

Now, from the cross-multiplication method,

⇒ \(\frac{x}{45-21}=\frac{y}{-21+15}=\frac{1}{-3+9}\)

⇒ \(\frac{x}{24}=\frac{y}{-6}=\frac{1}{6}\)

⇒ \(\frac{x}{4}=\frac{y}{-1}=\frac{1}{1}\)

⇒ x = 4, y = -1

Question 2.

1. For which values of a and b does the following pair of linear equations have an infinite number of solutions? 2x + 3y = 7, (a-b)x+(a+b)y = 3a+b-2
2. For which value of k will the following pair of linear equations have no solution? 3x- +y = 1, (2k-1)x+(k-1)y = 2k+1

Solution:

1. Here, 2x + 3y – 7 = 0

and (a – b)x + (a + b)y – (3a + b – 2) = 0

Now, a₁ = 2, b₁ = 3, c₁ = -7,

a₂= a – b, b₂ = a + b, c₂ =- (3a +b- 2)

For infinitely many solutions,

⇒ \(\frac{2}{a-b}=\frac{3}{a+b}=\frac{-7}{-(3 a+b-2)}\)

⇒ \(\left(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\right)\)

⇒ \(\frac{a-b}{2}=\frac{a+b}{3}=\frac{3 a+b-2}{7}\)

From the first and second expressions,

⇒ \(\frac{a-b}{2}=\frac{a+b}{3}\)

⇒ 3a-3b = 2a + 2b

⇒ a = 5b …(1)

From the second and third expressions,

⇒ \(\frac{a+b}{3}=\frac{3 a+b-2}{7}\)

⇒ 7a + 7b = 9a + 3b – 6

⇒ 4b = 2a -6

⇒ 2b = a-3 …(2)

From equations (1) and (2),

2b = 5b -3 ⇒ 6 = 1

Put 6= 1 in equation (1),

a = 5 x 1 ⇒ a = 5

2. Here, 3x+y- 1 =0 …..(1)

and (2k- 1)x + (k- l)y- (2k + 1) = 0 …(2)

Here, a₁ =3, b₁ = 1, c₁ = -1,

a₂ =2k-1, b₂ = k-1, c₂ = -(2k+1)

⇒ \(\frac{a_1}{a_2}=\frac{3}{2 k-1}, \frac{b_1}{b_2}=\frac{1}{k-1}, \frac{c_1}{c_2}=\frac{1}{2 k+1}\)

For no solution, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) i.e., \(\frac{3}{2 k-1}=\frac{1}{k-1} \neq \frac{1}{2 k+1}\)

⇒ \(\frac{3}{2 k-1}=\frac{1}{k-1} \neq \frac{1}{2 k+1}\)

⇒ 3(k-1) = 2k-1 and k-1 = 2k+1

⇒ 3k-3 = 2k-1 and k = -2

⇒ k = 2 and k = -2

Therefore, the given system will have no solution when k = 2

Question 3. Solve the following pair of linear equations by the substitution and cross-multiplication methods. 8x + 5y = 9, 3x + 2y = 4
Solution:

Given equations

8x + 5y = 9 …….(1)

3x- + 2y = 4 ……(2)

Substitution method

From equation (1),

⇒ 8x = 9 – 5y

⇒ \(x=\frac{9-3 y}{8}\) ……(3)

Put this value of x in equation (2),

⇒ \(\frac{3(9-5 y)}{8}+2 y=4\)

⇒ \(\frac{27-15 y+16 y}{8}=4\)

⇒ 27 +y = 32 = 5

Put the value of/ in equation (3)

⇒ \(x=\frac{9-5 \times 5}{8}=\frac{9-25}{8}=-2\)

⇒ x = -2, y = 5

From the cross-multiplication method,

8x + 5y- 9 = 0

3x + 2y- 4 = 0

⇒ \(\frac{x}{-20+18}=\frac{y}{-27+32}=\frac{1}{16-15}\)

⇒ \(\frac{x}{-2}=\frac{y}{5}=\frac{1}{1}\)

⇒ x = -2, y = 5

Here, both methods are appropriate.

Question 4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :

1. A part of monthly hostel charges are fixed and the remaining depends on the number of days one has taken food in the mess. When student A takes food for 20 days she has to pay ₹1000 as hostel charges whereas student B, who takes food for 26 days, pays ₹1180 as hostel charges. Find the fixed charges and the cost of food per day.
2. A fraction becomes when \(\frac{1}{3}\) is subtracted from the numerator and it becomes \(\frac{1}{4}\) when 8 is added to its denominator. Find the fraction.
3. Yash scored 40 marks on a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
4. Places’ A’ and B are 100 1cm apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
5. The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Solution:

Let fixed charge = ₹x

and the cost of food per day = ₹y

∴ x + 20y = 1000 ………(1)

and x + 26y = 1180 ………(2)

Subtracting equation (1) from equation (2),

⇒ y = 30

From equation (1),

x + 20×30 = 1000

⇒ x = 1000-600 = 400

∴ Fixed charge = ₹400

and cost of food per day = ₹30

2. Let, fraction = \(\frac{x}{y}\)

According to the problem,

⇒ \(\frac{x-1}{y}=\frac{1}{3}\)

⇒ 3x-3=y

⇒ 3x-y = 3 …….(1)

and \(\frac{x}{y+8}=\frac{1}{4}\)

⇒ 4x =y + 8 ….(2)

⇒ 4x -y = 8

Subtracting equation (2) from equation (1),

Put the value of x in equation (2),

4×5-y = 8 ⇒ y = 20-8 = 12

∴ Fraction = \(\frac{x}{y}=\frac{5}{12}\)

3. Let number of correct answers = x and number of incorrect answers =7.

According to the first condition,

3x -7 = 40 ……..(1)

According to the second condition,

4x-2y =50 ⇒ 2x -7 = 25 ……..(2)

Subtracting equation (2) from equation (1),

Put the value of x in equation (2),

⇒ 2(15) -7 = 25

⇒ 30 -7 = 25 ⇒ y = 30 – 25 = 5

Total number of questions in a test

= x +7 =15 + 5= 20

4. Let the speed of the car starting from A =xkm/hr and the speed of the car starting from B =                       7 km/hr

Case 1: When they move in the same direction:

Distance covered by car A = AP= x × 5 Km ( ∵ distance = speed x time)

Distance covered by carB = BP = y × 5 km

⇒ but AB = 100 km

⇒ AP-BP = 100

⇒ 5x -5y = 100 ⇒ x-y=20 …….(1)

Case 2: When they move in opposite directions :

Distance covered by car A = AQ=x × 1 km

Distance covered by carB = BQ =y × 1 km

AB = 100 km

AQ + BQ = 100 ⇒ x +y = 100 …(2)

Adding equations (1) and (2),

2x = 1 20 ⇒ x = 60 km/hr

Put x = 60 in equation (2)

y = 100 – 60 = 40 km/hr

∴ Speed Of car A = 60 km/hr and speed of cal- B = 40 km/hr.

5. Let the length of the rectangle be x and breadth y units.

∴ Area of rectangle = xy square units

According to the given conditions,

⇒ (x-5)(y + 3) =xy-9

⇒ xy+ 3x-5y- 15 = xy- 9

⇒ 3x- 5y = 6 …….(1)

and (x + 3)(y + 2) = xy + 67

⇒ xy + 2x + 3y + 6= xy + 67

⇒ 2x + 3y =61 ……..(2)

Multiplying equation (1) by 3 and equation (2) by 5

9x- 15y = 18 …(3)

10x + 15y = 305

Adding equations (3) and (4)

19x = 323 ⇒ x = 17

Put x = 17 in equations (1)

3 x 17 – 5y = 6

⇒ 51-5y = 6

⇒ -5y = -45 ⇒ y = 9

Therefore, the length of the rectangle = 17 units and the breadth = 9 units.

Linear Equations In Two Variables Exercise 3.6

Question 1. Solve the following pairs of equations by reducing them to a pair of linear equations :

1. \(\frac{1}{2 x}+\frac{1}{3 y}=2 \quad \frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6}\)
2. \(\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2 \quad \frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1\)
3. \(\frac{4}{x}+3 y=14 \quad \frac{3}{x}-4 y=23\)
4. \(\frac{5}{x-1}+\frac{1}{y-2}=2 \quad \frac{6}{x-1}-\frac{3}{y-2}=1\)
5. \(\frac{7 x-2 y}{x y}=5 \quad \cdot \quad \frac{8 x+7 y}{x y}=15\)
6. 6x + 3y = 6xy, 2x + 4y = 5xy
7. \(\frac{10}{x+y}+\frac{2}{x-y}=4\),\(\frac{15}{x+y}-\frac{5}{x-y}=-2\)
8. \(\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}\), \(\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{-1}{8}\)

Solution:

1. Given equations, \(\frac{1}{2 x}+\frac{1}{3 y}=2,\)

⇒ \(\frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6}\)

Let, \(\frac{1}{x}=a \text { and } \frac{1}{y}=b\)

∴ Given equations, \(\frac{a}{2}+\frac{b}{3}=2\)

⇒ 3a + 2b = 12 …….(1)

and \(\frac{a}{3}+\frac{b}{2}=\frac{13}{6}\)

⇒ 2a+3b = 13 ……….(2)

Multiplying equation (1) by 2 and equation (2) by 3 and subtracting,

⇒ b = 3

Put the value of b in equation (1),

⇒ 3a + 2 × 3 = 12

⇒ 3a = 12-6 = 6

⇒ a = 2

Now, \(\frac{1}{x}=a \quad \Rightarrow \quad x=\frac{1}{a}=\frac{1}{2}\)

and \(\frac{1}{y}=b \quad \Rightarrow \quad y=\frac{1}{b}=\frac{1}{3}\)

∴ \(x=\frac{1}{2}, y=\frac{1}{3}\)

2. Given equations, \(\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2\)

⇒ \(\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1\)

Let \(\frac{1}{\sqrt{x}}=A \text { and } \frac{1}{\sqrt{y}}=B\)

∴ Given equations, 2A + 3B =2 …(1)

4A-9B = -1 ……(2)

Multiplying equation (1) by 3 and adding in equation (2),

⇒ \(A=\frac{1}{2}\)

Put the value of A in equation (1),

⇒ \(2 \times \frac{1}{2}+3 B=2\)

35 = 2-1 = 1 = B = \(\frac{1}{3}\)

Now, \(\frac{1}{\sqrt{x}}=A \quad \Rightarrow \sqrt{x}=\frac{1}{A}\)

⇒ \(x=\frac{1}{A^2}=\frac{1}{\left(\frac{1}{2}\right)^2}=4\)

and \(\frac{1}{\sqrt{y}}=B \Rightarrow \sqrt{y}=\frac{1}{B}\)

⇒ \(y=\frac{1}{B^2}=\frac{1}{\left(\frac{1}{3}\right)^2}=9\)

∴ x = 4, y = 9

3. Given equations, \(\frac{4}{x}+3 y=14, \frac{3}{x}-4 y=23\)

Let \(\frac{1}{x}=A\)

∴ Given equations, 4A + 3y = 14 ………(1)

3A – 4y = 23 ………(2)

Multiplying equation (1) by 4 and equation (2) by 3 and adding

⇒ A = 5

Put the value of A in equation (1),

4 × 5 + 3y = 14

⇒ 3y = 14 – 20 = -6

⇒ y = -2

Now, \(\frac{1}{x}=A \Rightarrow x=\frac{1}{A}=\frac{1}{5}\)

⇒ \(x=\frac{1}{5}, y=-2\)

4. Given equations, \(\frac{5}{x-1}+\frac{1}{y-2}=2\)

and \(\frac{6}{x-1}-\frac{3}{y-2}=1\)

Let \(\frac{1}{x-1}=A \text { and } \frac{1}{y-2}=B\)

∴ Given equations, 5A + B = 2 …..(1)

6A-3B = 1 ……(2)

Multiplying equation (1) by 3 and adding in equation (2),

⇒ \(A=\frac{7}{21}=\frac{1}{3}\)

Put the value of A in equation (1),

⇒ \(5 \times \frac{1}{3}+B=2 \quad \Rightarrow B=2-\frac{5}{3}=\frac{1}{3}\)

Now, \(\frac{1}{x-1}=A\)

⇒ \(x-1=\frac{1}{A}=\frac{1}{1 / 3}=3\)

⇒ x = 4

and \(\frac{1}{y-2}=B\)

⇒ \(y-2=\frac{1}{B}=\frac{1}{1 / 3}=3\)

⇒ y = 5

∴ x = 4, y = 5

5. Given equations, \(\frac{7 x-2 y}{x y}=5\)

⇒ \(\frac{7 x}{x y}-\frac{2 y}{x y}=5\)

⇒ \(\frac{7}{y}-\frac{2}{x}=5\) …..(1)

and \(\frac{8 x+7 y}{x y}=15 \Rightarrow \frac{8 x}{x y}+\frac{7 y}{x y}=15\)

⇒ \(\frac{8}{y}+\frac{7}{x}=15\) ……(2)

Let \(\frac{1}{y}=A \quad \text { and } \quad \frac{1}{x}=B\)

∴ From equation (1), 7A-2B =5 …(3

From equation (2), 8A + 7B = 15 …(4)

Multiplying equation (3) by 7 and equation (4) by 2 and adding,

⇒ A = 1

Put the value of A in equation (4),

8 × 1 + 7B = 15

⇒ 7B = 15 – 8 = 7

⇒ B = 1

Now, \(\frac{1}{x}=B \Rightarrow x=\frac{1}{B}=\frac{1}{1}=1\)

and \(\frac{1}{y}=A \Rightarrow y=\frac{1}{A}=\frac{1}{1}=1\)

∴ x = 1, y =1

6. Given equations, 6x+3y = 6xy

⇒ \(\frac{6 x}{3 x y}+\frac{3 y}{3 x y}=\frac{6 x y}{3 x y} \Rightarrow \frac{2}{y}+\frac{1}{x}=2\) …….(1)

and 2x + 4y = 5xy

⇒ \(\frac{2 x}{x y}+\frac{4 y}{x y}=\frac{5 x y}{x y} \Rightarrow \frac{2}{y}+\frac{4}{x}=5\) ……..(2)

Let \(\frac{1}{y}=A \quad \text { and } \quad \frac{1}{x}=B\)

∴ From equation (1), 2A + B = 2 …..(3)

From equation (2), 2A + 4B = 5 …..(4)

Subtracting equation (3) from equation (4),

B=1

Put the value of B in equation’ (3),

2A + 1 = 2 ⇒ 2A = 2-1 = 1

⇒ \(A=\frac{1}{2}\)

Now, \(\frac{1}{y}=A\)

⇒ \(y=\frac{1}{A}=\frac{1}{1 / 2}=2\)

⇒ \(\frac{1}{x}=B \quad \Rightarrow x=\frac{1}{B}=\frac{1}{1}=1\)

∴ x = 1, y = 2

7. Given equations,

⇒ \(\frac{10}{x+y}+\frac{2}{x-y}=4\) …..(1)

⇒ \(\frac{15}{x+y}-\frac{5}{x-y}=-2\) ….(2)

Let \(\frac{1}{x+y}=A \text { and } \frac{1}{x-y}=B\)

∴ From equation (1), 10A + 25 = 4

From equation (2), 15A – 55 = -2

Multiplying equations (3) by 5 and equation (4) by 2 and adding,

⇒ \(A=\frac{16}{80}=\frac{1}{5}\)

Put the value of A in equation (3),

⇒ \(10 \times \frac{1}{5}+2 B=4 \Rightarrow 2+2 B=4\)

2B = 4-2 = 2 ⇒ B = 1

Now \(\frac{1}{x+y}=A=\frac{1}{5}\)

⇒ x+y = 5 …..(5)

and \(\frac{1}{x-y}=B=1\)

⇒ x-y = 1 …..(6)

Adding equation (5) and equation (6)

⇒ x= 3

Put the value of x in equation (5)

3 + = 5 ⇒ y = 2

∴ x = 3, y = 2

8. Given equations, \(\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}\) …….(1)

and \(\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=-\frac{1}{8}\) ….(2)

Let \(\frac{1}{3 x+y}=A \text { and } \frac{1}{3 x-y}=B\)

From equation (1)

⇒ \(\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}\) ……(1)

and \(\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=-\frac{1}{8}\) …..(2)

Let \(\frac{1}{3 x+y}=A \text { and } \frac{1}{3 x-y}=B\)

From equation (1), \(A+B=\frac{3}{4}\) ……(3)

From equation (2), \(\frac{A}{2}-\frac{B}{2}=-\frac{1}{8}\)

⇒ \(A-B=-\frac{2}{8} \Rightarrow A-B=-\frac{1}{4}\) ……(4)

Adding equations (3) and (4)

A = \(\frac{1}{4}\)

Put the value of A in equation (3),

⇒ \(\frac{1}{4}+B=\frac{3}{4}\)

⇒ \(B=\frac{3}{4}-\frac{1}{4}=\frac{2}{4}=\frac{1}{2}\)

Now \(\frac{1}{3 x+y}=A=\frac{1}{4}\)

⇒ 3x-y= 4 ….(5)

and \(\frac{1}{3 x-y}=B=\frac{1}{2}\)

⇒ 3x-y= 2 ….(6)

Adding equations (5) and (6),

⇒ x= 1

Put the value of x in equation (5),

⇒ 3 × 1 +y = 4

⇒ y = 4-3=1

∴ x = 1, y = 1

Question 2. Formulate the following problems as a pair of equations, and hence find their solutions:

1. Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
2. 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
3. Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Solution:

1. Let the speed of Ritu in still water = x km/hr

and speed of current =y km/hr

∴ speed in downstream = (x +y) km/hr

and speed in upstream = (x -y) km/hr

Now from \(\text { time }=\frac{\text { distance }}{\text { speed }}\)

⇒ \(\frac{20}{x+y}=2 \text { and } \frac{4}{x-y}=2\)

⇒ x + y = 10 …..(1)

and x-y = 2 ….(2)

Adding equations (1) and (2),

2x = 12 ⇒ x= 6

Put x = 6 in equation (1),

6 +y = 10 ⇒ y = 4

∴ Speed of Ritu in still water = 6 km/hr

and speed of current = 4 km/hr

2. Let 1 woman can finish the work in x days and 1 man can finish the work in y days.

∴ 1 day work of 1 woman = \(\frac{1}{x}\)

1 day work of 1 man = \(\frac{1}{y}\)

1-day work of 2 women and 5 men \(\frac{2}{x}+\frac{5}{y}=\frac{5 x+2 y}{x y}\)

∴ Time taken to finish the work = \(\frac{x y}{5 x+2 y}\)

Given that, \(\frac{x y}{5 x+2 y}=4\)

Similarly in second case, \(\frac{x y}{6 x+3 y}=3\)

Then, \(\frac{5 x+2 y}{x y}=\frac{1}{4} \text { and } \frac{6 x+3 y}{x y}=\frac{1}{3}\)

⇒ \(\frac{20}{y}+\frac{8}{x}=1 \text { and } \frac{18}{y}+\frac{9}{x}=1\)

Let \(\frac{1}{x}=u \text { and } \frac{1}{y}=v\)

∴ 20v + 8u = 1 …..(1)

and 18v+9u = 1 ….(2)

Multiplying equation (1) by 9 and equation (2) by 8,

180v-144v = 9-8

⇒ 36v = 1

⇒ v = \(\frac{1}{36}\)

Put v = \(\frac{1}{36}\) in equation (2),

⇒ \(18 \times \frac{1}{36}+9 u=1\)

⇒ 9u = \(1-\frac{1}{2}\)

⇒ 9u = \(\frac{1}{2}\)

⇒ u = \(\frac{1}{18}\)

Now, \(u=\frac{1}{18} \quad \text { and } \quad v=\frac{1}{36}\)

⇒ \(\frac{1}{x}=\frac{1}{18} \text { and } \frac{1}{y}=\frac{1}{36}\)

⇒ x= 18 and y = 36

Therefore, 1 woman can finish the work alone in 18 days and 1 man can finish the work alone in 36 days.

3. Let the speed of train = x km/hr and speed of bus =y km/hr

In the first case, Ruhi goes 60 km from the train and 240 km from a bus in 4 hours.

⇒ \(\frac{60}{x}+\frac{240}{y}=4\)

⇒ \(\frac{15}{x}+\frac{60}{y}=1\) …..(1)

Similarly in the second case, she goes 100 km from the train and 200 km from a bus in \(4 \frac{1}{6}\) hours.

⇒ \(\frac{100}{x}+\frac{200}{y}=4+\frac{1}{6}\) ,\(\left(10 \text { minutes }=\frac{1}{6} \mathrm{hr}\right)\)

⇒ \(\frac{100}{x}+\frac{200}{y}=\frac{25}{6}\)

⇒ \(\frac{24}{x}+\frac{48}{y}=1\) ….(2)

Let \(\frac{1}{x}=u \text { and } \frac{1}{y}=v\)

⇒ 15u + 60v = 1 ……(3)

and 24u + 48v = 1 …..(4)

Multiplying equation (3) by 24 and equation (4) by 15 and subtracting,

24(15u + 60v)- 15(24u + 48v) =24-15

⇒ 440v – 720v = 9

⇒ 20v = 9

⇒ v = \(\frac{9}{720}=\frac{1}{80}\)

Put v = \(\frac{1}{80}\)

⇒ \(15 u+60 \times \frac{1}{80}=1\)

⇒ \(15 u=1-\frac{3}{4}=\frac{1}{4}\)

⇒ u = \(\frac{1}{60}\)

Now, u = \(\frac{1}{60}\) and v = \(\frac{1}{80}\)

⇒ \(\frac{1}{x}=\frac{1}{60} \text { and } \frac{1}{y}=\frac{1}{80}\)

⇒ x = 60 and y = 80

Therefore, the speed of the train = 60 km/hr and speed of bus = 80 km/hr.

Linear Equations In Two Variables Exercise 3.7 (Optional)

Question 1. The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
Solution:

Given

The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years.

For convenience, let their ages be according to their names.

Now, A – B = ± 3 ……(1) (As we don’t know who is older)

and \(\left.\begin{array}{rl}
D=2 A \\
B=2 C \Rightarrow C=\frac{B}{2}
\end{array}\right\}\)

and D – C = 30

But we don’t know who is older, so D-C= ±30

⇒ \(2 A-\frac{B}{2}= \pm 30 \Rightarrow 4 A-B= \pm 60\) ……..(2)

Now from equations (1) and (2), four cases are possible:

Question 2. One says, “Give me a hundred, friend! f shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara 2]

[Hint : x + 100 = 2(y- 100),y + 10 = 6(x- 10)].

Solution:

Let them have ₹x and ₹y respectively.

∴ According to the given condition,

⇒ x+ 100 = 2 (y- 100)

⇒ x – 2y = -300 …..(1)

and 6(x- 10) = + 10 ⇒ 6x-y = 70 …(2)

Multiplying equation (2) by 2 and subtracting from equation (1),

x- 12x = -300 – 140 ⇒ -11x = -440

x = ₹40

Put in c(|untion (1),

40 – 2y = -300

2y – 340 ⇒ y = ₹170

So, they have ₹40 and ₹170 respectively.

Question 8. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Solution:

Let speed of train = x km/hr and distance covered = y km

Time taken at this speed = \(\frac{y}{x} \text { hours }\)

If speed = (x + 10) km/hr

then time taken = \(\frac{y}{x+10} \text { hours }\)

∴ \(\frac{y}{x+10}=\frac{y}{x}-2 \Rightarrow \frac{y}{x+10}=\frac{y-2 x}{x}\)

⇒ xy = (x+10)(y-2x)

⇒ xy = xy-2x²+10y-20x

⇒ -2x²+ 10y-20x = 0 …(1)

If speed = (x- 10) km/hr

then time taken = \(\frac{y}{x-10} \mathrm{hr}\)

∴\(\frac{y}{x-10}=\frac{y}{x}+3\)

⇒ \(\frac{y}{x-10}=\frac{y+3 x}{x}\)

⇒ xy = (x-10)(y+3x)

⇒ xy = xy + 3x²-10y-30x

⇒ 3x²– 10y-30x = 0 ……(2)

Adding equations (1) and (2),

x² – 50x = 0 ⇒ x(x-50) = 0

⇒ x = 0 or x-50 = 0

⇒ x = 0 or x = 50

x = 0 is not possible

∴ x = 50 km/hr

Put the value of x in equation (1),

-2(50)² + 10y- 20×50 = 0

⇒ -5000+ 10y- 1000 = 0

⇒ 10y = 6000

⇒ y = 6000km

∴ Distance covered = 600 km

Question 4. ‘The students of a class are made to stand in rows. If 3 students arc extra in a row, there would be 1 row. If 3 students are less in a and there would be 2 rows more. Find the number of students in the class.
Solution:

Given

‘The students of a class are made to stand in rows. If 3 students arc extra in a row, there would be 1 row. If 3 students are less in a and there would be 2 rows more.

Let the number of rows = x

and number of students in each row = y

Total number of students in the class = xy

According to the first condition,

(y + 3) (x- 1) = xy

⇒ xy-y + 3xr3 = xy

⇒ 3x-y = 3 …(1)

According to the second condition,

(y-3)(x + 2)=xy

⇒ xy + 2y-3x- 6 = xy

⇒ 3x-2y = -6 ……(2)

Subtracting equation (2) from equation (1),

y = 9

Put y = 9 in equation (1),

3x- 9 =3

⇒ 3x= 12

⇒ x = 4

∴ Total number of students in the class = xy = 4 x 9 = 36

Question 5. In ΔABC, ∠C = 3∠B = 2(∠A+∠B). Find the three angles.
Solution:

Given

In ΔABC, ∠C = 3∠B = 2(∠A+∠B).

Let ∠A=x° and ∠B=y°

∴ ∠C = 3∠B ⇒ ∠C = 3y°

and ∠C = 2(∠A + ∠B)

⇒ 3y° = 2(x° +y°)

⇒ 2x + 2y-3y =0

⇒ 2x-y = 0 ……(1)

∵ ∠A + ∠B + ∠C = 180° (sum of three angles of Δ)

⇒ x+y + 3y = 180

⇒ x + 4y = 180 …..(2)

Multiplying equation (2) by 2

2x+ 8y = 360 …….(3)

Subtracting equation (3) from (1),

-9y = -360 ⇒ y = 40°

Put y = 40° in equation (1),

2x- 40 = 0

⇒ 2x = 40 ⇒ x = 20°

So, ∠A = 20°, ∠B = 40°

and ∠C = 3 × 40° = 120°

Question 6. Draw the graphs of the equations 5x-y = 5 and 3x – y = 3. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis.
Solution:

5x-y = 5 y ⇒ 5x-5

Mark the points A(0, -5), B( 1, 0) and C(2, 5) and join them.

Again, 3x -y = 3 ⇒ y= 3x-3

Mark the points D(0, -3), B( 1, 0) and E(2, 3) and join them.

The coordinates of the vertices of ΔBDA formed by two lines and the y-axis are as follows :

B( 1, 0), A(0, -5) and D(0, -3)

Now area of ΔBAD = \(\frac{1}{2} \times A D \times B O\)

⇒ \(\frac{1}{2} \times 2 \times 1\)

= 1 sq. units

Question 7. Solve the following pair of linear equations :

1. px + qy -p – q , qx-py = p + q
2. ax + by = c, bx + ay = 1 + c
3. \(\), ax + by = a2 +b2
4. (a – b)x + (a + b)y = a2 – 2ab – b2, (a + b)(x +y) =a2 +b2
5. 152x – 378y = -74, -378x + 152y = -604

Solution:

1. Given equation,

px + qy = p-q ….(1)

and qx – py = p+q

Multiplying equation (1) by p and equation (2) by q and adding,

⇒ \(\left(p^2+q^2\right) x=p^2+q^2\)

⇒ x = 1

Put the value of x in equation (1),

p×1+qy = p-q

⇒ qy = -q

⇒ y = -1

∴ x = 1, y = -1

2. ax + by = c

⇒ ax + by -c = 0

⇒ bx + ay = 1 +c ⇒ bx + ay- (1+ c) = 0

Now,

∴ \(\frac{x}{-b(1+c)+a c}=\frac{y}{-b c+a(1+c)}=\frac{1}{a^2-b^2}\)

⇒ \(\frac{x}{a c-b-b c^y}=\frac{1}{a^2-b^2}\)

⇒ \(\frac{y}{-b c+a+a c}=\frac{1}{a^2-b^2}\)

⇒ \(x=\frac{a c-b-b c}{a^2-b^2}\)

⇒ \(y=\frac{a+a c-b c}{a^2-b^2}\)

3. \(\frac{x}{a}-\frac{y}{b}=0 \quad \Rightarrow b x-a y+0=0\)

⇒ ax+by = a²+b²

⇒ ax+by-(a²+b²) = 0

Now

∴ \(\frac{x}{a\left(a^2+b^2\right)-0}=\frac{y}{0+b\left(a^2+b^2\right)}=\frac{1}{b^2+a^2}\)

⇒ \(\frac{x}{a\left(a^2+b^2\right)}=\frac{y}{b\left(a^2+b^2\right)}=\frac{1}{a^2+b^2}\)

⇒ \(\frac{x}{a}=\frac{y}{b}=1\)

⇒ \(\frac{x}{a}=1 \text { and } \frac{y}{b}=1\)

⇒ x = a and y = b

4. (a-b)x+(a+b)y = a²-2ab-b²

⇒ (a-b)x+(a+b)y – (a² – 2ab -b²) = 0

(a + b) (x+y) = a² + b²

⇒ (a-b)x+(a+b)y – (a²-b²) = 0

Now x

⇒ \(\frac{x}{-(a+b)\left(a^2+b^2\right)+(a+b)\left(a^2-2 a b-b^2\right)}\)

⇒ \(\frac{y}{-(a+b)\left(a^2-2 a b-b^2\right)+(a-b)\left(a^2+b^2\right)}\)

⇒ \(\frac{1}{(a-b)(a+b)-(a+b)^2}\)

⇒ \(\frac{x}{(a+b)\left(-a^2-b^2+a^2-2 a b-b^2\right)}\)

⇒ \(\begin{gathered}\frac{y}{-\left(a^3-a^2 b-3 a b^2-b^3\right)} \\+\left(a^3-a^2 b+a b^2-b^3\right)\end{gathered}\)

⇒ \(\frac{1}{(a+b)(a-b-a-b)}\)

\(\frac{x}{(a+b)\left(-2 a b-2 b^2\right)}=\frac{y}{4 a b^2}=\frac{1}{(a+b)(-2 b)}\)

⇒ \(\frac{x}{-2 b(a+b)^2}=\frac{1}{-2 b(a+b)}\)

and \(\frac{y}{4 a b^2}=\frac{1}{-2 b(a+b)}\)

⇒ x = a + b and y = \(\frac{-2 a b}{a+b}\)

5. 152x-378y = -74 ….(1)

-378A + 152y = -604 …..(2)

Adding equations (1) and (2),

-226A- 226y = -678 ⇒ x+y= 3 ….(3)

Subtracting equation (2) from equation (1),

530x – 530y = 530 ⇒ x-y = 1 …(4)

Adding equation (3) and (4),

⇒ 2x = 4

⇒ x = 2

Put x = 2 in equation (3),

⇒ 2+y =3

⇒ y = 1

∴ x = 2, y = 1

Question 8. ABCD is a cyclic quadrilateral. Find the angles of the cyclic quadrilateral.
Solution:

Given

ABCD is a cyclic quadrilateral.

