Important Questions For CBSE Class 12 Maths Chapter 1 Relations And Functions

CBSE Class 12 Maths Chapter 1 Important Questions: Key Concepts and Solutions

Overview of Chapter 1 Relations And Functions

Chapter 1 of CBSE Class 12 Maths covers essential concepts such as relations and functions. Understanding these topics is crucial for solving problems effectively.

CBSE Class 12 Maths Chapter 1 Relations And Functions Important Questions

Question 1. Let set X = {1, 2, 3} and a relation R is defined in X as R = {(1, 3), (2, 2), (3,2)}, then minimum ordered pairs which should be added in relation R to make it reflexive and symmetric are:

  1. {(1,1), (2. 3), (1,2)}
  2. {(3, 3), (3, 1), (1,2)}
  3. {(1, 1), (3, 3), (3. 1), (2, 3)}
  4. {(1,1), (3, 3), (3, 1), (1,2)}

Solution: 3. {(1, 1), (3, 3), (3. 1), (2, 3)}

Given, R= {(1,3), (2,2), (3,2)} on set X = {1.2, 3}

Then: if R is reflexive and symmetric, then R= {(1, 1), (2, 2), (3, 3), (1,3), (3, 2), (3, 1), (2, 3)}

Hence: the ordered pairs to be added are: {(1, 1), (3, 3), (3, 1), (2, 3)}

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CBSE Class 12 Maths Chapter 1 Important Question 2. If R = {(x, y); x, y ∈ Z, x² + y² ≤ 4} is a relation in set Z. then the domain of R is :

  1. {0, 1,2}
  2. {-2,-1, 0, 1,2}
  3. {0,-1,-2}
  4. {-1,0, 1}

Solution: 2. {-2,-1, 0, 1,2}

Given: R = {(x, y) : x, y ∈ Z, x² + y² ≤ 4} is a relation in set Z.

Hence: according to the given relation.

Domain of R= {-2, —1, 0. 1.2}

Important Questions CBSE Class 12 Maths Chapter 1

Important Questions For Class 12 Maths Chapter 1

Question 3. Let X = {x²: x ∈ N} and the function f: N →X is defined by f(x) = x², x ∈ N. Then, this function is :

  1. Injective only
  2. Not objective
  3. Surjective only
  4. Bijective

Solution: 4. Bijective

f(x) = x²; x ∈ N; f : N → X where X = {x²: x ∈ N}

Now; f(x1) = f(x2)

⇒ x²1 = x²2

⇒ (x1 + x2) (x1 – x2) = 0

Now : x1 + x2 ≠ 0 (x1,x2 ∈ N)

⇒ x1 – x2 = 0 ⇒ x1 = x2

Hence; f is one-one.

Let y = f(x) ⇒ y = x²

⇒ x = √y

∀ y ∈ X, there exists x ∈ N

Hence, every element of the co-domain has pre-images in the domain.

⇒ Range = Co-domain

∴  f is onto.

∴ The given function is bijective.

Relations And Functions Class 12 Important Questions

Question 4. A function f: R → R defined by f(x) = 2 + x² is :

  1. Not one-one
  2. One-one
  3. Not onto
  4. Neither one-one nor onto

Solution: 4. Neither one-one nor onto

f(x) = 2 + x² f: R → R

One-one: Let x1 and x2 ∈ R such that

f(x1) = f(x2) => x²1 + 2 = x²2 = 2

⇒ x1 = ±x2 ⇒  f is not one-one

Onto: Let y = f(x) ⇒ y = x² + 2 ⇒ x \(=\sqrt{y-I} \in R\)

∴ y – 2 ≥ 0 ⇒ y ≥ 2 ⇒ y ∈ R [2, ∞)

Range 5≠ co-domain

Also: f is not onto as there is no pre-image for negative real numbers.

Hence; f is neither one-one nor onto.

Question 5. Write the smallest reflexive relation on set A = {a, b, c}.

Solution:

For reflexive relation, each and every element of the given set must be related to itself, at least i.e. the smallest reflexive relation on set A will be {(a, a), (b, b), (c, c)}.

Class 12 Maths Chapter 1 Important Questions With Solutions

Question 6. If f = {(1.2), (2, 4), (3, 1), (4, k)} is a one-one function from set A to A, where A = {1, 2, 3, 4}; then find the value of k.

Solution:

In a one-on-one function, different elements must have different images. So, according to the given function, f = {(1,2), (2. 4), (3, 1), (4, k)}, the value of k must be 3.

Important Questions CBSE Class 12 Maths Chapter 1 Relations And Functions One One Function

CBSE Class 12 Maths Chapter 1 Important Question 7.

  1. Check whether the relation R defined on the set {1, 2, 3, 4} as R = {(a, b); b = a + 1} is transitive. Justify your answer. Or
  2. If the relation R on the set A = {x : 0 ≤ x ≤12} given by R = {(a, b): a = b} is an equivalence relation, then find the set of all elements related to 1.

Solution:

1. Given. R = ((a, b): b = a + 1} defined on the set A = {1, 2, 3, 4}

⇒ R = {(1, 2), (2, 3), (3, 4)}

By definition, for transitive relation, if a R b and b R c ⇒ a R c ∀ a, b, c ∈ A

For 1,2, 3 ∈ A; (1,2) ∈ R and (2, 3) ∈ R but (1, 3) ∉ R

Therefore, R is not transitive.

Aliter: Given; R = {(a, b): b = a + 1} defined on the set A – {1,2, 3,4}

For transitive relation: Let (a, b) ∈ R and (b, c) ∈ R ∀ a, b, c ∈ A

⇒ b = a + 1 and c = b + 1

⇒ c = a + 2

or (a, c) ∉ R

∴ R is not transitive

or

2. The elements in R that are related to 1 will be those elements from set A which are equal to 1.

Hence, the set of elements related to 1 is {1}.

Aliter: Let x ∈ A which is related to 1 by the given relation R = {(a, b): a = b}

x R 1 ⇒ (x, 1) ∈ R ⇒ x = 1

∴ [1] = {1}

Relations And Functions Important Questions For Cbse Class 12

Question 8. The relation R in the set {1.2,3} given by R = {(1, 2), (2, 1), (1, 1)} is:

  1. Symmetric and transitive, but not reflexive
  2. Reflexive and symmetric, but not transitive
  3. Symmetric, but neither reflexive nor transitive
  4. An equivalence relation

Solution: 3. Symmetric, but neither reflexive nor transitive

Let A= {1,2, 3} and R = {(1, 2), (2, 1), (1, 1)}

R is not reflexive as (2, 2) and (3,3) ∉ R.

R is a symmetric relation as (a, b) ∈ R and (b, a) ∈ R ∀ a, b ∈ A.

R is not a transitive relation because (2, 1) ∈ R and (1,2) ∈ R but (2, 2) ∉ R.

Hence; R is symmetric, but neither reflexive nor transitive.

Question 9. Let A = {1,3, 5}, Then the number of equivalence relations in A containing (1,3) is

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

Solution: 2. 2

R1 = {(1, 1), (3, 3), (5, 5), (1, 3), (3, 1)}

and R2 = {(1, 1), (3. 3), (5, 5), (1, 3), (3, 1), (1,5), (5, 1), (3, 5), (5, 3)}

Hence, the number of equivalence relations in A containing (1, 3) is 2.

CBSE Class 12 Maths Chapter 1 Extra Questions

Question 10. Check whether the relation R in the set N of natural numbers given by R = {(a, b): a is divisor of b} is reflexive, symmetric or transitive. Also, determine whether R is an equivalence relation.

Solution:

The relation R in the set N is given by:

R = {(a, b): a is divisor of b}

⇒ aRb => \(\frac{b}{a}\) = K ∈ 1 ∀ a, b ∈ N,

Reflexive; aRa ∀ a∈N (By definition)

aRa ⇒ \(\frac{a}{a}\)= 1 ∈ I which is true.

So, R is reflexive relation.

Symmetric: we have

aRb ⇒ \(\frac{b}{a}\) = K ∈ I ∀ a, b ∈ N

⇒ \(\frac{a}{b}=\frac{1}{K} \notin I\)

⇒ b K a ∀ a, b∈ N

∴ If aRb ⇒ bRa ∀ a, b ∈ N, then R is not a symmetric relation.

Transitive: If aRb ⇒ \(\frac{b}{a}=K_1\) ∈ I ∀ a, b ∈ N ….(1)

bRc ⇒ \(\frac{c}{b}=K_2\) ∈ I ∀ b, c ∈ N ….(2)

from equation (1) and (2) \(\frac{b}{a} \times \frac{c}{b}=K_1 \times K_2 \Rightarrow \frac{c}{a}-\left(K_1 K_2\right) \in I \Rightarrow a R c\)

So, R is a transitive relation

∴ R is reflexive and transitive but not symmetric

R is not an equivalence relation on N.

Question 11. Check whether the relation R defined on the set A = {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric, or transitive.

Solution:

Given, R = {(a, b): b = a + 1} is defined on the set A = {1, 2, 3, 4, 5, 6}

⇒ R= {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

Reflexive: By definition, a R a ∀a, b ∈ A

Now, 1 ∈ A but (1, 1) ∉ R.

Therefore, R is not reflexive.

Symmetric: By definition, if a R b ⇒ b R a ∀ a, b ∈ A

For 1,2 ∈ A, It can be observed that (1,2) ∈ R but (2, 1) ∉ R.

Therefore, R is not symmetric.

Transitive: By definition, if a R b and b R c ⇒ a R c ∀ a, b, c ∈ A

For 1,2, 3 ∈ A, (1,2) ∈ R and (2, 3) ∈ R but (1, 3)  ∉ R

Therefore, R is not transitive.

Hence, R is neither reflexive nor symmetric nor transitive.

Aliter: Given; R = {(a, b): b = a + 1} defined on the set A = {1,2, 3, 4, 5, 6}

Reflexive: (a, a) ∈ R ∀ a ∈ A

⇒ a = a + 1; which is not true;

∴ R is not reflexive.

Symmetric: Let (a, b) ∈ R ∀ a, b ∈ A

⇒ b = a + 1

⇒ (b, a) ∉ R [a ≠ b + 1]

∴ R is not symmetric

Transitive: Let (a, b) ∈ R and (b, c) ∈ R ∀ a, b, c ∈ A

⇒ b = a + 1 and c = b + 1

⇒ c = a + 2 or (a, c ) ∉ R

∴ R is not transitive.

Hence; R is neither reflexive nor symmetric nor transitive.

Relations And Functions Previous Year Questions Class 12

Question 12. Let A = {x ∈ Z : 0 ≤ x ≤12}. Show that R = {(a, b): a, b ∈ A, |a – b| is divisible by 4} is an equivalence relation. Find the set of all elements related to 1. Also; write the equivalence class [2]. or, Show that the function f: R → R defined by \(f(x)=\frac{x}{x^2+1}\). ∀ x ∈R is neither one-one nor onto.

Solution:

Given, R = {(a, b): a, b ∈ A, |a – b| is divisible by 4}

Reflexive: Let a ∈ A

Now, |a — a| = 0. which is divisible by 4

So, (a, a) ∈ R ∀ a ∈ A

Hence, R is reflexive.

Symmetric: Let a, b ∈ A such that (a, b) ∈ R i.e. |a — b| is divisible by 4.

⇒ |-(b – a)| = |b — a| is also divisible by 4.

Hence; (b, a) ∈ R.

So, R is symmetric.

Transitive: Let a, b, c ∈ A such that (a, b), (b, c) ∈ R

i.e. |a – b| and |b – c| is divisible by 4.

Let \(|\mathrm{a}-\mathrm{b}|=4 \mathrm{k}_1\)

⇒ \(\mathrm{~b}-\mathrm{c}|=4 \mathrm{k}_2\)

⇒ \((\mathrm{a}-\mathrm{b})= \pm 4 \mathrm{k}_1\)….(1)

⇒ \((b-c)= \pm 4 k_2\)….(2)

Adding equations (1) and (2); \((\mathrm{a}-\mathrm{b})+(\mathrm{b}-\mathrm{c})= \pm 4 \mathrm{k}_1 \pm 4 \mathrm{k}_2= \pm 4\left(\mathrm{k}_1+\mathrm{k}_2\right)\)

⇒ a – c is divisible by 4.

⇒ |a – c| is divisible by 4.

Hence; (a, c) ∈ R

So, R is transitive.

Hence; R is an equivalence relation.

Further, let (x, 1) ∈ R ∀ x ∈ A ⇒ |x  – 1| is divisible by 4

⇒ x – 1 = 0, 4, 8, 12

⇒ x= 1, 5, 9 [x = 13 ∉ A]

Equivalence class of [1] = {1,5,9}

[The set of all elements related to 1 represents its equivalence class]

Now, we will find the equivalence class of [2]

Let (x, 2) ∈ R ∀ x ∈ A

⇒ |x — 2| = 0, 4, 8, 12

⇒ x = 2, 6, 10 [x = 14 ∉ A]

∴ Equivalence class of [2] = {2, 6, 10}. or,

f : \(R \rightarrow R ; f(x)=\frac{x}{x^2+1}\)

1. One-one function: we have \(\mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right), \forall, \mathrm{x}_1, \mathrm{x}_2 \in \mathrm{R}\)

⇒ \(\frac{x_1}{x_1^2+1}=\frac{x_2}{x_2^2+1}\)

⇒ \(x_1 x_2^2+x_1=x_2 x_1^2+x_2 \quad \Rightarrow\left(x_1 x_2^2-x_2 x_1^2\right)+\left(x_1-x_2\right)=0\)

⇒ \(x_1 x_2\left(x_2-x_1\right)+\left(x_1-x_2\right)=0 \Rightarrow\left(x_1-x_2\right)\left(-x_1 x_2+1\right)=0\)

⇒ \(x_1=x_2 \text { and } x_1 x_2=1 \text { or } x_1=\frac{1}{x_2}\)

One-one function Example: Let \(\mathrm{x}_1=2 \in \mathrm{R} ; \mathrm{x}_2=1 / 2 \in \mathrm{R}\)

Then, \(f\left(x_1\right)=f(2)=2 / 5\) and \(f\left(x_2\right)=f(1 / 2)=2 / 5\)

⇒ latex]x_1 \neq x_2[/latex] but \(f\left(x_1\right)=f\left(x_2\right)\)

∴ f is not a one-on-one function

2. Onto function: Let y = f(x)

⇒ y = \(\frac{x}{x^2+1} \Rightarrow x^2 y-x+y=0\)

⇒ x = \(\frac{1 \pm \sqrt{1-4 y^2}}{2 y} \in R \text { if } 1-4 y^2 \geq 0 \text { and } y \neq 0\)

⇒ \(4 y^2-1 \leq 0 \text { or }(2 y+1)(2 y-1) \leq 0\)

Relations And Functions Onto Function

⇒ \(x \in\left[-\frac{1}{2}, 0\right) \cup\left(0, \frac{1}{2}\right]\)

Thus, every element of the Co-domain does not have its pre-image in the domain.

Hence, f: R → R is not onto.

∴ f is neither one-one nor onto.

Important Questions For CBSE Class 12 Maths Chapter 2 Inverse Trigonometric Functions

CBSE Class 12 Maths Chapter 2 Important Questions: Key Concepts and Solutions

Overview of Chapter 2

Chapter 2 of CBSE Class 12 Maths focuses on Inverse Trigonometric Functions. Understanding these functions is crucial for solving various mathematical problems effectively.

CBSE Class 12 Maths Chapter 2 Inverse Trigonometric Functions Important Questions

Question 1. The principal value of \(\cos ^{-1}\left(\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{\sqrt{2}}\right)\) is

  1. \(\frac{\pi}{12}\)
  2. \(\pi\)
  3. \(\frac{\pi}{3}\)
  4. \(\frac{\pi}{6}\)

Solution: 1. \(\frac{\pi}{12}\)

⇒ \(\cos ^{-1}\left(\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\frac{\pi}{3}+\left(-\frac{\pi}{4}\right)\)

= \(\frac{4 \pi-3 \pi}{12}=\frac{\pi}{12}\)

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Question 2. The principal value of \(\tan ^{-1}\left(\tan \frac{9 \pi}{8}\right)\) is:

  1. \(\frac{\pi}{8}\)
  2. \(\frac{3 \pi}{8}\)
  3. \(-\frac{\pi}{8}\)
  4. \(-\frac{3 \pi}{8}\)

Solution: 1. \(\frac{\pi}{8}\)

⇒ \(\tan ^{-1}\left[\tan \left(\frac{9 \pi}{8}\right)\right]=\tan ^{-1}\left[\tan \left(\pi+\frac{\pi}{8}\right)\right]\)

= \(\tan ^{-1}\left[\tan \left(\frac{\pi}{8}\right)\right]=\frac{\pi}{8} \in\left(-\frac{\pi}{2} \cdot \frac{\pi}{2}\right)\)

Inverse Trigonometric Functions Class 12 Important Questions

Question 3. What is the domain of the function cos-1(2x-3)?

  1. [-1, 1]
  2. (1,2)
  3. (-1, 1)
  4. [1,2]

Solution: 4. [1,2]

For the given function: -1 ≤ 2x – 3 ≤ 1

⇒ 2 ⇒ 2x ⇒ 4

⇒ 1 ≤ x ≤ 2

Question 4. The principal value of [tan-1 √3 – cot-1 (-√3)] is

  1. π
  2. \(-\frac{\pi}{2}\)
  3. 0
  4. \(2 \sqrt{3}\)

Solution: 2. \(-\frac{\pi}{2}\)

⇒ \(\tan ^{-1}(\sqrt{3})-\cot ^{-1}(-\sqrt{3})=\frac{\pi}{3}-\left(\pi-\cot ^{-1}(\sqrt{3})\right)\)

=\(\frac{\pi}{3}-\left(\pi-\frac{\pi}{6}\right)=\frac{\pi}{3}-\frac{5 \pi}{6}=\frac{-3 \pi}{6}=\frac{-\pi}{2}\)

Important Questions For CBSE Class 12 Maths Chapter 2

Question 5. The range or the principal valurbrance of the function y = sec-1 x is ____

or,

The principal value of \(\cos ^{-1}\left(-\frac{1}{2}\right)\) is

Solution:

y = \(\sec ^{-1} x\) (given)

Range of \(\sec ^{-1} x\) is [0, π]-{π/2}

Or

Let \(\cos ^{-1}\left(-\frac{1}{2}\right)=y\)

⇒ \(\cos y=-\frac{1}{2} \Rightarrow \cos y=\cos \left[\pi-\frac{\pi}{3}\right]\)

⇒ \(\cos y=\cos \frac{2 \pi}{3} \Rightarrow y=\frac{2 \pi}{3} \in[0, \pi]\)

Class 12 Maths Chapter 2 Important Questions With Solutions

Question 6. Simplify \(\sec ^{-1}\left(\frac{1}{2 x^2-1}\right), 0<x<\frac{1}{\sqrt{2}} \text {. }\)

Solution:

Let y = \(\sec ^{-1}\left(\frac{1}{2 x^2-1}\right), 0<x<\frac{1}{\sqrt{2}}\)

Put x = \(\cos \theta\); then \(\theta=\cos ^{-1} x\)

y = \(\sec ^{-1}\left(\frac{1}{2 \cos ^2 \theta-1}\right)\)

= \(\sec ^{-1}\left(\frac{1}{\cos 2 \theta}\right)=\sec ^{-1}(\sec 2 \theta)=2 \theta=2 \cos ^{-1} x\)

Question 7. Prove that: \(3 \sin ^{-1} x=\sin ^{-1}\left(3 x-4 x^3\right), x \in\left[-\frac{1}{2}, \frac{1}{2}\right]\)

Solution:

To prove: \(3 \sin ^{-1} x=\sin ^{-1}\left(3 x-4 x^3\right), x \in\left[-\frac{1}{2}, \frac{1}{2}\right]\)

R.H.S = sin-1(3x – 4x³)

Put x = sinθ :  θ = sin-1x

⇒ R.H.S = sin-1 (3sinθ – 4sin³θ) – sin-1(sin3θ) = 3θ = 3sin-1x = L.H.S

Hence, proved.

CBSE Class 12 Maths Chapter 2 Conclusion

Focusing on these important questions will enhance your preparation for CBSE Class 12 Maths Chapter 2. Regular practice will build confidence and improve your problem-solving skills.

Important Questions For CBSE Class 12 Maths Chapter 4 Determinates

CBSE Class 12 Maths Chapter 4 Determinates Important Questions

Question 1. Three points P(2x, x + 3), Q(0, x) and R(x + 3, x + 6) are eollinear. then x is equal to

  1. 0
  2. 2
  3. 3
  4. 1

Solution: 4. 1

As per the given condition \(\left|\begin{array}{ccc}
2 x & x+3 & 1 \\
0 & x & 1 \\
x+3 & x+6 & 1
\end{array}\right|=0\)

[ar(ΔPQR) = 0]

⇒ 2x (x – x – 6) — 0(x + 3 – x — 6) + (x + 3) {x + 3 – x} =0

or -12x + 3x + 9 = 0 => -9x = -9 => x = 1

Read and Learn More CBSE Class 12 Maths Important Question and Answers

Question 2. If Cij denotes the cofactor of element pij of the matrix P = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & 2 & 4
\end{array}\right]\)
, then the value of C31, C23 is:

  1. 5
  2. 24
  3. -4
  4. -5

Solution: 1. 5

P = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & 2 & 4
\end{array}\right]\) (given)

⇒ C31, C23 = (3-4), {(-1)(2+3)} = 5

Determinants Class 12 Important Questions

Question 3. The system of linear equations 5x + ky = 5; 3x + 3y = 5 will be consistent if:

  1. k ≠ -3
  2. k = -5
  3. k = 5
  4. k ≠ 5

Solution: 4. k ≠ 5

The system of linear equations is given as:

5x + ky = 5 and 3x + 3y = 5

It can be written in matrix form as \(\left[\begin{array}{ll}
5 & k \\
3 & 3
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
5 \\
5
\end{array}\right]\)

or A X = B

So; the given system of linear equations is consistent if |A| ≠ 0

⇒ \(\left|\begin{array}{ll}
5 & \mathrm{k} \\
3 & 3
\end{array}\right|\) 0

⇒15 – 3 k ≠ 0

⇒ k ≠ 5

Important Questions For CBSE Class 12 Maths Chapter 4

Question 4. If, for the matrix \(A=\left[\begin{array}{cc}
\alpha & -2 \\
-2 & \alpha
\end{array}\right]\)
,\(\left|\mathrm{A}^2\right|=125\); then the value of α is:

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

Solution: 1. 3

Class 12 Maths Chapter 4 Important Questions With Solutions

Given \(\left|A^3\right|=125\)

⇒ \((|A|)^3=125\)

(because \(\left|A^n\right|=|A|^n\))

⇒ \(\left(\alpha^2-4\right)^3=125\)

⇒ \(\left(\alpha^2-4\right)=5\)

⇒ \(\alpha^2=9\) or \(\alpha= \pm 3\)

Question 5. Let matrix X = [xij] is given by X = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 4 & -5 \\
2 & -1 & 3
\end{array}\right]\)
, Then, the matrix Y = [mij]. where mij = Minor of Xij is:

  1. \(\left[\begin{array}{ccc}
    7 & -5 & -3 \\
    19 & 1 & -11 \\
    -11 & 1 & 7
    \end{array}\right]\)
  2. \(\left[\begin{array}{ccc}
    7 & -19 & -11 \\
    5 & -1 & -1 \\
    3 & 11 & 7
    \end{array}\right]\)
  3. \(\left[\begin{array}{ccc}
    7 & 19 & -11 \\
    -3 & 11 & 7 \\
    -5 & -1 & -1
    \end{array}\right]\)
  4. \(\left[\begin{array}{ccc}
    7 & 19 & -11 \\
    -1 & -1 & 1 \\
    -3 & -11 & 7
    \end{array}\right]\)

Solution: 

X = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 4 & -5 \\
2 & -1 & 3
\end{array}\right]\)

Now, Y = [mij]

⇒ Y = \(\left[\begin{array}{lll}
m_{11} & m_{12} & m_{13} \\
m_{21} & m_{22} & m_{23} \\
m_{31} & m_{32} & m_{33}
\end{array}\right] \Rightarrow Y=\left[\begin{array}{ccc}
7 & 19 & -11 \\
-1 & -1 & 1 \\
-3 & -11 & 7
\end{array}\right]\)

Determinants Important Questions CBSE Class 12

Question 6. If x =-4 is a root of \(\left|\begin{array}{lll}
x & 2 & 3 \\
1 & x & 1 \\
3 & 2 & x
\end{array}\right|\) = 0
, then the sum of the other two roots is:

  1. 4
  2. -3
  3. 2
  4. 5

Solution: 1. 4

⇒ \(\left|\begin{array}{lll}
x & 2 & 3 \\
1 & x & 1 \\
3 & 2 & x
\end{array}\right|\) = 0 (given)

⇒ \(x\left(x^2-2\right)-2(x-3)+3(2-3 x)=0\)

⇒ \(x^3-2 x-2 x+6+6-9 x=0\)

⇒ \(x^3-13 x+12=0\)

⇒ \((x+4)\left(x^2-4 x+3\right)=0\)

(because x=-4 is a root)

⇒ \((x+4)(x-1)(x-3)=0\)

Hence; the sum of other two roots = 1 + 3 = 4

Determinants Class 12 Important Questions

Question 7. The inverse of the matrix \(X=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right]\)

  1. \(24\left[\begin{array}{ccc}1 / 2 & 0 & 0 \\ 0 & 1 / 3 & 0 \\ 0 & 0 & 1 / 4\end{array}\right]\)
  2. \(\frac{1}{24}\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\)
  3. \(\frac{1}{24}\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right]\)
  4. \(\left[\begin{array}{ccc}1 / 2 & 0 & 0 \\ 0 & 1 / 3 & 0 \\ 0 & 0 & 1 / 4\end{array}\right]\)

Solution:

Given, \(X=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{array}\right]\)

|X| = \(2(12)-0+0=24 \neq 0\)

⇒ \(X^{-1}\) exists

Now, adj X = \(\left[\begin{array}{ccc}12 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 6\end{array}\right]\)

⇒ \(X^{-1}=\frac{1}{|X|} \cdot \mathrm{adj} X\)

⇒ \(X^{-1}=\frac{1}{24}\left[\begin{array}{ccc}
12 & 0 & 0 \\
0 & 8 & 0 \\
0 & 0 & 6
\end{array}\right]=\left[\begin{array}{ccc}
1 / 2 & 0 & 0 \\
0 & 1 / 3 & 0 \\
0 & 0 & 1 / 4
\end{array}\right]\)

Question 8. If A is square matrix of order 3 such that A(adj A) = \(\left[\begin{array}{ccc}-2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2\end{array}\right]\), then find |A|.

Solution:

Given, A (adj A) = \(\left[\begin{array}{ccc}-2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2\end{array}\right]\)

We know that A(adj A)= |A| I

Now, A(adj A)= \(-2\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] = -2I = |A| I\)

⇒ |A|=-2

CBSE Class 12 Maths Chapter 4 Extra Questions

Question 9. If A is a non-singular square matrix of order 3 such that A² = 3A, then the value of |A| is?

  1. -3
  2. 3
  3. 9
  4. 27

Solution:

Given, A² = 3A, |A| ≠ 0, order of A is 3

∴ |A²| = |3A|

⇒ |A|²= 3³|A|

(|A²| = |A|² and |KA| = Kn|A|)

or |A| = 27

Question 10. If A is a square matrix satisfying A’A = I, write the value of |A|.

Solution:

A’A = I (given)

⇒ \(\left|A^{\prime} A\right|=|I|\)

⇒ \(\left|A^{\prime}\right||A|=|I| \Rightarrow|A|^2=1\) (because \(\left|A^{\prime}\right|=|A|\))

⇒ \(|A|=1 \text { or }|A|=-1\)

Question 11. Given, \(A=\left[\begin{array}{cc}2 & -3 \\ -4 & 7\end{array}\right]\), compute A-1 and show that 2A-1 = 9I – A.

Solution:

Given, \(A=\left[\begin{array}{cc}2 & -3 \\ -4 & 7\end{array}\right]\)

|A| = \(14-(12)=2 \neq 0\)

Hence, A is invertible.

adj A = \(\left[\begin{array}{ll}
7 & 3 \\
4 & 2
\end{array}\right]\)

Hence, \(A^{-1}=\frac{1}{|A|} \cdot \mathrm{adj} \cdot(\mathrm{A})\)

⇒ \(\mathrm{A}^{-1}=\frac{1}{2}\left[\begin{array}{ll}
7 & 3 \\
4 & 2
\end{array}\right] \text { or } 2 \mathrm{~A}^{-1}=\left[\begin{array}{ll}
7 & 3 \\
4 & 2
\end{array}\right]\)….(1)

Now, R.H.S. = \(9 \mathrm{I}-\mathrm{A}\)

⇒ \(9\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]-\left[\begin{array}{cc}
2 & -3 \\
-4 & 7
\end{array}\right]=\left[\begin{array}{cc}
9 & 0 \\
0 & 9
\end{array}\right]-\left[\begin{array}{cc}
2 & -3 \\
-4 & 7
\end{array}\right]=\left[\begin{array}{ll}
7 & 3 \\
4 & 2
\end{array}\right]=2 A^{-1}\)= L.H.S. [from (1)]

Hence proved.

Question 12. If \(A=\left[\begin{array}{ccc}
3 & 4 & 2 \\
0 & 2 & -3 \\
1 & -2 & 6
\end{array}\right]\)
; find A-1.

Determinants Previous Year Questions Class 12

Hence solve the following system of equations: 3x + 4y + 2z = 8; 0x + 2y + 3z = 3 and x + 2y + 6z = -2 or,

OR

If \(A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$ and $B=\left[\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right]\) find \((A B)^{-1} \text {. }\)

Solution:

Given, \(A=\left[\begin{array}{ccc}
3 & 4 & 2 \\
0 & 2 & -3 \\
1 & -2 & 6
\end{array}\right]\)

⇒ |A| = 3(12 -6)-4(0 + 3) + 2(0-2)= 18- 12-4=2*0

Hence, A-1 exists.

Now, co-factors are given as:

C11 = 6, C12 = —3, C13 = -2,

C21=-28, C22 = 16, C23= 10,

C31= -16, C32 = 9, C33 = 6

Hence adj A = \(\left[\begin{array}{ccc}6 & -3 & -2 \\ -28 & 16 & 10 \\ -16 & 9 & 6\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ccc}6 & -28 & -16 \\ -3 & 16 & 9 \\ -2 & 10 & 6\end{array}\right]\)

⇒ \(A^{-1}=\frac{1}{|A|}(\mathrm{adj} A)=\frac{1}{2}\left[\begin{array}{ccc}
6 & -28 & -16 \\
-3 & 16 & 9 \\
-2 & 10 & 6
\end{array}\right]\)….(1)

The system of linear equations is given as:

3x + 4y + 2z = 8

0x + 2y —3 = 3

x – 2y + 6z = -2

This system is written in matrix form as

⇒ \(\left[\begin{array}{ccc}
3 & 4 & 2 \\
0 & 2 & -3 \\
1 & -2 & 6
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{r}
8 \\
3 \\
-2
\end{array}\right]\)

⇒ AX = B

⇒ X = \(A^{-1} \cdot B\)

⇒ X = \(\frac{1}{2}\left[\begin{array}{ccc}
6 & -28 & -16 \\
-3 & 16 & 9 \\
-2 & 10 & 6
\end{array}\right] \cdot\left[\begin{array}{l}
8 \\
3 \\
-2
\end{array}\right]=\frac{1}{2}\left[\begin{array}{l}
48-84+32 \\
-24+48-18 \\
-16+30-12
\end{array}\right]\)

[from (1) and (2)]

⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{2}\left[\begin{array}{r}
-4 \\
6 \\
2
\end{array}\right]\)

x=-2, y=3, z=1

Or

Given, \(A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]\)…(1)

and \(B=\left[\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right]\)

|B| = 1 (3 – 0) – 2(- 1 – 0) – 2(2 – 0) = 3 + 2 – 4 = 1 ≠ 0

Now, the co-factors of matrix B are given as

⇒ \(C_{11}=3, C_{12}=1, C_{13}=2\)

⇒ \(C_{21}=2, C_{22}=1, C_{23}=2,\)

⇒ \(C_{31}=6, C_{32}=2, C_{33}=5\)

∴ (adj)(B) = \(\left[\begin{array}{lll}
3 & 1 & 2 \\
2 & 1 & 2 \\
6 & 2 & 5
\end{array}\right]^1=\left[\begin{array}{lll}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5
\end{array}\right]\)

⇒ \(B^{-1}=\frac{1}{|B|}(\mathrm{adj} B)=\left[\begin{array}{lll}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5
\end{array}\right]\)….(2)

∴ (AB)-1 = B-1 A-1

= \(\left[\begin{array}{lll}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5
\end{array}\right] \cdot\left[\begin{array}{ccc}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]\)

=\(\left[\begin{array}{ccc}
9-30+30 & -3+12-12 & 3-10+12 \\
3-15+10 & -1+6-4 & 1-5+4 \\
6-30+25 & -2+12-10 & 2-10+10
\end{array}\right]\)

⇒ \((A B)^{-1}=\left[\begin{array}{ccc}
9 & -3 & 5 \\
-2 & 1 & 0 \\
1 & 0 & 2
\end{array}\right]\)

(from (1) and (2))

⇒ \((\mathrm{AB})^{-1}=\left[\begin{array}{ccc}
9 & -3 & 5 \\
-2 & 1 & 0 \\
1 & 0 & 2
\end{array}\right]\)

Question 13. If \(A=\left[\begin{array}{ccc}5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6\end{array}\right]\), find A-1 and use it to solve the following system of equations.

5x – y + 4z = 5

2x + 3y + 5z = 2

5x – 2y + 6z = -1

Solution:

Given, \(A=\left[\begin{array}{ccc}5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6\end{array}\right]\)

∴ |A| = \(\left|\begin{array}{ccc}
5 & -1 & 4 \\
2 & 3 & 5 \\
5 & -2 & 6
\end{array}\right|\) = 5(18+10) + (12 – 25) + 4 (-4 – 15)= 140 – 13 – 76 = 51 ≠ 0.

Hence: A-1 exists

Now co-factors of elements of A are:

⇒ \(A_{11}=28, A_{12}=13, A_{13}=-19\)

⇒ \(A_{21}=-2, A_{22}=10, A_{23}=5\)

⇒ \(A_{31}=-17, A_{32}=-17, A_{33}=17\)

∴ adj A = \(\left[\begin{array}{ccc}
28 & 13 & -19 \\
-2 & 10 & 5 \\
-17 & -17 & 17
\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ccc}
28 & -2 & -17 \\
13 & 10 & -17 \\
-19 & 5 & 17
\end{array}\right]\)

⇒ \(A^{-1}=\frac{1}{|A|}(\mathrm{adj} A)=\frac{1}{51}\left[\begin{array}{ccc}
28 & -2 & -17 \\
13 & 10 & -17 \\
-19 & 5 & 17
\end{array}\right]\)

Given system of equations are

5x – y + 4z = 5

2x + 3y + 5z = 2

5x – 2y + 6z = -1

This system is written in matrix form as

⇒ \({\left[\begin{array}{ccc}
5 & -1 & 4 \\
2 & 3 & 5 \\
5 & -2 & 6
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{r}
5 \\
2 \\
-1
\end{array}\right] }\)

⇒ AX = B ⇒ X=\(A^{-1}\) B

⇒ \({\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{51}\left[\begin{array}{ccc}
28 & -2 & -17 \\
13 & 10 & -17 \\
-19 & 5 & 17
\end{array}\right]\left[\begin{array}{r}
5 \\
2 \\
-1
\end{array}\right]}\)

or \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{51}\left[\begin{array}{l}
140-4+17 \\
65+20+17 \\
-95+10-17
\end{array}\right]\)

⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\frac{1}{51}\left[\begin{array}{r}
153 \\
102 \\
-102
\end{array}\right]=\left[\begin{array}{r}
3 \\
2 \\
-2
\end{array}\right]\)

⇒ x= 3, y = 2, z = -2

Question 14. Show that, for matrix A = \(\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]\)
, A³-6 A²+5 A+11 I=O, Hence, find \(A^{-1}\).

Or,

Using the matrix method, solve the following system of equations:

3x – 2y + 3z = 8

2x + y – z = 1

4x – 3y + 2z = 4

Important Question for Class 12 Maths Chapter 4

Solution:

Given A = \(\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]\)

To prove: A³ – 6A² + 5A + 11 I = O

⇒ \(A^2=A \cdot A=\left[\begin{array}{rrr}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right] \cdot\left[\begin{array}{rrr}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]=\left[\begin{array}{ccc}
1+1+2 & 1+2-1 & 1-3+3 \\
1+2-6 & 1+4+3 & 1-6-9 \\
2-1+6 & 2-2-3 & 2+3+9
\end{array}\right]\)

⇒ \(A^2=\left[\begin{array}{rrr}
4 & 2 & 1 \\
-3 & 8 & -14 \\
7 & -3 & 14
\end{array}\right]\)

and \(A^3= A^2 \cdot A=\left[\begin{array}{rrr}
4 & 2 & 1 \\
-3 & 8 & -14 \\
7 & -3 & 14
\end{array}\right] \cdot\left[\begin{array}{rrr}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]\)

–\(\left[\begin{array}{rrr}
4+2+2 & 4+4-1 & 4-6+3 \\
-3+8-28 & -3+16+14 & -3-24-42 \\
7-3+28 & 7-6-14 & 7+9+42
\end{array}\right]\)

⇒ \(A³=\left[\begin{array}{ccc}
8 & 7 & 1 \\
-23 & 27 & -69 \\
32 & -13 & 58
\end{array}\right]\)

L.H.S. = \(A^3-6 A^2+5 A+11I\)

= \(\left[\begin{array}{ccc}
8 & 7 & 1 \\
-23 & 27 & -69 \\
32 & -13 & 58
\end{array}\right]-6\left[\begin{array}{ccc}
4 & 2 & 1 \\
-3 & 8 & -14 \\
7 & -3 & 14
\end{array}\right]+5\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]+11\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)

= \(\left[\begin{array}{ccc}
8-24+5+11 & 7-12+5+0 & 1-6+5+0 \\
-23+18+5+0 & 27-48+10+11 & -69+84-15+0 \\
32-42+10+0 & -13+18-5+0 & 58-84+15+11
\end{array}\right]\)

= \(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]=\mathrm{O}\) Zero matrix

Now, A³-6A²+5A+11 I = O

A³A-1 – 6A²A -1+ 5AA-1 + 11 I A-1 = OA-1 (Post multiplying both sides by \(A^{-1}\)

A²(AA-1)-6A(AA-1)+5(AA-1)+11 I A-1 = OA-1

A²-6A +5I + 11A-1 = O

because AA-1=I and OA-1 =O

⇒ \(A^{-1}=\frac{-1}{11} \cdot\left(A^2-6 A+5I\right)=\frac{-1}{11}\left\{\left[\begin{array}{ccc}
4 & 2 & 1 \\
-3 & 8 & -14 \\
7 & -3 & 14
\end{array}\right]-6\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]+5\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\right\}\)

⇒ \(A^{-1}=\frac{-1}{11}\left[\begin{array}{ccc}
4-6+5 & 2-6+0 & 1-6+0 \\
-3-6+0 & 8-12+5 & -14+18+0 \\
7-12+0 & -3+6+0 & 14-18+5
\end{array}\right]\)

⇒ \(A^{-1}=\left[\begin{array}{ccc}
-3 / 11 & 4 / 11 & 5 / 11 \\
9 / 11 & -1 / 11 & -4 / 11 \\
5 / 11 & -3 / 11 & -1 / 11
\end{array}\right]\)

The given system of equations are:

3x -2y + 3 z = 8

2x + y – z = 1 and

4x – 3y + 2z = 4

Determinants Class 12 Questions With Answers

By using the matrix method; the given system of equations can be written as; AX = B

where \(A=\left[\begin{array}{ccc}3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]\) and \(B=\left[\begin{array}{l}8 \\ 1 \\ 4\end{array}\right]\)

Now; \(|A|=\left|\begin{array}{ccc}3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2\end{array}\right|\)=-3+16-30=-17 ≠0

Hence; \(\mathrm{A}^{-1}\) exists.