We know that the sum of opposite angles of a cyclic quadrilateral is 180°.

∴ ∠A + ∠C = 180°

⇒ 4y + 20+ (-4A) = 180°

⇒ -4x + 4y = 1 80°- 20° = 160°

⇒ -x+y = 40°…(1)

and ∠B + ∠D = 180°

⇒ (3y-5)+(-7x + 5) = 180°

⇒ -7x + 3y = 1 80° …..(2)

Multiplying equation (1) by 3 and subtracting from equation (2),

⇒ x = -15°

Put the value of A in equation (1),

⇒ -(-15°) +y = 40°

⇒ y = 40°- 15° = 25°

Now, ∠A = 4y + 20° = 4 × 25° + 20°

= 120°

∠B = 3y- 5

= 3 × 25°- 5° = 70°

∠C = -4A = -4(-15°)

= 60°

∠D = -7A + 5° =-7(-15°) + 5°

= 110°

Linear Equations In Two Variables Multiple Choice Questions

Question 1. The pair of equations 5x- 15y = 16 and 3x – 9y = \(\frac{48}{5}\) has/have :

1. One solution
2. No solution
3. Infinitely many solutions
4. None of these

Answer: 3. Infinitely many solutions

Question 2. If any pair of linear equations is consistent then the lines of their graphs will be :

1. Coincident or intersecting
2. Parallel
3. Always coincident
4. Always intersecting

Answer: 1. Coincident or intersecting

Question 3. The pair of equations x = 2,y = 3 represents those lines which are graphically:

1. Parallel
2. Coincident
3. Intersecting at (2, 3)
4. Intersecting at (3, 2)

Answer: 2. Coincident

Question 4. The sum of digits of a number of two digits is 9. If 9 is added to it, the digits of this number are reversed. The number is :

1. 27
2. 36
3. 45
4. 54

Answer: 3. 45

Question 5. If x + 3y = 8 and k x + 6y-16 = 0isa pair of coincident lines then the value of k is:

1. 1
2. 2
3. 3
4. 4

Answer: 2. 2

Question 6. If 3x + ky = 6 and 2x + 5y = 0 are parallel lines then the value of k is:

1. 7.5
2. -7.5
3. 2.5
4. -2.5

Answer: 1. 7.5

Question 7. Shyam has 25 coins of ₹5 and ₹10. If the total amount is ₹175, then the number of coins is:

1. 5
2. 10
3. 15
4. 20

Answer: 3. 15

Question 8. The solution of the equations x-y = 2 and x+y = 2 will be:

1. x = 0, y = 2
2. x = 2, y = 0
3. x = 4, y = 2
4. x = -2,y- 0

Answer: 2. x = 2, y = 0

NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Introduction

In daily life, there are so many examples which follow a certain pattern.

1. In a fixed deposit scheme, the amount becomes. 1 item of itself after every year. The maturity amount of ₹10,000 after 1, 2, 3 and 4 years will be ₹11,000, ₹12,100, ₹13,310,14.641
2. When a person is offered a job with a monthly salary of ₹18,000 and with an annual increment of ₹400, His salary for the succeeding years will be ₹18,000, ₹18,400, ₹ 18,800 per mon.
   • In the first example, the succeeding terms are obtained by multiplying with 1.1. In the second example, the succeeding terms are obtained by adding 400.
   • Now, we will discuss all those patterns in which the succeeding terms are obtained by adding a fixed number to the preceding terms.

NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Sequence

A number of things that come one after another form a sequence. It may be possible that we do not have a formula to find the nth term of the sequence, but still, we know about the next term.

Read and Learn More Class 10 Maths Solutions Exemplar

Sequence For example: 2, 3, 5, 7, 11,13, 17, 19, …is a sequence of prime numbers.

We all know about the next number of the sequence but we do not have any formula to calculate the next number.

• The numbers present in the sequence are called the terms of the sequence.
• The nth term of the sequence can be represented by T_n or a_n. It is called the general term.
• A sequence which has finite terms is called a finite sequence.

NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Progression

Those sequences whose terms follow certain patterns are called progressions. In progression, each term (except the first) is obtained from the previous one by some rule.

The difference between a progression and a sequence is that a progression has a specific formula to calculate its nth term, whereas a sequence can be based on a logical rule like a group of prime numbers, which does not have any formula associated with it.

The numbers 2, 4, 6, 8, 10, 12 … form a progression as its nth term T_n =2n while 2,3,5,7,11,13,17,19,23,29,31,… is a sequence of numbers, as there is no formula to find the next number. It is a group of prime numbers.

NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Series

If T₁, T₂, T₃,…, T_n form a progression, then T₁, T₂, T₃,…, T_n is called its corresponding series. In other words, if all the terms of a progression are connected by ‘+’ started

Series For example:1-2 + 3- 4+5-6 ……… is a series. The first term is 1, the second term is -2, the third term is 3, the fourth is -4 and so on. Its nth is n(-1) n+1

NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Series Solved Examples

Question 1. The nth of a sequence is an = 2n+1. Find its first four terms.
Solution:

Given

The nth of a sequence is an = 2n+1.

a_n=2n+1

put n= 1,2,3,4, we get

a₁ = 2×1+3 = 5

a₂ = 2×2+3 = 7

a₃ = 2×3+3 = 9

a₄ = 2×4+3 = 11

∴ The first four terms of the sequence are 5,7,9,11.

Question 2. The nth term of a sequence is an=n2+5. Find its first terms.
Solution:

Given

The nth term of a sequence is an=n²+5

a_n=n²+5

put n=1,2,3, we get

a₁ = 1² + 5 = 6

a₂ = 2² + 5 = 9

a₃ = 3² + 5 = 14

∴ The first four terms of the sequence are 6,9,14.

Question 3. Fibonacci sequence is defined as follows: a₁ = a₂ = 1 and a_n = a_n-2 + a_n-1 , where n > 2. Find third, fourth and fifth terms
Solution:

Given

Fibonacci sequence is defined as follows: a₁ = a₂ = 1 and a_n = a_n-2 + a_n-1 , where n > 2.

a₁ = a₁ = 1

⇒ \(a_n=a_{n-2}+a_{n-1}, n>2\)

put n = 3, we get

⇒ \(a_3=a_1+a_2=1+1=2\)

put n = 4, we get

⇒ \(a_4=a_2+a_3=1+2=3\)

put it = 5, we get

⇒ \(a_5=a_3+a_4=2+3=5\)

Question 4. A sequence is defined as follows: a₁ = 3, a_n = 2a_n-1 + 1, where n > 1. Find \(\frac{a_{n+1}}{a_n}\) for n=1,2,3.
Solution:

Given

A sequence is defined as follows: a₁ = 3, a_n = 2a_n-1 + 1, where n > 1.

a₁=3

⇒ \(a_n=2 a_{n-1}+1 \quad \text { where } n>1\)

put n= 2, we get

⇒ \(a_2=2 a_1+1=2 \times 3+1=7\)

put n = 3, we get

⇒ \(a_3=2 a_2+1=2 \times 7+1=15\)

put n = 4, we get

\(a_4=2 a_3+1=2 \times 15+1=31\)

Now, for n = 1

⇒ \(\frac{a_{n+1}}{a_n}=\frac{a_2}{a_1}=\frac{7}{3}\)

For n = 2,

⇒ \(\frac{a_{n+1}}{a_n}=\frac{a_3}{a_2}=\frac{15}{7}\)

For n = 3,

⇒ \(\frac{a_{n+1}}{a_n}=\frac{a_4}{a_3}=\frac{31}{15}\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression

An arithmetic progression (A.P.) is a list of numbers in which each term is obtained by adding a fixed number to the preceding term, except the first term.

or

If the difference of any two consecutive terms of a progression is the same (constant), it is called arithmetic progression.

For example: a, a + d, a + 2d, a + 3d, ….

NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression – Common Difference

The difference between two consecutive terms (i.c., any term — preceding term) of an arithmetic progression is called a common difference.

It is denoted by ‘d’

∴ d = a₂-a₁ = a₃-a₂= … = a_n-a_n-1.

NCERT Exemplar Solutions for Class 10 Maths Chapter 5 General Term Of Arithmetic Progression

Let the first term and common difference of an arithmetic progression be ‘a’ and ‘b’ respectively.

∴ Arithmetic progression is a, a +d, a + 2d, a + 3d,…..

Here, first term = a = a+(1-1)d

Second term = a+d = a+(2-1)d

Third term = a+2d = a+(3-1)d
.
.
.
.

nth term = a+(n-1)d

∴ a_n=a+(n-1)d

nth term of a progression is called its general term.

NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression nth Term From The End Of An Arithmetic Progression

Let the first term, common difference and last term of an arithmetic progression be a, d and / respectively.

Arithmetic progression is a, a + d, a + 2d,…….l – 2d, l-d,l

Here, the First term from the end -l = l – (1 -1)d

Second term from the end = l – d =l-(2-l)d

Third term from the end = l – 2d = l – (3 – 1 )d

.

.

.

nth term from the end =l-(n- 1 )d.

NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Solved Examples

Question 1. For the following A.P., write the first term and common difference: 3, 1,-1,-3, …
Solution:

3, 1,-1,-3,…

Here, first term a = 3

Common difference = 1 – 3 = -2

Question 2. Write the first four terms of the A.P., when the first term V and the common difference ‘d’ are given as follows:

a = 10, d = 5

Solution:

a = 10, d = 5

a₁ = 10

a₂ =10 + 5=15

a₃ =15 + 5 = 20

a₄ = 20 + 5 = 25.

∴ First 4 terms are 10, 15, 20, 25.

Question 3. Find the 18th term of the A.P. 4, 7, 10, …
Solution:

Given

A.P. 4, 7, 10

Here, a = 4,

d = 7 – 4 = 10-7 = 3,

n = 18

∴ a_n = a + (n – 1 )d

a₁₈ = 4 + (18 – 1) × 3 =4 + 51=55

∴ 18th term of the given A.P. = 55.

Question 4. What is the common difference of an A.P. in which a₂₁-a₇ = 84?
Solution:

Given

a₂₁-a₇ = 84

Let the first term of A.P. be ‘a’ and common difference be ‘d’.

Since, a₂₁-a₇ =84

∴ (a+20d) -(a-6d) = 84

⇒ 14d = 84

⇒ d= 6

Hence, the common difference is 6.

Question 5. Which term of the A.P. 3, 8, 13, 18,….is 88?
Solution:

Given

A.P. 3, 8, 13, 18,….is 88

Here, a = 3,

A = 8-3 = 13 -8 = 5

Let a_n = 88

⇒ 3 + (n – 1)5 = 88

⇒ 3 + 5K – 5 = 88

⇒ 5n – 2 = 88

⇒ 5n = 90

⇒ n= 18

The 18th term of the given A.P. is 88.

Question 6. Which term of the A.P. 90, 87, 84,…is zero?
Solution:

Given

90, 87, 84,…

Here, a = 90,

d = 87 – 90 = -3

Let a_n = 0

⇒ 90 + (n – 1) (-3) = 0

⇒ 90 – 3n + 3 = 0

⇒ -3n = -93

⇒ n= 31

∴ 31st term of the given A.P. is zero.

Question 7. If 2, a, b, c, d, e,f and 65 form an A.P., find the value of e.
Solution:

Given

2, a, b, c, d, e,f and 65 form an A.P.

Here, T₁ = 2, T₈ = 65, T₆ = e

Let the common difference be D

T₈ = 65 ⇒  T₁ + (8 – 1)D = 65

⇒ 2+7D = 65  ⇒ D = 9

∴ e = T₆ = T₁ + (6 – 1 )D = 2 + 5(9) = 47

Hence, the value of e is 47.

Question 8. Which term of the A.P. \(10,9 \frac{1}{3}, 8 \frac{2}{3}\) ….. is the first negative term?
Solution:

Here, a = 10,

⇒ \(d=9 \frac{1}{3}-10=8 \frac{2}{3}-9 \frac{1}{3}=-\frac{2}{3}\)

Let a_n<0

⇒ \(10+(n-1)\left(-\frac{2}{3}\right)<0\)

⇒ \(\frac{30-2 n+2}{3}<0\)

⇒ 32-2n<0

⇒  32 < 2n

⇒ 2n>32

⇒ n>16

⇒ n = 17, 18, 19, (all are negative terms)

∴ First negative term = 17th term

Now, a₁₇=a+(17-1) d

⇒ \(10+16\left(-\frac{2}{3}\right)=10-\frac{32}{3}=\frac{-2}{3}\)

Thus, 17th term which is \(-\frac{2}{3}\) is the first negative term.

Question 9. The 18th term of an A.P. exceeds its 10th term by 8. Find the common difference.
Solution:

The 18th term of an A.P. exceeds its 10th term by 8.

Let the first term and the common difference of the A.P. be ‘a’ and ‘d’ respectively.

Given that,

⇒ a₁₈ = a₁₀ + 8

⇒ a₁₈-a₁₀= 8

⇒ {a +(18-1)d}-{a + (10- 1)d} = 8

⇒ [a + 17d)-{a + 9d) = 8

⇒ a + 17d-a-9d = 8

⇒ 8d = 8

⇒ d = 1

∴ Common difference = 1.

Example 10. The 26th, 11th and last term of an A.P. are 0, 3 and \(-\frac{1}{5}\) respectively. Find the common difference and the number of terms.
Solution:

The 26th, 11th and last term of an A.P. are 0, 3 and \(-\frac{1}{5}\) respectively.

Let the first term, common difference and number of terms in the A.P. be a, d and n respectively.

Given that

⇒ a₂₆ = 0

⇒ a+25d = 0 ……(1)

⇒ a = -25d

⇒ a₁₁ = 3 a+10d = 3

⇒ -25d+10d = 3 [from (1)]

⇒ -15d = 3

⇒ d = \(-\frac{1}{5}\)

put d = \(-\frac{1}{5}\) in equation (1) we get

⇒ \(a=-25\left(-\frac{1}{5}\right)=5\) and \(a_n=-\frac{1}{5}\)

⇒ \(5+(n-1)\left(-\frac{1}{5}\right)=-\frac{1}{5}\)

⇒ \((n-1)\left(-\frac{1}{5}\right)=-\frac{1}{5}-5=-\frac{26}{5}\)

⇒ n- 1=26

⇒ n= 27

No. of terms in the A.P. = 27.

Question 11. If the 9th term of an A.P. is zero, prove that its 29th term is twice its 19th term.
Solution:

The 9th term of an A.P. is zero

Let the first term and the common difference of the A.P. be ‘a’ and ‘d’ respectively.

Given that,

a₉ = 0

a + 8d = 0

a = -8d

Now, a₂₉ = a+28d = -8d+28d

= 20d

T₁₉ = a+18d = -8d + 18d

= 10d

⇒ 2T₁₉ = 20d

∴ T₂₉ = 2T₁₉

Question 12. The nth term of a sequence is 3n + 5. Show that it is an A.P.
Solution:

Given

The nth term of a sequence is 3n + 5.

Here, a_n = 3n+5

a_n-1 = 3(n-1) +5

3n-3+5 = 3n+2

Now, a_n-a_n-1 = (3n+5)-(3n+2) = 3

which does not depend on n i.e., it is constant.

∴ Given sequence is in A.P.

Question 13. For what value of k will k + 9, 2k- 1 and 2k +7 are the consecutive terms of an A.P?
Solution:

Since k + 9, 2k- 1 and 2k + 7are in A.P.

So, there must be a common difference.

i.e., (2k- 1) – (k + 9) = (2k + 7) – (2k- 1)

⇒ k- 10 = 8

⇒ k= 18

The value of k= 18.

Question 14. Find how many integers between 200 and 500 are divisible by 8.
Solution:

Integers between 200 and 500 which are divisible by 8 are as follows :

∴ 208, 216, 224, 232, …, 496

This forms an A.P. whose first term = 208, common difference = 8 and last term = 496. Let there be n terms.

∴ a_n = 496

⇒ a+ (n -1)<d = 496

⇒ 208 + (n- 1)8 = 496

⇒ (n- 1)8 = 496-208 = 288

⇒ \(n-1=\frac{288}{8}\)

∴ n-1 = 36

∴ n = 37

Hence, 37 integers are between 200 to 500 which are divisible by 8.

Question 15. If m times the with term of an A.P. is equal to n times the mth term and m ≠ n, show that its (m + n)th term is zero.
Solution:

Let the first term and the common difference of the A.P. be ‘a and ‘d’ respectively.

Given that

m.a_m = n.a_n

⇒ \( m\{a+(m-1) d\}=n\{a+(n-1) d\}\)

⇒ \(a m+\left(m^2-m\right) d=a n+\left(n^2-n\right) d\)

⇒ \(a(m-n)+\left\{\left(m^2-n^2\right)-m+n\right\} d=0\)

⇒ \(a(m-n)+\{(m-n)(m+n)-1(m-n\} d=0\)

⇒ \(a(m-n)+(m-n)(m+n-1) d=0\)

⇒ \((m-n)\{a+(m+n-1) d\}=0\)

⇒ \(a+(m+n-1) d=0\) (…m=n)

⇒ \(a_{m+n}=0\)

∴ (m + n)th term of the given A.P. is zero.

Question 16. If the term of an A.P. is \(\frac{1}{n}\) and its nth term be \(\frac{1}{m}\) then shows that its (nm)th term is 1.
Solution:

Let the first term and the common difference of the A.P. be ‘a’ and ‘d’ respectively

∴ \(a_m=\frac{1}{n} \quad \Rightarrow \quad a+(m-1) d=\frac{1}{n}\) ….(1)

and \(a_n=\frac{1}{m} \quad \Rightarrow \quad a+(n-1) d=\frac{1}{m}\)

Subtract equation (2) from equation (1), we get

⇒ \((m-n) d=\frac{m-n}{n m}\)

⇒ \(d=\frac{1}{m n}\)

put \(d=\frac{1}{m n}\)

⇒ \( a+(m-1) \cdot \frac{1}{m n}=\frac{1}{n}\)

⇒ \(a+\frac{1}{n}-\frac{1}{m n}=\frac{1}{n}\)

⇒ \(a=\frac{1}{m n}\)

⇒ \(a_{m n}=a+(m n-1) d\)

⇒ \(\frac{1}{m n}+(m n-1) \cdot \frac{1}{m n}=\frac{1}{m n}+1-\frac{1}{m n}=1\)

∴ (mn)th term of the A.P. = 1

Question 17. Find the 7th term from the end of the A.P. : 3, 8, 13, 18,….98
Solution:

Here, l = 98,

d = 8-3 = 13-8 = 5,

n = 7

∴ 7th term from the end =l- (7 – 1)d

= 98 – 6 x 5

= 98 – 30

= 68

Alternative Method :

Write the A.P. in reverse order 98, …18, 13,8,3

Here, a = 98

d = 3-8 = 8- 13 = -5

a₇ = a + (7 – 1)d

= 98 + 6 x (-5) = 98- 30 = 68

which is the 7th term from the end of the given A.P.

NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Sum Of n Terms Of An A.P.

Let the first term and common difference of an A.P. be ‘a’ and ‘d’          respectively. Let the A.P. contain ‘n’ terms and the last term be ‘l’.

∴ l = a + (n-1)d ….(1)

Now. a sum of n terms

S_n = a + (a + d) + + (1-d)+l …..(2)

In reverse order

S_n = l + (l + d) + + (a-d)+a …..(3)

Adding equations (2) and (3), we get

2S_n = (a + l) + (a + l) + …… + (a+l) + (a +l) (no.of terms = n)

2S_n = n(a+l)

⇒ \(S_n=\frac{n}{2}(a+l)\)

⇒ \(S_n=\frac{n}{2}[a+a+(n-1) d]\)

From eq.(1)

⇒ \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Sum Of n Terms Solved Examples

Question 1. Find the sum of n terms of the series

⇒ \(\left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)+\ldots\)

Solution:

We are given the series

⇒ \(S=\left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)+\ldots \text { to } n \text { terms }\)

⇒ \((4+4+4+\ldots \text { to } n \text { terms })-\left(\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+\ldots \text { to } n \text { terms }\right)\)

⇒ \(4 n-\frac{1}{n}(1+2+3+\ldots \text { to } n \text { terms })\)

⇒ \(4 n-\frac{1}{n}\left[\frac{n}{2}\{2 \times 1+(n-1) \times 1\}\right] \quad( a=1, d=1)\)

⇒ \(4 n-\frac{1}{n}\left[\frac{n}{2}(n+1)\right]=4 n-\frac{n+1}{2}=\frac{8 n-n-1}{2}=\frac{1}{2}(7n-1)\)

Hence, the required sum of n terms is \(\frac{1}{2}(7 n-1) \text {.}\)

Question 2. Find the sum of the following series : 5 + (-41) + 9 +(-39) + 13 + (-37)+ 1 7 + … + (-5) + 81 + (-3)
Solution:

Let 5 = 5 + (-41) + 9 + (-39)+ 13 + (-37) + 17 + … + (-5) + 81 + (-3)

= (5 + 9+ 13 + 17 + … + 81) – (41 + 39 + 37 + … + 3)

= S₁-S₂

where, ,S₁ = 5 + 9+ I3 + 17 + … + 81

It is an A.P. with a₁ = 5,d = 9- 5 = 4

Let there be n terms.

∴ a_n = 81

⇒ a₁ + (n- 1)d = 81

⇒ 5 + (n – 1)4 = 81

⇒ (n-1)4 = 76

⇒ n-1= 19

⇒ n = 20

∴ S₁ = Sum of 20 terms with first term 5 and last term 81

⇒ \(\frac{20}{2}[5+81]=10 \times 86\)

=860

and S₂ = 41 + 39 + 37 + … + 3

It is also an A.P. with a1 = 41, d = 39 – 41 = 2

Let there be n terms.

∴ a_n = 3

⇒ a₁ + (n- 1)d = 3

⇒ 41 + (n – 1)(-2) = 3

⇒ (n- 1)(-2)=-38

⇒ n- 1 = 19

⇒ n=20

∴ S₂ = Sum of 20 terms with first term 41 and last term 3

⇒ \(\frac{20}{2}[41+3]\)

⇒ 10×44 = 440

From equation (1),

S = S₁ – S₂

⇒ S = 860- 440 = 420

Question 3. The first term of an A.P. is 5, the last term is 45 and the sum of all its terms is 150.

Find the number of terms and the common difference of the A.P.

Solution:

Given

The first term of an A.P. is 5, the last term is 45 and the sum of all its terms is 150.

Let a number of terms = n, the first term is a = 5 and the last term is l = 45.

We have, S_n = 150

⇒ \(\frac{n}{2}[a+l]=150\)

⇒ \(\frac{n}{2}(5+45)=150\)

n = 6

Now, l = 45 i.e., nth term = 45

⇒ a + (n-1)d = 45

⇒ 5 + (6- 1)d = 45

⇒ 5d = 40

⇒ \(d=\frac{40}{5}\)

= 8

Hence, number of terms = 6 and common difference = 8

Question 4. Solve for x: 5+ 13 + 21 +……. +x = 2139
Solution:

Given

5+ 13 + 21 +……. +x = 2139

Here, a = 5,d= 13-5 = 8

Let T_n=x

∴ S_n = 2139

∴ \(\frac{n}{2}[2 \times 5+(n-1) 8]=2139\)

⇒ n(5 + 4n – 4)= 2139

⇒ 4n² + n- 2139 = 0

⇒ 4n² – 92n + 93n -2139 = 0

⇒ (n – 23) (4n + 93) = 0

∴ n = 23 or \(n=-\frac{93}{4}\)

since n > 0, we have n = 23

x = T_n = a + (n- 1 )d = 5 + (23 – 1 )(8)

x = 181

Question 5. How many terms of an A.P. -6, \(\frac{-11}{2}\),-5 are needed to give the sum -25? Explain the double answer.
Solution:

Here, a= -6, \(d=\frac{-11}{2}-(-6)=\frac{1}{2}\)

Let -25 be the sum of it terms of this A.P. (n ∈ N)

Using \(S_n=\frac{n}{2}[2 n+(n-1) d]\)

⇒ \(-25=\frac{n}{2}\left[2(-6)+(n-1)\left(\frac{1}{2}\right)\right]\)

⇒ \(-50=n\left(-12+\frac{n-1}{2}\right)\)

⇒ \(-50=n\left(\frac{n-25}{2}\right)\)

⇒ -100 = n²– 25n

⇒ n²-25n+ 100 = 0

⇒ (n – 5) (n – 20) = 0

∴ n = 5, 20

Both the values of n are natural and therefore, admissible.

Explanation of Double Answer:

S₂₀ = \( -6-\frac{11}{2}-5-\frac{9}{2}-4-\frac{7}{2}-3-\frac{5}{2}-2-\frac{3}{2}-1-\frac{1}{2} +0+\frac{1}{2}+1+\frac{3}{2}+2+\frac{5}{2}+3+\frac{7}{2} \)

⇒ \(-6-\frac{11}{2}-5-\frac{9}{2}-4\)

⇒ S₅

Question 6. Find the sum of all odd numbers lying between 100 and 200.
Solution:

The series formed by odd numbers lying between 100 and 200 is 101 + 103 + 105 + … + 199

Here, a = 101,

d= 103- 101 = 105- 103 = 2

Let, a_n= 199

⇒ 101 + (n- 1)²= 199

⇒ (n- 1)² = 98

⇒ n-1 = 49

⇒ n = 50

Now, \(S_{50}=\frac{50}{2}(101+199)\)

=7500

The sum of all odd numbers lying between 100 and 200 =7500

Question 7. If a_n = 3 – 4n, show that a₁,a₂,a₃, … form an A.P. Also find S₂₀.
Solution:

Given

a_n = 3 – 4n

a_n = 3 – 4n

a_n-1 = 3-4(n-1)

= 3-4n+4=7-4n

∴ a_n-a_n-1 = (3-4n)-(7-4n) = -4n

Which does not depend on ‘n’ i.e., the difference of two consecutive terms is constant.

∴ Given sequence is in A.P.

Now, a₁ = a = 3 – 4(1 ) = -1 ,

d = -4

∴ \( S_{20}=\frac{20}{2}[2 a+(20-1) d]\)

⇒ \(=10[2(-1)+19(-4)]\)

=-780

Question 8. If the sum of the first 6 terms of an A.P. is 36 and that of the first 1 6 terms is 256, find the sum of the first 10 terms.
Solution:

Given

The sum of the first 6 terms of an A.P. is 36 and that of the first 1 6 terms is 256

Let the first term and common difference of A.P. be ‘a’ and ‘d’ respectively.

S₆ = 36

⇒ \(\frac{6}{2}[2 a+(6-1) d]=36\)

2a + 5d= 12 ……(1)

5,6 = 256

⇒ \(\frac{16}{2}[2 a+(16-1) d]=256\)

2a + 15d = 32 …..(2)

Subtract eq. (1) from eq. (2), we get

⇒ \(\begin{array}{r}
2 a+15 d=32 \\
2 a+5 d=12 \\
-\quad-\quad- \\
\hline 10 d=20
\end{array}\)

d = 2

put d = 2 in eq. (1), we get

2a + 5(2) = 12

⇒ 2a = 2

⇒ a = 1

Now, the sum of first 10 terms 510 = \(\frac{10}{2}[2 a+(10-1) d]=5[2(1)+9(2)]\)

= 100

The sum of first 10 terms = 100

Question 9. The sum of the first V terms of an A.P. whose first term is 8 and the common difference is 20, is equal to the sum of the first 2, i terms of another A.P. whose first term is -30 and the common difference is 8. Find ‘n’.
Solution:

Given

The sum of the first V terms of an A.P. whose first term is 8 and the common difference is 20, is equal to the sum of the first 2, i terms of another A.P. whose first term is -30 and the common difference is 8.

For first A.P., a = 8, d = 20

∴ \( S_n=\frac{n}{2}[2(8)+(n-1)(20)]\)

⇒ \(n(8+10 n-10)=10 n^2-2 n \)

For second A.P., a = -30, d = 8

⇒ \(S_{2 n}=\frac{2 n}{2}[2(-30)+(2 n-1)(8)]\)

⇒ \(n(-60+16 n-8)=16 n^2-68n\)

Question 10. If the pth term of an A.P. is x and the qth term is y. Show that the sum of first (p + q) terms is \(\frac{p+q}{2}\left\{x+y+\frac{x-y}{p-q}\right\}\)
Solution:

Given

The pth term of an A.P. is x and the qth term is y.

Let first term = n and common difference = d

Now, T_p = x a + (p- 1)d = x …(1)

and T_q + a(q- 1)d =y …(2)

Subtracting eq. (2) from eq. (1), we get

(p – q)d = x- y

⇒ \(d=\frac{x-y}{p-q}\)

Now, if you put the value of d in eq. (1 ) or (2), and find a then it will be a very tedious job. So, don’t find ‘a we will simplify especially:

⇒ \(S_{p+q}=\frac{p+q}{2}[2 a+(p+q-1) d]\)

Bifurcate the terms inside the bracket as

⇒ \(S_{p+q}=\frac{p+q}{2}[\{a+(p-1) d\}+\{a+(q-1) d\}+d]\)

⇒ \(S_{p+q}=\frac{p+q}{2}\left[x+y+\frac{x-y}{p-q}\right]\) [From (1) ,(2) and (3)]

Question 11. The sum of the first six terms of an A.P. is 42. The ratio of its 10th term to 30th term is 1 : 3. Calculate the first term and 1 3th term of the A.P.
Solution:

Given

The sum of the first six terms of an A.P. is 42. The ratio of its 10th term to 30th term is 1 : 3.

Let the first term and common difference of A.P. be V and ‘d’ respectively.

Given that,

⇒ \(\frac{a_{10}}{a_{30}}=\frac{1}{3}\)

⇒ \(\frac{a+9 d}{a+29 d}=\frac{1}{3}\)

⇒ 3a + 27d = a + 29d

⇒ 2a = 2d

⇒ a=d

and, S₆ = 42

⇒ \(\frac{6}{2}(2 a+5 d)=42\)

3(2 d+5 d)=42

d = 2

a = 2

a₁₃ = a+12d

= 2 + 12 × 2

= 26

The first term and 13th term of the A.P is 2 and 26.

Question 12. The ratio between the sum of fird=st n terms of two A.P.’s are in the ratio (7n-5): (5n+17). Find the ratio of their 10th terms.
Solution:

Given

The ratio between the sum of fird=st n terms of two A.P.’s are in the ratio (7n-5): (5n+17).

Let a₁ and d₁, be the first term and common difference of first A.P and let a₂ and d₂ be the first term and common difference of other A.P.

∴ \( \frac{\frac{n}{2}\left[2 a_1+(n-1) d_1\right]}{\frac{n}{2}\left[2 a_2+(n-1) d_2\right]}=\frac{7 n-5}{5 n+17}\)

⇒ \(\frac{2 a_1+(n-1) d_1}{2 a_2+(n-1) d_2} =\frac{7 n-5}{5 n+17}\)

⇒ \(\frac{a_1+\left(\frac{n-1}{2}\right) d_1}{a_2+\left(\frac{n-1}{2}\right) d_2}=\frac{7 n-5}{5 n+17}\)

Replace \(\frac{n-1}{2}\) by 9

⇒ \(\frac{n-1}{2}\)

⇒ n = 19

∴ From eq. (1)

⇒ \(\frac{a_1+9 d_1}{a_2+9 d_2}=\frac{7(19)-5}{5(19)+17}\)

⇒ \(\frac{T_{10}}{T_{10}^*}=\frac{128}{112}=\frac{8}{7}\)

∴ \(T_{10}: T_{10}^*=8: 7\)

Question 13. If the sum of the first ‘m’ terms of an A.P. be V and the sum of its first ‘n’ terms is ‘m’. then show that the sum of its first (m + n) terms is -(m +n).
Solution:

Given

The sum of the first ‘m’ terms of an A.P. be V and the sum of its first ‘n’ terms is ‘m’.

Let the first term and common difference of the A.P. be ‘a’ and ‘d’ respectively.

Given that,

S_m = n

⇒ \(\frac{m}{2}[2 a+(m-1) d] =n \)

⇒ \(2 a m+\left(m^2-m\right) d =2 n\) ….(1)

and S_n =m

⇒ \(\frac{n}{2}[2 a+(n-1) d] =m\)

⇒ \(2 a n+\left(n^2-n\right)d =2 m\) …..(2)

Subtract eq. (2) from eq. (1), we get

⇒ 2a(m – n) + {(m2 – m2) – (m – n))d = 2(n – m)

⇒ 2a(m – n) + {(m – n) (m + n) – (m – n)}d = -2(m – n)

⇒ 2d(m – n) + (m – n) (m + n – 1)d = -2(m – n)

⇒ 2a + (m + n – l)d = -2 …..(3)

Now, \(S_{m+n}=\frac{m+n}{2}\{2 a+(m+n-1) d\}\)

⇒ \(=\frac{m+n}{2}(-2)=-(m+n)\)

Hence proved.

Question 14. If the sum of first n, 2n and 3n terms of an A.P. be S1, S2 and S3 respectively, then prove that: S₃ = 3(S₂ – S₁)
Solution:

Given

The sum of first n, 2n and 3n terms of an A.P. be S1, S2 and S3 respectively,

Let the first term and common difference of the A.P. be V and ld’ respectively.

S₁ = sum of first ‘n’ terms

⇒ \(S_1=\frac{n}{2}[2 a+(n-1) d]\) ….(1)

S₂ = sum of first ‘2n terms

⇒ \(\frac{2 n}{2}[2 a+(2 n-1) d]\)

⇒ \(S_2=\frac{n}{2}[4 a+(4 n-2) d]\) ….(2)

and S₃ = sum of first ‘3n’ terms

⇒ \(S_3=\frac{3 n}{2}[2 a+(3 n-1) d]\) …..(3)

⇒ \(S_2-S_1=\frac{n}{2}[4 a+(4 n-2) d]-\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(\frac{n}{2}[4 a+(4 n-2) d-2 a-(n-1) d]=\frac{n}{2}[2 a+(3 n-1) d]\)

⇒ \(3\left(S_2-S_1\right)=\frac{3 n}{2}[2 a+(3 n-1) d]=S_3\)

Hence proved

Question 15. If the ratio of the sum of the first m and n terms of an A.P. is m²:n², show that the ratio of its mth and nth terms is (2m – 1) : (2n – 1).
Solution:

Given

The ratio of the sum of the first m and n terms of an A.P. is m²:n²

⇒ \(\frac{S_m}{S_n}=\frac{\frac{m}{2}[2 a+(m-1) d]}{\frac{n}{2}[2 a+(n-1) d]}=\frac{m^2}{n^2} \)

⇒ \(\frac{2 a+(m-1) d}{2 a+(n-1) d}=\frac{m}{n}\)

⇒ \(\frac{a+\left(\frac{m-1}{2}\right) d}{a+\left(\frac{n-1}{2}\right) d}=\frac{m}{n} \ldots\)

Caution

Some students write as :

put \(\frac{m-1}{2}=m-1 \quad \Rightarrow \quad m-1=2 m-2\)

m-2m = -2+1

m = 1

which is wrong.

Here has mixed up the m’s of \(\frac{m-1}{2} \text { and } m-1\)

We want, \(\frac{T_m}{T_n} \text { i.e., } \frac{a+(m-1) d}{a+(n-1) d}\)

So, we replace \(\frac{m-1}{2}\) as m-1

i.e., m-1 as 2(m – 1)

⇒ m – 1 as 2m – 2

⇒ m as 2m -2+1

i.e., replace m by 2m – 1

Similarly, replace n by 2n – 1 in eq. (1), we get

∴ \( \frac{a+(m-1) d}{a+(\dot{n}-1) d}=\frac{2 m-1}{2 n-1}\)

i.e., \(\frac{T_m}{T_n}=\frac{2 m-1}{2 n-1}\)

Question 16. An A.P. consists of 37 terms. The sum of the three middlemost terms is 225 and the sum of the last three terms is 429. Find the A.P.
Solution:

Given

An A.P. consists of 37 terms. The sum of the three middlemost terms is 225 and the sum of the last three terms is 429.

Total number of terms (n) = 37, which is odd

∴ Middle term = \(\frac{37+1}{2}\) th term = 19th term

So, 3 middle most terms are 18th, 19th and 20th

∴ T₁₈ +T₁₉ + = 225

⇒ a + 17d + a+ 18d + a+ 19d- 225

⇒ 3a + 54d = 225

⇒ a=18d = 75 ……(1)

Also, a sum of the last 3 terms = 429

T₃₅ + T₃₆ + T₃₇ = 429

⇒ a + 34d + a + 35d + a + 36d = 429

⇒ 3a + 105d = 429

⇒ a + 35d = 143 ……(2)

Solving, (1) and (2), we get d- 4 and a = 3

∴ Required A.P. is a, a + d, a + 2d, a + 3d,…

i.e., 3, 3 + 4, 3 + 2(4), 3 + 3(4)

i.e., 3, 7, 11, 15, …

The A.p = 3, 7, 11, 15, …

Question 17. 25 trees are planted in a straight line 5 metres apart from each other. To water them the gardner must bring water for each tree separately from a well 10 metres from the first tree in line with trees. Find the distance he will move in order to water all the trees beginning with the first if he starts from the well.
Solution:

Distance Covered by Gardner from well to well :

⇒ \(\underbrace{10+10}_{\text {well-1-well }}+\underbrace{(10+5)+(10+5)}_{\text {well-2-well }}+\underbrace{(10+2 \times 5)+(10+2 \times 5)}_{\text {well-3-well }}\)

⇒ \(+\underbrace{(10+23 \times 5)+(10+23 \times 5)}_{\text {well-24-well }}+\underbrace{(10+24 \times 5)}_{\text {well to } 25 \text { th tree }}\)

2[10 + (10 + 5) + (10 + 2×5) + ……+(10 + 23×5)] + (10 + 24×5)

⇒ \(2 \underbrace{[10+15+20+\ldots+125]}_{24 \text { terms }}+(10+120)\)

⇒ \(2 \times \frac{24}{2}[10+125]+130=24 \times 135+130=3240+130=3370 \mathrm{~m}\)

Question 18. A child puts one five-rupee coin of her savings in the piggy bank on the first day. She increases her savings by one five-rupee coin daily. If the biggy bank can hold 1 90 coins of five-rupees in all, find the number of days she can continue to put the five-rupee coins into it and find the total money she saved. Write your views on the habit of saving.
Solution:

Child’s savings day-wise are,

We can have at most 190 coins

i.e., 1+2 + 3+ 4 + 5 + …to n terms =190

⇒ \(\frac{n}{2}[2 \times 1+(n-1) 1]=190\)

⇒ n(n + 1) = 380

⇒ n² +n- 380 = 0

⇒ (n + 20) (n- 19) = 0

⇒ n = -20

or n = 19

But many coins cannot be negative

∴ n = 19

So, number of days =19

Total money she saved = 5+10+15+20 + … upto 19 terms

⇒ \(\frac{19}{2}[2 \times 5+(19-1) 5]\)

⇒ \(\frac{19}{2}[100]=\frac{1900}{2}\)

=950

So, number of days = 19

and total money she saved = ₹950

NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Arithmetic Mean Of Two Numbers

If three numbers are in A.P., then the middle term is called the arithmetic mean of the remaining two numbers.

Let A be the arithmetic mean of a and b.

∴ a, A, b are in A.P.

⇒ A – a = b – A

⇒ 2A = a +b

⇒ A= \(\frac{a+b}{2}\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Selection Of Continuous Terms Of A.P.

1. Three consecutive terms in A.P. a— d, a, a + d (common difference is d)
2. Four consecutive terms in A.P. a — 3d, a — d, a+ d, a + 3d (common difference is 2d)
3. Five consecutive terms in A.P. a — 2d, a — d, a, a + d, a + 2d (common difference is d)

NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Selection Of Continuous Terms Of A.P. Solved Examples

Question 1. Find the value of x for which (x + 2), 2x, (2x + 3) are three consecutive terms of A.P. 
Solution:

Given

(x + 2), 2x, (2x + 3) are three consecutive terms of A.P.

∴ 2x = \(\frac{(x+2)+(2 x+3)}{2}\)

⇒ 4x = 3x + 5

x = 5

The value of x = 5.

Question 2. Split 207 into three parts such that these are in A.P. and the product of the two smaller parts is 4623.
Solution:

Let three parts be a- d, a, a + d.

∴ a-d + a+ a+ d = 207

⇒ 3a = 207

⇒ a = 69

and (a-d)-a = 4623

⇒ (69-d)69 = 4623

⇒ 69 – d = 67

⇒ d = 2

∴ a-d = 69 – 2 = 67

a= 69

a+d = 692 = 71

⇒ three parts are 67,69,71

Question 3. The angles of a triangle are in A.P. The greatest angle is twice the least. Find all the angles of the triangle.
Solution:

Given

The angles of a triangle are in A.P. The greatest angle is twice the least.

Let angles of the triangle be a – d, a, a + d

∴ a – d+a+a + d = 180°

⇒ 3a = 1 80°

⇒ a = 60°

and a + d = 2(a – d)

⇒ 60° + W = 2(60°- d)

⇒ 60° + d = 120° -2d

⇒ 3d = 60°

⇒ d = 20°

∴ a- d = 60°- 20° = 40°

a +d = 60° + 20° = 80°

∴ Three angles of triangle are 40°, 60°, 80°

Question 4. The angles of a quadrilateral are in A.P. and their common difference is 10°. Find the angles.
Solution:

Given

The angles of a quadrilateral are in A.P. and their common difference is 10°.

Let the angles of the quadrilateral be

a, a + 10°, a + 20°, a + 30°   (∵ common difference is 10°)

∴ a + (a + 10°) + (a + 20°) + (a + 30°) = 360°

⇒ 4a + 60° = 360°

⇒ 4a = 300°

⇒ a = 75°

∴ a+ 10°= 75°+ 10° = 85°

a + 20° = 75° + 20° = 95°

a + 30° = 75° + 30°= 105°

Hence, the angles are 75°, 85°, 95°, 105°.

Alternative Method :

Let the four angles of a quadrilateral are

a – 3d, a – d, a + d and a + 3d

∴ Here the common difference is 2d (remember)

∴ (a – 3d) + (a-d) + (a + d) + (a + 3d) = 360°

4a = 360°

⇒ a = 90°

the common difference is given to be 10°

i.e., 2d = 10°

⇒ d = 5°

Four angles are a – 3d = 90°- 3(5°) = 75°,

a – d 90°- 5° = 85°,

a + d = 90° + 5° = 95°,

a + 3d = 90° + 3(5°) = 105°

Question 5. There are m arithmetic means between 5 and -16 such that the ratio of the 7th mean to the (m -7)th mean is 1: 4. Find the value of m.
Solution:

Given

There are m arithmetic means between 5 and -16 such that the ratio of the 7th mean to the (m -7)th mean is 1: 4

Let A₁, A₂,…….A_m be m arithmetic means between 5 and -16.

∴ 5,A₁, A₂, A₃,…..A_m, -16 are in A.P.

∴ \(T_{m+2}=-16\)

⇒ \(5+(m+1) d=-16\)

⇒ \(d=\frac{-21}{m+1}\)

⇒ \(\frac{A_7}{A_{m-7}}=\frac{1}{4}\)

⇒ \(\frac{T_8}{T_{m-6}}=\frac{1}{4}\)

⇒\(\frac{5+7 d}{5+(m-7) d}=\frac{1}{4}\)

⇒ \(\frac{5+7\left(\frac{-21}{m+1}\right)}{5+(m-7)\left(\frac{-21}{m+1}\right)} \frac{1}{4}\)

⇒ \(20-\frac{588}{m+1}=5-\frac{21(m-7)}{m+7}\)

⇒ 20m + 20- 588 = 5m +5- 21m + 147

⇒ 36m = 720

⇒ m = 20

The value of m = 20.

Question 6. The digits of a positive number of three digits are in A.P. and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.
Solution:

Given

The digits of a positive number of three digits are in A.P. and their sum is 15. The number obtained by reversing the digits is 594 less than the original number.

Let the 3 digits in A.P. at units, tens and hundredth places be a – d, a and a + d.

According to the first condition,

(a -d) +a + (a +d) = 15

⇒ 3a = 15

a = 5 …(1)

The number is (a – d) + 10a + 100 (a + d)….(1)

i.e., 111+ 99d …(2)

The number; on reversing the digits is

(a + d) + 1 0a + 100(A — d) i.e., 111+ 99d

∴ According to the 2nd condition,

111a- 99d = (11la + 99d) – 594

⇒ 198d = 594

⇒ d = 3 …(3)

∴ Required number = 111a + 99d [from (2)]

= 111(5) + 99(3) [from (1) and (3)]

= 555 + 297

= 852

Required number = 852

NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5.1

Question 1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

1. File taxi fare after each km when the fare is ₹15 for the first km and ₹8 for each additional km.
2. The amount of air present in a cylinder when a vacuum pump removes — the
   air remaining in the cylinder at a time.
3. The cost of digging a well after every metre of digging, when it costs ₹150 for the first metre rises by ₹50 for each subsequent metre.
4. The amount of money in the account every year, when ₹10000 is deposited at compound interest at 8% per annum.

Solution :

1. Fare of first kilometre = ₹15 

Fare of 2 kilolnetre = ₹(15 + 8) = ₹23

Fare of 3 kilometre = ₹(23 + 8) = ₹31

Fare of 4 kilometre = ₹(3 1 + 8) = ₹39

Now, a₁ = 15, a₂ = 23, a₃ = 31, a₄ = 39

a₂ -a₁ = 23 – 15 = 8

a₃-a₂ = 31 -23 = 8

a₄ – a₃ = 39 – 31 = 8

∴ The difference between two consecutive terms is constant.

∴ The taxi fare after each kilometre is in A.P

2. Let the initial volume of air in the cylinder = V

In the first pump,

air remove = \(\frac{V}{4}\)

Remaining air = \(V-\frac{V}{4}=\frac{3 V}{4}\)

In the second pump,

air remove = \(\frac{1}{4} \times \frac{3 V}{4}=\frac{3 V}{16}\)

Remaining air = \(\frac{3 V}{4}-\frac{3 V}{16}=\frac{9 V}{16}\)

Now, \(a_1=V, a_2=\frac{3 V}{4}, a_3=\frac{9 V}{16}\)

∴ \(a_2-a_1=\frac{3 V}{4}-V=-\frac{V}{4}\)

and \(a_3-a_2=\frac{9 V}{16}-\frac{3 V}{4}=-\frac{3 V}{16}\)

∵ a₂– a₁ ≠ a₃– a₂

∴ The volumes of air are not in A.P.

3. The cost of digging of first metre = ₹150

The cost of digging of 2 metres = ₹.(150 + 50)

= ₹200

The cost of digging of 3 metres

= ₹ (150 + 50 + 50)

= ₹250

The cost of digging of 4 metres

= ₹(150 + 50 + 50 + 50)

= ₹300

Now, a1=₹150, a2 = 200,a3 = ₹250, a4 = ₹300

∴ a₂-a₁ = ₹200-₹150 = ₹50

a₃-a₂= ₹250 -₹200 = ₹50

a₄-a₃= ₹300- ₹250 = ₹50

∵  The difference between two consecutive terms is constant.

∵ The costs of digging each metre are in A.P.

4. Principal P = ₹10,000;

rate of interest R = ₹8%

Amount after 1 year,

⇒ \(A_1=P\left(1+\frac{R}{100}\right)^1=10,000\left(1+\frac{8}{100}\right)^1\)

⇒ \(=10,000 \times \frac{108}{100}=₹ 10,800\)

Amount after 2 years,

⇒ \(A_2=P\left(1+\frac{R}{100}\right)^2=10,000\left(1+\frac{8}{100}\right)^2\)

⇒ \(=10,000 \times \frac{108}{100} \times \frac{108}{100}=₹ 11664\)

Amount after 3 years

⇒ \(A_3=P\left(1+\frac{R}{100}\right)^3=10,000\left(1+\frac{8}{100}\right)^3\)

⇒ \(=10,000 \times \frac{108}{100} \times \frac{108}{100} \times \frac{108}{100}\)

= ₹12597.12

Now, A₂-A₁ = 11664- 10800 = 864

A₃-A₂= 12597.12- 11664 = 933.12

∵ The difference between two consecutive terms is not the same.

∴ Amounts are not in A.P.

Question 2. Write the first four terms of the A.P., when the first term a and the common difference d are given as follows :

1. a= 10, d= 10
2. a =- 2, d = 0
3. a = 4,d = -3
4. a=-1, d=\(\frac{1}{2}\)

Solution:

1. a = 10, d = 10

First term = a = 10

Second term =a + d= 10 + 10 = 20

Third term = a + 2d = 10 + 2 × 10 = 30

Fourth term = a + 3d = 1 0 + 3 × 1 0 = 40

∴ The first four terms of A.P. are 10, 20, 30, 40

2. a = -2, d = 0

First term = a = -2

Second term = a + d = -2 + 0 = -2

Third term = a + 2d = -2 + 2 × 0 = -2

Fourth term = a + 3d = -2 + 3 × 0=-2

∴ The first four terms of A.P. are -2, -2, -2, -2

3. a= 4, d = -3

First term = a = 4

Second term = a + d = 4 + (-3) = 1

Third term = a + 2d = 4 + 2(-3) = -2

Fourth term = a + 3d = 4 + 3(-3) =-5

∴ The first four terms of A.P are 4, 1, -2, -5

4. a=-1, d = \(\frac{1}{2}\)

First term =a = -1

Second term = a + d = \(-1+\frac{1}{2}=-\frac{1}{2}\)

Third term= a + 2d = \(-1+2 \times \frac{1}{2}=0\)

Fourth term= a + 3d = \(-1+3 \times \frac{1}{2}=\frac{1}{2}\)

∴ First four terms of A.P. are = \(-1,-\frac{1}{2}, 0, \frac{1}{2}\)

5. a =- 1.25, d = -0.25

First term = a = -1.25

Second term = a+d =-1.25 + (-0.25) = -1.50

Third term =a + 2d = -1.25 + 2(-0.25)

= -1.75

Fourth term = a + 3d =- 1.25 + 3(-0.25.)

= -2.00

∴ The first four terms of A.P. are

= -1.25, -1.50, -1.75, -2.00

Question 3. For the following A.Ps., write the first term and the common difference :

1. 3, 1,-1, -3,…
2. -5, -1,3, 7,…
3. \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots\)
4. 0.6, 17,2.8,3.9,…

Solution:

1. 3, 1, -1, -3,…

First term a = 3

Common difference d = 1-3 = (-1)- 1 = -2

2. -5,-1, 3, 7,…

First term a = -5

Common difference d = -1-(-5) = 3-(-1) = 4

3. \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots\)

First term a = \(\frac{1}{3}\)

Common difference d = \(\frac{5}{3}-\frac{1}{3}=\frac{9}{3}-\frac{5}{3}=\frac{4}{3}\)

4. 0.6, 17,2.8,3.9,…

First term a = 0.6

Common difference d= 1.7 – 0.6

= 2.8-17=1.1

Question 4. Which of the following are A.P.s? If they form an A.P., find the common difference d and write three more terms.

1. \(2,4,8,16,\)
2. \(2, \frac{5}{2}, 3, \frac{7}{2},\)
3. \(-1.2,-3.2,-5.2,-7.2,\)
4. \(-10,-6,-2,2,\)
5. \(3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2},\)
6. \(0.2,0.22,0.222,0.2222,\)
7. \(0,-4,-8,-12,\)
8. \(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},\)
9. \(1,3,9,27,\)
10. \(a, 2 a, 3 a, 4 a,\)
11. \(a, a^2, a^3, a^4,\)
12. \(\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32},\)
13. \(\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12},\)
14. \(1^2, 3^2, 5^2, 7^2,\)
15. \(1^2, 5^2, 7^2, 73,\)

Solution:

1. 2,4, 8, 16,…

Here, a₁ = 2, = 4, = 8, = 16

a₂– a₁ = 4- 2 = 2

a₃-a₂ = 8- 4 = 4

∵ a₂-a₁ ≠ a₃-a₂

∴ Given sequence is not an A.P.

2. \(2, \frac{5}{2}, 3, \frac{7}{2},\)

Here, \(a_1=2, a_2=\frac{5}{2}, a_3=3, a_4=\frac{7}{2}\)

⇒ \(a_2-a_1=\frac{5}{2}-2=\frac{1}{2}\)

∴ \(a_3-a_2=3-\frac{5}{2}=\frac{1}{2}\)

⇒ \(a_4-a_3=\frac{7}{2}-3=\frac{1}{2}\)

∵ \(a_2-a_1=a_3-a_2=a_4-a_3=\frac{1}{2}\)

∴ Given sequence is an A.P. and d = \(\frac{1}{2}\)

Now, fifth term \(a_5=a_4+d=\frac{7}{2}+\frac{1}{2}=4\)

Sixth term \(a_6=a_5+d=4+\frac{1}{2}=\frac{9}{2}\)

Seventh term \(a_7=a_6+d=\frac{9}{2}+\frac{1}{2}=5\)

∴ Next three terms of A.P. = 4, \(\frac{9}{2}\),5

3. -1.2, -3.2, -5.2, -7.2, …

Here a₁ = -1.2, a₂ =-3.2, a₃ = -5.2, a₄ =-7.2

∴ a₂-a₁ =(-3.2) -(-1.2) =-3.2+ 1.2= -2

a₃ -a₂ = (-5-2) – (-3.2) =-5.2 + 3.2 =-2

a₄ – a₃ = (-7.2) – (-5.2) = -7.2 + 5.2 =- 2

∵ a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = -2

∴ Given sequence is A.P. and d = -2

Now, fifth term a₅= a₄ + d = -7.2 + (-2) = -9.2

Sixth term a₆ = a₅ + d = -9.2 + (-2) = -1 1 .2

Seventh term a₇ =a₆+ d = -11.2 + (-2) =-13.2

∴ Next three terms of A.P. =- 9.2, -1 1.2, -13.2

4. -10, -6, -2,2,…

Here a₁ = -10, a₂ =-6, a₃ =-2, a₄ = 2

∴ a₂-a₁=(-6)-(-10)=-6+ 10 = 4

a₃ – a₂ = (-2) – (-6) =-2 + 6 = 4

a₄ – a₃ = 2-(-2) = 2 + 2 = 4

∵ a₂-a₁ = a₃ – a₂ = a₄ – a₃ = 4

∴ Given sequence is A.P. and d = 4

Now, fifth term a₅ = a₄ + d = 2 + 4 = 6

Sixth term a₆ = a₅ + d = 6 + 4 = 10

Seventh term a₇ = + d = 1 0 + 4 = 14

∴ Next three-term of A.P. = 6, 10, 14

5. \(3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2},\)

Here, \(a_1=3, a_2=3+\sqrt{2}, a_3=3+2 \sqrt{2},\)

⇒ \(a_4=3+3 \sqrt{2}\)

⇒ \(a_2-a_1=(3+\sqrt{2})-3=\sqrt{2}\)

∴ \(a_3-a_2=(3+2 \sqrt{2})-(3+\sqrt{2})=\sqrt{2}\)

⇒ \(a_4-a_3=(3+3 \sqrt{2})-(3+2 \sqrt{2})=\sqrt{2}\)

∵ \(a_2-a_1=a_3-a_2=a_4-a_3=\sqrt{2}\)

∴ Given sequence is A.P. and d = \(\sqrt{2}\)

Now, fifth term \(a_5=a_4+d\)

⇒ \(3+3 \sqrt{2}+\sqrt{2}=3+4 \sqrt{2}\)

Sixth term \(a^6=a^5+d\)

⇒ \(3+4 \sqrt{2}+\sqrt{2}=3+5 \sqrt{2}\)

Seventh term \(a_7=a_6+d\)

⇒ \(3+5 \sqrt{2}+\sqrt{2}=3+6 \sqrt{2}\)

Next three terms of A.P.

⇒ \(3+4 \sqrt{2}, 3+5 \sqrt{2}, 3+6 \sqrt{2}\)

6. 0.2,0.22,0.222,0.2222, …

Here a₁ = 0.2, a₂ = 0.22, a₃ = 0.222, a₄ = 0.2222

∴ a₂-a₁=0.22-0.2 = 0.02

a₃-a₂ = 0.222 -0.22 = 0.002

∵ a₂ – a₁ = a₃ – a₂

∴ Given sequence is not an A.P.

7. 0, -4, -8, -12, …

Here, a₁ = 0, a₂ = -4, a₃ = -8, a₄ = -12

∴ a₂ – a₁ = -4 – 0 = – 4

a₃-a₂ =-8- (-4) =- 8 + 4 = -4

∵ a₄ – a₃ = -12 – (-8) =-12 + 8 =- 4

a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = -4

∴ Given sequence is A.P. and d =-4

Now, fifth term a₅ =a₄ + d =-12 + (-4) =-16

Sixth term a₆ = a₅ + d = -16 + (-4) = -20

Seventh term a₇ = a6 + d = -20 + (-4) = -24

∴ Next three terms of A.P. =-16, -20, -24

8. \(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \ldots\)

Here, \( a_1=a_2=a_3=a_4=-\frac{1}{2}\)

∴ \(a_2-a_1=\left(-\frac{1}{2}\right)-\left(-\frac{1}{2}\right)=0\)

⇒ \(a_3-a_2=\left(-\frac{1}{2}\right)-\left(-\frac{1}{2}\right)=0\)

⇒ \(a_4-a_3=\left(-\frac{1}{2}\right)-\left(-\frac{1}{2}\right)=0\)

∴ a₂– 1 = a₃ – a₂ = a₄ – a₃ = 0

∴ Given sequence is A.P. and d = 0

Now, fifth term \(a_5=a_4+d=\frac{-1}{2}+0=-\frac{1}{2}\)

Sixth term \(a_6=a_5+d=\frac{-1}{2}+0=-\frac{1}{2}\)

Seventh term \(a_7=a_6+d=\frac{-1}{2}+0=-\frac{1}{2}\)

∴ Next three terms of A.P. = \(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\)

9. 1,3, 9,27,…

Here, a₁ = 1, a₂ = 3, a₃ = 9, a₄ = 27

∴ a₂-a₁ = 3-1=2

a₃– a₂ = 9- 3 = 6

∵ a₂ -a₁ ≠ a₃– a₂

∴ Given sequence is not an A.P.