Now; \(A=\left[\begin{array}{ccc}3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2\end{array}\right]\)

Co-factors are given as: \(C_{11}=-1, \quad C_{12}=-8, C_{13}=-10, C_{21}=-5, C_{22}=-6, C_{23}=1, C_{31}=-1, C_{32}=9, C_{33}=7\)

Hence, (adj)(A)= \(\left[C_{i j}\right]^{\mathrm{T}}\)

adj(A)= \(\left[\begin{array}{ccc}
-1 & -5 & -1 \\
-8 & -6 & 9 \\
-10 & 1 & 7
\end{array}\right]\)

Now; AX = \(B \Rightarrow X=A^{-1} \cdot B\) = \(\frac{\mathrm{adj} A}{|A|} \cdot B\)

⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=-\frac{1}{17}\left[\begin{array}{ccc}
-1 & -5 & -1 \\
-8 & -6 & 9 \\
-10 & 1 & 7
\end{array}\right] \cdot\left[\begin{array}{l}
8 \\
1 \\
4
\end{array}\right]=-\frac{1}{17}\left[\begin{array}{c}
-8-5-4 \\
-64-6+36 \\
-80+1+28
\end{array}\right]=-\frac{1}{17}\left[\begin{array}{l}
-17 \\
-34 \\
-51
\end{array}\right]\)

⇒ \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)

x=1, y=2, z=3

Question 15. If \(A=\left[\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right]\)
, find A-1. Hence, using A-1, solve the system of equations:

2x – 3y + 5z = 11,

3x + 2y – 4z = -5,

x + y -2z = -3.

Solution:

Given, \(A=\left[\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right]\)

A is invertible if |A| ≠ 0

Now. |A| = 2(-4 + 4) + 3 (-6 + 4) + 5(3 – 2)= 0 – 6 + 5 =-1≠0

Co-factors are given as :

⇒ \(C_{11}=0, C_{12}=2, C_{13}=1\)

⇒ \(C_{21}=-1, C_{22}=-9, C_{23}=-5\)

⇒ \(C_{31}=2, C_{32}=23, C_{33}=13\)

⇒ (adj) \((\mathrm{A})=\left[\mathrm{C}_{1 \mathrm{ij}}\right]^{\mathrm{T}}\)

or (adj) \((\mathrm{A})=\left[\begin{array}{ccc}
0 & 2 & 1 \\
-1 & -9 & -5 \\
2 & 23 & 13
\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ccc}
0 & -1 & 2 \\
2 & -9 & 23 \\
1 & -5 & 13
\end{array}\right]\)

⇒ \(\mathrm{A}^{-1}=\frac{\mathrm{adj}(\mathrm{A})}{|\mathrm{A}|}=\left[\begin{array}{ccc}
0 & 1 & -2 \\
-2 & 9 & -23 \\
-1 & 5 & -13
\end{array}\right]\)

Now, given equations are 2x-3y + 5z = 11; 3x + 2y -4z = -5 and x + y – 2z = -3

⇒ \({\left[\begin{array}{ccc}
2 & -3 & 5 \\
3 & 2 & -4 \\
1 & 1 & -2
\end{array}\right]\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\left[\begin{array}{l}
11 \\
-5 \\
-3
\end{array}\right]}\)

⇒ \(\mathrm{AX}=\mathrm{B} \text { or } \mathrm{X}=\mathrm{A}^{-1} \mathrm{~B} \\
\Rightarrow \mathrm{X}=\left[\begin{array}{ccc}
0 & 1 & -2 \\
-2 & 9 & -23 \\
-1 & 5 & -13
\end{array}\right]\left[\begin{array}{l}
11 \\
-5 \\
-3
\end{array}\right]\)

or \(\mathrm{X}=\left[\begin{array}{c}
0 \times 11+(1) \times(-5)+(-2) \times(-3) \\
(-2) \times 11+9 \times(-5)+(-23) \times(-3) \\
(-1) \times 11+5 \times(-5)+(-13) \times(-3)
\end{array}\right] \\
\Rightarrow \quad\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)

⇒ \(\mathrm{x}=1, \mathrm{y}=2, \mathrm{z}=3\)

Important Questions For CBSE Class 12 Maths Chapter 3 Matrices

CBSE Class 12 Maths Chapter 3 Matrices Important Questions

Question 1. The number of all possible matrices of order 2 x 3 with each country 1 or 2 is :

  1. 16
  2. 6
  3. 64
  4. 24

Solution: 3. 6

Required number of possible matrices

= (Number of entries)order

= (2)2×3 = (2)6 = 64

Read and Learn More CBSE Class 12 Maths Important Question and Answers

Important Questions For Class 12 Maths Chapter 3

Question 2. If a matrix A is both symmetric and skewed, symmetric. then A is necessarily a

  1. Diagonal matrix
  2. Zero square matrix
  3. Square matrix
  4. Identity matrix

Solution: 2. Zero square matrix

Given; AT = A (symmetric matrix)

and -AT= A (skew-symmetric matrix)

⇒ 2A = O or A = O

Hence, A is necessarily a zero-square matrix

Important Questions CBSE Class 12 Maths Chapter 3

Question 3. If \(\left[\begin{array}{cc}
3 c+6 & a-d \\
a+d & 2-3 b
\end{array}\right]=\left[\begin{array}{cc}
12 & 2 \\
-8 & -4
\end{array}\right]\)
, then the value of ab-cd is

  1. 4
  2. 16
  3. -4
  4. -16

Solution: 1. 4

Given: \(\left[\begin{array}{cc}
3 c+6 & a-d \\
a+d & 2-3 b
\end{array}\right]=\left[\begin{array}{cc}
12 & 2 \\
-8 & -4
\end{array}\right]\)

On comparing both sides, we get

a – d = 2 and a + d = -8

⇒  2a = -6 Or a=-3

⇒ d = -5

Also; 3c + 6 = 12, 2-3b = -4

⇒ c = 2, b = 2

Hence, ab- cd = (-3)2 – 2(-5) = -6 + 10 = 4

Question 4. For two matrices \(P=\left[\begin{array}{cc}
3 & 4 \\
-1 & 2 \\
0 & 1
\end{array}\right] \text { and } Q^{\top}=\left[\begin{array}{ccc}
-1 & 2 & 1 \\
1 & 2 & 3
\end{array}\right]\),
P-Q is

  1. \(\left[\begin{array}{cc}2 & 3 \\ -3 & 0 \\ 0 & -3\end{array}\right]\)
  2. \(\left[\begin{array}{cc}4 & 3 \\ -3 & 0 \\ -1 & -2\end{array}\right]\)
  3. \(\left[\begin{array}{cc}4 & 3 \\ 0 & -3 \\ -1 & -2\end{array}\right]\)
  4. \(\left[\begin{array}{cc}2 & 3 \\ 0 & -3 \\ 0 & -3\end{array}\right]\)

Solution: 2. \(\left[\begin{array}{cc}4 & 3 \\ -3 & 0 \\ -1 & -2\end{array}\right]\)

⇒ \(Q^{\top}=\left[\begin{array}{ccc}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right] \Rightarrow Q=\left[\begin{array}{cc}-1 & 1 \\ 2 & 2 \\ 1 & 3\end{array}\right]\)

Hence, \(P-Q=\left[\begin{array}{cc}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]-\left[\begin{array}{cc}-1 & 1 \\ 2 & 2 \\ 1 & 3\end{array}\right]=\left[\begin{array}{cc}4 & 3 \\ -3 & 0 \\ -1 & -2\end{array}\right]\)

Matrices Class 12 Important Questions

Question 5. A matrix A = [aij]3×3 is defined by \(a_{i j}=\left\{\begin{array}{ccc}
2 i+3 j & ; & i<j \\
5 & ; & i=j \\
3 i-2 j & ; & i>j
\end{array}\right.\)
, The number of elements in A, which are more than 5, is

  1. 3
  2. 4
  3. 5
  4. 6

Solution: 2. 4

Given; A = [aij]3×3

where \(a_{i j}=\left\{\begin{array}{ccc}
2 i+3 j & ; & i<j \\
5 & ; & i=j \\
3 i-2 j & ; & i>j
\end{array}\right.\)

⇒ A= \(\left[\begin{array}{lll}
5 & 8 & 11 \\
4 & 5 & 13 \\
7 & 5 & 5
\end{array}\right]\)

Hence; required number =4

Question 6. For the matrix X = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\), (X²-X) is:

  1. 21
  2. 31
  3. 1
  4. 51

Solution: 1. 21

⇒ \(X^2=\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right] \cdot\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]=\left[\begin{array}{lll}
2 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 2
\end{array}\right]\)

⇒ \(X^2-X=\left[\begin{array}{lll}
2 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 2
\end{array}\right]-\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]=\left[\begin{array}{lll}
2 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2
\end{array}\right]=21\)

Question 7. Find the order of matrix A such that \(\left[\begin{array}{cc}2 & -1 \\ 1 & 0 \\ -3 & 4\end{array}\right]\), A = \(\left[\begin{array}{cc}-1 & -8 \\ 1 & -2 \\ 9 & 22\end{array}\right]\)

or,

If B = \(\left[\begin{array}{cc}
1 & -5 \\
0 & -3
\end{array}\right]\)
and A+2 B = \(\left[\begin{array}{cc}
0 & 4 \\
-7 & 5
\end{array}\right]\)
, find matrix A.

Solution:

Let \(\mathrm{B}=\left[\begin{array}{cc}2 & -1 \\ 1 & 0 \\ -3 & 4\end{array}\right]\) and \(\mathrm{C}=\left[\begin{array}{cc}-1 & -8 \\ 1 & -2 \\ 9 & 22\end{array}\right]\)

We have \(\left[\begin{array}{cc}2 & -1 \\ 1 & 0 \\ -3 & 4\end{array}\right]_{3 \times 2} \quad A=\left[\begin{array}{cc}-1 & -8 \\ 1 & -2 \\ 9 & 22\end{array}\right]_{3 \times 2}\)

Let the order of A be m x n

BA will be possible if several columns in matrix B should be equal to several rows in matrix A ⇒ m = 2.

and the order of BA is 3 x n

Since, BA = C.

⇒ Order of BA will be the same as that of matrix C

⇒ 3 x n = 3 x 2

⇒ n = 2

Then, the order of matrix A is 2 x 2

A+2 B \(\left[\begin{array}{cc}
0 & 4 \\
-7 & 5
\end{array}\right]\) (given)

⇒ A = \(\left[\begin{array}{cc}
0 & 4 \\
-7 & 5
\end{array}\right]-2 B=\left[\begin{array}{cc}
0 & 4 \\
-7 & 5
\end{array}\right]-2\left[\begin{array}{cc}
1 & -5 \\
0 & -3
\end{array}\right]=\left[\begin{array}{cc}
0 & 4 \\
-7 & 5
\end{array}\right]-\left[\begin{array}{cc}
2 & -10 \\
0 & -6
\end{array}\right]\)

⇒ A = \(\left[\begin{array}{cc}
-2 & 14 \\
-7 & 11
\end{array}\right]\)

Question 8. If A = \(\left[\begin{array}{lll}
1 & 0 & 4
\end{array}\right]\) and B = \(\left[\begin{array}{l}
2 \\
5 \\
6
\end{array}\right]\)
, find AB.

Solution:

AB= \(\left[\begin{array}{lll}
1 & 0 & 4
\end{array}\right]_{1 \times 3}\left[\begin{array}{l}
2 \\
5 \\
6
\end{array}\right]_{3 \times 1}\)

= \([1 \times 2+0 \times 5+4 \times 6]_{1 \times 1}=[2+0+24]=[26]\)

Class 12 Maths Chapter 3 Important Questions With Solutions

Question 9. Given, a skew-symmetric matrix \(A = \mathrm{A}=\left[\begin{array}{ccc}
0 & \mathrm{a} & 1 \\
-1 & \mathrm{~b} & 1 \\
-1 & \mathrm{c} & 0
\end{array}\right]\)
, then value of (a+b+c)² is

Solution:

Given \(A = \mathrm{A}=\left[\begin{array}{ccc}
0 & \mathrm{a} & 1 \\
-1 & \mathrm{~b} & 1 \\
-1 & \mathrm{c} & 0
\end{array}\right]\)

A is a skew-symmetric matrix, AT = -A

⇒ \(A^{\top}=\left[\begin{array}{ccc}
0 & -1 & -1 \\
a & b & c \\
1 & 1 & 0
\end{array}\right] \text { and }-A=\left[\begin{array}{ccc}
0 & -a & -1 \\
1 & -b & -1 \\
1 & -c & 0
\end{array}\right]\)

So, a = 1, b = 0 and c = -1

Now (a + b + c)² = (1 +0-1)² = 0

Question 10. If the matrices A = \(\left[\begin{array}{ccc}
0 & \mathrm{a} & -3 \\
2 & 0 & -\mathrm{t} \\
\mathrm{b} & 1 & 0
\end{array}\right]\)
 is skew symmetric, find the values of ‘a’ and ‘b’.

Solution:

A = \(\left[\begin{array}{ccc}
0 & \mathrm{a} & -3 \\
2 & 0 & -\mathrm{t} \\
\mathrm{b} & 1 & 0
\end{array}\right]\) (given)

A is skew-symmetric ⇒ AT = -A

⇒ \(\left[\begin{array}{ccc}
0 & 2 & b \\
a & 0 & 1 \\
-3 & -1 & 0
\end{array}\right]=\left[\begin{array}{ccc}
0 & -a & 3 \\
-2 & 0 & 1 \\
-b & -1 & 0
\end{array}\right]\)

⇒ a=-2, b=3

Question 11. If the matrix A = \(\left[\begin{array}{cc}
0 & 6-5 x \\
x^2 & x+3
\end{array}\right]\)
 is symmetric, find the values of x?

Solution:

A = \(\left[\begin{array}{cc}
0 & 6-5 x \\
x^2 & x+3
\end{array}\right]\) (given)

For a symmetric matrix, A = A1

⇒ \(\left[\begin{array}{cc}
0 & 6-5 x \\
x^2 & x+3
\end{array}\right]=\left[\begin{array}{cc}
0 & x^2 \\
6-5 x & x+3
\end{array}\right]\)….(1)

∴ 6-5x = x² [from (1)]

⇒ x² + 5x – 6 = 0

⇒ (x + 6) (x -1) = 0

⇒ x = – 6, 1

Question 12. If A = \(\left[\begin{array}{cc}
-3 & 2 \\
1 & -1
\end{array}\right] \text { and } I=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
, find scalar k such that A² + I = KA

Solution:

Given, A = \(\left[\begin{array}{cc}
-3 & 2 \\
1 & -1
\end{array}\right] \text { and } I=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)

⇒ \(\left[\begin{array}{cc}
-3 & 2 \\
1 & -1
\end{array}\right]\left[\begin{array}{cc}
-3 & 2 \\
1 & -1
\end{array}\right]+\left[\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right]=K\left[\begin{array}{cc}
-3 & 2 \\
1 & -1
\end{array}\right]\)

(because \(A^2+I=K A\))

⇒ \(\left[\begin{array}{cc}
11 & -8 \\
-4 & 3
\end{array}\right]+\left[\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right]\) = \(\left[\begin{array}{cc}
-3 \mathrm{~K} & 2 \mathrm{~K} \\
\mathrm{~K} & -\mathrm{K}
\end{array}\right]\)

∴ \(\left[\begin{array}{cc}
12 & -8 \\
-4 & 4
\end{array}\right]\) = \(\left[\begin{array}{cc}
-3 \mathrm{~K} & 2 \mathrm{~K} \\
\mathrm{~K} & -\mathrm{K}
\end{array}\right]\)

⇒ K = -4 (on comparing both sides)

Matrices Subjective And Objective Questions Class 12

Question 13. If A = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]\)
, show that (A-2I)(A-3I) = O.

Solution:

A = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]\) (given)

⇒ A-2I = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]-2\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
2 & 2 \\
-1 & -1
\end{array}\right]\)

and A-3I = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]-3\left[\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
1 & 2 \\
-1 & -2
\end{array}\right]\)

⇒ (A-2I)(A-3I) = \(\left[\begin{array}{cc}
2 & 2 \\
-1 & -1
\end{array}\right]\left[\begin{array}{cc}
1 & 2 \\
-1 & -2
\end{array}\right]=\left[\begin{array}{cc}
2-2 & 4-4 \\
-1+1 & -2+2
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]=O\)

Important Questions For CBSE Class 12 Maths Chapter 12 Linear Programming

CBSE Class 12 Maths Chapter 12 Linear Programming Important Questions

Question 1. A Linear programming problem is as follows: Minimize Z = 2x + y

Subject to the constraints:

  • x ≥ 3, x ≤ 9, y ≥ 0,
  • x-y ≥ 0, x + y ≤ 14

The feasible region has:

  1. 5 corner points including (0, 0) and (9, 5)
  2. 5 corner points including (7, 7) and (3, 3)
  3. 5 corner points including (14, 0) and (9, 0)
  4. 5 corner points including (3, 6) and (9, 5)

Solution: 2. 5 corner points including (7, 7) and (3, 3)

Given; Z = 2x + y subject to the constraints

x ≥ 3, x ≤ 9, y ≥ 0, x – y ≥ 0, x + y ≤ 14

Read and Learn More CBSE Class 12 Maths Important Question and Answers

Important Questions For CBSE Class 12 Maths Chapter 12 Linear Programming Graph For 5 Corner Points

Hence, the feasible region has 5 corner points including (7, 7) and (3, 3)

Important Questions For CBSE Class 12 Maths Chapter 12

Question 2. A Linear Programming Problem is as follows: Maximise/Minimise objective function Z = 2x – y + 5

Subject to the constraints:

  • 3x + 4y ≤ 60,
  • x + 3y ≤ 30,
  • x ≥ 0, y ≥ 0

Linear Programming Class 12 Important Questions

If the corner points of the feasible region are A(0, 10), B(12, 6), C(20, 0) and 0(0,0); then which of the following is true?