10. a, 2a, 3a, 4a

Here, a₁ = a, a₂ = 2a, a₂ = 3a, a₄ = 4a

∴ a₂-a₁ =2a – a = a

a₃ – a₂ = 3a – 2a =a

a₄ – a₃= 4a – 3a = a

a₂– a₁ = a₃ – a₂ =a₄– a₃ = a

∴ Given sequence is A.P. and d = a

Now, fifth term a₅ = a₄ + d = 4a + a = 5a

Sixth term a₆ = a₅ + d = 5a + a = 6a

Seventh term a₇ = a₆ + d = 6a + a = 7a

∴ Next three terms of A.P. = 5A, 6a, 7a

11. a, a², a³, a⁴ = ,…

Here, a₁ = a, a₂ = a², a₃ = a³, a₄ = a⁴

∴ a₂-a₁ = a² – a

a₃ – a₂ = a³-a²

a₂– a₁ ≠ a₃– a₂

∴ Given sequence is not an A.P.

12. \(\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \cdots\)

Here, \(=\sqrt{2}, 2 \sqrt{2}, 3 \sqrt{2}, 4 \sqrt{2}, \cdots\)

∴ \(a_1=\sqrt{2}, a_2=2 \sqrt{2}, a_3=3 \sqrt{2}, a_4=4 \sqrt{2}\)

⇒ \(a_2-a_1=2 \sqrt{2}-\sqrt{2}=\sqrt{2}\)

⇒ \(a_3-a_2=3 \sqrt{2}-2 \sqrt{2}=\sqrt{2}\)

⇒ \(a_4-a_3=4 \sqrt{2}-3 \sqrt{2}=\sqrt{2}\)

⇒ \(a_2-a_1=a_3-a_2=a_4-a_3\)

∴ Given sequence is A.P. and d = \(\sqrt{2}\)

Now, fifth term \(a_5=a_4+d=4 \sqrt{2}+\sqrt{2}=5 \sqrt{2}\)

Sixth term \(a_6=a_5+d=5 \sqrt{2}+\sqrt{2}=6 \sqrt{2}\)

Seventh term \(a_7=a_6+d=6 \sqrt{2}+\sqrt{2}=7 \sqrt{2}\)

∴ Next three terms of A.P. = \(5 \sqrt{2}, 6 \sqrt{2}, 7 \sqrt{2}\)

13. \(\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \ldots\)

Here, \(\quad a_1=\sqrt{3}, a_2=\sqrt{6}, a_3=\sqrt{9}, a_4=\sqrt{12}\)

⇒ \(a_2-a_1=\sqrt{6}-\sqrt{3}=\sqrt{3}(\sqrt{2}-1)\)

⇒ \(a_3-a_2=\sqrt{9}-\sqrt{6}=\sqrt{3}(\sqrt{3}-\sqrt{2})\)

⇒ \(a_2-a_1 \neq a_3-a_2\)

∴ Given sequence is not an A.P.

14. \(1^2, 3^2, 5^2, 7^2, \ldots\)

Here, \(\quad a_1=1^2, a_2=3^2, a_3=5^2, a_4=7^2\)

∴ \(a_2-a_1=3^2-1^2=9-1=8\)

⇒ \(a_3-a_2=5^2-3^2\)

⇒ \(=25-9=16\)

∵ \(a_2-a_1\neq a_3-a_2\)

∴ Given sequence is not an A.P.

15. \(1^2, 5^2, 7^2, 73, \ldots\)

Here, \(a_1=1^2, a_2=5^2, a_3=7^2, a_4=73\)

∴ \(a_2-a_1=5^2-1^2=25-1=24\)

⇒ \(a_3-a_2=7^2-5^2=49-25=24\)

⇒ \(a_4-a_3=73-7^2=73-49=24\)

∵ \(a_2-a_1=a_3-a_2=a_4-a_3=24\)

∴ Given sequence is not an A.P. and d = 24

Now, fifth term a₅ = a₄ + d = 73 + 24 = 97

Sixth term a₆= a₅+d = 97 + 24

= 121 = 11²

Seventh term a₇=a₆ + d= 121 + 24 = 145

∴ Next three terms of A.P. = 97, 11², 145

NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5.2

Question 1. Fill in the blanks in the following table given the first term. d the common difference and an nth term of the A.P. :

Solution:

1. Here, a = 7, d = 3. n= 8

∴ a_n – a + (n -1)d

= 7 + (8 – 1) 3

= 7 + 7 × 3=28

Therefore, a_n =28

2. a =-18, n = 10, a_n = 0

∴ a + (n – 1)d = a_n

⇒ -18 + (10- 1)d = 0

⇒ 9d = 18

⇒ d = 2

3. d =-3, n= 1 8, a_n =-5

∴ a + (n – 1 )d = a_n

⇒ a + (18 – 1) (-3) =-5

⇒ a-51 =- 5

⇒ a = -5 + 51 =46

4. a = -18.9, d = 2.5, a_n = 3.6

∴ a + (n -1)d = an

-18.9 + (n -1) × 2.5 = 3.6

⇒ (n- 1) × 2.5 =3.6+18.9 = 22.5

⇒ \(n-1=\frac{22.5}{2.5}=9\)

⇒ n = 9+1 = 10

5. a = 3.5, d = 0, n= 105

∴ a_n = a + (n -1 )d

= 3.5 + (105 – 1) × 0

= 3.5

Question 2. Choose the correct choice in the following and justify :

1. 30th term of the A.P.: 10, 7, 4, …, is
   1. 97
   2. 77
   3. -77
   4. -87
2. 11th term of the A.P.: -3, 2 …, is
   1. 28
   2. 22
   3. -38
   4. \(-48 \frac{1}{2}\)

Solution:

1. Given A.P.: 10,7,4…

Mere n= 10,d= 7- 10 = 4- 7 = -3, n =30

∴ a_n =a( n – 1 )d

⇒  a₃₀ = 10 + (30 – 1) (-3) = 10 – 87 – -77

2. Given A.P. : \(-3,-\frac{1}{2}, 2, \ldots\)

Here a = \(a=-3, d=-\frac{1}{2}-(-3)=2-\left(-\frac{1}{2}\right)=\frac{5}{2}\)

n = 11

∴ a_n = a + (n – 1 )d

⇒ \(a_{11}=-3+(11-1) \frac{5}{2}=-3+25=22\)

Question 3. In the following A.Ps., find the missing terms in the boxes :

Solution:

1. Here, first term a = 2

Third term = 26

⇒ a + (3-1)d = 26 2 + 2d = 26

⇒ 2d = 24

⇒ d= 12

∴ Second term = a + d = 2 + 12 = 14

∴ Term of the box = 14

2. Second term =13

⇒ a + (2-1)d = 13 (where a = first term and d = common difference)

⇒ a+d =13 ….(1)

and Fourth term = 3

⇒ a + (4 – 1 )d= 3 ⇒ a + 3d = 3 ….(2)

Subtracting equation (1) from equation (2),

⇒ d = -5

Put the value of d in equation (1)

a + (-5) = 13 ⇒ a = 13+5

= 18

Third term a₂ + d = 13 + (-5) = 8

∴ Term of the boxes = 18. 8 respectively.

3. Here, first term a = 5

Fourth term \(a_4=9 \frac{1}{2}\)

⇒ \(a+3 d=\frac{19}{2}\)

⇒ \(5+3 d=\frac{19}{2}\)

⇒ \(3d=\frac{19}{2}-5=\frac{9}{2} \Rightarrow d=\frac{3}{2}\)

Now, second term \(a_2=a+d=5+\frac{3}{2}=\frac{13}{2}\)

Third term \(a_3=a_2+d=\frac{13}{2}+\frac{3}{2}=\frac{16}{2}=8\)

∴ Term of the boxes = \(\frac{13}{2}, 8\)

4. Here, first term a = -4

Sixth term = 6

⇒ a + 5d = 6 ⇒ -4 + 5d = 6

⇒ 5d = 6+4= 10

⇒ d =2

∴ Second term a₂ = a + d = -4 + 2 = -2

Third term a₃ = a₂ + d = -2 + 2 = 0

Fourth term a₄ = a₃+d = 0 + 2 = 2

Fifth term a₅= a₄ + d = 2 + 2 = A

∴ Term of the boxes = -2, 0, 2, 4 respectively.

5. Here, second term a₂ = 38

⇒ a + d= 38

Sixth term a₅ = -22

⇒ a + 5d = -22

Subtracting equation (1) from (2),

d = -15

Put the value of d in equation (1),

a + (-15) =38

⇒ a = 38+ 15=53

Third term a₃ = a₂ + d = 38 + (-15) = 23

Fourth term a₄ = a₃+d = 23 + (-15) = 8

fifth term a₅ = a₄ + d = 8 + (-15) -7

∴ Term of the boxes = 53, 23, 8, -7 respectively.

Question 4. Which term of the A.P. : 3, 8, 13, 18,… is 78?
Solution:

Given A.P. : 3, 8, 13, 18,…

Here a = 3, d = 8- 3 = 13-8 = 5

Let a_n = 78 ⇒ a + (n – 1 )d = 78

⇒ 3 + (11 – 1)5 = 78

⇒ (n- 1)5 =78-3 = 75

⇒ \(n-1=\frac{75}{5}=15\)

⇒ n= 15+ 1 = 16

∴ The 16th term is 78.

Question 5. Find the number of terms in each of the following A.Ps. :

1. 7, 13, 19,…, 205
2. 18, \(15 \frac{1}{2}\), 13…..-47

Solution:

Given A.P.: 7, 13, 19,…, 205

a=7

d= 13-7= 19- 13 = 6

Let an = 205

⇒ a + (n- l)d = 205

⇒ 7 + (n – 1)6 = 205

⇒ (n- 1)6 = 205-7 = 198

⇒ \(n-1=\frac{198}{6}=33\)

⇒ n = 33 + 1 = 34

∴ Number of terms in given A.P. = 34

2. Given A.P.:18, \(15 \frac{1}{2}\), 13…..-47

Here, a= 18

⇒ \(d=15 \frac{1}{2}-18=13-15 \frac{1}{2}=-2 \frac{1}{2}=\frac{-5}{2}\)

Let \(a_n=-47\)

⇒ \(a+(n-1) d=-47\)

⇒ \(18+(n-1)\left(\frac{-5}{2}\right)=-47\)

⇒ \((n-1)\left(\frac{-5}{2}\right)=-47-18=-65\)

⇒ \(n-1=(-65)\left(-\frac{2}{5}\right)=26\)

⇒ n = 26 + 1 = 27

∴ Number of terms in given A.P. = 27

Question 6. Check whether -150 is a term of the A.P. 1 1,8, 5. 2….
Solution:

Given A.P: 1 1, 8, 5, 2,…

⇒ a= 11, d=8- 11 =5- 8 = -3

Let a_n = -150

⇒ a + (n- 1 )d = -150

⇒ 11 + (n – 1) (-3) = -150

⇒ 11 – 3n + 3 = -150

⇒ 14 + 150 = 3

⇒ \(n=\frac{164}{3}=54 \frac{2}{3}\)

∵ The value of n is not a whole number.

∴ -150 is not a term of a given A.P.

Question 7. Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.
Solution:

Given

In AP 11th term is 38 and the 16th term is 73

Let a be the first term and d the common difference of the A.P.

Now, a₁₁ = 38

⇒ n +(11 – 1)d = 38

⇒ a+ 10 = 38 ….(1)

and a16 = 73

⇒ a+ (16- 1)d = 73

⇒ a+15d=73 …..(2)

Subtracting equation (1) from (2),

Put the value of d in equation (1),

a+ 10 × 7 = 38

⇒ a=38- 70 =-32

Now, the 31st term of A.P.

a₃₁ = n + (31 – 1)d

=- 32 + 30 × 7

= -32 + 210= 178

The 31st term of an A.P = 178.

Question 8. An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:

Given

An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106.

Let a be the first term and d the common difference of A.P.

∴ a₃ = 12 ⇒ a + (3 -1)d = 12

⇒ a+2d = 12 ….(1)

and last term= 50th term = 106

⇒ a + (50-1)d = 106

⇒ a + 49 d = 106 …..(2)

Subtracting equation (1) from (2)

⇒ d = 2

Put the value of d in equation (1),

a + 2 × 2 = 12

a = 12-4 = 8

Now, 29th term = a + (29 – 1)d = 8 + 28 × 2

= 8 + 56 = 64

The 29th term = 64

Question 9. If the 3rd and the 9th terms of an A.P. are 4 and- 8 respectively, which term of this A.P. is zero?
Solution:

Given

The 3rd and the 9th terms of an A.P. are 4 and- 8 respectively,

Let a be the first term and d the common difference of A.P.

a₃ = 4 ⇒ a + (3-1)d = 4

a + 2d = 4…..(1)

a9 = -8 ⇒ a + (9- 1 )d = -8

a + 8d = -8 ……(2)

Subtracting equation (1) from (2),

⇒ d = -2

Put the value of d in equation (1),

⇒ a + 2(-2) = 4

⇒ a = 4 + 4 =8

Now, let a_n = 0 a + (n- 1)d =0

⇒ 8 + (n- l)(-2) =0

⇒ 8- 2n + 2 = 0

⇒ -2n =-10

⇒ n =5

∴ 5th term of the progression is zero.

Question 10. The 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.
Solution:

Given

The 17th term of an A.P. exceeds its 10th term by 7.

Let a be the first term and d the common difference of A.P.

∴ a₁₇ =a₁₀ + 7

⇒ a + (17- 1)d = a + (10- 1)d + 7

16d-9d = 7 ⇒ 7d =7

⇒ d=1

Common difference of progression = 1

Question 11. Which term of the A.P. : 3, 15, 27, 39, … will be 132 more than its 54th term?
Solution:

Given A.P. : 3, 15, 27, 39…

a = 3, d=15-3 = 27-15=12

∴ a₅₄ = a+ (54 -1)d = 3 + 53 × 12

= 3 + 636 = 639

Let a_n =a₅₄ + 132

⇒ a + (n – 1)d =639+ 132

⇒ 3 + (n -1)12 =771

⇒ (n- 1)12 =771 -3 = 768

⇒ \((n-1)=\frac{768}{12}=64\)

⇒ n = 64 + 1 = 65

∴ Required term = 65th term.

Question 12. Two A.P.s have the same common difference. The difference between their 100th terms is 1 00, what is the difference between their 1,000th terms?
Solution:

Given

Two A.P.s have the same common difference. The difference between their 100th terms is 1 00

Let the first term be a and the common difference be d of first A.P.

Let the first term be A and the common difference be D of the second A.P.

100th term of first progression = + (100 -1)d

= a + 99d

100th term of the second progression

=A + (100- 1)d

=A + 99d

Difference of 100th terms of two progression

= 100

⇒ (a + 99d) – (A + 99d) = 100

⇒ a + 99d-A-99d = 100

⇒ a-A = 100 ……(1)

Again, the 1000th term of the first progression

= a + (1000-1)d

= a + 999d

1000th term of the second progression

=A + (1000-1)d

=A + 999d

Difference of 1000th terms of two progressions

= (a + 999d) – (A + 999d)

= a + 999d-A-999d

= a – A = 100 [from equation (1)]

Question 13. How many three-digit numbers by 12. are divisible by 7?
Solution:

Three digit numbers: 100, 101 102, …, 999

Three-digit numbers divisible by 7: 105, 112, 119,…, 994

Here, a = 105, d= 1 12 – 105 = 1 19 – 1 12 = 7

Let a_n = 994

⇒ a + (n-1)d =994 ⇒ 105 + (n -1)7 = 994

⇒ (n- 1)7 =994- 105 = 889

⇒ \(n-1=\frac{889}{7}=127\)

⇒ n= 127+ 1 = 128

∴ Number of 3-digit numbers divisible by 7.

= 128

Question 14. How many multiples of 4 lie between 10 and 250?
Solution: The multiples of 4 between 10 and 250 are :

12, 16, 20, …,248

Here a=12,d= 16-12 = 20-16 = 4

a_n =248

⇒ a + (n- 1)d =248

⇒12 + (n- 1)4=248

⇒ (n- 1)4 =248- 12 = 236

⇒ n-1 = 59

⇒ n = 59 + 1 = 60

∴ Multiples of 4 between 10 and 250 = 60

Question 15. For what value of n, are the nth terms of two A.Ps?: 63,65,67,… and 3, 10, 17,… equal?
Solution:

First A.P.: 63, 65, 67…

Here a = 63

d = 65 – 63 = 67- 65 = 2

a_n = a + (n – 1 )d

= 63 + (n – 1)2 = 63 + 2n – 2

= 2n + 61

Second A.P. 3, 10, 17, …

Here A =3,D= 10-3= 17- 10 = 7

A_n =A + (n -1)D = 3 + (n – 1)7

= 3 + 7n-7 = 7n-4

According to the problem, an =An

⇒ 2n + 61 = 7n – 4

⇒ n -7n = -4 – 61

⇒ -5n = -65

⇒ n = 13

∴ The 13th terms of given progressions are equal.

Question 16. Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution:

Let the first term be a and common difference d of the A.P.

According to the problem,

a₇ – a₅ =12

⇒ (a + 6d) – (a + 4d) = 12

⇒ a+6d-a-4d = 12

⇒ 2d = 12

⇒ d = 6

⇒ a₃ = 16

⇒ a + (3-1)d = 16

⇒ a + 2 × 6 = a + (3-1)d = 16

⇒ a + 2 × 6 = 16

⇒ a = 16 – 12 = 4

Now A.P.: 4, 4 + 6, 4 + 2 × 6,…

= 4, 10, 16, …

Question 17. Find the 20th term from the last term of the A.P. : 3, 8, 13, …. 253.
Solution:

Given A.P. : 3, 8, 13,…, 253

Here, last term l = 253, = 8- 3 = 13-8 = 5

20th term from the end =l-(n-1)d

= 253 – (20- 1) × 5

= 253- 19 × 5 = 253-95 = 158

∴ 20th term from the end of the progression

= 158

The 20th term from the last term of the A.P = 158

Question 18. The sum of the 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the
Solution :

Given

The sum of the 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44.

Let the first term be a and common difference d of the A.P.

∴ a₄ + a₈ =24 ⇒ a + 3d + a + 7d = 24

⇒ 2a + 10d =24

⇒ a+2d = 12 …(1)

and a₆ + a₁₀ = 44 ⇒ a + 5d + a + 9d = 44

⇒ 2a+14d =44

⇒ a + 7d =22

Subtracting equation (1) from (2),

⇒ d = 5

Put the value of d in equation (1),

a + 5 × 5= 12

⇒ a= 12-25 =-13

∴ Second term = a + d = -13 + 5 =-8

Third term-a + 2d =-13+2×5 = -3

So, the first three terms of A.P. are -13, -8,-3.

Question 19. Subba Rao started work in 1995 at an annual salary of ₹5000 and received an increment of ₹200 each year. In which year did his income reach ₹7000?
Solution:

Given

Subba Rao started work in 1995 at an annual salary of ₹5000 and received an increment of ₹200 each year.

Salary in first year = ₹5000 Salary in second year = ₹5000 + ₹200 = ₹5200 Salary in third year = ₹5200 + ₹200 = ₹5400 Progression formed from the salary of each year ₹5000, ₹5200, ₹5400, …

Here, a₂-a₁ = 5200 -5000 = 200

a₃-a₂= 5400 -5200 = 200

a₂ – a₁ = -a₂

⇒ The above progression is an A.P.

a = ₹5000, d = ₹200

Let in the nth year, the salary becomes ₹7000

Let in the nth year, the salary becomes ₹7000.

∴ a_n =7000

⇒ a + (n-1)d =7000

⇒ 5000 + (n- 1)200 =7000

⇒ (n- 1)200 =7000-5000

⇒ (n- 1)200 =2000

⇒ \(n-1=\frac{2000}{200}=10\)

⇒ n= 10+ 1 = 11

∴ In the 11th year, the salary of Subba Rao will be ₹7000.

Question 20. Ramkali saved ₹5 in the first week of the year and then increased her weekly savings by ₹1.75. If in the nth week, her weekly savings become ₹20.75, find n.
Solution:

Given

Ramkali saved ₹5 in the first week of the year and then increased her weekly savings by ₹1.75. If in the nth week, her weekly savings become ₹20.75

Saving of the first week = ₹5

∵ ₹1.75 is increasing in the savings of every week. The saving of every week from an A.P, in which.

a = ₹5 and d = ₹1.75

Let the saving in the nth week = ₹20.75

⇒ a + (n – 1) d = 20.75

⇒ 5 + (w- 1) (1.75) = 20.75

⇒ (n-1) (1.75) =20.75 -5 = 15.75

⇒ \(n-1=\frac{15.75}{1.75}=9\)

⇒ n=9 + 1 = 10

n= 10

NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5.3

Question 1. Find the sum of the following A.Ps.:

1. 2, 7, 12, …, to 10 terms.
2. -37, -33, -29, …, to 12 terms
3. 0.6, 1.7, 2.8, …, to 100 terms.
4. \(\frac{1}{15}, \frac{1}{12}, \frac{1}{10},\)…, to 11 terms.

Solution:

1. 2, 7, 12…..to 10th term

Here, a = 2,d = 7 -2= 12-7 = 5,n= 10

From the formula \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{10}=\frac{10}{2}[2 \times 2+(10-1) \times 5]\)

= 5(4 + 45) = 5×49 = 245

2. -37, -33, -29,…, to 12 terms

Here a =-37

d = -33- (-37) =-29- (-33) = 4, n = 12

From the formula \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{12}=\frac{12}{2}[2 \times(-37)+(12-1) \times 4]\)

= 6(-74 + 44) = 6 x (-30) = -180

3. 0.6, 1.7, 2.8,…, to 100 terms

Here a = 0.6, d = 1.7 – 0.6 = 2.8 – 1.7 = 1.1 n= 100

From the formula \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{100}=\frac{100}{2}[2 \times 0.6+(100-1) \times 1.1]\)

= 50[1.2 + 108.9] = 50×110.1

= 5505

4. \(\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots, \text { to } 11 \text { terms }\)

Here, \(a=\frac{1}{15}, d=\frac{1}{12}-\frac{1}{15}=\frac{1}{10}-\frac{1}{12}=\frac{1}{60}, n=11\)

From the formula \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \( S_n=\frac{11}{2}\left[2 \times \frac{1}{15}+(11-1) \times \frac{1}{60}\right]\)

⇒ \(=\frac{11}{2}\left[\frac{8+10}{60}\right]=\frac{11}{2} \times \frac{18}{60}=\frac{33}{20}\)

Question 2. Find the sums given below :

1. 7 +\(10 \frac{1}{2}\) + 14+ … + 84
2. 34 + 32 + 30 + … + 10
3. -5 + (-8) + (-11) + …(-230)

Solution:

1. 7 +\(10 \frac{1}{2}\) + 14+ … + 84

Here a = 7, d = \(10 \frac{1}{2}\) = \(14-10 \frac{1}{2}\)

⇒ \(3 \frac{1}{2}=\frac{7}{2}\)

Let a_n = 84

⇒ a + (n-1)d = 84 => \(7+(n-1) \frac{7}{2}=84\)

⇒ \( (n-1) \frac{7}{2}=84-7=77\)

⇒ \(n-1=77 \times \frac{2}{7}=22\)

⇒ \(n=22+1=23\)

From the formula, \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \( S_{23}=\frac{23}{2}\left[2 \times 7+(23-1) \times \frac{7}{2}\right]\)

⇒ \(\frac{23}{2}[14+77]=\frac{23 \times 91}{2}\)

⇒ \(\frac{2093}{2}\)

⇒ \(1046 \frac{1}{2}\)

2. 34 + 32 + 30 + … + 10

Here a = 34, = 32- 34 = 30 – 32 = -2

Let an = 10 ⇒ a + (n – 1)d = 10

⇒ 34 + (n-1)(-2) =10

⇒ (n- 1) (-2) = 10-34

⇒ (n-1) (-2) =-24

⇒ n-1 =12

⇒ n=12+1 = 13

From the formula, \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{13}=\frac{13}{2}[2 \times 34+(13-1)(-2)]\)

⇒ \(\frac{13}{2}[68-24]\)

⇒ \(\frac{13}{2} \times 44=286 \quad\)

3. -5 + (_8) + (-11)… + (-230)

Here a =-5, d =-8- (-5) =-11 – (-8) =-3

Let an = -230 ⇒ a + (n-1)d = -230

⇒ -5 + (n – 1) (-3) = -230

⇒ (n- 1) (-3) =-230 + 5

⇒ (n – 1)(-3) = -225

⇒ n- 1 = 75

⇒ n = 75 + 1 = 76

From the formula \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{76}=\frac{76}{2}[2 \times(-5)+(76-1)(-3)]\)

= 38(-10 – 225)

= 38 × (-235) = -8930

Question 3. In an A.P. :

1. Given a = 5, d = 3, a_n = 50, find ii and S_n.
2. Given a = 7, a₁₃ = 35, find d and S₁₃.
3. Given a₁₂ = 37, d = 3, find a and S₁₂.
4. Given a₃ = 15, S₁₀ = 125, findd and a₁₀.
5. Given d = 5, S₉ = 75, find a and a₉.
6. Given a = 2, d = 8, S_n = 90, find n and an.
7. Given a = 8, an = 62, S_n = 210, find n and d.
8. Given a_n = 4, n = 2, S_n = -14, find n and a.
9. Given a = 3, n = 8, S = 192, find d.
10. Given l = 28, S = 144, and there are a total 9 terms. Find a

Solution:

1. a = 5,d = 3, a_n = 50

⇒ a + (n -1)d = 50

⇒ 5 + (n-1)3 =50

⇒  (n-1)3 = 45

⇒ a-1= 15

⇒ n = 16

Now, \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(\frac{16}{2}[2 \times 5+(16-1) \times 3]\)

⇒ \(8(10+45)=8 \times 55=440\)

n = 16 and Sn = 440

2. a = 7

a₁₃= 35

⇒ a + (13 – 1)d = 35

⇒ 7 + 12d = 35

⇒ 12d =35-7 = 28

⇒ \(d=\frac{28}{12}=\frac{7}{3}\)

and, from \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{13}=\frac{13}{2}\left[2 \times 7+(13-1) \times \frac{7}{3}\right]\)

⇒ \(\frac{13}{2}[14+28]=\frac{13}{2} \times 42=273\)

∴ \(d=\frac{7}{3} \text { and } S_{13}=273 \quad\)

3. d = 3

a₁₂ = 37

a+(12-1)d = 37

a + 11 × 3 = 37

a = 37 – 33 = 4

and, from \( S_n=\frac{n}{2}[2 n+(n-1) d]\)

⇒ \(S_{12}=\frac{12}{2}[2 \times 4+(12-1) \times 3]\)

⇒ \(6[8+33]=6 \times 41=246\)

∴ \(a=4, S_{12}=246\)

4. Let the first term be a and the common difference is d.

a₃ = 15 ⇒ a + (3 -1)d= 15

⇒ a+2d = 15 …(1)

and S₁₀= 125

⇒ \(\frac{10}{2}[2 a+(10-1) d]=125\)

⇒ 5[2a + 9d] = 125

⇒ 2a + 9d = 25 …(2)

Multiply equation (1) by 2 and subtracting from equation (2),

d = -1

Put the value of d in equation (1)

⇒ a + 2 × – 1 = 15

⇒ a-2= 15

⇒ a = 15+2= 17

a₁₀ = a+ (10-1)d

= 17 + 9(-1) = 17-9 = 8

∴ d = -1, a₁₀ = 8

5. d = 5

and S₉ = 75

⇒ \(\frac{9}{2}[2 a+(9-1) \times 5]=75\)

⇒ \( 2 a+40=\frac{75 \times 2}{9}\)

⇒ \(a +20=\frac{75}{9} \Rightarrow a+20=\frac{25}{3}\)

⇒ \(a=\frac{25}{3}-20=\frac{25-60}{3}\)

⇒ \(a=\frac{-35}{3}\)

and \(a_9=a+8 d=\frac{-35}{3}+8 \times 5=\frac{-35}{3}+40\)

⇒ \(\frac{-35+120}{3}=\frac{85}{3}\)

∴ \(a=\frac{-35}{3} \text { and } a_9=\frac{85}{3}\)

6. a = 2, d= 8

S_n = 90

⇒ \( \frac{n}{2}[2 a+(n-1) \cdot d]=90\)

⇒ \(\frac{n}{2}[2 \times 2+(n-1) \cdot 8]=90\)

⇒ \(\frac{4n}{2}[1+(n-1) \cdot 2]=90\)

⇒ \( n(2 n-1)=\frac{90 \times 2}{4}\)

⇒ \(2 n^2-n=45\)

⇒ \(2 n^2-n-45=0\)

⇒ \(2 n^2-10 n+9 n-45=0\)

⇒ \(2 n(n-5)+9(n-5)=0\)

⇒ \((n-5)(2 n+9)=0\)

⇒ \(n-5=0 \quad \text { or } 2 n+9=0\)

⇒ \(n =5 \quad \text { or } \quad n=-\frac{9}{2}\)

⇒ \(n =-\frac{9}{2} \text { is not possible.}\)

∴ n = 5

Now, an = a + (n-1)d

a₅ = 2 + (5 – 1) × 8 = 2 + 32 = 34

n = 5 and a_n = 34

7. a = 8, a_n = 62

S_n= 210

⇒ \(\frac{n}{2}\left(a+a_n\right)=210\)

⇒ \(\frac{n}{2}(8+62)=210\)

⇒ \(n=\frac{210 \times 2}{70}=6\)

⇒ \(a_n=62\)

⇒ \(a+(n-1) d=62\)

⇒ 8 + (6-1)d= 62 => 5d = 62 – 8 = 54

⇒ \(d=\frac{54}{5}\)

∴ \(n=6, d=\frac{54}{5}\)

8. a_n = 4, d = 2 and S_n=-14

a_n = 4

⇒ a + (n -1)d = 4

⇒ a + (n – 1).2 = 4

⇒ a + 2n – 2 =4

⇒ a + 2n = 6

and S_n = -14 …(1)

⇒ \( \frac{n}{2}\left(a+a_n\right)=-14\)

⇒ \(\frac{n}{2}(a+4)=-14\)

⇒ \(\frac{n}{2}(6-2 n+4)=-14 \quad \text { from eqn. (1) }\)

⇒ \(\frac{n}{2}(10-2 n)=-14\)

⇒ \(n(5-n)=-14\)

⇒ \(5 n-n^2=-14\)

⇒ \(0=n^2-5 n-14\)

⇒ \(n^2-7 n+2 n-14=0\)

⇒ \( n(n-7)+2(n-7)=0\)

⇒ \((n-7)(n+2)=0\)

⇒ \(n-7=0 \text { or } n+2=0\)

⇒ \(n=7 \text { or } \quad n=-2\)

⇒ \(n=-2 \text { is not possible.}\)

∴ n=7

From equation (1)

a + 2 × 7 = 6

a = 5

a = 6 – 14= -8

∴ n = 7 and a = -8

9. \(a=3, n=8 \text { and } S=192\)

⇒ \(S=192\)

⇒ \(\frac{n}{2}[2 a+(n-1) d]=192\)

⇒ \(\frac{8}{2}[2 \times 3+(8-1) d]=192\)

⇒ \(4(6+7 d)=192\)

⇒ \(24+28 d=192\)

⇒ \(28 d=192-24=168\)

⇒ \(d=\frac{168}{28}=6\)

∴ d=6

10. l=28, S =144, n=9

S=144

⇒ \(\frac{n}{2}(a+l)=144 \Rightarrow \frac{9}{2}(a+28)=144\)

⇒ \(a+28=\frac{144 \times 2}{9}=32\)

⇒ a=32-28=4

∴ a=4

Question 4. How many terms of the A.P.: 9, 17, 25,… must be taken to give a sum of 636?
Solution:

Given A.P: 9. 17. 25….

Here a = 9,d= 17-9 = 25-17 = 8

Let S_n=636

⇒ \(\frac{n}{2}[2 a+(n-1)] d=636\)

⇒ \(\frac{n}{2}[2 \times 9+(n-1) \cdot 8]=636\).

⇒ \(n[9+(n-1) \cdot 4]=636\)

⇒ \(n(9+4 n-4)=636\)

⇒ \(n(4 n+5)=636\)

⇒ \(4 n^2+5 n-636=0\)

⇒ \(4 n^2+53 n-48 n-636=0\)

⇒ \( n(4 n+53)-12(4 n+53)=0\)

⇒ \((4 n+53)(n-12)=0\)

⇒ \(4 n+53=0 \text { or } n-12=0\)

⇒ \(n=-\frac{53}{4} \text { or } n=12\)

but \(n=-\frac{53}{4}\) is not possible.

∴ n = 12

Therefore, the number of terms = 12

Question 5. The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:

Here a = 5

Let the number of terms = n

⇒ \(a_n=45 \text { and } S_n=400\)

⇒ \(S_n=400 \Rightarrow \frac{n}{2}\left(a+a_n\right)=400\)

⇒ \(\frac{n}{2}(5+45)=400\)

⇒ \(n=\frac{400 \times 2}{50}=16\)

⇒ \(a_n=45\)

⇒ \(a+(n-1) d=45\)

⇒ \(5+(16-1) d=45\)

⇒ \(15 d=45-5=40\)

⇒ \(d=\frac{40}{15}=\frac{8}{3}\)

∴ \(n=16 \text { and } d=\frac{8}{3}\)

The number of terms and the common difference 16 and\(\frac{8}{3}\)

Question 6. The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Solution:

Given

The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9

Let the number of terms = n

a= 17, d = 9

an = 350 ⇒ a + {n- l)d = 350

⇒ \( 17+(n-1) \cdot 9=350\)

⇒ \((n-1)9=350-17=333\)

⇒ \(n-1=\frac{333}{9}=37\)

⇒ \(n=37+1=38\)

Now, from \( S_n=\frac{n}{2}\left(a+a_n\right)\)

⇒ \(S_{38}=\frac{38}{2}(17+350)\)

⇒ \(19 \times 367=6973\)

∴ n=38 and S_n=6973

Question 7. Find the sum of the first 22 terms of an A.P. in which d = 7 and the 22nd term is 149.
Solution:

d =7

a₂₂=149

⇒ \(a+(22-1) d=149\)

⇒ \(a+21 \times 7=149\)

⇒ \(a+147=149\)

⇒ \(a=149-147=2\)

⇒ \(S_n=\frac{n}{2}\left(a+a_n\right)\)

⇒ \(S_{22}=\frac{22}{2}\left(a+a_{22}\right)=11(2+149)\)

= 11 × 151 = 1661

The sum of the first 22 terms of an A.P. = 1661

Question 8. Find the sum of the first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively.
Solution :

Given

Second and third terms of an A.P are 14 and 18 respectively

Let the first form of A.P. be a and the common difference be d.

Now, \(a_2=14 \quad \Rightarrow \quad a+d=14 \quad \ldots(1)\)

and \(a_3=18 \quad \Rightarrow a+2 d=18 \quad \ldots(2)\).

Subtracting equation (1) from (2)

Put the value of d in equation (l),

a+4 = 14

a = 14-4 = 10

Now, from the formula \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{51}=\frac{51}{2}[2 \times 10+(51-1) \times 4]\)

⇒ \(\frac{51}{2}(20+200)\)

⇒ \(\frac{51}{2} \times 220=5610\)

∴ The sum of 51 terms = 5610

Question 9. If the sum of first 7 terms of an A.P is 49 and that of 1 7 terms is 289, find the sum of first n terms.
Solution:

Given

The sum of first 7 terms of an A.P is 49 and that of 1 7 terms is 289

Let the first term of A.P. be a and common difference be d.

S₇ = 49

⇒ \( \frac{7}{2}[2 a+(7-1) d]=49 \Rightarrow \frac{1}{2}[2 a+6 d]=7\)

⇒ \(a+3 d=7\)

⇒ \(S_{17}=289\)

⇒ \(\frac{17}{2}[2 a+(17-1) d]=289\)

⇒ \(\frac{1}{2}[2 a+16 d]=17 \Rightarrow a+8 d=17 \ldots(2)\)

Subtracting equation (1) from (2),

d = 2

Put the value of d in equation (1),

a + 3×2=7

⇒ a + 6=7

⇒ a =7 -6=1

Now, \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(\frac{n}{2}[2 \times 1+(n-1) \cdot 2]\)

⇒\(n(1+n-1)=n^2\)

∴ The sum of n terms of A.P. = n²

Question 10. Show that a₁, a₂, an A.P. where a„ is defined as below :

1. a_n = 3 + 4n
2. a_n = 9 – 5n

Also, find the sum of the first 15 terms in each case.

Solution:

the nth term of the sequence,

a_n=3+4n

Put n = 1\(a_1=3+4 \times 1=3+4=7\)

Put n = 2\(a_2=3+4 \times 2=3+8=11\)

Put n = 3\(a_3=3+4 \times 3=3+12=15\)

Now, \(a_2-a_1=11-7=4\)

⇒ \(a_3-a_2=15-11=4\)

⇒ \( a_2-a_1=a_3-a_2=4\)

∴ The difference between two consecutive terms of the sequence is constant.

So, the sequence is A.P.

Now, d = 4 and a = 7

A sum of first 15 terms

⇒ \(frac{n}{2}[2 a+(n-1) d]\)

⇒ \(\frac{15}{2}[2 \times 7+(15-1) \times 4]\)

⇒ \(\frac{15}{2}[14+56]=\frac{15}{2} \times 70=525\)

2. \(a_n=9-5 n\)

Put n = 1, \(a_1=9-5 \times 1=9-5=4\)

Put n = 2, \(a_2=9-5 \times 2=9-10=-1\)

Put n = 3, \(a_3=9-5 \times 3=9-15=-6\)

Now, \(a_2-a_1=-1-4=-5\)

⇒ \(a_3-a_2=6-(-1)=-6+1=-5\)

⇒ \(a_2-a_1=a_3-a_2=-5\)

∴ The difference between two consecutive terms of the sequence is constant.

So, the sequence is A.P.

Now d =-5,a = 4

∴ The sum of the first 15 terms

⇒ \(\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(\frac{15}{2}[2 \times 4+(15-1)(-5)]\)

⇒ \(\frac{15}{2}(8-70)=\frac{15}{2} \times(-62)=-465\)

Question 11. If the sum of the first n terms of an A.P. is 4n – n², what is the first term (that is S₁)? What is the sum of the first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
Solution:

Here S_n = 4n – n²

Put n = 1

⇒ \(S_1=4 \times 1-(1)^2=4-1=3\)

Put n = 2

⇒ \(S_2=4 \times 2-2^2=8-4=4\)

Second term \(a_2=S_2-S_1=4-3=1\)

Put n = 3

⇒ \(S_3=4 \times 3-3^2=12-9=3\)

⇒ \(a_3=S_3-S_2=3-4=-1\)

Put n = 9

⇒ \(S_9=4 \times 9-9^2=36-81=-45\)

Put n = 10

⇒ \( S_{10}=4 \times 10-10^2=40-100=-60\)

⇒ \(a_{10}=S_{10}-S_9=(-60)-(-45)\)

= -60+45 = -15

Replace n by (n – 1)

⇒ \(S_{n-1}=4(n-1)-(n-1)^2\)

⇒ \(4 n-4-\left(n^2-2 n+1\right)\)

⇒ \(4 n-4-n^2+2 n-1=6 n-n^2-5\)

⇒ \(a_n=S_n-S_{n-1}\)

⇒ \(\left(4 n-n^2\right)-\left(6 n-n^2-5\right)\)

⇒ \(4 n-n^2-6 n+n^2+5\)

⇒ 5-2 n

Question 12. Find the sum of the first 40 positive integers divisible by 6.
Solution:

The progression formed the positive integers divisible by 6

6, 12, 1 8, 24, … to 40 terms

Here a = 6, = 6, n = 40

From the formula, \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{40}=\frac{40}{2}[2 \times 6+(40-1) \times 6]\)

= 20(12 + 234) = 20 x 246 = 4920

The sum of 40 terms = 4920

Question 13. Find the sum of the first 15 multiples of 8.
Solution:

The first 15 multiples of 8 are 8, 16, 24, …, to 15 terms

Here, a = 8, d = 16 — 8 = 24 — 16 = 8, n= 15

From the formula \( S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{15}=\frac{15}{2}[2 \times 8+(15-1) \times 8]\)

⇒ \(\frac{15}{2} \times 8[2+14]\)

= 60×16 = 960

The sum of the first 15 multiples of 8 = 960

Question 14. Find the sum of the odd numbers between 0 and 50.
Solution:

Odd numbers between 0 and 50 are 1,3,5…..49.

Here, a= 1,d =3- 1 =5-3=2

Let, an = 49

⇒ 1+(n-1).2 = 49

⇒ 1 + 2n – 2 = 49

⇒ 2n – 2 = 49

⇒ 2n = 49 + 1 = 50

⇒ n = 25

Now, from the formula \(S_n=\frac{n}{2}\left(a+a_n\right)\)

⇒ \(S_{25}=\frac{25}{2}(1+49)\)

⇒ \(\frac{25}{2} \times 50=625\)

The sum of the odd numbers between 0 and 50 is 625.

Question 15. A contract on a construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, etc., the penalty for each succeeding day being? 50 more than for the preceding day. How much money does the contractor have to pay as a penalty if he has delayed the work by 30 days?
Solution:

Penalty for delay of first day = ₹200

Penalty for delay of second day = ₹250

Penalty for delay of third day = ₹300

This progression is an A.P.

Here a = 200, d = 250 – 200 = 300 – 250 = 50, n = 30

From the formula \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \( S_{30}=\frac{30}{2}[2 \times 200+(30-1) \times 50]\)

=15(400+1450)

=15×1850=27750

The contractor will pay the penalty of ₹27750.

Question 16. A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each of the prizes.
Solution:

Let first prize = ₹a

Common difference d =- 20, n = 7

Given, S₇ = 700

⇒ \(\frac{7}{2}[2 a+(7-1)(-20)]=700\)

⇒ \({[2 a-120]=200}\)

⇒ 2a =200+ 120

⇒ 2a =320

⇒ a = 160

∴ a₂ =a + d= 160-20= 140

⇒ \(a_3=a_2+d=140-20=120\)

⇒ \(a_4=a_3+d=120-20=100\)

⇒ \(a_5=a_4+d=100-20=80\)

⇒ \(a_6=a_5+d=80-20=60\)

⇒ \(a_7=a_6+d=60-20=40\)

Prizes are ₹160, ₹140, ₹120, ₹100, ₹80, ₹ 60 and ₹40.

Question 17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, would be the same as the class, in which they are studying, There are three sections of each class. How many trees will be planted by the students?
Solution:

The sequence so formed : 3, 6, 9, … is an A.P.

Here, a = 3,d = 6- 3 = 9- 6 = 3

and n= 12

Now, from the formula \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{12}=\frac{12}{2}[2 \times 3+(12-1) \times 3]\)

= 6(6 + 33) = 6 x 39 = 234

Total trees planted = 234

Question 18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with the centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm,…. What is the total length of such a spiral made up of thirteen consecutive semicircles? ( Take \(\pi=\frac{22}{7}\))

Solution:

The radius of first semicircle r₁ = 0.5 cm.

The radius of second semicircle r₂ =1.0 cm

The radius of third semicircle r₃ = 1.5 cm

.
.
.
.
.
.
.

This sequence is an A.P.

Here a = 0.5 and d = 1.0 – 0.5 = 0.5, n = 13

Now, the length of the spiral is made of 1 3 consecutive semicircles.

⇒ \(\pi r_1+\pi r_2+\pi r_3+\ldots \text { to } 13 \text { terms }\)

⇒ \(\pi\left[r_1+r_2+r_3+\ldots 13\right.\)

⇒ \(\frac{22}{7} \times \frac{13}{2}[2 a+(13-1) \cdot d]\)

⇒ \(\frac{143}{7}[2 \times 0.5+12 \times 0.5]\)

⇒ \(\frac{143}{7} \times 7=143 \mathrm{~cm}\)

The total length of such a spiral made up of thirteen consecutive semicircle 143cm.

Question 19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 1 9 in the next row, 1 8 in the row next to it and so on (see figure). In how many rows are the 200 logs placed and how many logs are in the top row?

Solution:

No. of logs in lowest row = 20.

Starting from the bottom.

Logs in first row = 20

Logs in the second row =19

Logs in the third row = 18

This sequence is an A.P. in which

a = 20, d = 19-20 =-l

Let no. of rows = n

∴ S_n=200

⇒ \(\frac{n}{2}[2 a+(n-1) \cdot d]=200\)

⇒ \(\frac{n}{2}[2 \times 20+(n-1) \cdot(-1)]=200\)

⇒ \(n(40-n+1)=400\)

⇒ n(41-n)=400

⇒ \(41 n-n^2=400\)

⇒ \(0=n^2-41 n+400\)

⇒ \(n^2-25 n-16 n+400=0\)

⇒ \(n(n-25)-16(n-25)=0\)

⇒ \((n-25)(n-16)=0\)

⇒ \(n-25=0 \text { or } n-16=0\)

⇒ \(n=25 \text { or } n=16\)

∴ n = 25th

a₂₅ =n + (25-1)d = 20 + 24(-l)

= -4 which is not possible.

∴ n= 16

n₁₆ =n + (16-1)d = 20 + 1 5(-1)

= 20-15 = 5

So total rows = 1 6

and no. of logs in upper row = 5

Question 20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket drops it In, and continues in the same way until all the potatoes arc in the bucket. What is the total distance the competitor has to run?
Solution:

Distance of the first potato from the first bucket = 5 m

Distance of the second potato from a bucket

= 5 + 3 = 8 m

Distance of the third potato from the bucket

= 8 + 3 = 11 cm

Once start from a bucket, pick up a potato and run back with it, drop it in the bucket.

Distance covered to drop the potatoes in the bucket.

= 2×5m,2×8m,2×11m, …

= 10 m, 16 m, 22 m,

Here a = 10, d = 16 – 10 = 22 – 16 = 6, n = 10

Distance covered to drop n potatoes in a bucket

⇒ \(\frac{n}{2}[2 a+(n-1) d]\)

Distance covered to drop 10 potatoes in a bucket

⇒ \(\frac{10}{2}[2 a+(10-1) d]\)

⇒ \(5[2 \times 10+9 \times 6]=5(20+54)\)

⇒ \(5 \times 74=370 \mathrm{~m}\)

The total distance the competitor has to run 370 meters,

NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Exercise 5.4 (Optional)

Question 1. Which term of the A.P.: 121, 117, 113…..is its first negative term?

[Hint: Find n for an < 0]

Solution:

Given, A.P.: 121, 117, 113, …

Here a= 121,d= 117- 121 = 113- 117 = -4

Let \(a_n<0 \quad \Rightarrow a+(n-1) d<0\)

⇒ \(121+(n-1)(-4)<0\)

⇒ \(121-4 n+4<0\)

⇒ \(-4 n<-125\)

⇒ \(n>\frac{125}{4} \quad \Rightarrow \quad n>31 \frac{1}{4}\)

n = 32, 33, 34…

∴ The first negative term = 32nd term.

Question 2. The sum of the third and the seventh terms of an A.P. is 6 and their product is 8. Find the sum of the first sixteen terms of the A.P.
Solution:

Let the first term of A.P. be a and the common difference be d.

∴ Third term a3 = a + 2d

and Seventh term a7 = a + 6d

According to the problem,

⇒ \( a_3+a_7=6\)

⇒ \(a_3 a_7=8\)

⇒ \(a_3\left(6-a_3\right)=8 \text { [from equation }(1)]\)

⇒ \(6 a_3-a_3^2=8\)

⇒ \(0=a_3^2-6 a_3+8\)

⇒ \(a_3^2-4 a_3-2 a_3+8=0\)

⇒ \(a_3\left(a_3-4\right)-2\left(a_3-4\right)=0\)

⇒ \(\left(a_3-4\right)\left(a_3-2\right) =0\)

⇒ \(a_3-4=0 \text { or } \quad a_3-2=0\)

⇒ \(a_3=4 \text { or } \quad a_3=2\)

If a₃ = 4 then from equation (1)

Now, \( a_7=6-4=2\)

a+2 d=4 ….(2)

a+6 d=2 ….(3)

On subtracting

-4d = 2

⇒ \(d=-\frac{1}{2}\)

Put the value of d in equation (2),

⇒ \(a+2\left(-\frac{1}{2}\right)=4\)

a-1=4 ⇒ a = 4+1=5

⇒ \(S_{16}=\frac{16}{2}[2 a+(16-1) d]\)

⇒ \(8\left[2 \times 5+15\left(-\frac{1}{2}\right)\right]\)

⇒ \(8\left(10-\frac{15}{2}\right)\)

⇒ \(8 \times \frac{5}{2}=20\)

If a₃ = 2 then from equation (1), a₇ = 4

Now, a + 2d = 2 …..(4)

a + 6d = 4 …..(5)

On subtracting

-4d =-2

d = \(\frac{1}{2}\)

Put the value of d in equation (4)

⇒ \(a+2 \times \frac{1}{2}=2 \Rightarrow a=2-1=1\)

⇒ \(S_{16}=\frac{16}{2}[2 a+(16-1) d]\)

⇒ \(8\left[2 \times 1+15 \times \frac{1}{2}\right]\)

⇒ \(8\left(2+\frac{15}{2}\right)\)

⇒ \(8 \times \frac{19}{2}=76\)

The sum of the first sixteen terms of the A.P = 76.

Question 3. A ladder has rungs 25 cm apart. (see figure). The rungs decrease uniformly in length from 45cm at the bottom to 25cm at the top. If the top and the bottom rungs are 2\(\frac{1}{2}\) m apart, what is the length of the wood required for the rungs?

[Hint : Number of rungs = \(\frac{250}{25}+1\)]

Solution:

Horizontal distance between first and last rung = 2\(\frac{1}{2}\) m = 250cm

and distance between two consecutive rungs

Number of rungs in ladder \(=\frac{250}{25}+1=10+1=11\)

Now, the length of the first rung a = 25 cm

Length of last rung l = 45 cm

Length of wood used in 11 rungs

⇒ \(\frac{11}{2}(a+l)=\frac{11}{2}(25+45)\)

⇒ \(11 \times 35=385 \mathrm{~cm}\)

The length of the wood required for the rungs 385cm.

Questionfrom4. The1 tohouses49. Showofa row that is numbered there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

Hint: \(S_{x-1}=S_{49}-S_x\)

Solution:

Numbers mark on houses: 1, 2, 3, …47, 48, 49

x is a number such that Sum of the numbers before x = sum of the numbers after x

1 +2 + 3 + …+(x-1)

= (x+ 1) + (x + 2) + … + 49

⇒ \(S_{x-1}=S_{49}-S_x\)

⇒ \( \frac{x-1}{2}[1+x-1]=\frac{49}{2}[1+49]-\frac{x}{2}(1+x)\)

⇒ \(\frac{x^2-x}{2}=1225-\frac{x^2+x}{2}\)

⇒ \(\frac{x^2-x}{2}+\frac{x^2+x}{2}=1225\)

⇒ \(\frac{x^2-x+x^2+x}{2}=1225\)

⇒ \(x^2=(35)^2\)

x=35

The value of x =35.

Question 5. A small terrace at a football ground comprises of 1 5 steps each of which is 50 m long and built of solid concrete.

Each step has a rise of \(\frac{1}{4}\)m and a tread of \(\frac{1}{2}\)m (see figure). Calculate the total volume of concrete required to build the terrace.

[Hint: Volume of concrete required to build first step = \(\frac{1}{4} \times \frac{1}{2} \times 50 \mathrm{~m}^3\)]

Solution:

Given, the length of each step=50m and breadth is \(\frac{1}{2}\)m

The number of steps are 15 and the height of each step from the ground from an A.P. is as follows:

⇒ \(\frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{4}{4}, \frac{5}{4}, \frac{6}{4}, \frac{15}{4}\)

So, the volume of concrete used in the first step

⇒ \(50 \times \frac{1}{2} \times \frac{1}{4}=\frac{50}{8} \mathrm{~m}^3\)

The volume of concrete used in the second step

⇒ \(50 \times \frac{1}{2} \times \frac{2}{4}=\frac{100}{8} \mathrm{~m}^3\)

The volume of concrete used in the third step

⇒ \(50 \times \frac{1}{2} \times \frac{3}{4}=\frac{150}{8} \mathrm{~m}^3\)

The volume of concrete used in the fourth step

⇒ \(50 \times \frac{1}{2} \times \frac{4}{4}=\frac{200}{8} \mathrm{~m}^3\)

So, the volume of total concrete

⇒ \(\frac{50}{8}+\frac{100}{8}+\frac{150}{8}+\frac{200}{8}+\ldots+\text { to } 15 \text { term }\)

⇒ \(a=\frac{50}{8}\)

⇒ \(d=\frac{100}{8}-\frac{50}{8}=\frac{50}{8} \text { and } n=15\)

Therefore, the total volume of concrete

⇒ \(V=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(\frac{15}{2}\left[2 \times \frac{50}{8}+(15-1) \frac{50}{8}\right]\)

⇒ \(\frac{15}{2} \times \frac{50}{8}[2+15-1]\)

⇒ \(\frac{15}{2} \times \frac{50}{8} \times 16=15 \times 50=750 \mathrm{~m}^3\)

Therefore, the volume of concrete used in the terrace = 750 m3

NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression Multiple Choice Questions

Question 1. The sum of the first 6 multiples of 3 is :

1. 55
2. 60
3. 63
4. 65

Answer: 3. 63

Question 2. The sum of 10 terms of the progression 5, 11, 17, … is :

1. 300
2. 320
3. 280
4. 240

Answer: 2. 320

Question 3. 8 times the 8th term of an A.P. is equal to 12 times the 12th term. Its 20th term is

1. 20
2. 0
3. -20
4. None of these

Answer: 2. 0

Question 4. The first two terms of an A.P. are 2 and 7. Its 18th term is :

1. 87
2. 92
3. 82
4. None of these

Answer: 1. 87

Question 5. How many terms are there in A.P. 42, 63, 84 … 210?

1. 7
2. 8
3. 10
4. 9

Answer: 4. 9

Question 6. In an A.P., d =-4,n = 7,an = 7, then the value of a is:

1. 6
2. 7
3. 28
4. 30

Answer: 3. 28

Question 7. The common difference between the two arithmetic progressions are same. If their first terms are 2 and 10 respectively, then the difference between their 5th terms is :

1. 8
2. 2
3. 10
4. 6

Answer: 1. 8

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Quadratic Equation

A quadratic equation In the variable x Is the equation of the form ax²+bx+c=0, where a, b, c are real numbers, a ≠ 0. For example.,

1. 3x² + 5x -1=0
2. 3x-x² + 1 =0

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Roots Of Quadratic Equation

A real number a is called a root of the quadratic equation ax² + bx + c = 0, a= 0 if
aα² + bα + c = 0

i.e., x = α satisfies the equation ax² + bx + c = 0

or x = α is a solution of the equation ax² + bx + c = 0

The roots of a quadratic equation ax² + bx + c = 0 are called zeroes of the polynomial ax² + bx + c.

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Solution Of A Quadratic Equation By Factorisation Method

Consider the quadratic equation ax² + bx + c = 0, a≠0

Let it be expressed as a product of two linear expressions (factors) namely (px + q) and (rx + s) where p, q, r, s are real numbers and p ≠ 0, r ≠ 0, then

ax² + bx + c = 0 ⇒ (px + q)(rx + s) = 0

⇒ px + q = 0  or  rx + s = 0

Read and Learn More Class 10 Maths Solutions Exemplar

⇒ \(x=-\frac{q}{p}\)  or  \(x=-\frac{s}{r}\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Solution Of A Quadratic Equation Solved Examples

Question 1. Which of the following are the solutions of 2x² – 5x – 3 = 0?

1. x = 2
2. x = 3
3. \(x=-\frac{1}{2}\)

Solution:

The given equation is 2x² – 5x – 3 = 0

1. On substituting x = 2 in the given equation

L.H.S. = 2×2² – 5×2-3 = 8-10-3 = -5 ≠ R.H.S.

∴ x = 2 is not a solution of 2x² – 5x – 3 = 0

2. On substituting x = 3 in the given equation

L.H.S. = 2×3²– 5×3-3 = 18 -15- 3 = 0 = R.H.S. 2

∴ x = 3 is a solution of 2ײ – 5x – 3 = 0

3. On substituting x = \(-\frac{1}{2}\) in the given equation

L.H.S = \(2 \times\left(\frac{-1}{2}\right)^2-5 \times\left(\frac{-1}{2}\right)-3\)

⇒ \(2 \times \frac{1}{4}+5 \times \frac{1}{2}-3\)

⇒ \(\frac{1}{2}+\frac{5}{2}-3=\frac{1+5-6}{2}\)

=0 = R.H.S

∴ x = \(-\frac{1}{2}\) is a solution of 2×2-5x-3 = 0

Question 2. If x = 2 and x = 3 are roots of the equation 3x²-mx+ 2n = 0, then find the values of m and n.
Solution:

Since, x = 2 is a solution of 3x² – mx + 2n = 0

∴ 3 × (2)² -m×2 + 2n = 0

⇒ 12 -2m + 2n=0

⇒ -m + n = -6 ……(1)

Again, x = 3 is a solution of 3x² – mx + 2n = 0

∴ 3 × (3)² – m × 3 + 2n = 0

⇒ 27 – 3m + 2n = 0

⇒ -3m + 2n = -27 ……(2)

On multiplying equation (1) by 2 and subtracting from (2) we get

-m =-15 or m= 15

On substituting the value of m in equation (1) we get

-15 + n =-6

⇒ n= -6+15

⇒ n=9

Hence,The values of m = 15 and n = 9

Question 3. Solve the following quadratic equation : (3x-5)(2x + 3) = 0
Solution:

Given equation is (3x – 5)(2x + 3) = 0

⇒ 3x -5=0 or 2x + 3 = 0

⇒ 3x = 5 or 2x = -3

⇒ \(x=\frac{5}{3}\)  or  \(x=-\frac{3}{2}\)

Here, x = \(\frac{5}{3}\) and x = \(-\frac{3}{2}\) are the solutions.