  1. The maximum value of Z is 40
  2. The minimum value of Z is -5
  3. The difference between the maximum and minimum values of Z is 35
  4. At two corner points, the value of Z is equal

Solution: 2. The minimum value of Z is -5.

Important Questions For CBSE Class 12 Maths Chapter 12 Linear Programming Corner Points Of The Feasible Region

Question 3. The corner points of the feasible region determined by a set of constraints (linear inequalities) are P(0, 5), Q(3, 5), R(5, 0), and S(4, 1) and the objective function is Z = ax + 2by; where a, b > 0. The condition on ‘a’ and ‘b’ such that the maximum of Z occurs at Q and S is:

  1. a-5b=0
  2. a-3b = 0
  3. a – 2b = 0
  4. a – 8b = 0

Solution: 4. a – 8b = 0

Given points arc P(0, 5), Q(3, 5), R(5, 0) and S(4, 1) and Z = ax + 2by

Since the maximum of Z occurs at Q and S;

⇒ 3a + 10b = 4a + 2b

⇒ 8b = a ⇒ a – 8b = 0

Class 12 Maths Chapter 12 Important Questions With Solutions

Question 4. For an L.P.P.. the objective function is Z = 4x + 3y and the feasible region is determined by a set of constraints (linear inequations) as shown in the graph.

Important Questions For CBSE Class 12 Maths Chapter 12 Linear Programming Graph For Set Of Constraints Linear Equations

Which one of the following statements is true?

  1. The maximum value of Z is at R
  2. The maximum value of Z is at Q
  3. The value of Z at R is less than the value of P
  4. The value of Z at Q is less than the value of R

Solution: 2. Maximum value of Z is at Q

Z = 4x + 3y

Important Questions For CBSE Class 12 Maths Chapter 12 Linear Programming Maximum Value At Q

A maximum value of Z is at Q.

Linear Programming Important Questions CBSE

Question 5. Solve the following problem graphically: Maximize Z = 3x + 9y

Subject to the constraints:

  • x + 3y ≤ 60,
  • x + y ≥ 10,
  • x ≤ y,
  • x ≥ 0, y ≥ 0

Solution:

First of all, let us graph the feasible region of the system of linear inequalities given above. The feasible region ABCD is shown.

Note that the region is bounded. The coordinates of the comer points A, B, C, and D are (0, 10), (5, 5), (15, 15), and (0, 20) respectively.

Important Questions For CBSE Class 12 Maths Chapter 12 Linear Programming Graph For Feasible Regio Of The Ssytem Of Linear Inequalities

Important Questions For CBSE Class 12 Maths Chapter 12 Linear Programming Line Segments Joining

The maximum value of Z is 180. which occurs at every point of the line segment joining the points C and D.

Important Questions For CBSE Class 12 Maths Chapter 13 Probability

CBSE Class 12 Maths Chapter 13 Probability Important Questions

Question 1. If A and B are two independent events and P(A) = \(\frac{1}{3}\), P(B) = \(\frac{1}{2}\); find \(\mathrm{P}(\overline{\mathrm{A}} \mid \overline{\mathrm{B}})\).

Solution:

Given, A and B are two independent events and P(A) = \(\frac{1}{3}\), P(B) = \(\frac{1}{2}\)

⇒ P(A∩B) = P(A).P(B) = \(\frac{1}{3}\) x \(\frac{1}{2}\) = \(\frac{1}{6}\)

⇒ P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{\frac{6}{1}}{\frac{1}{2}}=\frac{1}{3}\)

⇒ \(\mathrm{P}(\overline{\mathrm{A}} / \overline{\mathrm{B}})=\mathrm{P}(\overline{\mathrm{A} / \mathrm{B}})=1-\mathrm{P}(\mathrm{A} / \mathrm{B}) \Rightarrow \mathrm{P}(\overline{\mathrm{A}} / \overline{\mathrm{B}})=1-\frac{1}{3}=\frac{2}{3}\)

Read and Learn More CBSE Class 12 Maths Important Question and Answers

Important Questions For CBSE Class 12 Maths Chapter 13

Question 2. A coin is tossed once. If the head comes up, a die is thrown; blit if the tail comes up, the coin is tossed again. Find the probability of obtaining a head and number 6.

Solution:

S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, H), (T, T)}

Let event A denote “obtaining head” and B denote “obtaining number 6”

⇒ P(A) = \(\frac{1}{2}\), P(B) = \(\frac{1}{6}\)

Hence, the probability of obtaining head and number 6

= P[{(H, 6)}]= P(A∩B) = P(A).P(B) = \(\frac{1}{2}\) x \(\frac{1}{6}\) = \(\frac{1}{12}\)

Probability Class 12 Important Questions

Question 3. Two cards are drawn successively with replacement from a well-shuffled pack of 52 cards. Find the probability distribution of the number of spade cards.

Solution:

Let random variable X denote the number of spade cards; then the possible values of X are 0, 1 or 2.

P(X = 0) = P(no spade and no spade) = \(\frac{39}{52} \times \frac{39}{52}=\frac{3}{4} \times \frac{3}{4}=\frac{9}{16}\)

P(X = 1) = P(spade and no spade or no spade and spade)

= \(\left(\frac{13}{52} \times \frac{39}{52}\right)+\left(\frac{39}{52} \times \frac{13}{52}\right)=\left(\frac{1}{4} \times \frac{3}{4}\right)+\left(\frac{3}{4} \times \frac{1}{4}\right)=\frac{3}{8}\)

P(X = 2) = P(spade and spade) = \(\left(\frac{13}{52} \times \frac{13}{52}\right)=\left(\frac{1}{4} \times \frac{1}{4}\right)=\frac{1}{16}\)

Hence, the probability distribution of X is:

Important Questions For CBSE Class 12 Maths Chapter 13 Probability Probability Distribution Of Number Of Spade Cards

Question 4. A pair of dice is thrown and the sum of the numbers appearing on the dice is observed to be 7. Find the probability that the number 5 has appeared on at least one die.

Or,

The probability that A hits the target is \(\frac{1}{3}\) and the probability that B hits it is \(\frac{2}{5}\). If both try to hit the target independently, find the probability that the target is hit.

Solution:

When a pair of dice is thrown, the sample space S contains 36 outcomes.

Let E: Event that number 5 has appeared on at least one die.

F: Event that the sum of the numbers on the dice is 7

⇒ E = {(5,1), (5,2), (5,3), (5, 4), (5, 5), (5,6), (1,5), (2,5), (3,5), (4,5), (6,5)}, F = {(1, 6), (2,5), (3,4), (4,3), (5,2), (6,1)}

∴ E ∩ F = {(5, 2), (2,5)}

Class 12 Maths Chapter 13 Important Questions With Solutions

Now; P(E∩F) = \(\frac{2}{36}\),P(F) = \(\frac{6}{36}\)

Hence, required probability = \(P(E / F)=\frac{P(E \cap F)}{P(F)}=\frac{(2 / 36)}{(6 / 36)}=\frac{1}{3}\)

Given; P(E1) = \(\frac{1}{3}\), P(E2) =\(\frac{2}{5}\)

⇒ \(P\left(\overline{E_1}\right)=\frac{2}{3} \cdot P\left(\overline{E_2}\right)=\frac{3}{5}\)

⇒ Required Probability = P(target is hit)

= \(1-\mathrm{P}\left(\overline{\mathrm{E}}_1 \overline{\mathrm{E}}_2\right)=1-\mathrm{P}\left(\overline{\mathrm{E}}_1\right) \cdot \mathrm{P}\left(\overline{\mathrm{E}}_2\right)=1-\left(\frac{2}{3} \times \frac{3}{5}\right)=1-\frac{2}{5}=\frac{3}{5}\)

Or,

Let E1: Event that A hits the target

Let E2: Event that B hits the target

Given; \(P\left(E_1\right)=\frac{1}{3}, P\left(E_2\right)=\frac{2}{5} \Rightarrow P\left(\overline{E_1}\right)=\frac{2}{3} \cdot P\left(\overline{E_2}\right)=\frac{3}{5}\)

⇒ Required Probability = P (target is hit)

= \(1-\mathrm{P}\left(\overline{\mathrm{E}}_1 \overline{\mathrm{E}}_2\right)=1-\mathrm{P}\left(\overline{\mathrm{E}}_1\right) \cdot \mathrm{P}\left(\overline{\mathrm{E}}_2\right)=1-\left(\frac{2}{3} \times \frac{3}{5}\right)=1-\frac{2}{5}=\frac{3}{5}\)

Question 5. A random variable X has the probability distribution:

Important Questions For CBSE Class 12 Maths Chapter 13 Probability A Random Variable Of Probability Distribution

Find the value of K and P(X ≤ 2).

Solution:

We know that ∑ P(X) = 1 (for probability distribution)

⇒ 0 + K + 4K + 3K + 2K = 1

⇒ 10 K = 1

⇒ K = \(\frac{1}{10}\)…..(1)

∴ P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

= 0 + k + 4K = 5K = \(\frac{5}{10}\) = \(\frac{1}{2}\) (from (1))

Probability Important Questions CBSE

Question 6. A purse contains 3 silver and 6 copper coins and a second purse contains 4 silver and 3 copper coins. If a coin is drawn at random from one of the two purses, find the probability that it is a silver coin.

Solution:

Let E1: Event that first purse is selected. E2: Event that second purse is selected and A: Event that silver coin is drawn

⇒ \(P\left(E_1\right)=\frac{1}{2}, P\left(E_2\right)=\frac{1}{2}, P\left(A / E_1\right)=\frac{3}{9}, P\left(A / E_2\right)=\frac{4}{7}\)

∴ \(P(A)-P\left(E_1\right) P\left(A / E_1\right)+P\left(E_2\right) \cdot P\left(A / E_2\right)\) (by total probability theorem)

= \(\left(\frac{1}{2} \times \frac{3}{9}\right)+\left(\frac{1}{2} \times \frac{4}{7}\right)=\frac{1}{6}+\frac{2}{7}=\frac{19}{42}\)

Question 7. A coin is tossed 5 times. What is the probability of getting

  1. 3 heads,
  2. At most 3 heads?

Or,

Find the probability distribution of X, the number of heads in a simultaneous loss of two coins.

Solution:

1. P 3 Heads= \({ }^5 \mathrm{C}_3\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^2=10 \times \frac{1}{8} \times \frac{1}{4}=\frac{5}{16}\)

2. P (at most 3 Heads) =\(1-\mathrm{P}(4)-\mathrm{P}(5)\)

= \(1-{ }^3 C_4\left(\frac{1}{2}\right)^4\left(\frac{1}{2}\right)-{ }^3 C_5\left(\frac{1}{2}\right)^5=1-\frac{5}{32}-\frac{1}{32}=\frac{26}{32}=\frac{13}{16}\)

Or,

Let X denotes the number of heads in a simultaneous toss of two coins, then the possible values of X are 0, 1 or 2.

P(X=0)=P(TT)= \(\left(\frac{1}{2} \cdot \frac{1}{2}\right)=\frac{1}{4}\),

P(X=1)=P(HT)+P(T H)=\(\left(\frac{1}{2} \cdot \frac{1}{2}\right)+\left(\frac{1}{2} \cdot \frac{1}{2}\right)=\frac{2}{4}\),

P(X=2)=P(HH)=\(\left(\frac{1}{2} \cdot \frac{1}{2}\right)=\frac{1}{4}\)

Hence; the probability distribution of X is given as :

Important Questions For CBSE Class 12 Maths Chapter 13 Probability Probability Distribution Of Number Of Heads In A Toss Of Two Coins

Question 8. If P(not A) = 0.7, P(B) = 0.7 and P(B/A) = 0.5, then find P( A/B)

Solution:

Given; \(\mathrm{P}(\overline{\mathrm{A}})\) = 0.7 ⇒ 1 – P(A) = 0.7 ⇒ P(A) = 0.3 and P(B) = 0.7

Also, P(B/A) = 0.5; P(A/B) = ?

We know that: P(B/A) = \(\frac{P(B \cap A)}{P(A)} \Rightarrow 0.5=\frac{P(B \cap A)}{0.3}\)

⇒ P(B ∩ A) = 0.5 x 0.3 = 0.15

Now, P(A ∩ B) = P(B ∩ A) = 0.15

⇒ P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{0.15}{0.70}=\frac{15}{70}=\frac{3}{14}\)

Question 9. A black and a red die are rolled together. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

Solution:

Let E: Event of obtaining the sum 8 on the dice.

F: Event that red die resulted in a number less than 4, and let first and second die represent the black and red die respectively.

⇒ E = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}

⇒ n(E) = 5,

F = {(1,1),(1,2),(1,3),(2,1),(2,2),(2,3)}(3,1),(3,2),(3,3),(4,1),(4,2),(4,3),(5,1),(5,2),(5,3),(6,1),(6,2),(6,3)}

⇒ n(F) = 18

and E ∩ F = {(5, 3), (6, 2)}

⇒ n(E ∩F) = 2

Here, \(\mathrm{P}(\mathrm{F})=\frac{18}{36}=\frac{1}{2}$ and $\mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{2}{36}=\frac{1}{18}\)

∴ Required probability = \(P(E / F)=\frac{P(E \cap F)}{P(F)}=\frac{1 / 18}{1 / 2}=\frac{1}{9}\)

Probability Previous Year Questions Class 12

Question 10. A shopkeeper sells three types of flower seeds A1, A2, A3. They are sold in the form of a mixture, where the proportions of these seeds are 4:4:2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively. Based on the above information, answer the following questions

  1. Calculate the probability that a randomly chosen seed will germinate.
  2. Calculate the probability that the seed is of type A,, given that a randomly chosen seed germinates.

Solution:

Given, A1: A2: A3 = 4 : 4 : 2

⇒ \(P\left(A_1\right)=\frac{4}{10}, P\left(A_2\right)=\frac{4}{10}\) and \(P\left(A_3\right)=\frac{2}{10}\)

Let E be the event that a seed germinates.

∴ \(P\left(\frac{E}{A_1}\right)=\frac{45}{100} \cdot P\left(\frac{E}{A_2}\right)=\frac{60}{100}\) and \(P\left(\frac{E}{A_7}\right)=\frac{35}{100}\)

1. \(\mathrm{P}(\mathrm{E})=\mathrm{P}\left(\mathrm{A}_1\right) \cdot \mathrm{P}\left(\mathrm{E} / \mathrm{A}_1\right)+\mathrm{P}\left(\mathrm{A}_2\right) \cdot \mathrm{P}\left(\mathrm{E} / \mathrm{A}_2\right)+\mathrm{P}\left(\mathrm{A}_3\right) \cdot \mathrm{P}\left(\mathrm{E} / \mathrm{A}_3\right)\)….(1)

= \(\left(\frac{4}{10} \cdot \frac{45}{100}\right)+\left(\frac{4}{10} \cdot \frac{60}{100}\right)+\left(\frac{2}{10} \cdot \frac{35}{100}\right)=\frac{490}{1000}=\frac{49}{100}=0.49\)

2. \(P\left(A_2 / E\right)=\frac{P\left(A_2\right) \cdot P\left(E / A_2\right)}{P\left(A_1\right) \cdot P\left(E / A_1\right)+P\left(A_2\right) \cdot P\left(E / A_2\right)+P\left(A_3\right) \cdot P\left(E / A_3\right)}\)

= \(\left(\frac{\frac{4}{10} \times \frac{60}{100}}{\frac{49}{100}}\right)\) (from (1))

= \(\frac{24}{49}=0.48\)

Question 11. A student either knows or guesses or copies the answer to a multiple-choice question with four choices. The probability that he makes a guess is 1/3 and the probability that he copies the answer is 1/6. The probability that his answer is correct given that he copied, is 1/8. Let E1, E2 and E3 are the events that the student guesses, copies or knows the answer respectively and A is the event that the student answers correctly.

1. What is the probability that he answers correctly, given that he knows the answer?

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

2. What is the probability that he answers correctly, given that he knows the answer?

  1. 1
  2. 0
  3. 1/4
  4. 1/8

3. What is the probability that he answers correctly, given that he had made a guess?

  1. 1/4
  2. 0
  3. 1
  4. 1/8

4. What is the probability that he knew the answer to the question, given that he answered it correctly?

  1. 24/29
  2. 4/29
  3. 1/29
  4. 3/29

5. \(\sum_{k=1}^3 P\left(E_1 \mid A\right) \text { is : }\)

  1. 0
  2. 1/3
  3. 1
  4. 11/8

Solution:

Given that:

E1 = Event that the student guesses the answer.

E2 = Event that the student copies the answer.

E3 = Event that the student knows the answer,

and A = event that the student answers correctly

Now, \(P\left(E_1\right)=\frac{1}{3}\) and \(P\left(E_2\right)=\frac{1}{6}\)

∴ \(P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)=1 \Rightarrow \frac{1}{3}+\frac{1}{6}+P\left(E_3\right)=1\)

or \(P\left(E_3\right)=1-\frac{3}{6}=\frac{1}{2}\)….(1)

Also, \(\mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)=\frac{1}{8}\) (given), \(\mathrm{P}\left(\mathrm{A} / \mathrm{E}_3\right)=1\)….(2)

(because his answer is correct, given that he knows it]

Chapter 13 Maths Class 12 Important Questions

and \(P\left(A / E_1\right)=\frac{1}{4}\)…..(3)

(his answer is correct, given that he guesses and the multiple choice question has four choices)

1. (2)Required probability = P(E3) = 1/2 (from eq(1))

2. (1)Required probability = P( A/E3) = 1 (from eq(2))

3. (1) Required probability = P( A/E1) = 1/4 (from eq(3))

4. (1) Required probability = P(E3/A)

= \(\frac{P\left(E_3\right) \cdot P\left(A / E_3\right)}{P\left(E_1\right) \cdot P\left(A / E_1\right)+P\left(E_2\right) \cdot P\left(A / E_2\right)+P\left(E_3\right) \cdot P\left(A / E_3\right)}\) (By Baye’s Theorem)

= \(\frac{\left(\frac{1}{2} \times 1\right)}{\left(\frac{1}{3} \times \frac{1}{4}\right)+\left(\frac{1}{6} \times \frac{1}{8}\right)+\left(\frac{1}{2} \times 1\right)}=\frac{\frac{1}{2}}{\left(\frac{1}{12}+\frac{1}{48}+\frac{1}{2}\right)}\)

= \(\frac{\frac{1}{2}}{\left(\frac{4+1+24}{48}\right)}=\frac{1}{2} \times \frac{48}{29}=\frac{24}{29}\)

5. (3) \(\sum_{k=1}^3 \mathrm{P}\left(\mathrm{E}_2 \mid \mathrm{A}\right)\)

= \(\mathrm{P}\left(\mathrm{E}_1 / \mathrm{A}\right)+\mathrm{P}\left(\mathrm{E}_2 / \mathrm{A}\right)+\mathrm{P}\left(\mathrm{E}_2 / \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \cdot \mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)+\mathrm{P}\left(\mathrm{E}_3\right) \cdot \mathrm{P}\left(\mathrm{A} / \mathrm{E}_3\right)}{\mathrm{P}\left(\mathrm{E}_1\right) \cdot \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)+\mathrm{P}\left(\mathrm{E}_2\right) \cdot \mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)+\mathrm{P}\left(\mathrm{E}_3\right) \cdot \mathrm{P}\left(\mathrm{A} / \mathrm{E}_3\right)}\)

[By applying Baye’s theorem and then taking L.C.M.] =1

Probability MCQ Questions Class 12

Question 12. Suppose a girl throws a die. If she gets 1 or 2, she tosses a coin three times and notes the number of tails. If she gets 3, 4. 5 or 6. she tosses a coin once and notes whether a ‘head’ or ‘tail’ is obtained. If she obtained exactly one ‘tail1, what is the probability that she threw 3, 4.. 5 or 6 with the die?