Question 4. Find the roots of the following quadratic equation by factorization: 2z² + az – a² = 0
Solution:

Given equation is 2z² + az – a² =0

⇒ 2z² + (2a – a) z -a² = 0

⇒ 2z² + 2az – az-a² = 0

⇒ 2z(z + a) -a(z + a) = 0

⇒ (z + a)(2z – a) = 0

⇒ z + a = 0 or 2z-a = 0

when z + a = 0 ⇒ z = – a

and 2z – a = 0 ⇒ z = \(\frac{a}{2}\)

Hence, the roots of the equation are -a and \(\frac{a}{2}\)

Question 5. Solve the following quadratic equations by factorization:

1. 4-11x = 3x²
2. \(x^2-\frac{11}{4} x+\frac{15}{8}=0\)

Solution:

1. Given equation is 4 -11x = 3x²

⇒ 3x²+11x-4 = 0

⇒ 3x²+(12-1)-4 = 0

⇒ 3x²+12x-x-4 = 0

⇒ 3x(x+4)(x+4)

⇒ (3x-1)(x+4)=0  or x+4 = 0

when 3x+1 = 0

⇒ \(x=\frac{1}{3}\)

and x + 4 = 0

x = -4

Hence,\(\frac{1}{3}\) and- 4 are roots of equation

2. Given equation is \(x^2-\frac{11}{4} x+\frac{15}{8}=0\)

Multiplying both sides by 8, we get

⇒ 8x²– 22x + 15 = 0

⇒ 8x² – (12 + 10)x+ 15 = 0

⇒ 8x²– 12x- 10x+ 15 = 0

⇒ 4x(2x- 3) – 5(2x- 3) = 0

⇒ (2x-3)(4x-5) = 0

∴ either 2x- 3 = 0  or 4x- 5 = 0

2x = 3 or 4x = 5

⇒ \(x=\frac{3}{2}\) or \(x=\frac{5}{4}\)

Hence \(x=\frac{3}{2}\) and \(x=\frac{5}{4}\) are the roots of given equation.

Question 6. Solve the following quadratic equation :

⇒ \(x^2-(1+\sqrt{2}) x+\sqrt{2}=0\)

Solution:

Given equation is

⇒ \(x^2-(1+\sqrt{2}) x+\sqrt{2}=0\)

⇒ \(x^2-x-\sqrt{2} x+\sqrt{2}=0\)

⇒ \(x(x-1)-\sqrt{2}(x-1)=0\)

⇒ \((x-1)(x-\sqrt{2})=0\)

x-1 = 0 or \(x-\sqrt{2}=0\)

when x-1 = 0 ⇒ x = 1

and \(x-\sqrt{2}=0\) ⇒ \(\sqrt{2}\)

Hence, 1 and \(\sqrt{2}\) are roots of the equation.

Question 7. Solve the following quadratic equation: a²b²x² + b²x- a²x-1 = 0
Solution:

Given equation is

a²b²x² + b²x- a²x-1=0

b²x(a2x + 1)-1 (a²x + 1 ) = 0

(a²x+1)(b²x- 1) = 0

a²x + 1 = 0 or b²x-1=0

when a²x+1=0 ⇒ \(x=-\frac{1}{a^2}\)

and b²x- 1=0 ⇒ \(x=\frac{1}{b^2}\)

Hence, \(-\frac{1}{a^2} \text { and } \frac{1}{b^2}\) are roots equation.

Question 8. Solve the following quadratic equation: 4x²– 2(a² + b²)x + a²b²=0
Solution.

Given equation is

4x² -2(a² + b²)x + a²b² = 0

4x²– 2a²x- 2b²x + a²b² = 0

2x(2x- a²)-b²(2x -a²) = 0

(2x-2)(2x-b²) = 0

2x-a² = 0

or 2x-b² = 0

when 2x-a² = 0 ⇒ \(x=\frac{a^2}{2}\)

and 2x-b² = 0 ⇒ \(x=\frac{b^2}{2}\)

Question 9. Solve the following equation :

⇒ \(\frac{2 x}{x-4}+\frac{2 x-5}{x-3}=\frac{25}{3},(x \neq 4,3)\)

Solution:

⇒ \( \frac{2 x}{x-4}+\frac{2 x-5}{x-3}=\frac{25}{3}\)

⇒ \(\frac{2 x(x-3)+(x-4)(2 x-5)}{(x-4)(x-3)}=\frac{25}{3}\)

⇒ \(\frac{2 x^2-6 x+2 x^2-5 x-8 x+20}{x^2-3 x-4 x+12}=\frac{25}{3}\)

⇒ \(\frac{4 x^2-19 x+20}{x^2-7 x+12}=\frac{25}{3}\)

⇒ \(3\left(4 x^2-19 x+20\right)=25\left(x^2-7 x+12\right)\)

⇒ \(12 x^2-57 x+60=25 x^2-175 x+300\)

⇒ \(25 x^2-175 x+300-12 x^2+57 x-60=0\)

⇒ \(13 x^2-118 x+240=0\)

⇒ \(13 x^2-(78+40) x+240=0\)

⇒ \(13 x^2-78 x-40 x+240=0\)

⇒ 13x(x-6)-40(x-6) = 0

⇒ (x-6)-409x-6) = 0

⇒ x-6 = 0 or 13x-40 = 0

⇒ x-6 0  or \(x=\frac{40}{13}\)

Hence, 6 and \(\frac{40}{13}\) are roots of the equation.

Question 10. Solve the following equation:

⇒ \(2\left(\frac{2 x-1}{x+3}\right)-3\left(\frac{x+3}{2 x-1}\right)=5,(x \neq-3,1)\)

Solution:

Given equation is

⇒ \(2\left(\frac{2 x-1}{x+3}\right)-3\left(\frac{x+3}{2 x-1}\right)=5\)…..(1)

Let \(\frac{2 x-1}{x+3}=y\)

Hence, \(\frac{x+3}{2 x-1}=\frac{1}{y}\)

Now from equation (1)

⇒ \(2 y-\frac{3}{y}=5\)

⇒ \(2 y^2-3=5 y\)

⇒ \(2 y^2-5 y-3=0\)

⇒ \(2 y^2-(6-1) y-3=0\)

⇒ \(2 y^2-6 y+1 y-3=0\)

⇒ 2y(y-3)+1(y-3)=0

⇒ (2y+1)(y-3)=0

⇒ 2y+1=0 or y-3=0

when 2y + 1 = 0 ⇒ y = \(-\frac{1}{2}\)

and y-3 = 0 ⇒ y = 3

Substituting values of y in equation (2)

when y = \(-\frac{1}{2}\)

⇒ \(\frac{2 x-1}{x+3}=-\frac{1}{2}\)

⇒ 2(2x-1) = -1(x + 3)

⇒ 4x- 2 = -x- 3

⇒ 5x = -1

⇒ x = \(-\frac{1}{5}\)

when y = 3

⇒ \(\frac{2 x-1}{x+3}=3\)

⇒ 2x- 1 = 3(x + 3) ⇒ 2x- 1 = 3x+9

⇒ -x = 10        ⇒           x =-10

Hence, x = -10 or x = \(-\frac{1}{5}\) are roots of the equation.

Question 11. Solve the equation:

⇒\(\frac{a}{x-b}+\frac{b}{x-a}=2 \quad(x \neq b, a)\)

Solution:

Given equation is \(\frac{a}{x-b}+\frac{b}{x-a}=2\)

⇒ \(\frac{a}{x-b}+\frac{b}{x-a}=1+1\)

⇒ \(\frac{a}{x-b}-1+\frac{b}{x-a}-1=0\)

⇒ \(\frac{a-x+b}{x-b}+\frac{b-x+a}{x-a}=0\)

⇒ \((a+b-x)\left(\frac{1}{x-b}+\frac{1}{x-a}\right)=0\)

a+b -x= 0 or \(\frac{1}{x-b}+\frac{1}{x-a}=0\)

when a +b-x= 0 ⇒  x=a +b

and when  \(\frac{1}{x-b}+\frac{1}{x-a}=0\) ⇒ \(\frac{x-a+x-b}{(x-b)(x-a)}=0\)

⇒ 2x-a-b = 0

⇒ 2x= a + b

⇒ \(x=\frac{a+b}{2}\)

Alternatively,

⇒ \(\frac{1}{x-b}=-\frac{1}{x-a}\)

⇒ \(x-b=a-x\)

⇒ \(2 x=a+b\)

⇒ \(x=\frac{a+b}{2}\)

Hence, x = a + b and \(x=\frac{a+b}{2}\) arc roots of the equation.

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Quadratic Formula

The roots of the quadratic equation ax² + bx+ c= 0 where a ≠ 0 can be obtained by using the formula.

⇒ \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

Proof: Given ax2 + bx + c = 0

Dividing each term by a, we get

⇒ \(x^2+\frac{b}{a} x+\frac{c}{a}=0\)   (∵ a ≠ 0)

⇒ \(x^2+\frac{b}{a} x=-\frac{c}{a}\)  (transposing the constant)

Adding \(\left(\frac{1}{2} \text { coefficient of } x\right)^2 \text { i.e. }\left(\frac{b}{2 a}\right)^2\) on both sides, we get

⇒ \(x^2+\frac{b}{a} x+\left(\frac{b}{2 a}\right)^2=\left(\frac{b}{2 a}\right)^2-\frac{c}{a}\)

⇒ \(\left(x+\frac{b}{2 a}\right)^2=\frac{b^2}{4 a^2}-\frac{c}{a}\)

⇒ \(\left(x+\frac{b}{2 a}\right)^2=\frac{b^2-4 a c}{4 a^2}\)

Taking square root on both sides, we get

⇒ \(x+\frac{b}{2 a}= \pm \frac{\sqrt{b^2-4 a c}}{2 a}\)  (try to understand ‘±’)

⇒ \(x=\frac{-b}{2 a} \pm \frac{\sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-b+\sqrt{b^2-4 a c}}{2 a}\) or ⇒ \(x=\frac{-b-\sqrt{b^2-4 a c}}{2 a}\)

This is known as “Sridharacharya Formula” or “quadratic formula”.

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Discriminant

For the quadratic equation ax² + bx + c = 0, the expression D = b²– 4ac is called the discriminant. Roots of ax² + bx + c = 0 are real, only when b² – 4ac > 0, otherwise they are imaginary (not real).

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Sum Of Roots And Product Of Roots

We know that the two roots of a quadratic equation ax² + bx + c = 0 are

⇒ \(\alpha=\frac{-b+\sqrt{D}}{2 a} \quad \text { and } \quad \beta=\frac{-b-\sqrt{D}}{2 a}\) where D=b2- 4acis called the discriminant.

∴ Sum of roots :

⇒ \(\alpha+\beta=\frac{-b+\sqrt{D}}{2 a}+\frac{-b-\sqrt{D}}{2 a}=\frac{-b+\sqrt{D}-b-\sqrt{D}}{2 a}=\frac{-2 b}{2 a}=\frac{-b}{a}\)

∴ Sum of roots = \(\frac{-b}{a}=-\frac{\text { Coeff. of } x}{\text { Coeff. of } x^2}\)

Product of roots :

⇒ \(\alpha \beta=\frac{-b+\sqrt{D}}{2 a} \times \frac{-b-\sqrt{D}}{2 a}\)

⇒ \(\frac{(-b)^2-(\sqrt{D})^2}{4 a^2}=\frac{b^2-D}{4 a^2}=\frac{b^2-\left(b^2-4 a c\right)}{4 a^2}=\frac{4 a c}{4 a^2}=\frac{c}{a}\)

∴ Product of roots : \(\frac{c}{a}=\frac{\text { Constant term }}{\text { Coeff. of } x^2}\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Sum Of Roots And Product Of Roots Solved Examples

Question 1. Find the roots of the following quadratic equation, if they exist by the method of completing the square.

⇒ \(3 x^2+4 \sqrt{3} x+4=0\)

Solution:

Given equation is

⇒ \(3 x^2+4 \sqrt{3} x+4=0\)

Dividing both sides by 3

⇒ \(x^2+\frac{4 \sqrt{3}}{3} x+\frac{4}{3}=0\)

⇒ \(x^2+\frac{4}{\sqrt{3}} x=\frac{-4}{3}\)

Adding \(\left(\frac{\text { coefficient of } x}{2}\right)^2 \text { i.e., }\left(\frac{2}{\sqrt{3}}\right)^2=\frac{4}{3}\) on both sides

⇒ \(x^2+\frac{4}{\sqrt{3}} x+\frac{4}{3}=-\frac{4}{3}+\frac{4}{3}\)

⇒ \(\left(x+\frac{2}{\sqrt{3}}\right)^2=0\)

⇒ \(\left(x+\frac{2}{\sqrt{3}}\right)=0 \quad \text { and } \quad\left(x+\frac{2}{\sqrt{3}}\right)=0\)

⇒ \(x=\frac{-2}{\sqrt{3}} \quad \text { and } \quad x=\frac{-2}{\sqrt{3}}\)

Hence, roots of the equation are \(\frac{-2}{\sqrt{3}} \text { and } \frac{-2}{\sqrt{3}}\)

Question 2. Find roots of the following quadratic equations by using the quadratic formula, if they exist.

1. 3x² +x-4 = 0
2. 3x² +x+ 4 = 0

Solution:

1. Given equation is 3x² +x-4 = 0

On comparing with ax² + bx+ c = 0, we get

a = 3, b = 1 and c = -4

∴ Discriminant, D = b² – 4ac

D = (1)² – 4 × 3 × (-4)

D = 1 +48

D = 49 > 0

Hence, the given equation has two real roots.

∴ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{-1 \pm \sqrt{49}}{6}=\frac{-1 \pm 7}{6}=\frac{6}{6}\)

⇒ \(\text { or } \frac{-8}{6}\)

⇒ \(1 \text { or } \frac{-4}{3}\)

⇒ \(x=1, \frac{-4}{3}\) are roots of the equation.

2. Given equation is 3x² +x+ 4 = 0

On comparing with ax² + bx + c = 0, we get

a = 3, b = 1 and c = 4

∴ Discriminant, D – b² – 4ac

D = (1)² – 4 × 3 × 4

D = 1 – 48

D = -47 < 0

Hence, the equation has no real roots.

Question 3. Find roots of the equation by quadratic formula : x+x -(a + 2) (a+1) = 0
Solution:

The given equation is x² + x- (a + 2) (a +1) = 0

Comparing it with Ax² + Bx + C = 0, we get

A = 1, B = 1 and C = -(a + 2) (a+1)

∴ \(x=\frac{-B \pm \sqrt{B^2-4 A C}}{2 A}\)

⇒ \(x=\frac{-1 \pm \sqrt{1^2-4 \times 1 \times[-(a+2)(a+1)]}}{\cdots}\)

⇒ \(x=\frac{-1 \pm \sqrt{1+4\left(a^2+3 a+2\right)}}{2}\)

⇒ \(x=\frac{-1 \pm \sqrt{1+4 a^2+12 a+8}}{2}\)

⇒ \(x=\frac{-1 \pm \sqrt{4 a^2+12 a+9}}{2}\)

⇒ \(x=\frac{-1 \pm \sqrt{(2 a+3)^2}}{2}\)

⇒ \(x=\frac{-1 \pm(2 a+3)}{2}\)

⇒ \(x=\frac{-1+2 a+3}{2} \text { and } \frac{-1-2 a-3}{2}\)

⇒ \(x=\frac{2 a+2}{2} \text { and } \frac{-2 a-4}{2}\)

⇒ x = (a + 1) and -(a + 2) are roots of the equation.

Question 4. Solve the following equation by the method of completing the square :

⇒ \(4 \sqrt{3} x^2+5 x-2 \sqrt{3}=0\)

Solution:

We have, \(4 \sqrt{3} x^2+5 x-2 \sqrt{3}=0\)

Dividing each term by \(4 \sqrt{3}\) we get

⇒ \(x^2+\frac{5}{4 \sqrt{3}} x-\frac{2 \sqrt{3}}{4 \sqrt{3}}=0\)

⇒ \(x^2+\frac{5}{4 \sqrt{3}} x=\frac{1}{2}\)

Adding \(\left(\frac{1}{2} \text { coefficient of } x\right)^2 \text { i.e., }\left(\frac{5}{8 \sqrt{3}}\right)^2\) to both sides, we get

⇒ \(x^2+\frac{5}{4 \sqrt{3}} x+\left(\frac{5}{8 \sqrt{3}}\right)^2=\left(\frac{5}{8 \sqrt{3}}\right)^2+\frac{1}{2}\)

⇒ \(\left(x+\frac{5}{8 \sqrt{3}}\right)^2=\frac{25}{192}+\frac{1}{2} \Rightarrow\left(x+\frac{5}{8 \sqrt{3}}\right)^2=\frac{25+96}{192}\)

⇒ \(\left(x+\frac{5}{8 \sqrt{3}}\right)^2=\frac{121}{192}\)

Taking the square root of both sides, we get

⇒ \(x+\frac{5}{8 \sqrt{3}}= \pm \frac{11}{8 \sqrt{3}}\)

∴ \(x=-\frac{5}{8 \sqrt{3}} \pm \frac{11}{8 \sqrt{3}}=\frac{-5 \pm 11}{8 \sqrt{3}}\)

⇒ \(x=\frac{-5+11}{8 \sqrt{3}} \quad \text { or } \quad x=\frac{-5-11}{8 \sqrt{3}}\)

⇒ \(x=\frac{3}{4 \sqrt{3}} \quad \text { or } \quad x=\frac{-2}{\sqrt{3}}\)

Hence x = \(\frac{3}{4 \sqrt{3}} \text { or } x=\frac{-2}{\sqrt{3}}\) are the solutions of given equation.

Question 5. Solve: x² + x- (a + 2) (a + 1) = 0 by

1. Factorisation
2. Method of completing the square

Solution:

1. By factorisation :

We have x² +x- (a + 2) (a + 1) = 0

⇒ x²+x× 1 -{a + 2) (a+ 1) = 0

⇒ x² +x[(a + 2) — (a + 1)] — (a + 2) (n + 1) = 0

⇒ x² +x(a + 2) -x(a + 1) – (a + 2) (a + 1) = 0

⇒ x[x+ (a + 2)] – (a + 1)[x+ (a + 2)] = 0

⇒ [x+(a + 2)] [x-(a+ 1)] = 0

∴ either x+ (a + 2) = 0  or x- (a + 1) = 0

⇒ x = -(a + 2) or x=(a+ 1)

2. By the method of completing the square:

We have x² + x- (a + 2) (a + 1) = 0

⇒ x² +x=(a + 2) (a + 1)

Adding \(\left(\frac{1}{2}\right)^2\) on both sides, we get

⇒ \(x^2+x+\left(\frac{1}{2}\right)^2=a^2+3 a+2+\left(\frac{1}{2}\right)^2\)

⇒ \(\left(x+\frac{1}{2}\right)^2=\frac{4 a^2+12 a+9}{4}\)

⇒ \(\left(x+\frac{1}{2}\right)^2=\left(\frac{2 a+3}{2}\right)^2\)

Taking the square root of both sides, we get

⇒ \(x+\frac{1}{2}= \pm \frac{2 a+3}{2}\)

∴ \(x=\frac{-1}{2} \pm \frac{2 a+3}{2}=\frac{-1 \pm(2 a+3)}{2}\)

∴ \(x=\frac{-1+2 a+3}{2}\text { or }x=\frac{-1-(2 a+3)}{2}\)

⇒ \(x=\frac{2(a+1)}{2}\text { or }x=\frac{-2(a+2)}{2}\)

⇒ \(x=(a+1)\text { or }x=-(a+2)\)

Question 6. Let f(x) = 3x² – 5x- 1. Then solve f(x) = 0 by

1. Factoring the quadratic
2. Using the quadratic formula
3. Completing the square and then rewrite f(x) in the formv4(x±B)² ± C.

Solution:

f(x) = 3x²– 5x- 1

f(x) = 0

3x²-5x- 1=0

1. The given quadratic equation cannot be fully factorised using real integers. So, it is better to solve this equation by any other method.

2. 3x-5x- 1 =0

Compare it with ax² +bx + c = 0, we get

a = 3, b = -5,c =-1

∴ Let two roots of this equation are

⇒ \(\alpha=\frac{-b+\sqrt{b^2-4 a c}}{2 a}\)

⇒ \(\frac{-(-5)+\sqrt{(-5)^2-4(3)(-1)}}{2 \times 3}\)

⇒ \(\frac{5+\sqrt{37}}{6}\)

and

⇒ \(\beta=\frac{-b-\sqrt{b^2-4 a c}}{2 a}\)

⇒ \(\frac{-(-5)-\sqrt{(-5)^2-4(3)(-1)}}{2 \times 3}\)

⇒ \(\beta=\frac{5-\sqrt{37}}{6}\)

∴ Two values of x are \(\frac{5+\sqrt{37}}{6} { and } \frac{5-\sqrt{37}}{6}\)

3. \(3 x^2-5 x-1=0\)

⇒ \(x^2-\frac{5}{3} x-\frac{1}{3}=0\)

⇒ \(x^2-\frac{5}{3} x+\ldots \ldots=\frac{1}{3}+\ldots \ldots\)

⇒ \(x^2-\frac{5}{3} x+\left(\frac{5}{6}\right)^2\)

⇒ \(\frac{1}{3}+\left(\frac{5}{6}\right)^2\)

⇒ \(\left[\text { adding }\left(\frac{\text { Coeff. of } x}{2}\right)^2 \text { on both sides }\right]\)

⇒ \(\left(x-\frac{5}{6}\right)^2=\frac{1}{3}+\frac{25}{36} \Rightarrow\left(x-\frac{5}{6}\right)^2=\frac{12+25}{36}\)

∴ \(\left(x-\frac{5}{6}\right)^2=\left(\frac{\sqrt{37}}{6}\right)^2\)

⇒ \(x-\frac{5}{6}= \pm \frac{\sqrt{37}}{6}\)

∴ Two values of x are \(\frac{5+\sqrt{37}}{6} \text { and } \frac{5-\sqrt{37}}{6} \text {. }\)

Now, f(x) = \(3 x^2-5 x-1=3\left(x^2-\frac{5}{3} x\right)-1\)

⇒ \(3\left(x^2-\frac{5}{3} x+\frac{25}{36}-\frac{25}{36}\right)-1\)

⇒ \(3\left(x-\frac{5}{6}\right)^2-\frac{25}{12}-1=3\left(x-\frac{5}{6}\right)^2-\frac{37}{12}\)

which is of the form A(x- B)²– C, where A = 3, B = \(\frac{5}{6}\), C = \(\frac{37}{12}\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Nature Of Roots Of A Quadratic Equation

The nature of roots of a quadratic equation ax² +bx + c = 0 depends on the value of its discriminant (D). i.e., upon b² – 4ac.

If a, b, and c are real numbers and a ≠ 0 then discriminant D = b²-4ac.

The value of discriminant affects the nature of roots in the following ways :

The roots of a quadratic equation are :

1. Real: When D > 0 i.e. (D > 0 or D = 0) (when quadratic equation can be expressed as in real linear factors, D > 0)
2. No Real (Imaginary): When D < 0
3. Real and Distinct: When D > 0
4. Real and Equal (Coincident): When D = 0

In this case each equal root will \(\left(\frac{-b}{2 a}\right)\)

Remember:

1.  ax-b>0 ⇒ \(x>\frac{b}{a} \text {, if } a>0 \text { and } x<\frac{b}{a} \text {, if } a<0\)
2. x²– a² > 0 ⇒ x<-a or x>a
3. x²– a² = 0 ⇒ x= —a or x =  a
4. x²– a² < 0 ⇒ x<a or x> -a ⇒ -a< x < 0
5. (x-a) (x-b)>0,a<b ⇒ x<a or x>b
6. (x- a) (x- b) < 0, a < b ⇒ a<x<b

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Nature Of Roots Of A Quadratic Equation Solved Examples

Question 1. Find the value of k so that the equation 2x² – 5x + k = 0 has two equal roots.
Solution:

The given equation is 2x² – 5x + K = 0

Comparing with ax² + bx + c = 0, we get

a = 2,b = -5 and c = k

The equation will have two equal roots if

D = 0

D = b²– 4ac = 0

or (-5)² -4×2×K = 0

25 – 8K = 0

⇒ \(k=\frac{25}{8}\)

The value of k =\(\frac{25}{8}\)

Question 2. The equation 3x² – 12x + (n – 5) = 0 has repeated roots. Find the value of n.
Solution:

Given equation is 3x²– 12x + (n – 5) = 0

Comparing with ax² +bx + c = 0, we get

a = 3, b = -12 and c = n- 5

The equation will have repeated (two equal) roots if

Discriminant (D) = 0

∴ D =b²-4ac=0

or (-12)²– 4 × 3 × (n- 5) = 0

⇒ 144 -12n + 60 = 0

⇒ 204- 12n = 0

⇒ -12n = -204

⇒ n = 17

Hence, the value of n is 17

Question 3. Find the value of k for which the equation 2 + k(2x +k- l) + 2 = 0 has real and equal
roots.
Solution:

Given equation is

x² + k(2x + k – 1) + 2 = 0

x² + 2kx +k(k- 1) + 2 = 0

x² + 2kx + (k² -k + 2) = 0

Comparing with ax² + bx + c = 0,we get

a = 1, b = 2k and c = k² – k + 2

For real and equal roots,

Discriminant (D) = 0

∴ D =b²-4ac=0

or (2k)² – 4(1)(k² -k + 2)= 0

4k² – 4(k² -k + 2) = 0

k²– (k² -k + 2) = 0

K-2 = 0

k = 2.

Hence, the value of k is 2

Question 4. Find the value of p, for which one root of the quadratic equation px² – 14x + 8 = 0 is 6 times the other.
Solution:

Given equation is px² – 14x + 8 = 0

Let a and p be two roots of quadratic equation px²– 14x + 8 = 0, such that β = 6a

∴ Sum of roots = \(\frac{-b}{a}\)

⇒ \(\alpha+6 \alpha=-\frac{(-14)}{p}\)

⇒ \(7 \alpha=\frac{14}{p} \Rightarrow \alpha=\frac{2}{p}\)….(1)

Product of roots = \(\frac{c}{a}\)

⇒ \(\alpha \cdot 6 \alpha=\frac{8}{p}\)

⇒ \(6 \alpha^2=\frac{8}{p} \Rightarrow \alpha^2=\frac{8}{6 p}\) ….(2)

From equations (1 ) and (2), we get

⇒ \(\left(\frac{2}{p}\right)^2=\frac{8}{6 p} \Rightarrow \frac{4}{p^2}=\frac{4}{3 p}\)

P² = 3P

p² – 3p = 0

p(p-3) = 0 (don’t cancel p on both sides)

Either p = 0 or p = 3.

But p = 0 is not possible, as on putting, p = 0 in the given equation, we don’t have a quadratic equation and therefore we cannot get two roots.

Hence, P = 3

Alternatively,

Let one root of the quadratic equation px² – 14x + 8 = 0 is a.

∴ pα² – 14α + 8 =0 ….(1)

∴ Other root of the equation will be 6a.

p(6α)²– 14(6α) + 8 =0

36pα²– 84α + 8=0

9pα² – 21α + 2 =0 ….(2)

Solving equations (1) and (2) by cross-multiplication method.

⇒ \(\frac{\alpha^2}{-14(2)-8(-21)}=\frac{\alpha}{8(9 p)-2 p}=\frac{1}{p(-21)-9 p(-14)}\)

⇒ \(\frac{\alpha^2}{-28+168}=\frac{\alpha}{70 p}=\frac{1}{105 p}\)

⇒ \( \frac{\alpha^2}{140}=\frac{1}{105 p}\text { and }\frac{\alpha}{70 p}=\frac{1}{105 p}\)

⇒ \(\alpha^2=\frac{140}{105 p}=\frac{4}{3 p}\text { and }\alpha=\frac{70 p}{105 p}=\frac{2}{3}\)

⇒ \(\left(\frac{2}{3}\right)^2=\frac{4}{3 p} \quad \Rightarrow \quad \frac{4}{9}=\frac{4}{3 p} \quad \Rightarrow \quad 3 p=9\)

p = 3

The value of p = 3

Question 5. The equation x²+2(m-1)x+ (m + 5) 0 has real and equal roots. Find the value of m.
Solution:

Given equation is x²+ 2 (m – 1 )x + (m+ 5) = 0

Comparing with ax² + bx + c = 0, we get

a = 1, b = 2(m -1), c = m + 5

The equation will have two real and equal roots if

Discriminant (D) = 0

∴ D = b² – 4ac= 0

or [2(m-1)]2-4 × 1 × (m + 5) = 0

4(m² + 1 – 2m) – 4(m + 5) = 0

4m² + 4- 8m – 4m – 20 = 0

4m²– 12m, – 16 = 0

m²– 3m, -4 = 0

m² – 4m, + m, – 4 = 0

m(m, – 4) + 1 (m – 4) = 0

(m + 1) (m – 4) = 0

m = -1 or m = 4

Hence, the value(s) of m are -1 and 4.

Question 6. If -4 is a root of the equation x² + px- 4 = 0 and the equation x² + px + q = 0 has coincident roots, find the values of p and q.
Solution:

Since -4 is a root of x² + px- 4 = 0

Hence, (-4) will satisfy the equation.

Therefore, (-4)² +p(-4) -4=0

16 – 4p – 4 = 0

-4p +12 = 0

-4p = -12 .

p = 3 …….(1)

Given that, x² + px + q = 0 has coincident roots.

D = b² – 4ac = 0

D = p²-4×1×q=0

p²-4q = 0

3² -4q =0 [from (1)]

9 – 4q = 0

-4q = -9

q = \(\frac{9}{4}\)

Hence, the values of p = 3 and q = \(\frac{9}{4}\)

Question 7. Prove that both roots of the equation (x -a) (x- b) + (x- b) (x- c) + {x- c) (x-a) = 0 are real but they are equal only when a =b =c.
Solution:

The given equation may be written as

3x² – 2(a + b + c)x+ (ab + bc + ac) = 0

∴ Discriminant D = B² – 4AC

D = \(4(a+b+c)^2-4 \times 3(a b+b c+a c)\)

D = \(4\left(a^2+b^2+c^2+2 a b+2 b c+2 a c\right)-12(a b+b c+a c)\)

D = \(4\left(a^2+b^2+c^2-a b-b c-a c\right)\)

D = \(2\left(2 a^2+2 b^2+2 c^2-2 a b-2 b c-2 a c\right)\)

D = \(2\left[a^2+b^2-2 t b+b^2+c^2-2 b c+c^2+a^2-2 a c\right]\)

D = \(2\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \geq 0\)

∵ \((a-b)^2 \geq 0,(b-c)^2 \geq 0 \text { and }(c-a)^2 \geq 0\)

Hence, both roots of the equation are real.

For equal root, we must have D = 0

⇒ (a – b)² + (b- c)² + (c- a)² = 0

⇒ a-b =0, b-c = 0, c-a = 0

⇒ a = b,b = c,c = a

⇒ a – b =c

Hence, roots are equal only when a=b = c

Question 8. Find the positive values of k for which the equations x² + kx + 64 = 0 and x² – 8x + k = 0 nail both have real roots:
Solution:

Given equations are

x² + kx + 64 = 0 …(1)

and x²-8x + k= 0 …(2)

Let D₁ and D₂ be discriminants of equations (1) and (2) respectively, then

D₁ = k² – 4 x 64 or  D₁ = k² – 656

and D₂ = (-8)² – 4k or D₂ = 64-4K

Both equations will have real roots, if

D₁ ≥ 0 and D₂ ≥ 0

⇒ K² -256 ≥ 0 and 64- 4k ≥ 0

⇒ k² ≥ 256 and64 ≥ 4k

⇒ k ≥ 16 and K ≤ 16

k = 16

Hence, both equations will have real roots, when k = 16.

The positive values of k = 16.

Question 9. Find the value(s) of k for which the given quadratic equations have real and distinct roots:

1. 2x² + kx + 4 = 0
2. 4x²-3kx+ 1=0
3. kx² + 6x + 1 = 0
4. x²-kx+ 9 = 0

Solution:

1. The given equation is 2x² + kx + 4 = 0

Comparing with ax² +bx + c = 0, we get

a-2, b=k and c = 4

∴ D = b²– 4ac ≥ 0 for real and distinct roots.

Therefore, D = k²– 4 × 2 × (4) ≥ 0

⇒ k² – 32 ≥ 0

⇒ k² ≥ 32

⇒ \(k \leq-4 \sqrt{2} \text { and } k \geq 4 \sqrt{2}\)

2. The given equation is 4x² – 3kx + 1 – 0

Comparing with ax² + bx + c = 0, wc get

a = 4, b = -3k and c = 1

∴ D = b² – 4ac ≥ 0 for real and distinct roots

Therefore, D = (-3k)² – 4 × 4 × 1 ≥ 0

9k² -16>0 ⇒ 9k² > 16

⇒ \(k^2 \geq \frac{16}{9}\)

⇒ \(k \leq-\frac{4}{3} \quad \text { and } \quad k \geq \frac{4}{3}\)

3.  The given equation is kx² + 6x+1 = 0

Comparing with ax² +bx + c = 0, we get

a = k, b = 6 and c = 1

D = b² – 4ac > 0 for real and distinct roots

Therefore, D = (6)²– 4 × k × 1 ≥ 0

36-4K ≥ 0

36 ≥ 4K

k ≤ 9

4. The given equation is

x² -kx + 9 = 0

Comparing with ax² + bx + c = 0, we get

a = 1, b = -k and c = 9

∴ D = b² – 4ac ≥ 0 for real and distinct roots

Therefore, D = (-k)²– 4 × 1 × 9 ≥ 0

⇒ K² – 36 ≥ 0

⇒ k ≤ -6 or K ≥ 6

Question 10. If roots of the equation (1 + m²)x² + 2mcx + (c² – a²) = 0 are equal, prove that : c² = a²(1+m²)
Solution:

We have,

(1 + m²)x² + 2mcx + (c² – a²) = 0

It has equal roots, if

D = 0

B²-4AC =0

⇒ (2mc)² – 4(1 + m²) (c² – a²) =0

⇒ 4m²c² – 4(c² – a² + m²c²– m²a²) = 0

⇒ m²c² -c² + a² – m²c² + m²a² = 0

⇒ c² = a²+ m²a² = a²(1 +m²)

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Word Problems Based On Quadratic Equations

To solve the word problem, first translate the words of the problem into an algebraic equation, then solve the resulting equation.

For solving a word problem based on a quadratic equation adopt the following steps:

Step 1: Read the statement of the problem carefully.

Step 2: Represent the unknown quantity of the problem by a variable.

Step 3: Translate the given statement to form an equation in terms of variables.

Step 4: Solve the equation.

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Quadratic Equations Solved Examples

Question 1. The sum of a number and its reciprocal is \(\frac{10}{3}\), find the number(s).
Solution:

Let the number be x

∴ According to a given statement

⇒ \(x+\frac{1}{x}=\frac{10}{3}\)

⇒ \(\frac{x^2+1}{x}=\frac{10}{3}\)

⇒ \(3 x^2+3=10 x\)

⇒ \(3 x^2-10 x+3=0\)

⇒ \(3 x^2-(9+1) x+3=0\)

⇒ \(3 x^2-9 x-x+3=0\)

⇒ 3x(x-3)- l(x-3) = 0

⇒ (3x-1) (x-3) = 0

when 3x- 1 = 0  \(x=\frac{1}{3}\)

and when x-3 = 0,  x = 3

Hence, the number(s) are 3 and \(\frac{1}{3}\)

Question 2. Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164.
Solution:

Let the larger part be x. Then, the smaller part = 16 – x

According to a given statement

⇒ 2x²-(16-x)² = 164

⇒ 2x² – (256 +x² – 32x) = 164

⇒ 2x² – 256- x² + 32x- 164 = 0

⇒ x² + 32x- 420 = 0

⇒ x² + 42x- 10x- 420 = 0

⇒ x(x + 42)- 10(x + 42) = 0

⇒ (x + 42) (x- 10) = 0

⇒ x = -42 or x =10

⇒ x = 10

Hence, the required parts are 10 and 6.

Question 3. The sum of squares of three consecutive natural numbers is 149, Find the numbers.
Solution:

Given

The sum of squares of three consecutive natural numbers is 149,

Let three consecutive natural numbers be a-, (x +1) and (x +2) respectively.

According to the given condition

⇒ \(x^2+(x+1)^2+(x+2)^2=149\)

⇒ \(x^2+\left(x^2+1+2 x\right)+\left(x^2+4+4 x\right)=149\)

⇒ \(3 x^2+6 x+5=149\)

⇒ \(3 x^2+6 x-144=0\)

⇒ \(x^2+2 x-48 =0\)

⇒ \(x^2+8 x-6 x-48=0\)

⇒ x(x+8)-6(x+8)=0

⇒ (x+8)(x-6)=0

⇒ x = -8 and x = 6

⇒ x = 6 (x = -8 is not a natural number)

Hence, the required natural numbers are 6, (6 + 1), (6 + 2) = 6, 7, 8 respectively.

Question 4. A two-digit number is such that the product of its digits is 8. When 63 is subtracted from the number digits interchange their places. Find the number.
Solution:

Given

A two-digit number is such that the product of its digits is 8. When 63 is subtracted from the number digits interchange their places.

Let the digit at the unit place be x and the digit at ten’s place be y

∴ Number = x × 1 + 10 × y

= x+ 10y.

After reversing the order of digits, the reversing number =y + 10x

According to the first condition,

xy = 8 ….(1)

According to the second condition,

⇒ (x+10y) – 63 -y + 10ar

⇒ 9y- 9x = 63

⇒ y -x = 7

⇒ y =x + 7 ….(2)

∴ from (1) and (2), we get

x (x + 7) = 8

⇒ x² +7x- 8 = 0

⇒ (x + 8) (x- 1) = 0

∴ x= 1 or x = -8

If \(\left.\begin{array}{l}
x=1 \\
y=1+7=8
\end{array}\right\} \quad \Rightarrow \text { number }=1+10(8)=81\)

If \(\left.\begin{array}{rl}
x=-8 \\
y=-8+7=-1
\end{array}\right\}\) not possible as digits cannot be negative.

So, required number = 81

Question 5. The denominator of a fraction is one more than twice the numerator. If the sum of the 16 fractions and its reciprocal is \(2 \frac{16}{21}\), find the fraction.
Solution:

Given

The denominator of a fraction is one more than twice the numerator. If the sum of the 16 fractions and its reciprocal is \(2 \frac{16}{21}\)

Let, the fraction be \(\frac{x}{y}\) where numerator is.v, then denominator = 2x + y

According to a given statement

⇒ \(\frac{x}{2 x+1}+\frac{2 x+1}{x}=2 \frac{16}{21}\)

Now, let \(\frac{x}{2 x+1}=a\)

⇒ \(a+\frac{1}{a}=\frac{58}{21} \Rightarrow \frac{a^2+1}{a}=\frac{58}{21}\)

⇒ \(21 a^2+21=58 a\)

⇒ \(21 a^2-58 a+21=0\)

⇒ \(21 a^2-49 a-9 a+21=0\)

⇒ \(7 a(3 a-7)-3(3 a-7)=0\)

⇒ \((7 a-3)(3 a-7)=0\)

⇒ \(a=\frac{3}{7} \quad \text { or } \quad a=\frac{7}{3}\)

when, \(a=\frac{3}{7} \Rightarrow \frac{x}{2 x+1}=\frac{3}{7}\)

⇒ 7x = 6x + 3

⇒ x = 3

when, \(a=\frac{7}{3} \Rightarrow \frac{x}{2 x+1}=\frac{7}{3}\)

⇒ 3x = 14x + 7

⇒ -11x = 7

⇒ \(\frac{-7}{11}\)

x = 3   \(\left(x=\frac{-7}{11} \text { is not a natural number }\right)\)

Hence, required fraction is \(\frac{x}{2 x+1}=\frac{3}{7} \text {.}\)

Question 6. The hypotenuse of a right-angled triangle is 6 meters more than twice the shorter side. If the third side is 2 meters less than the hypotenuse, find the sides of the triangle.
Solution:

Given

The hypotenuse of a right-angled triangle is 6 meters more than twice the shorter side. If the third side is 2 meters less than the hypotenuse

Let, the length of the shortest side be x meters

then hypotenuse = (2x + 6) metres

the third side = (2x + 6 – 2) metres

= (2x + 4) metres

Now, using Pythagoras theorem (2x+6)²=x² + (2x + 4)²

⇒ (4x² + 24x + 36)=x² + (4x²+ 16x + 16)

⇒ x² – 8x- 20 = 0

⇒ x² – 10x+2x- 20 = 0

⇒ x(x-10)+2(x+2) = 0

⇒ (x+10)(x+2) = 0

⇒ x = 10 or x = -2

⇒ x = 10

So, the length of the shortest side = 10 meters

length of the hypotenuse = (2 × 10 + 6) = 26 metres

and length of the third side = (2 × 10 + 4) = 24 metres

Hence, the sides of the triangle are 10 m, 24 m, and 26 m.

Question 7. The age of a father is twice the square of the age of his son. Eight years hence, the age of his father will be 4 years more than 3 times the age of the son. Find their present ages.
Solution:

Given

The age of a father is twice the square of the age of his son. Eight years hence, the age of his father will be 4 years more than 3 times the age of the son.

Let, the present age of the son be x years.

Hence, the age of father = 2×2

8 years hence, the age of son = (x + 8) years

and the age of father = (2x² + 8) years

According to a given statement

⇒ \(2 x^2+8=3(x+8)+4\)

⇒ \(2 x^2+8=3 x+24+4\)

⇒ \(2 x^2-3 x-20=0\)

⇒ \(2 x^2-8 x+5 x-20=0\)

⇒ \(2 x(x-4)+5(x-4)=0\)

⇒ (2 x+5)(x-4)=0

⇒ \(x=\frac{-5}{2}\) and x = 4

⇒ x = 4

Therefore, the present age of the son is 4 years and present age of the father is 2 × 4² = 32 years.

Question 8. Two taps running together can fill a tank in 3 \(3 \frac{1}{13}\) hours. If one tap takes 3 hours more 13 than the other to fill the tank, then how much time will each tap take to fill the tank?
Solution:

Given

Two taps running together can fill a tank in 3 \(3 \frac{1}{13}\) hours. If one tap takes 3 hours more 13 than the other to fill the tank

Let the time taken by TapI to fill a tank = x hrs

∴ The time taken by TapII to fill a tank = (x + 3) hrs

and time taken by both to fill a tank =  \(3 \frac{1}{13}=\frac{40}{13} \mathrm{hrs}\)

∴ Tap 1’s 1 hr work = \(\frac{1}{x}\)

Tap 2’s 1 hr work = \(\frac{1}{x+3}\)

and (Tap1 + Tap 2)’s 1hr work = \(\frac{13}{40}\)

⇒ \(\frac{1}{x}+\frac{1}{x+3}=\frac{13}{40}\)

⇒ \(\frac{x+3+x}{x(x+3)}=\frac{13}{40}\)

⇒ \(40(2 x+3)=13 x(x+3)\)

⇒ \(80 x+120=13 x^2+39 x\)

⇒ \(13 x^2-41 x-120=0\)

⇒ \(13 x^2-65 x+24 x-120=0\)

⇒ \(13 x(x-5)+24(x-5)=0\)

⇒ (x-5)(13 x+24)=0

∴ Either x-5 = 0 or 13x+ 24 = 0

x = 5 or \(x=\frac{-24}{13}\)

But time cannot be negative, so we reject x = \(\frac{-24}{13}\)

x = 5

Hence, time taken by Tap 1 = 5 hrs

and time taken by Tap 2 = (5 + 3) hrs = 8 hrs.

Question 9. A takes 6 hours less than B to complete a work. If together they complete the work in 13 hours 20 minutes, find how much time B alone takes to complete the work.
Solution:

Given

A takes 6 hours less than B to complete a work. If together they complete the work in 13 hours 20 minutes,

Let, B alone complete the work in* hours, then A alone will complete the work in (x- 6) hours.

⇒ \(\frac{1}{x-6}+\frac{1}{x}=\frac{3}{40}\)    \(\left(13 \mathrm{hr}+20 \mathrm{~min}=\frac{40}{3} \mathrm{hrs} .\right)\)

⇒ \(\left(13 \mathrm{hr}+20 \mathrm{~min}=\frac{40}{3} \mathrm{hrs} .\right)\)

⇒ \(\frac{x+x-6}{(x-6) x}=\frac{3}{40}\)

⇒ \(3 x^2-18 x=80 x-240\)

⇒ \(3 x^2-98 x+240=0\)

⇒ \(3 x^2-90 x-8 x+240=0\)

⇒ \(3 x(x-30)-8(x-30)=0\)

⇒ \((x-30)(3 x-8)=0\)

x = 30

⇒ \(x=\frac{8}{3}\)

x = 30   \(\left(\text { if } x=\frac{8}{3} \text {, then } x-6 \text { in negative }\right)\)

B alone will take 30 hours to complete the work.

Question 10. An airplane takes 1 hour less for a journey of 1200 km if its speed is increased by 100 km/hr from its usual speed. Find its usual speed.
Solution:

Given

An airplane takes 1 hour less for a journey of 1200 km if its speed is increased by 100 km/hr from its usual speed.

Let the usual speed of an airplane be x km/hr. Given, distance = 1200 km

Time taken for journey of 1200 km = \(\frac{1200}{x} \text { hours }\)

⇒ \(\left(\text { Time }=\frac{\text { Distance }}{\text { Speed }}\right)\)

When the speed is increased by 100 km/hr, time taken for the same journey = \(\frac{1200}{x+100} \text { hours. }\)

According to the given condition

⇒ \(\frac{1200}{x}-\frac{1200}{x+100}=1\)

⇒ \(\frac{1200(x+100)-1200 x}{x(x+100)}=1\)

⇒ \(1200 x+120000-1200 x=x^2+100 x\)

⇒ \(x^2+100 x-120000=0\)

⇒ \(x^2+400 x-300 x-120000=0\)

⇒ \(x(x+400)-300(x+400)=0\)

⇒ \((x+400)(x-300)=0\)

⇒ x = -400 or x = 300

⇒ x = 300 km/hr

∴ The usual speed = 300 km/hr.

Question 11. A motor boat whose speed is 15 km/hr in still water goes 30 Ion downstream and comes back in a total of 4 hours 30 minutes. Determine the speed of water.
Solution:

Given

A motor boat whose speed is 15 km/hr in still water goes 30 Ion downstream and comes back in a total of 4 hours 30 minutes.

Let, the speed of water be x km/hr.

Given the speed of a motor boat in still water is 15 km/hr.

Therefore, its speed downstream is (15 + x) km/hr and the speed upstream is (15- x) lon/hr.

Time taken for going 30 Ion downstream = \(\frac{30}{15+x} \text { hours. }\)

Time taken for going 30 Ion upstream = \(\frac{30}{15-x} \text { hours. }\)

According to the given condition

⇒ \(\frac{30}{15+x}+\frac{30}{15-x}=4+\frac{30}{60}\)

⇒ \(\frac{30(15-x)+30(15+x)}{(15+x)(15-x)}=\frac{9}{2}\)

⇒ \(\frac{450-30 x+450+30 x}{225-x^2}=\frac{9}{2}\)

⇒ \(900 \times 2=9\left(225-x^2\right)\)

⇒ \(1800=2025-9 x^2\)

⇒ \(9 x^2=225 \text { or } x^2=25\)

⇒ \(x= \pm 5\)

⇒ x=5

Hence, the speed of water is 5 Km/hr.

Question 12. A dealer sells an article for ₹24 and gains as much percent as the price of the article. Find the cost price of the article.
Solution:

Given

A dealer sells an article for ₹24 and gains as much percent as the price of the article.

Let, the C.P. of article be ₹x

Then, gain = x %

∴ \(\text { S.P. }=\frac{100+\text { gain } \%}{100} \times \text { C.P. } \quad\left(\text { or S.P. }=\text { C.P. }+ \text { C.P. } \times \frac{x}{100}\right)\)

⇒ \(24=\frac{100+x}{100} \times x\)

⇒ \(2400=100 x+x^2\)

⇒ \(x^2+100 x-2400=0\)

⇒ \(x^2+120 x-20 x-2400=0\)

⇒ \(x(x+120)-20(x+120)=0\)

⇒ \((x-20)(x+120)=0\)

⇒ x = 20 or x = -120

⇒ x = 20

Hence, the cost of the article is ₹20.

Question 13. A piece of cloth costs ₹200. If the piece was 5 m longer and each meter of cloth cost 2 less, the cost of the piece would have remained unchanged. How long is the piece and what is the original rate per meter?
Solution:

Given

A piece of cloth costs ₹200. If the piece was 5 m longer and each meter of cloth cost 2 less, the cost of the piece would have remained unchanged.

Let, the length of the piece be A meters

Since the cost of A meters of cloth = ₹200

⇒ Cost of each metre of cloth = \(₹ \frac{200}{x}\)

New length of cloth = (x + 5)m

New cost of each metre of cloth = \(₹ \frac{200}{x+5}\)

Now, given \(\frac{200}{x}-\frac{200}{x+5}=2\)

⇒ \(\frac{200(x+5)-200 x}{x(x+5)}=2 \quad \Rightarrow \quad \frac{200 x+1000-200 x}{x^2+5 x}=2\)

⇒ \(1000=2 x^2+10 x\)

⇒ \(2 x^2+10 x-1000=0\)

⇒ \(x^2+5 x-500=0\)

⇒ \(x^2+25 x-20 x-500=0\)

⇒ \(x(x+25)-20(x+25)=0\)

⇒ \((x-20)(x+25)=0\)

x = 20 or x = -25

Since length cannot be negative

Hence, x = 20 m

and the original rate per metre = \(₹ \frac{200}{20}=₹ 10\)

Question 14. Some students planned a picnic. The budget for the food was ₹480. As eight of them failed to join the picnic, the cost of the food for each member increased by ₹10. Find how many students went for a picnic.
Solution:

Given

Some students planned a picnic. The budget for the food was ₹480. As eight of them failed to join the picnic, the cost of the food for each member increased by ₹10.

Let, no. of students who planned the picnic = x

Given, the budget for food = ₹480

∴ Share of each student = \(₹ \frac{480}{x}\)

Since eight of them failed to join the picnic

∴ No. of students went for picnic = (x- 8)

∴ Share of each student = \(₹ \frac{480}{x-8}\)

According to the given condition

⇒ \(\frac{480}{x-8}-\frac{480}{x}=10\)

⇒ \(\frac{480 x-480(x-8)}{(x-8) x}=10\)

⇒ \(\frac{480 x-480 x+3840}{x^2-8 x}=10\)

⇒ \(10 x^2-80 x=3840\)

⇒ \(x^2-8 x-384=0\)

⇒ \(x^2-24 x+16 x-384=0\)

⇒ \(x(x-24)+16(x-24)=0\)

⇒ (x- 24) (x + 16) =0 ⇒  x = 24    or  x = -16

Since no. of students cannot be negative

Hence, x = 24

No. of students who went for picnic = x- 8 = 24-8=16 students.

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Quadratic Equation Exercise 4.1

Question 1. Check whether the following are quadratic equations:

1. \((x+1)^2=2(x-3)\)
2. \(x^2-2 x=(-2)(3-x)\)
3. (x-2)(x+1) =(x-1)(x+3)
4. \((x-3)(2 x+1)=x(x+5)\)
5. \((2 x-1)(x-3)=(x+5)(x-1)\)
6. \(x^2+3 x+1=(x-2)^2\)
7. \((x+2)^3=2 x\left(x^2-1\right)\)
8. \(x^3-4 x^2-x+1=(x-2)^3\)

Solution:

1. \((x+1)^2=2(x-3)\)

⇒ x² + 2x + 1 = 2x- 6

⇒ x² + 7 =0

The highest power of the variable x in it is 2.

∴ The given equation is a quadratic equation.

2. \(x^2-2 x=(-2)(3-x)\)

x² -2x =- 6 + 2x

x²-2x-2x+6 =0

x² -4x + 6 =0

The highest power of the variable x in it is 2.

∴ Given equation is a quadratic equation.

3. (x-2)(x+1) =(x-1)(x+3)

⇒ x(x + 1 ) -2 (x + 1 ) = x(x + 3) – 1 (x + 3)

⇒ x² +x-2x- 2 =x² + 3x-x-3

⇒ x² – x- 2 =x² + 2x- 3

⇒ x² +x-2- x² +3 = 0

⇒ -3x+1 = 0

The highest power of the variable x is not 2, in it.

∴ Given equation is not a quadratic equation.

4. (x-3) (2x+ 1) = x(x + 5)

x(2x+1)-3 (2x + 1) =x² + 5X

2x² + x- 6x- 3 – x² – 5x =0

x²– 10x – 3 =0

The highest power of the variable x in it is 2.

∴ Given equation is a quadratic equation.

5. (2x-1)(x-3)=(x +5)(x-1)

2x(x-3)-1(x-3) = x(x-1) + 5(x- 1)

2x² – 6x-x + 3 = x² -x+ 5x- 5

2x² – 7x + 3 = x² + 4x- 5

2x²– 7x + 3 -x² – 4x + 5 = 0

x²– 11x+ 8 = 0

The highest power of variable x is 2 in it.

∴ Given equation is a quadratic equation.

6. x² + 3x + 1 = (x- 2)²

x² + 3x + 1 =x² – 4x + 4

x² + 3x+ 1 -x² + 4x- 4 = 0

7x – 3 = 0

The highest power of the variable x is not 2 in it.

∴ Given equation is not a quadratic equation.

7. (x + 2)³ = 2x(x² – 1 )

⇒ x³ + 3.x.2 .(x + 2) + 2³ = 2x³ – 2x

⇒ x³+ 6x² + 12x + 8 – 2x³ + 2x =0

⇒ -x³+6x²+14x+8 = 0

The highest power of the variable x is not 2 in it.

∴ Given equation is a quadratic equation.

8. \(x^3-4 x^2-x+1=(x-2)^3\)

⇒ \(x^3-4 x^2-x+1=x^3-3 x \cdot 2(x-2)-2^3\)

⇒ \(x^3-4 x^2-x+1=x^3-6 x^2+12 x-8\)

⇒ \(x^3-4 x^2-x+1-x^3+6 x^2-12 x+8=0\)

⇒ \(2 x^2-13 x+9=0\)

Question 2. Represent the following situations in the form of quadratic equations :

1. The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of tire plot.
2. The product of two consecutive positive integers is 306. We need to find the integers.
3. Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
4. A train travels a distance of 480 1cm at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Solution:

1. Let the breadth of the plot =x meter

∴ Length of plot = (2x + 1) metre

Now, from length x breadth = area

(2x + 1) × x =528

⇒ 2x² + x =528

⇒ 2x² + x- 528 =0

which is the required quadratic equation.

2. Let two consecutive positive integers are x and x+ 1

∵ The product of two consecutive positive integers = 306

∴ x(x+1) = 360

⇒ x²+x = 306

⇒ x²+x-306 = 0

which is the required quadratic equation.

3. Let the present age of Rohan = x years

∴ Present age of Rohan’s mother

= (x + 26), years

After 3 years,

Rohan’s age = (x + 3) years

The age of Rohan’s mother = (x + 26 + 3) years

= (x + 29) years

According to the problem,

After 3 years, the product of their ages = 360

(x + 3) (x + 29) = 360

x(x + 29) + 3(x + 29) = 360

x² + 29x + 3x + 87 = 360

x² + 32x + 87 – 360 = 0

x² + 32x- 273 = 0

which is the required quadratic equation.

4. Let the speed of train = A km/hr

Distance = 480 1cm

∴ Time taken to cover a distance of 480 km

⇒ \(\frac{480}{x} \mathrm{hrs}\)

If, the speed of the train = (x- 8) Km/hr  then the time is taken to cover 480 Km distance

⇒ \(\frac{480}{x-8} \mathrm{hrs}\)

According to the problem,

⇒ \(\frac{480}{x-8}-\frac{480}{x}=3\)

⇒ \(\frac{480 x-480(x-8)}{x(x-8)}=3\)

⇒ 480x- 480x + 3840 = 3x(x- 8)

⇒ 3840 = 3(x² – 8x)

⇒ 1280 =x²– 8x

⇒ 0= x² -8x -1280

⇒ x² -8x- 1280 = 0

which is the required quadratic equation.

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Quadratic Equation Exercise 4.2

Question 1. Find the roots of the following quadratic equations by factorisation:

1. x²-3x-10 = 0
2. 2x²+x-6 = 0
3. \(\sqrt{2} x^2+7 x+5 \sqrt{2}=0\)
4. \(2 x^2-x+\frac{1}{8}=0\)
5. 100x²– 20x + 1=0

Solution:

1. x²-3x-10 = 0

⇒ x²-5 x+2 x-10=0

⇒ x(x-5)+2(x-5)=0

⇒ (x-5)(x+2)=0

⇒ 5=0 or x+2=0

⇒ x=5 or x=-2

∴ Roots of given quadratic equation = 5, -2

2. 2 x^2+x-6=0

⇒ 2 x^2+4 x-3 x-6=0

⇒ 2 x(x+2)-3(x+2)=0

⇒ (x+2)(2 x-3)=0

⇒ x+2 =0 or 2x-3 = 0

⇒ x = -2 or 2x = 3

⇒ x =-2 or x = \(\frac{3}{2}\)

∴ Roots of given quadratic equation = -2, \(\frac{3}{2}\)

3. \(\sqrt{2} x^2+7 x+5 \sqrt{2}=0 \)

⇒ \(\sqrt{2} x^2+5 x+2 x+5 \sqrt{2}=0\)

⇒ \((\sqrt{2} x+5)+\sqrt{2}(\sqrt{2} x+5)=0\)

⇒ \((\sqrt{2} x+5)(x+\sqrt{2})=0\)

⇒ \(\sqrt{2} x+5=0 \quad \text { or } x+\sqrt{2}=0\)

⇒ \(\sqrt{2} x=-5 \text { or }x=-\sqrt{2}\)

⇒ \(x=\frac{-5}{\sqrt{2}} \text { or }x=-\sqrt{2}\)

∴ Roots of given quadratic equation

⇒ \(\frac{-5}{\sqrt{2}},-\sqrt{2} \text {. }\)

4. \(2 x^2-x+\frac{1}{8}=0\)

⇒ \(16 x^2-8 x+1=0\)

⇒ \(16 x^2-4 x-4 x+1=0\)

⇒ \(4 x(4 x-1)-1(4 x-1)=0\)

⇒ (4x-1)(4 x-1)=0

⇒ 4x-1 = 0  4x- 1 = 0

⇒ 4x= 1 or 4x= 1

⇒ \(x=\frac{1}{4} \quad \text { or } \quad x=\frac{1}{4}\)

∴ Roots of given quadratic equation = \(\frac{1}{4}, \frac{1}{4} \text {.}\)

5. \(100 x^2-20 x+1=0\)

⇒ \(100 x^2-10 x-10 x+1=0\)

⇒ 10x(10 x-1)-1(10 x-1)=0

⇒ (10x-1)(10 x-1)=0

⇒ 10x-1=0 or 10 x-1=0

⇒ 10x=1 or 10 x=1

⇒ \(x=\frac{1}{10} \text { or } \quad x=\frac{1}{10}\)

∴ Roots of given quadratic equation \(\frac{1}{10}, \frac{1}{10}\)

Question 2. Represent the following situations mathematically:

1. John and Jiwanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
2. A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹750. We would like to find out the number of toys produced on that day.