Solution:

When a die is thrown then the sample space contains 6 outcomes i.e. S= {1,2, 4, 5, 6}

Let E1: Event that she gets 1 or 2 on the die,

E2: Event that she gets 3, 4, 5 or 6 on the die.

A: Event that she acts exactly one tail

Here, \(\mathrm{P}\left(\mathrm{E}_1\right)=\frac{2}{6}=\frac{1}{3}, P\left(\mathrm{E}_2\right)=\frac{4}{6}=\frac{2}{3} \text { and } \mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)=\frac{3}{8}, \mathrm{P}\left(\mathrm{A} / \mathrm{E}_2\right)=\frac{1}{2}\)

When she gets 1 or 2, then she throws a coin three times, and the outcomes are {HHH, TTT, HHT, THH, HTH, TTH, HTT, THT}

⇒ Required probability = P(E2/A)

= \(\frac{P\left(E_2\right) \cdot P\left(A / E_2\right)}{P\left(E_1\right) P\left(A / E_1\right)+P\left(E_2\right) \cdot P\left(A / E_2\right)} \text { (Using Baye’s Theorem) } \)

= \(\frac{\frac{2}{3} \times \frac{1}{2}}{\left(\frac{1}{3} \times \frac{3}{8}\right)+\left(\frac{2}{3} \times \frac{1}{2}\right)}=\frac{\frac{1}{3}}{\left(\frac{1}{8}+\frac{1}{3}\right)}=\frac{8}{11}\)

Question 13. Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find the probability distribution of the random variable X. and hence find the mean of the distribution. [CBSE 2018|

Solution:

S = {1, 2, 3, 4, 5, 6}

X denotes the larger of the two numbers (given)

If X = 2, then favourable cases are {(1, 2),(2,1)}

If X = 3, then favourable cases are {(1, 3),(2, 3), (3, 1),(3, 2)}

If X = 4, then favourable cases are {(1, 4),(2, 4),(3, 4),(4, 1),(4. 2),(4, 3)}

If X = 5, then favourable cases are {(1, 5),(2, 5),(3, 5),(4, 5),(5,1),(5, 2),(5, 3),(5, 4)}

If X = 6, then favourable cases are {(1,6),(2,6),(3,6),(4,6),(5,6),(6, 1),(6,2),(6,3),(6,4),(6,5)}

Important Questions For CBSE Class 12 Maths Chapter 13 Probability Two Numbers At Random Positive Integers

⇒ Mean = \(\Sigma X . P(X)=\frac{4+12+24+40+60}{30}=\frac{140}{30}=\frac{14}{3}\)

Question 14. A bag contains 5 red and 4 black balls, a second bag contains 3 red and 6 black balls. One of the two bags is selected at random and two balls are drawn at random (without replacement), both of which are found to be red. Find the probability that the balls are drawn from the second bag.

Solution:

Let E1: Event that first bag is selected,

E2: Event that the second bag is selected,

E: Event that both drawn balls are red

∴ Required probability = \(\mathrm{P}\left(\mathrm{E}_2 / \mathrm{E}\right)\)

Probability Subjective And Objective Questions

= \(\frac{P\left(E_2\right) \cdot P\left(E_{/} / E_2\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right)}\)(Using Bayes’ Theorem)

= \(\frac{\frac{1}{2} \times \frac{1}{12}}{\frac{1}{2} \times \frac{5}{18}+\frac{1}{2} \times \frac{1}{12}}=\left(\frac{\frac{1}{12}}{\frac{5}{18}+\frac{1}{12}}\right)=\frac{3}{13}\)

Important Questions For CBSE Class 12 Maths Chapter 8 Application Of Integrals

CBSE Class 12 Maths Chapter 8 Application Of Integrals Important Questions

Question 1. Using integration, find the area bounded by the curve y ²= 4x, y-axis, and y = 3.

Or,

Using integration, find the region’s area bounded by the line 2y = – x + 8, x-axis. x = 2 and x = 4.

Solution:

Given curve is y² = 4x ….(1)

and given line is y = 3 …..(2)

From equations (1) and (2):

Point of intersection is B(9/4, 3)

Read and Learn More CBSE Class 12 Maths Important Question and Answers

Important Questions For CBSE Class 12 Maths Chapter 8 Application Of Integrals Integrals Area Bounded By The Curve

⇒ Required Area = \(\int_0^3\) x dy

= \(\frac{1}{4} \int_0^3 y^2, d y=\frac{1}{4}\left[\frac{y^3}{3}\right]_0^3=\frac{1}{4}\left[\frac{3^3}{3}-\frac{0}{3}\right]=\left(\frac{1}{4}, \frac{27}{3}\right)=\frac{9}{4} \text { sq. units }\)

Or,

Given lines are 2y + x = 8, x = 2 and x = 4

Important Questions For CBSE Class 12 Maths Chapter 8 Application Of Integrals Integrals Area Region Bounded By The Curve

⇒ Required area = \(\int_2^4 y \mathrm{dx}=\int_2^4\left(\frac{8-\mathrm{x}}{2}\right) \mathrm{dx}\)

= \(\int_2^4\left(4-\frac{1}{2} \mathrm{x}\right) \mathrm{dx}=\left[4 \mathrm{x}-\frac{1}{2}\left(\frac{\mathrm{x}^2}{2}\right)\right]_2^4\)

= \(\left[4 \times 4-\frac{1}{4}(4)^2\right]-\left[4 \times 2-\frac{1}{4}(2)^2\right]\)

=12-7=5

∴Required Area = 5 sq. units

Application Of Integrals Class 12 Important Questions

Question 2. Using integration, find the area bounded by the circle x² + y² = 9.

Solution:

The whole area enclosed by the given circle will be 4 times the area of the region AOBA bounded by the curve, x-axis, and the ordinates x = 0 and x = 3 [as the circle is symmetrical about both the x-axis and y-axis]

Important Questions For CBSE Class 12 Maths Chapter 8 Application Of Integrals Are Bounded By The Circle

⇒ Required area = \(4 \int_0^3 y d x=4 \int_0^3 \sqrt{3^2-x^2} d x \quad\left[x^2+y^2=3^2 \text { gives } y= \pm \sqrt{3^2-x^2}\right]\)

As the region AOBA lies in the first quadrant, v is taken as positive. Integrating, we get the whole area enclosed by the given circle

⇒ Required area = \(4\left[\frac{x}{2} \sqrt{3^2-x^2}+\frac{3^2}{2} \sin ^{-1}\left(\frac{x}{3}\right)\right]_0^3\)

= \(4\left[\left(\frac{3}{2} \times 0+\frac{3^2}{2} \sin ^{-1}(1)\right)-0\right]=4\left(\frac{3^2}{2}\right)\left(\frac{\pi}{2}\right)=9 \pi \text { sq. units }\)

Question 3. Find the area of the region bounded by curve 4x²=y and the line y = 8x + 12. using integration.

Solution:

Given curve is 4x²= y….(1)

and given line is y = 8x + 12…..(2)

From equation (1) and (2), we get:

4x² – 8x – 12 = 0

⇒ x² – 2x – 3 = 0

⇒ (x-3) (x + 1) = 0 =3 x = 3, -1

Important Questions For CBSE Class 12 Maths Chapter 8 Application Of Integrals Are Region Bounded By The Curve And Line

From equation (1); when x = 3, y = 36 and when x = – 1 ⇒ y = 4.

So, point of intersection of the curve and line tire (3, 36) and (-1,4).

⇒ Required Area = \(\int_{-1}^3\left\{(8 x+12)-4 x^2\right\} d x=\left[\frac{8 x^2}{2}+12 x-\frac{4 x^3}{3}\right]_{-1}^3\)

= \((36+36-36)-\left(4-12+\frac{4}{3}\right)=36+\frac{20}{3}=\frac{128}{3} \text { sq. units }\)

Important Questions For CBSE Class 12 Maths Chapter 8

Class 12 Maths Chapter 8 Important Questions With Solutions

Question 4. Using integration, find the area of the region bounded by the curves x² + y² = 4, x =√3y, and the x-axis lying in the first quadrant.

Solution:

Given curve x² + y² = 4 is a circle with center (0, 0) and radius 2.

And line is x = √3y

Now, for the point of intersection of the line and circle, we have:

⇒ 4y² = 4 ⇒ y = ± 1

For y = 1; x = √3

So. point C is (3, 1)

Important Questions For CBSE Class 12 Maths Chapter 8 Application Of Integrals Area Bounded By The Curve And x axis Lying In First Quardant

Required area = area of OACO + Area of ABCA

= \(\int_0^{\sqrt{3}} y_{\text ({line })} d x+\int_{\sqrt{3}}^2 y_{\mid \text {circle } \mid} d x=\frac{1}{\sqrt{3}} \int_0^{\sqrt{3}} x d x+\int_{\sqrt{3}}^2 \sqrt{4-x^2} d x\)

= \(\frac{1}{2 \sqrt{3}}\left[x^2\right]_0^{\sqrt{3}}+\left[\frac{x}{2} \sqrt{4-x^2}+\frac{4}{2} \sin ^{-1}\left(\frac{x}{2}\right)\right]_3^{-2}\)

= \(\frac{1}{2 \sqrt{3}}(3)+\left\{2 \sin ^{-1}(1)-\frac{\sqrt{3}}{2}-2 \sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)\right\}=\frac{\sqrt{3}}{2}+\left(2 \times \frac{\pi}{2}\right)-\frac{\sqrt{3}}{2}-2 \times \frac{\pi}{3}=\frac{\pi}{3} \text { sq.units }\)

Application Of Integrals Important Questions CBSE

Question 5. Using the method of integration, find the area of the triangle ABC. coordinates of whose vertices are A(2. 0), B(4, 5) and C(6, 3).

Solution:

Vertices of ΔABC are A(2,0), B(4,5) and C (6,3)

Equation of line AB: y = 5/2(x – 2)

Equation of line BC: y = 9-x

Equation of line AC: y= 3/4(x-2)

Important Questions For CBSE Class 12 Maths Chapter 8 Application Of Integrals Area Of The Triangle

⇒ Required Area of ΔABC = \(\int_2^4(\text { line } \mathrm{AB}) \mathrm{dx}+\int_4^6(\text { line } \mathrm{BC}) \mathrm{dx}-\int_2^6(\text { line } \mathrm{AC}) \mathrm{dx}\)

= \(\frac{5}{2} \int_2^4(x-2) d x+\int_4^6(9-x) d x-\frac{3}{4} \int_2^6(x-2) d x\)

= \(\frac{5}{2}\left[\frac{x^2}{2}-2 x\right]_2^4+\left[9 x-\frac{x^2}{2}\right]_4^6-\frac{3}{4}\left[\frac{x^2}{2}-2 x\right]_2^6\)

= \(\frac{5}{2}[(8-8)-(2-4)]+[(54-18)-(36-8)]-\frac{3}{4}[(18-12)-(2-4)]\)

= \(\left(\frac{5}{2} \times 2\right)+8-\left(\frac{3}{4} \times 8\right)=5+8-6=7 \text { sq. units }\)

Question 6. Using the method of integration, find the area of a triangle whose vertices arc (1, 0), (2, 2), and (3, 1).

Solution:

Important Questions For CBSE Class 12 Maths Chapter 8 Application Of Integrals Area Of ThE Triangle With Vertices

Equation of line AB is y – 0 = \(\frac{(2-0)}{(2-1)}(x-1) \Rightarrow y=2(x-1)\)

Equation of line BC is y – 2 = \(\frac{(1-2)}{(3-2)}(x-2) \Rightarrow y=(-x+1)\)

Equation of line AC is y – 0 = \(\frac{(1-0)}{(3-1)}(x-1) \Rightarrow y=1/2(x-1)\)

⇒ Required Area = \(\int_1^2(\text { line } A B) d x+\int_2^3(\text { line } B C) d x-\int_1^3(\text { line } A C) d x\)

= \(\int_1^2 2(\mathrm{x}-1) \mathrm{dx}+\int_2^3(-\mathrm{x}+4) \mathrm{dx}-\int_1^3 \frac{1}{2}(\mathrm{x}-1) \mathrm{dx}\)

= \(2\left(\frac{\mathrm{x}^2}{2}-\mathrm{x}\right)_1^2+\left(\frac{-\mathrm{x}^2}{2}+4 \mathrm{x}\right)_2^3-\frac{1}{2}\left(\frac{\mathrm{x}^2}{2}-\mathrm{x}\right)_1^3\)

= \(2\left[(2-2)-\left(\frac{1}{2}-1\right)\right]+\left[\left(\frac{-9}{2}+12\right)-(-2+8)\right]-\frac{1}{2}\left[\left(\frac{9}{2}-3\right)-\left(\frac{1}{2}-1\right)\right] \)

= \(\left(2 \times \frac{1}{2}\right)+\left(\frac{-9}{2}+6\right)-\frac{1}{2}(4-2)=1+\frac{3}{2}-1=\frac{3}{2} \text { sq. units }\)

CBSE Class 12 Maths Chapter 8 Extra Questions

Question 7. Using integration, find the region’s area in the first quadrant enclosed by the x-axis, the line y = x, and the circle x² + y² = 32.

Solution:

Given line is y = x ….(1)

and given circle is x² + y² = 32….(2)

From equations (1) and (2); we have

2x² = 32 ⇒ x = ± 4

∴ y = ± 4

Important Questions For CBSE Class 12 Maths Chapter 8 Application Of Integrals Area Of The Region In First Quadrant

Now; (4, 4) lies in 1st quadrant

⇒ Required Area  = \(\int_0^4 x d x+\int_1^{4 \sqrt{2}} \sqrt{32-x^2} d x\)

= \(\left(\frac{x^2}{2}\right)_0^4+\left[\frac{x}{2} \sqrt{32-x^2}+\frac{32}{2} \sin ^{-1}\left(\frac{x}{4 \sqrt{2}}\right)\right]_{-4}^{4 \sqrt{2}}\)

= \((8-0)+\left[\left(0+16 \times \frac{\pi}{2}\right)-\left(8+16 \times \frac{\pi}{4}\right)\right]=4 \pi \text { sq. units }\)

NCERT Class 6 Science Chapter 4 Getting To Know Plants Question And Answers

NCERT Class 6 Science Chapter 4 Getting To Know Plants Long Question And Answers

Question 1. Correct the following statements and rewrite them in your notebook.

  1. Stem absorbs water and minerals from the soil
  2. Leaves hold the plant upright
  3. Roots conduct water to the leaves
  4. The number of petals and sepals In a flower is always equal
  5. If the sepals of a flower are joined together, Its petals are also joined together
  6. If the petals of a flower are joined together, then the pistil is joined to the petal

Answer:

  1. Correct statements are as follows
  2. Roots absorb water and minerals from the soil.
  3. Roots hold the plant upright.
  4. The stem conducts water to the leaves.
  5. The number of petals and sepals in a flower can be equal or different.
  6. If the sepals of a flower are joined together, its petals are not necessarily joined together.
  7. If the petals of a flower are joined together, then the pistil may or may not be necessarily joined together.

Read and Learn More Class 6 Science Question And Answers

Question 2. Can you find a plant in your house or In your neighbourhood, which has a long, but weak stem? Write its name. In which category will you place it?
Answer: Yes, we can find money plants in our house as well as in our neighbourhood, which have a long, but weak stem. It is classified as a climber. It readily climbs up by taking support from neighbouring structures such as a tree or a rod, etc.

NCERT Class 6 Science Chapter 4 Getting To Know Plants Questions And Answers

Question 3. What is the function of a stem?
Answer: The functions of a stem in a plant are as follows

  1. It conducts water from the roots to the leaves and other parts of the plants.
  2. It conducts food from leaves to other parts of the plants.
  3. It bears leaves, flowers and fruits.
  4. It provides support to the plant.

Question 4. Which of the following leaves have reticulate venation? Wheat, Tulsi, Maize, Grass, Coriander (dhania), China rose.
Answer: Leaves of tulsi, coriander and China rose have reticulate venation.

Question 5. If a plant has fibrous roots, what type of venation do its leaves have?
Answer: If a plant has a fibrous root, its leaves have parallel venation.

Question 6. If a plant has leaves with reticulate venation, what kind of roots will it have? If a plant has leaves with reticulate venation, then the plant is likely to have tap roots. Is it possible for you to find out whether a plant has tap roots or fibrous roots by looking at the impression of its leaf on a sheet of paper?
Answer: Yes, it is possible to find whether a plant has a taproot or fibrous root by looking at the impression of its leaf on paper. If the leaf has parallel venation, the roots ofthe plant will be fibrous root the leaf has reticulate venation, the root will be tap root.

Question 7. What are the parts of a flower?
Answer: The parts of a flower are as follows

  1. Petals
  2. sepals
  3. Stamens
  4. Pistil

Question 8. From the following plants, which of them have flowers?
Grass, maize, wheat, chilli, tomato, tulsi, peepal, sheesham, banyan, mango, jamun, guava, pomegranate, papaya, banana, lemon, sugarcane, potato, groundnut.
Answer: All ofthe given plants produce flowers. But in some plants such as peepal, sugarcane, grass, etc. the flowers are very small The plants which produce prominent flowers are

Class 6 Science Chapter 4 Getting To Know Plants The Following Plants And Flowers

Question 9. Name the part of the plant which produces food. Name this process.
Answer: The part plant which produces food is the leaf. This process is known as photosynthesis.

Question 10. In which part of a flower, you will find the ovary?
Answer: Ovary is found in the lowermost part of the pistiL

Question 11. Name two plants in which one has joined sepals and the other has separate sepals.
Answer: Flowers with joined sepals are found in Hibiscus, Datura, cotton, tomato (anyone). Flowers with separate sepals are found in mustard, lotus, lily, jasmine, and rose (anyone).

Question 12. The table lists four different groups of plants and their features

Class 6 Science Chapter 4 Getting To Know Plants The Different Groups Of Plants And their features

The Picture Shows A Garden Plant.

Class 6 Science Chapter 4 Getting To Know Plants A Garden Plant

1. Which group does the garden plant belong to?

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

Answer: 2. Group 2

2. Which Group Of Plants Most likely needs support to grow?

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

Answer: 1. Group 1

Question 13. Three similar potted plants were taken to conduct an activity to determine the conditions essential for plant growth. Plant A was kept in sunlight, but not watered. Plant B was kept in sunlight and watered. Plant C was watered and kept in a dark room.

  1. Which plant will grow best and which plant(s) will not show proper growth?
  2. The above experiment lists two main factors necessary for plant growth
  3. Leaves were taken from each of the plants and boiled to remove the green colour. They then had an iodine solution placed on the leaves. Describe and explain what would be observed for leaves A and B.

Answer: Plant TV will grow best and plants ‘A‘ and ‘C’ will not show proper growth.

The two main factors necessary for plant growth are sunlight and water.

The result will not be the same as the leaf from plant ‘A’ will show no change In colour due to the absence of starch, but the leaf from plant ‘B’ will turn bluish-black because of the presence of starch.

Question 14. Ajit wants to test If plants need sunlight to make food. He keeps a potted plant in sunlight, After five days, he tests for the presence of starch In the leaves.

Class 6 Science Chapter 4 Getting To Know Plants If Plants Need Sunlight To Make Food

How can Ajit improve his test?