Solution : 

1. Let the number of marbles initially with John =x

∴ Initially, the number of marbles with Jiwanti = 45-x

After losing 5 marbles each,

Remaining marbles with John =x- 5

Remaining marbles with Jiwanti = 45 -x- 5

= 40-x

According to the problem,

Product of remaining marbles with them =124

(x- 5) (40-x) = 124

⇒ x(40-x) -5(40-x) = 124

⇒ 40x-x²– 200 + 5x= 124

⇒ -x² + 45x- 200 = 124

⇒ 0= 124 +x²– 45x + 200

⇒ x² – 45x + 324 = 0

⇒ x²– 36x -9x + 324 = 0

⇒ x(x- 36) -9(x- 36) = 0

⇒ (x-36) (x-9) = 0

⇒ x-36 = 0 or x-9 = 0

⇒ x = 36 or x = 9

If x = 36 then 45-x = 45-36 = 9

If x = 9 then 45-x = 45-9 = 36

∴ Marbles with John = 36

and marbles with Jiwanti = 9

or

Marbles with John = 9

and marbles with Jiwanti = 36

2. Let the number of toys = x

Cost of each toy =  ₹(55 -x)

Cost of x toys = ₹(55 – x) x-

₹(55x-x2)

According to the problem,

⇒ 55x -x² = 750

⇒ 0 =x²-55X+750

⇒ x² – 30x- 25x + 750 =0

⇒ x(x- 30) -25 (x-30) =0

⇒ (x-30) (x-25) =0

⇒ x-30 = 0 or x- 25 = 0

⇒ x = 30 or x = 25

∴ Number of toys produced = 25 or 30.

Question 3. Find two numbers whose sum is 27 and whose product is 182.
Solution:

Let one number = x

∴ Second number =27 -x

According to the problem,

Product of two numbers = 182

⇒ x (27-x) = 182

⇒ 27x -x² = 182

⇒ 0 =x² – 27x+ 182

⇒ x²– 13x- 14x+ 182 =0

⇒ x(x- 13) -14 (x- 13) =0

⇒ (x- 13) (x- 14) =0

⇒ x-13 = 0 or x-14 =0

⇒ x= 13 or x = 14

If x = 13, then 27-x = 27-13 = 14

If x = 14, then 27 -14 = 13

Therefore,numbers=(13 and 14)or(14and 13).

Question 4. Find two consecutive positive integers, a sum of whose squares is 365.
Solution:

Given

The sum of whose squares is 365

Let two consecutive positive integers be x and x + 1.

According to the problem,

x² + (x+1)² =365

x² + x² + 2x + 1 =365

2x² + 2x + 1 – 365 = 0

2x² + 2x- 364 = 0

x²+x- 182 =0

x²+14x-13x-182 = 0

=x (x + 14) (x- 13) =0

x + 14=0 or x-13=0

x = -14 or x = 13

x is a positive integer,

∴ neglecting x = -14,

x = 13

x+1 = 13+1 = 14

Therefore, required positive integers = 13 and 14.

Question 5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Solution:

Given

The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm

Let the base of the right triangle = x cm

∴ Its height = (x- 7) cm

Given, the hypotenuse of the right triangle = 13cm

From Pythagoras theorem, in a right triangle (base)² + (height)² = (hypotenuse)²

⇒ x² + (x- 7)² = 132

⇒ x²+x²– 14x + 49 = 169

⇒ 2x² – 14x + 49 — 169 =0

⇒ 2x² -14x- 120 =0

⇒ x² – 7x- 60 =0

⇒ x²– 12x + 5x -60 =0

⇒ x(x- 12) + 5(x- 12) =0

⇒ (x- 12) (x+ 5) =0

⇒ x-12 =0 or x + 5= 0

⇒ x = 12 or x = -5

but the value of x cannot be negative.

∴ Neglecting

x = -5

x = 12

⇒ x-7 =12-7 = 5

Therefore, the other two sides of the triangle =12 cm and 5 cm.

Question 6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹90, find the number of articles produced and the cost of each article.
Solution:

Given

A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹90,

Let the number of pottery articles produced in a day = x

∴ Cost of each article = ₹(2x+ 3)

⇒ Cost of x articles = ₹(2x + 3)x

According to the problem,

(2x + 3) x = 90

⇒ 2x² + 3x = 90

⇒ 2x² + 3x-90 =0

⇒ 2x² + 15x- 12x- 90 = 0

⇒ x(2x+ 15) -6(2x+ 15) =0

⇒ (2x+ 1 5) (x -6) =0

⇒ 2x + 15 = 0 or x -6 = 0

⇒ \(x=-\frac{15}{2} \text { or } \quad x=6\)

but x = \(-\frac{15}{2}\) is not possible

∴ x = 6

⇒ 2x + 3 = 2×6 + 3=15

Therefore, the number of pottery articles in a day = 6, and the cost of each article =₹15.

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Quadratic Equation Exercise 4.3

Question 1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square :

1. 2x² – 7x + 3 = 0
2. 2x² + x- 4 = 0
3. \(4 x^2+4 \sqrt{3} x+3=0\)
4. 2x² +x + 4 = 0

Solution:

1. 2x² – 7x + 3 = 0

⇒ \(x^2-\frac{7}{2} x+\frac{3}{2}=0\)

⇒ \( {\left[x^2-2 x \cdot \frac{7}{4}+\left(\frac{7}{4}\right)^2\right]+\frac{3}{2}-\left(\frac{7}{4}\right)^2 }=0\)

⇒ \(\left(x-\frac{7}{4}\right)^2+\frac{24-49}{16}=0\)

⇒ \(\left(x-\frac{7}{4}\right)^2-\frac{25}{16}=0\)

⇒ \(\left(x-\frac{7}{4}\right)^2-\left(\frac{5}{4}\right)^2=0\)

⇒ \(\left(x-\frac{7}{4}-\frac{5}{4}\right)\left(x-\frac{7}{4}+\frac{5}{4}\right)=0\)

⇒ \((x-3)\left(x-\frac{1}{2}\right)=0\)

⇒ \(x-3=0\text { or }x-\frac{1}{2}=0\)

⇒ \(x=3\text { or }x=\frac{1}{2}\)

∴ Roots of given equation = \(3, \frac{1}{2}\)

2. \(2 x^2+x-4=0\)

⇒ \(x^2+\frac{1}{2} x-2=0\)

⇒ \(x^2+2 x \cdot \frac{1}{4}+\left(\frac{1}{4}\right)^2-2-\left(\frac{1}{4}\right)^2=0\)

⇒ \(\left(x+\frac{1}{4}\right)^2-\left(\frac{32+1}{16}\right)=0\)

⇒ \(\left(x+\frac{1}{4}\right)^2-\frac{33}{16}=0\)

⇒ \(\left(x+\frac{1}{4}\right)^2-\left(\frac{\sqrt{33}}{4}\right)^2=0\)

⇒ \(\left(x+\frac{1}{4}-\frac{\sqrt{33}}{4}\right)\left(x+\frac{1}{4}+\frac{\sqrt{33}}{4}\right)=0\)

⇒ \(x+\frac{1}{4}-\frac{\sqrt{33}}{4}=0 \text { or } x+\frac{1}{4}+\frac{\sqrt{33}}{4}=0\)

⇒ \(x=\frac{\sqrt{33}}{4}-\frac{1}{4} \quad \text { or } \quad x=-\frac{\sqrt{33}}{4}-\frac{1}{4}\)

⇒ \(x=\frac{\sqrt{33}-1}{4} \quad \text { or } \quad x=-\left(\frac{\sqrt{33}+1}{4}\right)\)

∴ Roots of a given equation

⇒ \(\frac{\sqrt{33}-1}{4},-\left(\frac{\sqrt{33}+1}{4}\right)\)

3. \(4 x^2+4 \sqrt{3} x+3=0\)

⇒ \(x^2+\sqrt{3} x+\frac{3}{4}=0\)

⇒ \(x^2+2 x \cdot \frac{\sqrt{3}}{2}+\left(\frac{\sqrt{3}}{2}\right)^2=0\)

⇒ \(\left(x+\frac{\sqrt{3}}{2}\right)^2=0\)

⇒ \(x+\frac{\sqrt{3}}{2}=0 \text { or }x+\frac{\sqrt{3}}{2}=0\)

⇒ \(x=-\frac{\sqrt{3}}{2}\text { or }x=-\frac{\sqrt{3}}{2}\)

∴ Roots of given equation = \(-\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}\)

4. \(2 x^2+x+4=0\)

⇒ \(x^2+\frac{1}{2} x+2=0\)

⇒ \(x^2+2 x \cdot \frac{1}{4}+\left(\frac{1}{4}\right)^2+2-\left(\frac{1}{4}\right)^2=0\)

⇒ \(\left(x+\frac{1}{4}\right)^2+\frac{32-1}{16}=0\)

⇒ \(\left(x+\frac{1}{4}\right)^2=\frac{-31}{16}\)

⇒ \(x+\frac{1}{4}=\sqrt{-\frac{31}{16}}\) which is an imaginary number.

∴ Roots of given equation does not exist.

Question 2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.
Solution:

1. 2x² – 7x + 3 = 0

On comparing with ax² +bx + c,

a = 2,b=-7,c = 3

Now, \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-(-7) \pm \sqrt{(-7)^2-4(2)(3)}}{2(2)}\)

⇒ \(\frac{7 \pm \sqrt{49-24}}{4}=\frac{7 \pm \sqrt{25}}{4}=\frac{7 \pm 5}{4}\)

⇒ \(x=\frac{7+5}{4}\text { or }x=\frac{7-5}{4}\)

⇒ \(x=3\text { or }x=\frac{1}{2}\)

∴ Roots of equation = \(3, \frac{1}{2}\)

2. 2x² +x- 4 = 0

On comparing with ax² + bx + c = 0

a = 2,b = 1,c = -4

Now, \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-1 \pm \sqrt{(1)^2-4(2)(-4)}}{2 \times 2}=\frac{-1 \pm \sqrt{1+32}}{4}\)

⇒ \(\frac{-1 \pm \sqrt{33}}{4}\)

⇒ \(x=\frac{-1+\sqrt{33}}{4} \text { or } \frac{-1-\sqrt{33}}{4}\)

∴ Roots of equation = \(\frac{-1+\sqrt{33}}{4}, \frac{-1-\sqrt{33}}{4}\)

3. \(4 x^2+4 \sqrt{3} x+3=0\)

On comparing with ax² +bx + c = 0

a = 4, b = \(4 \sqrt{3}\), c = 3

Now, \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-4 \sqrt{3} \pm \sqrt{(4 \sqrt{3})^2-4 \times 4 \times 3}}{2 \times 4}\)

⇒ \(\frac{-4 \sqrt{3} \pm \sqrt{48-48}}{8}=\frac{-4 \sqrt{3} \pm \sqrt{0}}{8}\)

⇒ \(-\frac{4 \sqrt{3}}{8}=-\frac{\sqrt{3}}{2}\)

Number of roots of a quadratic equation = 2

∴ Roots of given equation = \(-\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}\)

4. 2x² + x + 4 = 0

On comparing with ax², + bx + c = 0

a = 2,b=1,c = 4

Now, \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-1 \pm \sqrt{(1)^2-4(2)(4)}}{2 \times 2}\)

⇒ \(\frac{-1 \pm \sqrt{1-32}}{4}=\frac{-1 \pm \sqrt{-31}}{4}\)

⇒ \(\sqrt{-31}\)  is an imaginary number,

∴ the values for are imaginary.

So, the real roots of the given equation do not exist.

Question 3. Find the roots of the following equations:

1. \(x-\frac{1}{x}=3, x \neq 0\)
2. \(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}, x \neq-4,7\)

Solution:

1. \(x-\frac{1}{x}=3, x \neq 0\)

⇒ \(\frac{x^2-1}{x}=3\)

⇒ \(x^2-1=3 x \Rightarrow x^2-3 x-1=0\)

On comparing with ax² + bx + c = 0

a= 1, b = -3, c =-1

x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-1)}}{2(1)}=\frac{3 \pm \sqrt{9+4}}{2}\)

⇒ \(\frac{3 \pm \sqrt{13}}{2}\)

⇒ \(x=\frac{3+\sqrt{13}}{2} \text { or } \quad x=\frac{3-\sqrt{13}}{2}\)

Therefore, the roots of given equations

⇒ \(\frac{3+\sqrt{13}}{2}, \frac{3-\sqrt{13}}{2} \text {. }\)

2. \(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}, x \neq-4,7\)

⇒ \(\frac{(x-7)-(x+4)}{(x+4)(x-7)}=\frac{11}{30}\)

⇒ \(\frac{x-7-x-4}{x(x-7)+4(x-7)}=\frac{11}{30}\)

⇒ \(\frac{-11}{x^2-7 x+4 x-28}=\frac{11}{30}\)

⇒ \(\frac{-1}{x^2-3 x-28}=\frac{1}{30}\)

⇒ \(x^2-3 x-28=-30\)

⇒ \(x^2-3 x-28+30=0\)

⇒ \(x^2-3 x+2=0\)

⇒ \(x^2-2 x-x+2=0\)

⇒ x(x-2)-1(x-2)=0

⇒ (x-2)(x-1)=0

⇒ x-2 = 0 or x-1 = 0

⇒ x = 2 or x = 1

∴ Roots of given equation = 2,1

Question 4. The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is \(\frac{1}{3}\). Find his present age.
Solution:

Given

The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is \(\frac{1}{3}\).

Let the present age of Rehman = x years

3 years before, Rehman’s age = (x- 3) years

After 5 years, Rehman’s age = (x + 5) years

According to the problem, \(\frac{1}{x-3}+\frac{1}{x+5}=\frac{1}{3}\)

⇒ \(\frac{(x+5)+(x-3)}{(x-3)(x+5)}=\frac{1}{3}\)

⇒ \(\frac{2 x+2}{x(x+5)-3(x+5)}=\frac{1}{3}\)

⇒ \(\frac{2 x+2}{x^2+5 x-3 x-15}=\frac{1}{3}\)

⇒ \(\frac{2 x+2}{x^2+2 x-15}=\frac{1}{3}\)

⇒ \(x^2+2 x-15=6 x+6\)

⇒ \(x^2+2 x-15-6 x-6=0\)

⇒ \(x^2-4 x-21=0\)

⇒ \(x^2-7 x+3 x-21=0\)

⇒ x(x-7)+3(x-7)=0

⇒ (x-7)(x+3)=0

⇒ x-7=0 or  x+3=0

⇒ x=7 or  x=-3

but the age cannot be negative.

∴ x = 7

⇒ Rehman’s age = 7 years.

Question 5. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
Solution:

Given

In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210.

Let marks obtained by Shefali in Mathematics =x

∴ Her marks in English = (30- x)

If, marks in Mathematics = x + 2

marks in English =30-x-3 = 27-x

then (x + 2) (27 -x) =210

⇒ x(27 -x) + 2(27 -x) =210

⇒ 27x -x² + 54 – 2x = 210

⇒ 25x-x² + 54 =210

⇒ 0 = x²– 25x- 54 + 210

⇒ x² – 25x + 156 =0

⇒ x²– 13x- 12x+ 156 =0

⇒ x(x -13) – 12(x – 13) =0

⇒ (x- 13) (x- 12) =0

⇒ x -13 = 0 or x-12=0

⇒ x -13 or x-12

If x = 13 then 30-x = 30-13 = 17

If x = 12 then 30-x = 30-12 = 18

So, for Shefali

Maries in Mathematics = 13

and marks in English = 17

or

Maries in Mathematics = 12

and marks in English = 18

Question 6. The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.
Solution:

Given

The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side,

Let the smaller side of the rectangular field =xmetre = breadth

Larger side = (x + 30) metre = length

diagonal = (x + 60) metre

Now, (length) + (breadth) = (diagonal)

⇒ x² + (x+ 30)² = (x+60)²

⇒ x² +x² + 60x + 900 =x² + 120x + 3600

⇒ x² +x² + 60x + 900 -x² – 120x- 3600 = 0

⇒ x²– 60x -2700 = 0

⇒ x²– 90x + 3x -2700 =0

⇒ x(x- 90) + 30(x- 90) = 0

⇒ (x-90) (x + 30) =0

⇒ x-90 =0 or x+30 = 0

⇒ x = 90 or x = -30

but the side cannot be negative.

⇒ x = 90

⇒ x+30 = 90+30 = 120

Therefore, sides of the rectangular field = 120 m and 90 m.

Question 7. The difference of the squares of two numbers is 1 80. The square of the smaller number is 8 times the larger number. Find the two numbers.
Solution:

Given

The difference of the squares of two numbers is 1 80. The square of the smaller number is 8 times the larger number.

Let smaller number =x

∴ Larger number × 8 = x²

⇒ Larger number = \(\frac{x^2}{8}\)

According to the problem, (larger number)² – (smaller number)² =180

⇒ \(\left(\frac{x^2}{8}\right)^2-x^2=180\)

⇒ \(\left(\frac{y}{8}\right)^2-y=180\)

⇒ \(\text { where } x^2=y \text { (say) }\)

⇒ \(\frac{y^2}{64}-y=180\)

⇒ \(y^2-64 y=11\)

⇒ \(y^2-64 y-11520=0\)

⇒ \(y^2-144 y+80 y-11520=0\)

⇒ \(y(y-144)+80(y-144)=0\)

⇒ \((y-144)(y+80)=0\)

⇒ \(y-144=0 \text { or } y+80=0\)

⇒ \(y=144 \text { or } y=-80\)

⇒ \(x^2=144 \text { or } x^2=-80\)

x² =- 80 is not possible.

∴ \(x^2=144\)

⇒ \(x= \pm 12\)

⇒ \(x=12 \quad \text { or } \quad x=-12\)

⇒ \(\frac{x^2}{8}=\frac{144}{8}=18\)

Therefore, a number are 12, 18 or -12, 18.

Question 8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:

Given

A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey.

Let the speed of the train =x km/hr

∴ Time taken to cover 360 km distance = \(\frac{360}{x} \mathrm{~km}\)

If the speed of train = (x+ 5) km/hr

then time taken to cover 360 km distance = \(\frac{360}{x+5} \mathrm{hr}\)

According to the problem, \(\frac{360}{x}-\frac{360}{x+5}=1\)

⇒ \(\frac{360(x+5)-360 x}{x(x+5)}=1\)

⇒ \(\frac{360 x+1800-360 x}{x^2+5 x}=1\)

⇒ \(\frac{1800}{x^2+5 x}=1\)

⇒ \(x^2+5 x=1800\)

⇒ \(x^2+5 x-1800=0\)

⇒ \(x^2+45 x-40 x-1800=0\)

⇒ x(x+45)-40(x+45)=0

⇒ (x+45)(x-40)=0

⇒ x+45=0  or x-40=0

⇒ x=-45  or  x=40

but speed cannot be negative.

∴ Speed of train = 40 km/hr.

Question 9. Two water taps together can fill a tank in \(9 \frac{3}{8}\)hours. The tap of a larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Solution:

Given

Two water taps together can fill a tank in \(9 \frac{3}{8}\)hours. The tap of a larger diameter takes 10 hours less than the smaller one to fill the tank separately.

Let the time taken to fill the tank by the tap of smaller diameter =x hours

Time taken to fill the tank by the tap of larger diameter = (x- 10) hours

Now, work done by tap of smaller diameter in 1 hour = \(\frac{1}{x}\)

and work done by tap of larger diameter in 1hour = \(\frac{1}{x}+\frac{1}{x-10}\)

Work done by two taps in 1 hour = \(\frac{1}{x}+\frac{1}{x-10}\)

According to the problem, both taps fill the Time taken by the passenger train to cover 132 1cm tank in \(9 \frac{3}{8}=\frac{75}{8} \)

∴ \(\left(\frac{1}{x}+\frac{1}{x-10}\right) \times \frac{75}{8}=1\)

⇒ \(\frac{1}{x}+\frac{1}{x-10}=\frac{8}{75}\)

⇒ \(\frac{x-10+x}{x(x-10)}=\frac{8}{75}\)

⇒ \(\frac{2 x-10}{x^2-10 x}=\frac{8}{75}\)

⇒ \(8 x^2-80 x=150 x-750\)

⇒ \(8 x^2-80 x-150 x+750=0\)

⇒ \(8 x^2-230 x+750=0\)

⇒ \(4 x^2-115 x+375=0\)

⇒ \(4 x^2-100 x-15 x+375=0\)

⇒ 4 x(x-25)-15(x-25)=0

⇒ (x-25)(4 x-15)=0

⇒ x-25=0  or  4 x-15=0

⇒ x-25 or \(x=\frac{15}{4}\)

but \(x=\frac{15}{4}\) is not possible because both taps fill the tank in \(9 \frac{3}{4}\) hours.

⇒ x = 25 and x- 10 = 25 – 10= 15

Therefore, the tap with a smaller diameter fills the tank in 25 hours and the tap with a larger diameter fills the tank in 15 hours.

Question 10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.
Solution:

Given

An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train

Let the average speed of passenger train =x km/hr.

∴ Average speed of express train = (x+11) km/hr

Time taken by passenger train to cover 1321cm distance = \(\frac{132}{x} \mathrm{hr}\)

and time taken by express train to cover 1321cm distance = \(\frac{132}{x+11} \mathrm{hr}\)

According to the problem,

⇒ \(\frac{132}{x}-\frac{132}{x+11}=1\)

⇒ \(\frac{132(x+11)-132 x}{x(x+11)}=1\)

⇒ \(\frac{132 x+1452-132 x}{x^2+11 x}=1\)

⇒ \(x^2+11 x=1452\)

⇒ \(x^2+11 x-1452=0\)

⇒ \(x^2+44 x-33 x-1452=0\)

⇒ x(x+44)-33(x+44)=0

⇒ (x+44)(x-33)=0

⇒ x+44=0  or  x-33=0

⇒ x=-44  or  x=33

but the speed cannot be negative.

∴ x = 33

⇒ x + 11 =33+ 11 =44

Average speed of passenger train = 33 km/hr

an average speed of express train = 44 km/hr

Question 11. The sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.
Solution:

Given

The sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m

Let side of a square = x metre

Perimeter of this square = 4x metre

∴ Perimeter of second square = (4x + 24) metre

⇒ Side of second square = \(\frac{4 x+24}{4}\) = (x + 6) metre

Now, area of first square =x2 m2 and area of second square = (x+ 6)2 m2

According to the problem,

⇒ \((x+6)^2+x^2=468\)

⇒ \(x^2+12 x+36+x^2-468=0\)

⇒ \(2 x^2+12 x-432=0\)

⇒ \(x^2+6 x-216=0\)

⇒ \(x^2+18 x-12 x-216=0\)

⇒ x(x+18)-12(x+18)=0

⇒ (x+18)(x-12)=0

⇒ x+18=0   or   x-12=0

⇒ x=-18   or x=12

but the side of a square cannot be negative.

∴ x= 12

⇒ x+ 6 = 12 + 6= 18

Therefore, sides of squares = 12 m and 18 m

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Quadratic Equation Exercise 4.4

Question 1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them :

1. \(2 x^2-3 x+5=0\)
2. \(3 x^2-4 \sqrt{3} x+4=0\)
3. \(2 x^2-6 x+3=0\)

Solution:

1. \(2 x^2-3 x+5=0\)

On comparing with ax² + bx + c = 0

a = 2, b=-3,c = 5

Discriminant D = b² – 4ac = (-3)² – 4(2) (5)

= 9-40 =-31

∵ D is negative

∴ Roots of the equation are imaginary.

Here, the real roots of the equation do not exist.

2. \(3 x^2-4 \sqrt{3} x+4=0\)

On comparing with ax² + bx + c = 0

a = 3,b =\(-4 \sqrt{3}\), c= 4

Discriminant D = b²– 4ac

⇒ \((-4 \sqrt{3})^2-4(3)(4)\)

= 48-48 = 0

The roots of given equation are real and equal

Now, \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{4 \sqrt{3} \pm 0}{2 \times 3}=\frac{2}{\sqrt{3}}\)

Roots of given equation = \(\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)

3. 2x² – 6x + 3=0

On comparing with ax² + bx + c = 0

a = 2, b = – 6, c = 3

∴ Discriminant D = b²– 4ac = (-6)²– 4 (2) (3)

= 36-24 = 12 > 0

∵ D is positive

Roots of the given equation are real and unequal

Now, \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{6 \pm \sqrt{12}}{2 \times 2}=\frac{6 \pm 2 \sqrt{3}}{4}\)

⇒ \(x=\frac{3 \pm \sqrt{3}}{2}\)

⇒ \(\frac{3+\sqrt{3}}{2} \text { or } \frac{3-\sqrt{3}}{2} \text {. }\)

Question 2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.

1. 2x²+kx+ 3 = 0
2. k x(x- 2) + 6 = 0

Solution:

2x²+kx+ 3 = 0

On comparing with ax² +bx + c = 0

a = 2, b=k, c = 3

DiscriminantD = b²– 4ac = k² – 4 x 2 x 3

= k² -24

Given that, the roots are equal.

∴ D = 0  ⇒ k² – 24 = 0

⇒ k² = 24

⇒ K = \(\pm \sqrt{24}= \pm 2 \sqrt{6}\)

2.  kx(x -2) + 6 = 0

kx² – 2kx+6 = 0

On comparing with a² + bx + c = 0

a = k, b = -2k, c = 6

∴ Discriminant D = b²– 4ac = (-2k)² – 4k(6)

= 4k²– 24k = 4k(k- 6)

Given that the roots are equal.

∴ D = 0

⇒ 4k(k – 6) = 0

⇒ k – 6 = 0

⇒ k = 6

Question 3. Is It possible to design a rectangular mango grove whose length Is twice its breadth, and the area is 800 m2? If so, find its length and breadth.
Solution:

Let the breadth of mango grove

= x metre

∴ Length = 2x metre

Area of grove = length x breadth

= (2x)(x) = 2x²

According to the problem,

⇒ 2x² = 800

⇒ x² = 400

⇒ x = ± 20

but the breadth cannot be negative.

∴ x = 20

2x = 2 × 20 = 40

Therefore, a grove is possible, and its length = 40m and breadth = 20 m.

Question 4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their age in years was 48.
Solution:

Let the present age of one friend =  x years

∴ Present age of second friend = (20- x) years

4 years ago, the age of first friend = (x- 4) years

The age of second friend = 20 – x- 4

= (16- x) years

According to the problem,

(x-4) (16 -x) =48

⇒ x(1 6 – 4) -4(16 – x) =48

⇒ 16x – x²– 64 + 4x = 48

⇒ 20x-x² -64 = 48

⇒ 0 = 48 – 20x + x² + 64

x²-20x+ 112=0 ….(1)

On comparing with ax² + bx + c = 0

a = 1, b = -20, c = 112

Now, discriminant D=b² – 4ac

= (- 20)² – 4 (1) (112)

= 400 -448 =-48 <0

Roots of equation (1) are imaginary.

Therefore, the given condition is not possible.

Question 5. Is it possible to design a rectangular park of perimeter 80 m and an area of 400 m2? If so, find its length and breadth.
Solution:

Let the length of the park = x meter

Now, 2 (length + breadth) = perimeter

2(x + breadth) = 80 metre

x+ breadth = 40 metre

breadth = (40 – x) metre

According to the problem,

Area of park= 400 m²

x(40 – x) = 400

40x-x² = 400

0 = x² – 40x + 400

x² – 40x + 400 = 0

On comparing with ax² +bx+ c = 0

a = 1, b = -40, c = 400

Discriminant D = b² – 4ac

= (-40)² – 4(1)(400)

= 1600- 1600 = 0

⇒ Roots are equal

Now, \(x=\frac{-b \pm \sqrt{D}}{2 a}=\frac{40 \pm 0}{2 \times 1}=20\)

So, such a park is possible.

Its length = breadth = 20m.

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Quadratic Equation Multiple Choice Questions

Question 1. Which of die following equations has the root 3?

1. \(x^2+x+1\)
2. \(x^2-4 x+3=0\)
3. \(3 x^2+x-1=0\)
4. \(x^2+9=0\)

Answer: 2. \(x^2-4 x+3=0\)

Question 2. The sum of roots of the equation 5x² – 3x + 2 = 0 is

1. \(\frac{3}{5}\)
2. \(-\frac{3}{5}\)
3. \(\frac{2}{5}\)
4. \(-\frac{2}{5}\)

Answer: 1. \(\frac{3}{5}\)

Question 3. The quadratic equation \(2 x^2-\sqrt{5} x+1=0\) has

1. Two distinct real roots
2. Two equal roots
3. No real root
4. More than two real roots

Answer: 3. No real root

Question 4. One root of the equation x² + k x + 4 = 0 is -2. The value of k is :

1. -2
2. 2
3. -4
4. 4

Answer: 4. 4

Question 5. The value of k for which die roots of equation 2kx² – 6x + 1 = 0 is equal, is :

1. \(-\frac{9}{2}\)
2. \(\frac{9}{2}\)
3. 9
4. -9

Answer: 2. \(\frac{9}{2}\)

Question 6. Which of the following is a quadratic equation?

1. (x + 2)² =x²– 5x + 3
2. x³ +x² = (x- 1)³
3. 3x² + 1 = (3x- 2) (x + 5)
4. 5x- 7 = 1 + x

Answer: 2. x³ +x² = (x- 1)³

Question 7. If the product of roots of the equation 5×2- 3x + /{ = 0 is 2, then the value of k is:

1. 1
2. 2
3. 5
4. 10

Answer: 4. 10

Question 8. The discriminant of quadratic equadon 3xx³ +x² = (x- 1)³– 6x + 4 = 0 is :

1. 12
2. 13
3. -12
4. \(3 \sqrt{6}\)

Answer: 3. -12

Question 9. The roots of the quadratic equation x²– 4 = 0 are :

1. ± 0.2
2. ±1
3. ± 2
4. ±4

Answer: 3. ± 2

Question 10. The discriminant of equation \(3 x^2-2 x+\frac{1}{3}=0\) will be:

1. 3
2. 2
3. 1
4. 0

Answer: 4. 0

Question 11. If the roots of the quadratic equation 3x² – 12x + m = 0 are equal, then the value of m will be :

1. 4
2. 7
3. 9
4. 12

Answer: 4. 12