  1. He should choose a plant with larger green leaves.
  2. He should test another plant without leaves under the Sun.
  3. He should test a similar plant kept in the dark for five days,
  4. He should cover the plant with a transparent glass box to keep it warm.
  5. Which of the following statements Is true? Write ‘Yes’ or ‘No’ to mark your responses.

Class 6 Science Chapter 4 Getting To Know Plants The Following Statements Is True or not

Answer:

  1. He should test a similar plant kept in the dark for five days.
  2. No, Yes, Yes

Question 15. Identify the Incorrect statements and correct them.

  1. Anther Is a part of the pistil.
  2. The visible part of a bud is the petals.
  3. Lateral roots are present In a tap root.
  4. Leaves perform the function of transpiration only

Answer:

  1. It is an incorrect statement as another is a part of the stamen.
  2. It is an incorrect statement as the visible parts ofthe bud are sepals.
  3. It is a correct statement.
  4. It is an incorrect statement as leaves also perform photosynthesis along with transpiration.

Question 16. Read the functions of parts of a plant given below

  1. Fixes plant to the soil
  2. Prepares starch
  3. Takes part in reproduction
  4. Supports branches and bears flowers.
  5. In the diagram given in the below figure, write the names of the parts whose function you have just read in the appropriate space.

Class 6 Science Chapter 4 Getting To Know Plants plant

Answer: The functions of parts of a plant are

Class 6 Science Chapter 4 Getting To Know Plants Funtions And Parts Of A Plant

Question 17. Observe the figure and attempt the questions that follow it.

Class 6 Science Chapter 4 Getting To Know Plants leaf

  1. Label the parts 1, 2, 3 and 4 in the figure.
  2. What type of venation does the leaf have?
  3. What type of venation is seen in grass leaves

Answer: 2. In the given figure

  1. Petiole
  2. Midrib
  3. Lamina
  4. Vein

The leaf has reticulate venation, and the veins in the leaf occur in an irregular way forming a net-like pattern. Grass leaves have parallel types of venation. In which the veins run parallel to each other on both sides of the midrib.

Question 18. Draw the veins of leaves given in the figure below and write the type of venation.

Class 6 Science Chapter 4 Getting To Know Plants Types Of Venation

Answer:

The veins of leaves and their type of venation is

Class 6 Science Chapter 4 Chapter 4 Getting To Know Plants Types Of Venations

Question 19. Sudha fills two beakers with equal amounts of water she places a plant in beaker 1 with the roots of the plant dipped in water. She tightly covers the mouth of both beakers with plastic sheets. She leaves the beakers in sunlight and notes the amount of water in each beaker after 3 days

Class 6 Science Chapter 4 Getting To Know Plants Amount of water in the beaker at the start of the activity

What Is Sudha trying to find out?

  1. Do plants need water to live?
  2. Do plant roots absorb water?
  3. Does water help plants to stand straight?
  4. Does water evaporate faster in sunlight?

Suppose the beakers are not covered with plastic sheets. Will the amount of water in the beakers remain the same as shown in the table? Explain your answer.

Answer:

  1. Do plant roots absorb water?
  2. The amount of water in both beakers will not remain the same and will reduce because of the evaporation.

Question 20. The pictures show four different plants.

Class 6 Science Chapter 4 Getting To Know Plants The Four different plants

Which plants have the same type of roots?

  1. Only plant 1 and plant 2
  2. Only plant 2 and plant 3
  3. Plant 1, plant 2 and plant 3.
  4. Plant 2, plant 3 and plant 4
  5. Why is the root of plant1 thick and round?

Answer:

  1. Plant 1, plant 2 and plant 3
  2. The root of plant 1 is thick and round because these roots store food for the plant.

Question 21. The picture shows different parts of a flower.

Class 6 Science Chapter 4 Getting To Know Plants Types Of A Flower

  1. How many petals can be seen in the picture?
  2. What Is Label X?

Answer:

  1. 3 petals can be seen in the picture.
  2. Label X is the filament.

Question 22. Observe the picture of an activity given as a figure, carried out with leaves of plants and a polythene bag.

Class 6 Science Chapter 4 Getting To Know Plants Plants And Polythene Bag

  1. Now answer the following
  2. Which process is demonstrated in the activity?
  3. When will this activity show better results on a bright sunny day or a cloudy day?
  4. What will you observe in the polythene bag after a few hours of setting up the activity?
  5. Mention any one precaution you must take, while performing this activity.

Answer: The process demonstrated in the activity is transpiration.

  1. The activity will show the best results on a bright sunny day because transpiration is maximum in sunlight.
  2. After a few hours of setting up the activity, one observes small droplets of water inside the polythene bag.
  3. A major precaution one must take while performing the activity is that the polythene bag should be cleaned and its mouth should be sealed properly. Also, the twig should be fresh with 10-12 leaves.

NCERT Class 6 Science Chapter 4 Getting To Know Plants Short Question And Answers

Question 1. What kind of stem do the money plant, beanstalk, gourd plants and grapevines have?
Answer: All these above-mentioned plants, i.e. money plants, beanstalk, gourd plants and grapevines have soft, green and weak stems and are climbers.

Question 2. Do all the leaves have petioles?
Answer: No, all the leaves do not have petioles. In some plants, leaves are attached directly to the plant stem.

Question 3. Can you name the process that makes water drops appear on the polythene cover?
Answer: The process that makes water droplets appear inside the polythene cover is known as transpiration. This occurs due to the loss of water from the leaves.

Question 4. Look at the figure below. Who do you think is watering their plant correctly? Paheli or Boojho? Why

Class 6 Science Chapter 4 Chapter 4 Getting To Know Plants Watering Their Plants

Answer: I think that Paheli is watering the plants correctly. This is because she is sprinkling water on the roots, which helps in the upward conduction of water. How are the types of roots and leaf venation in a plant related to each other? Fill in the table below to justify your answer.

Class 6 Science Chapter 4 Getting To Know Plants The types of roots and leaf venation.

Answer: The table below shows the relation between the roots and leaf venation types in a plant.

Getting To Know Plants The types of roots and leaf venation.

Question 5. Write the name of the material that goes up in the stem and that which comes down.
Answer: The stem is like a street with two-way traffic. Through it, water and minerals go up while it conducts food from the leaves to other parts of the plant.

Question 6. Write down two examples of trees, shrubs, herbs and creepers growing near your area.
Answer: Two examples of trees – Oak and banyan (barged). Two examples of shrubs- the China rose and jasmine. Two examples of herbs are tomato and wheat. Two examples of creepers are pumpkin and watermelon.

Question 7. Can the stem of a plant be compared with a street with two-way traffic? Give reason.
Answer: Yes, the stem of a plant can be compared with a street with two ways traffic because

It carries water and minerals from the roots to the leaves and other parts ofplantin an upward direction.

It takes the food prepared by the leaves to other parts of the plant.

Question 8. Is it right to call the leaf as food factory of the plant? Justify your answer.
Answer: Yes, the leafs called as food factory of the plant. It is because the main function of a leaf is to synthesise food by the process of photosynthesis.

Question 9. Roots are necessary to keep the plants healthy and alive. Explain.
Answer: Roots absorb minerals and water from the soil. Both of these are needed for the manufacture of food from plant leaves. So, roots are necessary to keep the plant healthy and alive.

Question 10. Taproot is different from the fibrous root. Explain how.
Answer: The differences between tap and fibrous roots are

Class 6 Science Chapter 4 Getting To Know Plants The differents between tap root and fibrous root

Question 11. The type of leaf venation and root in plants are related interestingly. Explain the statement.
Answer: The relation between the type of leaf venation and the type of roots is as follows

  1. The plant having leaves with reticulate venation has tap roots, for Example sunflower plant.
  2. The plant having leaves with parallel venation has fibrous roots, Example wheat.

Question 12. Will a leaf taken from a potted plant kept in a dark room for a few days turn blue-black when tested for starch? Give a reason for your answer.
Answer: No, a leaf from a potted plant kept in the dark will not turn blue-black when tested for the presence of starch. This is because all the stored starch would have been used up by the plant. No fresh starch would be synthesised due to the non-availability of sunlight.

Question 13. Boojho wanted to test the presence of starch in leaves. He performed the following steps

  1. He took a leaf and boiled it in water.
  2. He placed the leaf in a Petri dish and poured some iodine over it
  3. He did not get the expected result. Which step did he miss? Explain.

Answer: Boojho did not get the expected results in his experiment because he missed an important step in the procedure. He did not boil the leaf in spirit to remove chlorophyll. It is necessary to remove chlorophyll because it interferes with the test for starch. It is also essential to remove chlorophyll from leaves so that the leaves get decolourised.

Question 14. Suggest the type of root system In grass. Also, explain that root system.
Answer: In grass, a fibrous root system is present. In such types of roots, there is no main root, instead many roots arise from one region. These grow horizontally in the soil and make a bushy/clustered appearance.

NCERT Class 6 Science Chapter 4 Getting To Know Plants Assertion-Reason Questions

Question 1. The following questions consist of two statements. Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below

  1. Both A and R are true and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A
  3. A is true, but R is false
  4. A is false, but R is true

1. Assertion (A) The Lamina of the leaf is green and helps in photosynthesis.
Reason (R) It is green due to the presence of green-coloured pigment.

2. Assertion (A) Roots are aerial parts of the plants.
Reason (R) Their main function is to absorb water and minerals from the soil.

3. Assertion (A) In tap root, the main root is present.
Reason (R) The smaller roots arising from the main root are called lateral roots.

4. Assertion (A) Transpiration helps in the movement of water from roots to leaves.
Reason (R) Transpiration is the loss of water from the stomata on leaves. As water evaporates from leaves, it creates a suction force that pulls water from the roots through the plant.

Answer:

Both A and R are true and R is the correct explanation of A.

A is false, but R is true. A can be corrected as Shoot is the aerial part ofthe plant whereas roots are present deep in the soil (underground)

Both A and R are true, but R is not the correct explanation of A. It can be corrected as In tap root one main root is present which grows vertically downward into the soil.

Both A and R are true and R is the correct explanation of A.

Question 2. Assertion (A) Plants with weak stems that cannot stand upright but spread on the ground are called creepers.
Reason (R) Plants that take support and climb up are called climbers.
Answer: Plants that take support and climb up are called climbers.

Question 3. Assertion (A) The Female part of the flower is called a pistil.
Reason (R) Transfer of pollen grains from the anther to the stigma is called pollination.
Answer: 2. Transfer of pollen grains from the anther to the stigma is called pollination.

 

NCERT Class 6 Science Chapter 4 Getting To Know Plants Very Short Questions and Answers

Question 1. Plants are classified into three main categories. Explain briefly.
Answer: Based on the size and nature of the stem, die plants are classified into three main categories, i.e. herbs, shrubs and trees.

Question 2. What is the usual name of medium-sized plants with hard and woody stems whose many branches arise just above the ground?
Answer: Shrubs

Question 3. All plants are of the same size. Do you agree?
Answer: No, all plants are of different sizes. According to their size, they are classified as herbs, shrubs and trees.

Question 4. Herb is different from a shrub. Explain.
Answer: The herb is a small plant with a tender and green stem. Whereas, the shrub is bigger than an herb and has a strong and thicker stem.

Question 5. Give the term for each of the following.

  1. These are small plants with green, soft, tender stems.
  2. These are bigger than herbs with thick and hard stems and branching at the base.

Answer: Herbs Shrubs

Question 6. It Is given that a plant Is very tall and has a hard, thick stem with branching on the upper part of the plant. Based on the given characteristics, categorise the plant.
Answer: Based on the given characteristics, it can be concluded that the given plant is a tree.

Question 7. Money plant Is an example of a creeper. Do you agree? Explain.
Answer: No, a money plant is not an example of a creeper. It is an example of a climber as it grows or climbs up by taking the support of neighbouring structures.

Question 8. Differentiate between vein and midrib.
Answer: A large number of linear structures that spread to all parts of the leaf are called veins. Whereas, the main vein (thick vein) in the middle of the leaf is called midrib.

Question 9. The leaves of grass are different from those of mango. Explain.
Answer: The leaves of grass show parallel venation whereas mango leaves have reticulate venation.

Question 10. leaves when treated with iodine give a blue-black colour. Give the reason.
Answer: The leaf gives a blue-black colour due to the presence of starch in it.

Question 11. In some plants, roots do not have any main root but all the roots are similar. Give some examples of such fibrous roots.
Answer: Examples of plants with fibrous roots are maize, wheat, rice, etc.

Question 12. Name the part of a plant which helps in holding the plant to the soil.
Answer: The roots anchor the plant to the soil.

Question 13. Roots absorb water and minerals from the soil. Why are these needed?
Answer: Roots absorb water and minerals from the soil. These are needed for the manufacture of food from plant leaves.

Question 14. Roots are helpful in the conservation of soil. Explain.
Answer: Roots help in holding the soil together. In this way, roots prevent the soil from being blown away by wind or washed away by water and thus, help in the conservation of soil.

Question 15. Differentiate between sepals and petals.
Answer: The small, green-coloured leaf-like structures seen in flowers are called sepals. Whereas, petals are the big, brightly coloured leaf-like structures seen in flowers.

Question 16. Plants also have reproductive organs like animals. Name the reproductive part of plants.
Answer: Stamens and pistil are the male and female reproductive parts of plants, respectively

NCERT Class 6 Science Chapter 4 Getting To Know Plants Fill In The Blanks 

1. The type of venation found in pea is Reticulate

2. Water comes out of the leaf by a process called transpiration

3. The small green leaves at the base of flowers are known as Sepals

4. The swollen basal part of the pistil is the Ovary, which bears the

5. Stamen has two parts called Anther, filaments.

6. The young unopened flower is termed a bud

7. Sepals, Petals stamens and pistils are the parts of a flower. Stamen Is made up of anther and filament and It represents the male part of the flower. The female part of the flower is called the pistil The basal, swollen part of the pistil is called the ovary which contains the ovules.

8. Loss of water through leaves is called Transpiration

9. The unwanted plants that grow in a field are called weeds

10. Mango leaves have venation reticulate

NCERT Class 6 Science Chapter 4 Getting To Know Plants True Or False

1. Stem bears branches, flowers and fruits. True

2. The arrangement of veins in the leaf lamina is called venation. True

3. Leaves are generally multicoloured. False, leaves are generally green in colour

4. Lateral roots are present in a tap root. True

5. Anther is a part of the pistil. False, another is a part of the stamen

6. Pollen grains are present in the anther of the pistil. False

7. Wheat has parallel venation and fibrous roots. True

NCERT Class 6 Science Chapter 4 Getting To Know Plants Match The Columns

Question 1. Match the Column 1 with Column 2.

Class 6 Science Chapter 4 Getting To Know Plants Match the coloumn 1 and column 2

Answer: A-2, B-1, C-4, D-3

Question 2. Match the Column 1 with Column 2.

Class 6 Science Chapter 4 Getting To Know Plants Match the coloumn 1 and column 2.2

Answer: A-3, B-2, C-4,D-5

Question 3. Match the Column 1 with Column 2.

Class 6 Science Chapter 4 Getting To Know Plants Match the column 1 and column 2.3

Answer: A-2, B-1, C-4, D-3

NCERT Class 6 Science Chapter 5 Body Movement Question And Answers

NCERT Class 6 Science Chapter 5 Class 6 Science Body Movement Long Questions And Answers

Question 1. Plants do not move from one place to another. Do they show any other kind of movement? What types of movements are shown by plants?
Answer: Plants do not move from one place to another, but they show other kinds of movements. For example, the movement of sunflowers toward the Sun, the movement of roots toward the water, and the wrinkling of leaves of touch-me-not plants when touched, etc.

Question 2. Fill in the table given below to explain how animals move from place to place.

The Living Organisms How Animals Move From Palce To Place

Answer: The table given below explains the movement of animals from place to place

The Living Organisms How Animals Move From Palce To Place.

Question 3. Why are there so many differences In the way that animals move from place to place?
Answer: There are so many differences in the way that animals move from place to place because of the differences in their body structure and also the different types of environments in which they live.

Read and Learn More Class 6 Science Question And Answers

Question 4. Fill In the table given below showing different types of movement in our body.

Class 6 Science Chapter 5 Body Movements Movements In Our Body

Answer:

Class 6 Science Chapter 5 Body Movements Movements In Our Body.

Question 5. Why is it that we can move a few parts of our body easily in various directions and some only in one direction? Why are we unable to move some parts at
Answer: We can move a few parts ofour body easily in various directions and some only in one direction. Some parts are fixed, i.e. they do not move at all. This is because the movement depends upon the type of joint that joins two or more bones together.

For example, the body parts having ball and socket joints move in all directions, while the body parts having hinge joints move only in one direction. The body parts having fixed joints are unable to move at all.

NCERT Class 6 Science Chapter 5 Body Movements Question And Answers

Question 6. How do we bend our elbows?
Answer: The elbow is not one long bone from the upper arm to our wrist. It consists of different bones joined together at the elbow. The joint present at the elbow is the hinge joint that helps in bending

Question 7. How many bones does the middle finger have?
Answer: The middle finger has three bones.

Question 8. Are muscles and bones always required for mo Pages
Answer: Yes, muscles and bones are always required for movement.

Question 9. Do all animals have bones?
Answer: No, all animals do not have bones. For example, earthworms.

Question 10. How does an earthworm fix parts of its body to the ground?
Answer: Earthworms possess a large number of tiny projections called bristles. These bristles are connected with muscles. They help to get a grip on the soil and thereby, help the earthworm in fixing parts of its body to the ground.

Question 11. Is the movement of a snail slow or fast as compared to an earthworm?
Answer: The movement snail is slow as compared to an earthworm.

Question 12. Why do different animals have the body parts that they do have and how do these body parts help animals to move the way they do?
Answer: Use of the differences in the way of their movement. For example, a snail has a thick muscular foot for movement Cockroach has three pairs of legs that help in walking.

Question 13. What are the differences in body parts between different animals used for movement? 
Answer: Animals of different kinds of body structures move in different ways using different body parts such as legs, wings, fins, tails, etc. These help them to walk, jump, fly, swim, run, etc.

Question 14. Why are there two legs for humans and four for cows and buffaloes?
Answer: Technically, humans do have four legs however when we evolved, the muscles in our legs grew stronger and longer, allowing us to stand upright. However, cows did not evolve with time and they still need four legs to balance themselves and around.

Question 15. Why is the bending of our legs different from that of our arms?
Answer: The bending ofour legs is different from that ofour arms because of the differences in their position and role in our body. Although they both are hinge joints, they bend in opposite directions.

Question 16. Distinguish between the following

  1. Movable joint and fixed joint
  2. Ball and socket joint and hinge joint

Answer: Differences between movable joints and fixed joints are as follows

Class 6 Science Chapter 5 Body Movements Movable Joint And Fixed Joint

Differences between Ball and socket joint and hinge joint are as follows-

Class 6 Science Chapter 5 Body Movements Ball And Socket Joint Hinge Joint

Question 17. Draw a diagram to show the Joint In the hand and answer the following questions.

  1. Can you bend your finger at every joint?
  2. How many bones are there in a human skull?
  3. Is your wrist flexible?
  4. What would happen if your hand had only one bone?

Answer:

  • We can bend our fingers at every joint.
  • The human skull has 22 bones
  • Yes, our wrist is flexible.
  • If our hand had only one bone, then we would not be able to bend our fingers and other parts of our hand

Question 18. How is the skeleton of a bird well-suited for flying?
Answer: The skeleton of a bird is well-suited for flying because

  • Bones are hollow and light in weight.
  • Bones of hindlimbs are used for walking and perching.
  • The bones of the forelimbs are modified as wings.
  • Shoulder bones are strong and helpful in flying.
  • Breast bones hold flight muscles and are used to move the wings up and down for flying.

Question 19. In the figure given below, there are two snakes of the same size, slithering on sand. Can you identify which of them would move faster and why?

Class 6 Science Chapter 5 Body Movements Two snakes of the same size slithering on sand

Answer:

A snake forms loops in its body while slithering. Each loop of the snake gives it a forward push by pressing against the ground.

The snake with a larger number of loops will move much faster than the snake with a lesser number of loops. Thus, snake A will move faster than snake B as we can see that loops in snake A are more than in snake B.

Question 20. Joints are the locations in the human body where two bones are connected. The picture shows three types of joints. The arrows show the movement of the bones in each Joint

Class 6 Science Chapter 5 Body Movements The Movement Of The Bones In each Joint

  1. In which joint can the pair of bones move in all directions?
  2. Which two joints shown in the picture are of the same type?
  3. Mention the name of that type of Joint.
  4. Some bone Joints are fixed and the bones at these joints cannot move. Which of these are fixed Joints?
  • Joints in the toe
  • Joints in the neck
  • Joints in the wrist
  • Joints in the skull.

Answer: The pair of bones can move in all directions in joint 3 as it is a ball and socket joint.

Joint 1 and joint 2, shown in the picture are the two joints that are of the same type. These are hinge joints.

Joints in the skull

21.

Class 6 Science Chapter 5 Body Movements Postion 1 And Postion 2

  1. In which position are the biceps most contracted?
  2. Which of these statements is true?
  3. The largest muscles of the human body are located in the arms
  4. The bending of the anus is controlled by muscle
  5. The muscles located in the arm have a feed shape
  6. Bending of arms involves the contraction and relaxation of a pair of muscles

Answer: The bending of arms involves the contraction and relaxation of a pair of muscles.

NCERT Class 6 Science Chapter 5 Body Movement Short Question And Answers

Question 1. Given below is a list of different types of movements in animals: Running, jumping, walking, slithering, crawling, flying swimming creeping. Write the types of movements seen in each animal.

  1. Duck 
  2. Horse 
  3. Kangaroo 
  4. Snail
  5. Snake
  6. Fish
  7. Human being
  8. Cockroach 

Answer: The types of movements seen in each animal are-

  1. Duck — Flying, walking, and swimming
  2. Horse — Running, jumping, and walking
  3. Kangaroo — Jumping
  4. Snail — Creeping
  5. Snake — Slithering
  6. Fish — Swimming
  7. Human being — Walking and running
  8. Cockroach — Walking and flying

Question 2. Which type of movement would have been possible if

  1. Our elbow had a fixed Joint.
  2. We were to have a ball and socket joint between our neck and head

Answer: If our elbow had a fixed joint, we would not be able to bend or fold our arms. A fixed joint does not allow any movement.

If we were to have a ball and socket joint between our neck and head, we would be able to rotate our head 360°. A ball and socket joint allows movement in all directions.

Question 3. Bones are hard structures and cannot be bent, but we can still bend our elbows, knees, etc. How is this possible?
Answer: The elbow and knee are not made up of a single bone rather they are made up of two or more bones that are joined to each other at a joint, i.e. hinge joint. This joint along with the muscles helps us to bend the elbow and knee.

Question 4. Study the classification chart given below

Class 6 Science Chapter 5 Body Movements Joints

What are A and B? Why?

Answer: A is the movable joint, which is further classified into

  • Ball and socket joint-e.g. hip joint and shoulder joint.
  • Pivotal joint – example between neck joint and head joint.
  • Hinge joint – Example elbow joint and knee joint.
  • B is the immovable or fixed joint. It is located within the skull. The upper jaw is fixed to the skull bones with the help of fixed joints.

Question 5. Write the type of joint which is used for each of the following movements

  1. A cricket bowler bowls the ball.
  2. A girl moves her head in the right and left direction.
  3. A person lifts weights to build up his biceps.

Answer:

  1. Ball and socket joint
  2. Pivotal joint
  3. Hinge joint

Question 6. Mention the number of bones present In the human skeletal system. Give its functions.
Answer: The human skeletal system is made up of 206 bones.

The functions of the skeletal system are

  1. To provide shape and support to the body.
  2. To protect internal organs.
  3. To make movements possible

Question 7. There are 12 pairs of curved bones called the Xln chest region. One end of bone X is joined to backbone and the other end is joined to bone Tin the front to form a box-like structure Z. This strong, box-like structure protects the delicate organs P, Q, and Rour body.

  1. What are (a) X, (b), and (c) Z?
  2. Name the organ
    • P
    • Q
    • R

Answer:

  1. Ribs (X)
  2. Breast bone (Y)
  3. Ribcage (Z)
  4. Heart (P)
  5. Lungs
  6. Liver (R)

Question 8. State the consequence if the backbone had only one long bone instead of many vertebrae.
Answer: Our backbone is comprised of 33 small bones called vertebrae. These are joined together with soft tissue in the form of cartilage discs. This arrangement makes these bony structures have some limited movements of the body.

If there is only one long bone instead of many in the backbone, the organism will not be able to avail the limited movements that are necessary for the organism for its survival.

Question 9. Write two ways by which we may know the shape of the human skeleton.
Answer: The two ways are given below

  • We can determine the shape of the human skeleton by feeling the bones ofour body.
  • We can also take X-ray images skeleton by using X-ray imaging

Question 10. Write the differences between bone and cartilage.
Answer: The differences between bone and cartilage are as follows

Class 6 Science Chapter 5 Body Movements The difference between Bone And Cartilage

Question 11. The skeleton is made up of bones, but it has some cartilage too. Do you agree?
Answer: Yes, cartilages are the additional parts of the skeleton that are not as hard as the bones and which can bent. So, we can say that though most of the skeleton is made up of bones, it has some cartilage too.

Question 12. Provide one-word answers to the statements given below.

  1. Joint which allows movement In all directions
  2. The hard structure that forms the skeleton
  3. Part of the body with a fixed joint.
  4. Help in the movement of the body through contraction and relaxation
  5. Bones that Join with the chest bone at one end and to the backbone at the other end
  6. Framework of bones which gives shape to our body
  7. Bones which enclose the organs of our body that lie below the abdomen our
  8. A joint where our neck joins the head
  9. Part of the skeleton that forms the earlobe

Answer:

  1. Ball and socket joint
  2. Bones
  3. Upper jaw with the skull
  4. Muscles
  5. Rib
  6. Skeleton
  7. Pelvic bones
  8. Pivotal joint
  9. Cartilage

Question 13. We need two muscles together to move a bone. Why?
Answer: A muscle can only pull, it cannot push. Thus, two muscles are required to work together to move a bone. When one muscle contracts, the bone is pulled. When another muscle of the pair pulls, it brings the bone back to its original position.

Question 14. Bhoojho fell off a tree and hurt his ankle. On examination, the doctor confirmed that the ankle was fractured. How was it detected?
Answer: The doctor must have observed swelling around his ankle and must have taken an X-ray of his ankle. X-ray images confirm any type of injury or fractures in the bones.

Question 15. The gait differs in different animals. Explain.
Answer: Different animals use different organs to move from one place to another. So, the manner of movement, i.e. the gait also differs in different animals.

Question 16. Earthworms are known as farmer’s friends why?
Answer: An earthworm is called a farmer’s friend as it eats its way through the soil thereby, loosening the soil. Also, the earthworm’s body excretes the undigested materials that increase the fertility of the soil.

Question 17. Can you imagine how an earthworm fixes parts of its body to the ground?
Answer: Under the earthworm’s body, it has a large number of tiny bristles projecting out. The bristles are connected with muscles and help to get a good grip on the ground.

Question 18. The skeleton of a cockroach is called an exoskeleton. Do you agree? Explain.
Answer: The body of a cockroach is covered with a hard outer protective covering called an exoskeleton. The skeleton of a cockroach is called the exoskeleton because it is present outside the body.

Question 19. The body of the fish is similar to that of the boat. Explain, how fish move in the water.
Answer: The streamlined shape of fish helps it to move in the water. The skeleton and muscles present on the front side of fish move on one side, while the tail muscles move the body on the other side. This makes a jerk and pushes the body forward. In this way, fish moves in water.

Question 20. Name the organisms that show these movements.

  • A thick muscular foot of the organism helps in dragging the shell.
  • The organism curves its body into various loops.
  • Their loops push the body forward.

Answer:

  1. Snail
  2. Snake

Question 22. The table shows how different animals move

Class 6 Science Chapter 5 Body Movements Animals How Does The Animal Move

Which animal uses all of its body parts to move?

  1. Frog
  2. Mouse
  3. Dragon
  4. Earthworm

Answer: 4. Earthworm

What helps a frog hop?

  1. Strong leg muscles
  2. Short body length
  3. Two pairs of legs
  4. Absence of a tail

Answer: 1. Strong leg muscles

Question 2. Rishi was playing in a garden. It was the rainy season. He observed an organism moving on the soil Based on his observation of the manner of movement, which organism is likely to be seen?

  1. Ant
  2. Cockroach
  3. Earthworm
  4. Snake

Answer: 3. Earthworm.

NCERT Class 6 Science Chapter 5 Body Movement Assertion-Reason Questions

The following questions consist of two statements. Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below.

  • Both A and R are true and R is the correct explanation of A
  • Both A and R are true, but R is not the correct explanation of A
  • A is true, but R is false
  • A is false, but R is true

Assertion (A) We can bend our body parts only at the joints.
Reason (R) Joints are those points where two or more bones are joined together.

Assertion (A) Pivot joint allows movement in all directions.
Reason (R) In a pivot joint, the rounded surface of one bone fits into a ring formed by the other bone.

Assertion (A) The bodies of the birds are well suited for flying.
Reason (R) The Presence of hollow bones makes the bones of birds very light

Answers

  1. Both A and R are true and R is the correct explanation of A.
  2. A is false, but R is true. A can be corrected as the Pivot joint allows only rotational movement.
  3. Both A and R are true and R is the correct explanation of A.

2. Direction The following questions consist of two statements. Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below.

  • Both A and R are true and R Is the correct explanation of A
  • Both A and R are true, but R is not the correct explanation of A
  • A is true, but R is false
  • A is false, but R is true

1. Assertion (A) An earthworm does not have bones. Reason (R) The body of an earthworm is made up of many rings, joined end to end.
Answer: 1. Both A and R are true and R Is the correct explanation of A

2. Assertion (A) Cockroaches can walk, climb and fly upto a short distance.
Reason (R) They have three pairs of jointed legs and two pairs of wings attached to the breast region which help them to walk and fly, respectively.
Answer: 1. Both A and R are true and R Is the correct explanation of A.

NCERT Class 6 Science Chapter 5 Body Movement Very Short Answer Type Questions

Question 1. Name the parts of our body responsible for the movements.
Answer: Bone and muscles ofour body are responsible for the movements.

Question 2. If there are no joints, then will it be possible to move? Explain.
Answer: No, movement in the body of an animal is possible in the presence of joints only.

Question 3. Can we bend our bodies at every part?
Answer: No, we can bend our body at every part, but only at joints.

Question 4. The arm rotates at a specific part of the body. Name the part at which the arms rotate.
Answer: Shoulders are that part of the body (or skeleton) to which our arms are attached and at which the arms rotate.

Question 5. Name the type of joints of your hand that help you
Answer: The hinge joint ofour hand helps us to grasp a badminton racquet.

Question 6. What do you mean by skeleton?
Answer: The bones and cartilage in our body form a framework that supports the whole body and gives a shape to it. This framework is called the skeleton.

Question 7. Name three components of the skeletal system.
Answer: A skeleton is made up of many bones, joints, and cartilage.

Question 8. Name the type of joints present mainly in the skull.
Answer: The skull bones are joined together by fixed joints.

Question 9. The hard and strong skull protects a delicate organ of our body. Name that organ.
Answer: The hard and strong skull protects a delicate organ of our body called the brain.

Question 10. Name the organs that are protected by the ribcage
Answer: The heart and lungs are protected by the ribcage.

Question 11. What would have happened if our backbone was
Answer: If our backbone was made up of one single bone, then we would not have been able to bend our body (from the waist).

Question 12. If an earthworm does not have bones or legs, then how does it move?
Answer: An earthworm does not have bones or legs. It has muscles which help it to lengthen and shorten the body to move.

Question 13. Comparing an earthworm and a snail, explain which of the two moves faster.
Answer: Earthworm moves faster than snails.

Question 14. Cockroaches can fly as well as walk. What helps the cockroach to walk?
Answer: Three pairs of legs help cockroaches to walk.

Question 15. Write any two adaptations of birds for flying in the air.
Answer: The following adaptations are seen in the body of birds.

  • Bones are hollow making the body lightweight for flying.
  • Forelimbs are modified into wings.

Question 16. Name three animals having streamlined bodies.
Answer: Fish, birds, and snakes have streamlined bodies.

Question 17. Some animals like fishes, birds, and snakes move due to their streamlined bodies. What do you mean by streamlined?
Answer: If the body tapers at both ends, then such a body shape is said to be streamlined.

Question 18. What is a ball and socket joint?
Answer: A joint in which the rounded end of one bone fits into the hollow space of another bone is called a ball and socket joint It allows movements in all directions. Joints between the upper arm and shoulder, thigh and hip are examples of ball and socket joints.

Question 19. Which of the skull bones are movable?
Answer: In the skull, only the lower jaw is movable. All the other bones of the skull are immovable.

Question 20. Why can our elbow not move backward?
Answer: Our elbow cannot move backward because the elbow has a hinge joint that allows movement in only one direction.

NCERT Class 6 Science Chapter 5 Body Movement Fill In The Blanks

1. Unscramble the jumbled words and write them in the blank spaces provided

  • Neosb-Bones
  • Tnemevom-Movement
  • Iontcaront-Contraction
  • Isecsum-Muscles
  • Arctigeal-Cartilage
  • Ephas-Shape
  • Inerlant sangro-Internal organs
  • laxaeriont-Relaxation.

2. Read the following paragraph and fill in the blanks using the word you unscrambled.

Bones and cartilage from the skeleton of the human body. they provide the framework, give shape to the body, and help in movement. They protect the internal organs. The bones are moved by alternate contraction and relaxation of two sets of muscles attached to them.

6. Contraction of muscles pulls the bone during movement.

7. The shoulder joint is an example of a ball and socket joint.

8. Joints of the bones help in the Movement of the body.

9. A combination of bones and cartilage forms the Skeleton of the body.

10. The bones at the elbow are joined by a Hinge joint.

11. The contraction of the Muscle pulls the bones during movement

NCERT Class 6 Science Chapter 5 Body Movement True Or False

1. All joints in our body are similar. False, joints are of mainly two types, i.e. fixed and
movable.

2. The upper jaw is movable. False, the lower jaw bone is the only movable bone in
the skull.

3. The ribcage protects vital internal organs like the heart, lungs, and liver. True

4. The joint of the upper arm and shoulder blade is a hinge joint. False, the joint of the upper arm and shoulder blade is a ball and socket joint.

5. Minute movable bristles in earthworms help in making a grip on the ground. True

6. An earthworm has a thick muscular foot. False

7. The body of a cockroach is covered with an exoskeleton. True

8. The body of a snail is streamlined. False

9. The movement and locomotion of all animals is the same. False, the movement and locomotion of animals differ.

10. The cartilage is harder than bones. False, the cartilage is not harder than bones

11. The finger bones do not have joints. False, the finger bones have joints.

12. The forearm has two bones. True

13. Cockroaches have an outer skeleton. True

NCERT Class 6 Science Chapter 5 Body Movement Match The Columns

Question 1. Match the column 1 with column 2 

Class 6 Science Chapter 5 Body Movements Match The Column 1 And 2

Answer: A-3, B-5, C-4, D-2, E-1

Question 2. Match the column 1 with column 2 

Class 6 Science Chapter 5 Body Movements Match The Column 1 And 2.

Answer: A-3, B-5; C-4, D-1, E-2

Question 3. Match the column 1 with column 2 

Class 6 Science Chapter 5 Body Movements Match The Column 1 And 2.3

Answer: A-4, B-1 or 7, C-5, D-6 or 2, E-2 or 3

NCERT Class 6 Science Chapter 3 Separation Of Substances Question And Answers

NCERT Class 6 Science Chapter 3 Separation Of Substances Long Question And Answers

Question 1. When separating sand and sawdust or powdered leaves, do both components fall in the same place?
Answer: When sand and sawdust are separated by winnowing, sawdust falls far away from the sand. This is because sand is heavy, while sawdust is light.

Question 2. Which method of separating tea leaves from prepared tea is better, decantation or filtration?
Answer: Tea leaves are better separated from prepared tea using the filtration method. It is because if the decantation method is used, it helps a little, but we still get a few leaves in our tea.

Read and Learn More Class 6 Science Question And Answers

Question 3. Many substances dissolve in water and form a solution. What happens if we go on adding more and more of these substances to a fixed quantity of water?
Answer: If we go on adding more and more of this substance to a fixed quantity of water, some of it remains undissolved in the solution and settles at the bottom of the vessel. This is called a saturated solution.

NCERT Class 6 Science Chapter 3 Separation Of Substance Questions And Answers

Question 4. What is sieving? Where is it used?
Answer: Sieving is a method by which fine particles are separated from bigger impurities by using a sieve. It is used in a flour mill, our homes, and at construction sites. In a flour mill, impurities like husk and stones are removed from wheat before grinding. At construction sites, pebbles and stones are removed from sand by sieving.

Question 5. Is it possible to separate sugar mixed with wheat flour? If yes, how will you do it?
Answer: Yes, it is possible to separate sugar mixed with wheat flour by sieving. The size of sugar particles is larger than that ofthe wheat flour. So, sugar will stay on a sieve and wheat flour will pass through the holes of the sieve.

Question 6. How will you separate sand and water from their mixture?
Answer: We will separate sand and water from their mixture by the process of sedimentation and decantation. Sedimentation During this process, we leave the mixture for some time. Heavier sand gets settled down at the bottom.

Decantation After the sedimentation process, water is poured into another vessel, and sand is left undisturbed in the first vessel. Thus, sand and water will be separated from their mixture.

Question 7. How would you obtain clear water from a sample of muddy water?
Answer: We will obtain clear water from a sample of muddy water by the process of filtration.

In this process, we use a piece of cloth to filter muddy water. Because in a piece of cloth, small holes or pores remain in between the woven threads. These pores in a cloth can be used as a filter.

If the water is still muddy, impurities can be separated by a filter paper that has even smaller pores.

A filter paper folded in the form of a cone is fixed onto a funnel. The mixture is then poured on the filter paper. Solid particles in the mixture do not pass through it and remain on the filter paper and in this way, clear water is obtained.

Question 8. Lemonade is prepared by mixing lemon juice and sugar in water. You wish to add ice to cool it. Should you add ice to the lemonade before or after dissolving sugar? In which case would it be possible to dissolve more sugar?
Answer: We should add ice to lemonade after dissolving sugar. Without ice, sugar dissolves easily in lemon juice because solubility depends on the temperature. When the temperature is high, then more sugar can be dissolved. After mixing ice, it gets cooled down so less sugar will dissolve in it.

Question 9. Both Sarika and Mohan were asked to make salt solutions. Sarika was given a teaspoonful of salt and half a glass of water, whereas, Mohan was given twenty teaspoonful of salt and half a glass of water.

  1. How would they make salt solutions?
  2. Who would be able to prepare a saturated solution

Answer:

They will mix salt with a certain amount of water to make a salt solution.

Saturation is a point at which no more substances can be dissolved in the solvent and the substance gets settled down. As Mohan has more quantity of salt than Sarika, hence, he would be able to prepare a saturated solution.

Question 10. Paheli was feeling thirsty, but there was only a pot of water at home which was muddy and unfit for drinking. How do you think Paheli would have made this water fit for drinking if the following materials were available to her? Tub, muslin cloth, gas stove, thread, alum, pan, and lid of the substance.
Answer: Paheli can make this water fit for drinking by working on the following sequence

At first, she needs to pour the water through a muslin cloth for filtration.

Next, the water has to be heated in a container on a gas stove until it vaporizes. The vapors are allowed to come in contact with a cold lid so that condense back into water which will be collected in a tub. Impuritiesifany will remain behind in the container.

Then she has to hang a piece of alum with a thread and submerge it in the water, keeping it undisturbed for some time. This allows the impurities to settle down as sediment.

Next, she has to pour off the undisturbed water in a tub by decantation.

Now, the water is fit for drinking.

Question 11. Read the story titled ‘Wise Farmer’ and tick the correct option from the underlined section to complete the story. A farmer was

  1. Sad/happy to see his healthy wheat crops ready for harvest. He harvested the crops and left them under the
  2. Sun/rain to dry the stalks. To separate the seeds from the bundles of the stalk, he
  3. Handpicked/threshed them. After gathering the seed grains, he wanted to separate the stones and husk from them. His wife
  4. Winnowed/ threshed them to separate the husk and later
  5. Sieved/ handpicked to remove stones from it. She grounded the wheat grains and
  6. Sieved/filtered the flour. The wise farmer and his wife got a good price for the flour. Can you tell me why?

Answer:

  1. Happy
  2. Sun
  3. Threshed
  4. Winnowed
  5. Handpicked
  6. Sieved

They got a good price as they used appropriate methods of separation to get good quality flour (atta).

Question 12. You are provided with a mixture of salt, sand, oil, and water. Write the steps involved in the separation of salt, sand, and oil from the mixture by giving an activity along with the diagram.
Answer: In the solution of salt, sand, oil, and water. Oil, being lighter will float on the solution forming a distinct layer, slowly oil is allowed to flow into another container and is separated from the water. This process is called decantation. Now, we are left with salt, sand, and water, to separate this.

Question 13. Pragati was helping her mother in the making of butter. She was amazed to see how easily the butter got separated from milk with the help of a churner.

Class 6 Science Chapter 3 Separation Of Substances Churching

1. What is the process that is being used above?

  1. Grinding
  2. Mashing
  3. Churning
  4. Threshing

Answer: 3. Churning

2. The principle behind the process of churning is

  1. Lighter particles of a solid get separated from liquid
  2. Heavier particles get settled at the bottom
  3. Heavier particles form heaps and lighter particles get swept away by the wind
  4. Particles of different sizes get separated

Answer: 1. Lighter particles of a solid get separated from liquid

Question 14. Saroj collected muddy water from a pond. He separated the water from the mud in three steps as shown in the pictures below.

Class 6 Science Chapter 3 Saroj collected muddy water from a pond. He Separated The Water From The Mud In Three Steps

1. Which separation method did Saroj use at each step? Select the correct row

Class 6 Science Chapter 3 Separation Of Substances Evaporation And Sedimentation Decantation

Answer: 1. Evaporation Sedimentation Decantation

2. Which of these properties did Saroj use to separate mud from water?

  1. It floats on water
  2. It dissolves in water
  3. Water turns cloudy due to mud
  4. It is heavier than water

Answer: 2. It dissolves in water

Question 15. Nidhi was doing an activity at home. She dissolved some sugar in a glass of water. Later she realised she had to add salt and not sugar. She wants to use the same water. Answer the following questions to help Nidhi with her activity.

  1. She can remove sugar from water by
  2. Filtering and then evaporating it
  3. Condensing and then evaporating it
  4. Evaporating and then condensing it
  5. Filtering and then condensing it

Answer: 1. Ice water

Water can be separated from sugar solution by first evaporating the water and then condensing it in another vessel to be used again with salt

2. In which of the following added sugar would be dissolved the least?

  1. Tap water
  2. Filtered water
  3. Ice water
  4. Boling water

Answer: 2. Filtered water

Ice water has the lowest temperature among all and hence would dissolve less amount of sugar as solubility decreases with a decrease in the temperature.

Question 16. Two beakers A and B are shown below. The components of each have been labeled.

Class 6 Science Chapter 3 Separation Of Substances Two beakers A And B

Paheli wants to mix the components of A and B beakers. However, she wants to first separate as many components as possible. Answer the following questions for her activity.

The order of separation of components from beaker A would be

  1. Oil layer > water = sand
  2. Sand = water > oil layer
  3. Sand > oil layer > water
  4. Water > oil layer > sand

Answer: 1. Oil layer > water = sand

The oil layer rests on the top of the solution hence it should be removed first. Sand is removed either by decantation or filtration. Hence, both are collected together separately. The correct order is Oil layer > water = sand.

Paheli can remove layers of oil by

  1. Filtration
  2. Decantation
  3. Handpicking
  4. Evaporation

Answer: 2. Decantation

Decantation is the process used to remove a layer of oil from the water-sulf solution as it forms a separate layer on top.

Question 17. Ashraf’s grandmother is a diabetic patient. Her doctor advised him to control her sugar level by taking less sugar and less fat content in her diet. So, she always prefers to take lassi instead of milk. Ashraf asked her mother when she explained the process of making lassi by removing cream from the curd. She told him this method is known as churning. Read this passage and answer the following questions.

  1. Explain the process of churning.
  2. What is the purpose of this method?
  3. What values are shown by Ashraf?

Answer:

  1. The process of separation of the lighter particles of a solid from a liquid is known as churning. It is also known as centrifugation.
  2. The purpose of churning is to remove butter or cream from the curd or milk. So in lassi, there is less fat content as compared to milk.
  3. Ashraf is a curious and intelligent boy.

Question 18. Rohit and Raman are best friends. They study in different classes, but in the same standard, Class 6. One day, they were playing in the school. They were throwing sand on each other. Suddenly, Raman asked Rohit if we mix salt in sand, then how will you separate the mixture of sand and salt? Rohit was quite intelligent, his science teacher had completed the chapter Separation of Substances last week. So, he immediately explained the method of separation. Rohit told him that his teacher had shown this activity in the class and the same method is applicable for the separation of the mixture of sugar and sand.

Read the passage carefully and answer the following questions.

  1. How is a mixture of sand and salt separated?
  2. Name the method(s) that are applicable in the separation of this mixture.
  3. What values are shown by Rohit?

Answer:

  1. This is done as follows
  2. Some water is added to the mixture of sand and salt in a beaker and stirred. Salt dissolves in water to form a salt solution, whereas sand remains undissolved. It is then filtered.
  3. On filtering, sand is obtained as a residue on the filter paper, and salt solution is obtained as a filtrate.
  4. The filtrate is evaporated and pure salt is left behind.
  5. The mixture of sand and salt has been separated into its components, i.e. sand and salt by doing two methods
  6. Filtration
  7. Evaporation
  8. Rohit is a knowledgeable and intelligent boy

Question 19. A construction worker poured a bag full of sand on the slanting sieve. All the sand particles filtered through the pores and the dust particles, stones, pebbles, and husk remain Name the process described above and state its principle.

  1. Why is it necessary to remove pebbles and dust particles from sand used for construction?
  2. You have been given a glass of muddy water, an empty glass, and a muslin cloth. Describe a method to obtain clean water.
  3. Identify the method of separation used to
  4. Separate chaff from the grain.
  5. Separate tea leaves from the water.

Class 6 Science Chapter 3 Separation Of Substances Pebbles and stones are removed from sand by sieving

Answer:

  1. The process used above is sieving. Principle – A mixture of components having different sizes gets separated with the help of a sieve
  2. A sieve has fine pores in it which lets the small-sized components pass through but holds back larger ones. Sand is sieved so that it can be used for construction work without any impurities.

Question 20.

Class 6 Science Chapter 3 Separation Of Substances Sepration of soild and liquid mixture A glass of muddy water An empty glass and a muslin cloth

  1. You have been given a glass of muddy water, an empty glass, and a muslin cloth. Describe a method to obtain clean water
  2. Identify the method of separation used to
  3. Separate chaff from the grain.
  4. Separate tea leaves from the water.

Answer:

  1. Filtration is done to obtain clean water from a glass of muddy water.
  2. Winnowing
  3. Filtration

Question 21. 

Class 6 Science Chapter 3 Separation Of Substances A group of salt solution The respective componets

  1. A group of students were given a salt solution and were asked to separate the respective components. They set up the apparatus as shown in the given figure.
  2. Name the solute and solvent for the given solution.
  3. Identify the processes for which the apparatus has been set up in the given figures.
  4. Identify the error in the set-up of the apparatus for the separation of a salt solution.

Answer:

  1. Solute-salt, solvent-water
  2. Filtration, condensation.
  3. Salt cannot be separated from water by filtration.

Question 22. Anu adds sugar and salt to two separate glasses of water. She stirs the mixture in each glass every time she adds salt or sugar to it.

Class 6 Science Chapter 3 Separation Of Substances Anu adds sugar and salt to two separate glasses of water

She notes her findings in a table

Class 6 Science Chapter 3 Separation Of Substances She Notes Her Findings

  1. What can Anu conclude from her activity?
  2. Sugar and salt dissolve equally in water
  3. Sugar is more soluble than salt in water
  4. Stirring helps in dissolving all substances in a liquid
  5. Sugar forms a saturated solution in water, but salt does not
  6. Anu used two spoons of the same size to add sugar and salt to the water. What else did Anu need to keep the same for her activity?
  7. Anu repeats the activity using hot water. Will the results in the table remain the same? Explain your answer.
  8. Which of these processes could be used to separate the sugar dissolved in the glass of water?
    1. Filtration
    2. Evaporation
    3. Condensation
    4. Sedimentation

Answer:

  1. Sugar is more soluble than salt in water
  2. Anu needs to keep the amount of water fixed to determine the solubility of a substance.
  3. No, results would not remain the same as temperature has a direct effect on solubility. With the increase in temperature, the solubility of both sugar and salt increases, but in different amounts.
  4. Evaporation

Question 23. A beaker contains a mixture of salt, sand, and water. The mixture is filtered using filter paper.

Class 6 Science Chapter 3 Separation Of Substances A beaker Contains a mixture of salt

  1. What is substance X?
  2. Write a single separation method by which water can be separated from a mixture of salt, sand, and water.

Answer:

  1. The substance ‘X sand as the process of filtration is used to separate insoluble solids, i.e. sand from its liquid mixture.
  2. By evaporating the mixture of salt, sand, and water, water will evaporate and separate. But sand and salt will remain in the vessel.

NCERT Class 6 Science Chapter 3 Separation Of Substances Short Question And Answer

Question 1. What is the principle of threshing?
Answer: Threshing is the method of separation based on the fact that the stalk of the crops and the chaff are soft materials, whereas the grains themselves are hard. Being soft, stalk and chaff can be broken into pieces on beating, but the grains remain unaffected.

Question 2. When a hot saturated solution is heated, crystals will be formed. Is It correct? If not, write the correct statement.
Answer: It is not correct as when a hot saturated solution is cooled, crystals will be formed.

Question 3. Name and describe briefly a method that can help separate a mixture of husk from grains. What is the principle of this method? NCERT Exemplar
Answer: Winnowing is used to separate grains from impurities like husk by dropping them vertically from a height. All light impurities are separated by a blowing air or wind and hence, husk from grains are separated easily.

Question 4. You have been given a solution of substance X. How will you test whether it is saturated or unsaturated? Give one factor affecting the solubility of a substance in solution.
Answer: When no more substance can be dissolved in the solution at a given temperature then the solution is saturated. Temperature affects the solubility of the substance.

Question 5. What happens when sand and powdered leaves are dropped at the same place in the wind?
Answer: Since, sand is heavier, while powdered leaves are lighter. When we drop these at the same place, blowing wind separates them as sand falls near that place vertically, while leaves are blown away. This process is known as winnowing.

Question 6. The best way to carry out the filtration of the mixture of water and sand is to use filter paper, not cloth, why?
Answer: Since, clothes may have small holes, but filter papers do not have any holes. It has very fine pores in it. Hence, sand particles settle down on the filter paper, while water after pouring reaches the bottom of the vessel.

NCERT Class 6 Science Chapter 3 Separation Of Substances Assertion-Reason Questions

Question 1. The following questions consist of two statements. Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below.

  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true, but R is not the correct explanation of A
  3. A is true, but R is false
  4. A is false, but R is true

1. Assertion (A) Husk and flour can be separated by the process of sieving.
Reason (R) Sieving is used when the particle sizes of the two components in the mixture differ.
Answer: Both A and R are true and R is the correct explanation of A.

2. Assertion (A) Evaporation can be used to separate a solid (like salt, or sugar) dissolved in a liquid. Reason (R) Evaporation is the process in which a solid gets converted to vapor.
Answer: A is true, but R is false. R can be corrected as Evaporation is the process in which liquid gets converted to vapor form and thus this process can be used to separate solids dissolved in a liquid.

3. Assertion (A) A substance gets dissolved more in hot water as compared to cold water. Reason (R) Solubility decreases with an increase in temperature.
Answer: A is true, but R is false. R can be corrected as Solubility increases with an increase in temperature

Question 2. Direction The following questions consist of two statements. Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below.

  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true, but R is not the correct explanation of
  3. A is true, but R is false
  4. A is false,

1. Assertion (A) The mixture of mud and water can be separated by using the process of filtration. Reason (R) The process of separating insoluble substances from a liquid using a filter is called filtration.
Answer: Both A and R are true and R is the correct explanation of A

2. Assertion (A) Filtration is used to separate chalk powder from water. Reason (R) Chalk powder is insoluble in water.
Answer: Both A and R are true and R is the correct explanation of A

NCERT Class 6 Science Chapter 3 Separation Of Substances Very Short Answer Type Questions

Question 1. Write one purpose of separating substances from a mixture.
Answer: To separate harmful or non-useful substances and to obtain the useful component.

Question 2. When is the process of handpicking used?
Answer: Handpicking is used to separate sightly large-sized undesirable components when present in small amounts.

Question 3. Which method is used to separate the pieces of stones from grains?
Answer: Handpicking is the method which is used to separate the pieces of stones from grains.

Question 4. Winnowing is based on which property?
Answer: Winnowing is based on the difference in the weight of particles of the different components.

Question 5. Give one example of sieving used in everyday life.
Answer: The separation of bran (choker) from flour is an example of sieving.

Question 6. A method is used to separate the components of a mixture that are of different sizes. Name it.
Answer: Sieving.

Question 7. What is the use of a strainer while preparing tea?
Answer: From prepared tea, we separate tea leaves by using a strainer as a filter. Tea leaves are bigger than the holes in the mesh.

Question 8. 1 kg wheat grains are mixed with 1 kg mustard seeds Write a suitable method to separate the mustard seeds from wheat grains.
Answer: The suitable method used to separate the mustard seeds from wheat grains is sieving.

Class 6 Science Chapter 3 Separation Of Substances Wheat Grains And Mustard Seeds

Question 9. Define the term condensation.
Answer: The process of conversion of water vapor into its liquid form by cooling is called condensation.

Question 10. Water is separated from rice and pulses after washing them. How?
Answer: By the process of decantation, water is separated from rice and pulses after washing them.

Question 11. How saturated solution can become an unsaturated solution?
Answer: When a saturated solution is heated, it becomes unsaturated.

Question 12. Temperature affects the solubility of solids in liquids. How?
Answer: The solubility of solids in liquids increases with increase in temperature.

Question 13. How chalk powder can be separated from a mixture of chalk powder and water?
Answer: A mixture of chalk powder and water can be separated by a filtration process.

Question 14. Sheela, Saima, and Ravi have to dissolve the maximum amount of sugar in the same amount of milk, to win in a game. Ravi took hot boiling milk, while Saima took ice-cold milk. Sheela managed to get milk at room temperature. Who do you think would win the game and why?
Answer: Milk at higher temperatures would dissolve more amount of sugar as solubility increases with temperature. Ravi took hot boiling milk so, he would win the game.

Question 15. Which method of separation is used to catch fish in the sea using a net?
Answer: A fishing net acts as a filter because it allows water to pass and holds fish in it. Thus, catching fishing using the net in the sea is an example of filtration.

Question 16. What will happen if a high quantity of substance is forced to dissolve in a fixed amount of water?
Answer: After the addition of a certain amount, substances remain undissolved in water as they become saturated.

Question 17. Does water dissolve equal amounts of different soluble substances?
Answer: No, water dissolves unequal amounts of different soluble substances.

Question 18. Write four materials that can be used for the process of filtration.
Answer: The materials that can be used for the process of filtration are a wire mesh, a piece of cotton, a piece of fine cloth, and a strainer.

NCERT Class 6 Science Chapter 3 Separation Of Substances Fill In The Blanks

1. The method of separating the seeds of paddy from its stalks is called Thereshing

2. When milk, cooled after boiling, is poured onto a piece of cloth, the cream (malai) is left behind on it. This process of separating cream from milk is an example of Filtration

3. Salt is obtained from seawater by the process of Evaporation

4. Impurities settled at the bottom when muddy water was kept overnight in a bucket. The clear water was then poured off from the top. The process of separation used in this example is called Sedimentation And Decantation

5. Insects can be separated from wheat by Handpicking

6. Crushing paddy under Cattle’s feet to separate grains is called Threshing

7. Husk from wheat flour is generally removed by Sieving

8. Slanting sieve Used at construction sites to separate pebbles and stones from sand.

9. The process of settling heavier particles in a solution is called Sedimentation.

NCERT Class 6 Science Chapter 3 Separation Of Substances True Or False.

1. A mixture of milk and water can be separated by filtration. False, we cannot separate the mixture of milk and water by filtration because milk is completely soluble in water.

2. A mixture of powdered salt and sugar can be separated by the process of winnowing. The false, winnowing process is used to separate heavier and lighter components of a mixture. Here, particles of size sugar and salt are almost of same size.

3. Separation of sugar from tea can be done by filtration. False, sugar cannot be separated from tea by filtration because sugar is completely soluble in tea.

4. Grain and husk can be separated by the process of decantation. False, grain and husk are separated by the process of winnowing.

5. Separation of components of a mixture is a useful process. True

6. A mixture of iron filings and rice flour can be separated by handpicking. False, iron filling can be separated from rice flour using a magnet

7. A mixture of wheat grains and rice flakes can be separated by winnowing. True

8. A mixture of oil and water can be separated by filtration. False, a mixture of oil and water can be separated by decantation

9. A mixture of tea leaves and milk can be separated by filtration. True

NCERT Class 6 Science Chapter 3 Separation Of Substances Match The Columns

Question 1. Match the following separation processes with their purposes and the way separated components are used.

Class 6 Science Chapter 3 Separation Of Substances The Following Separation process with their purpose

Answer: The correct match is 1.(b)-(i), 2.-(a)-(iii),3.-(c)-(ii)

Question 2. Match the Column 1 with Column 2

Class 6 Science Chapter 3 Separation Of Substances Match The Column 1 And Column 2

Answer: A-3, B-4,C-5,D-2, E-1

Question 3. Match The Column 1 And Column 2

Class 6 Science Chapter 3 Separation Of Substances Match The Column 1 And Column 2.

Answer: A-3,B-1,C-5,D-4, E-2