NCERT Exemplar Solutions for Class 10 Maths Chapter 1 Real Numbers

NCERT Exemplar Solutions for Class 10 Maths Chapter 1 Real Numbers Introduction

We have studied real numbers. Now we will discuss two important results, namely

  1. Euclid’s division lemma
  2. Fundamental theorem of arithmetic

NCERT Exemplar Solutions For Class 10 Maths Chapter 1 Real Numbers

Difference Between Algorithm And Lemma

Algorithm: An algorithm is a series of well-defined steps, that gives a procedure for solving a problem.

Lemma: It is a proven statement used to prove another statement.

Read and Learn More Class 10 Maths Solutions Exemplar

NCERT Exemplar Solutions for Class 10 Maths Chapter 1 Real Numbers Euclid’s Division Lemma

Real Numbers Euclids Division Lemma

Note: Remainder is always less Dividend than the divisor. It is greater = Divisor x Quotient+ Remainder than or equal to zero.

i.e., \(0 \leq r<\text { divisor }\)

We think that all of you are familiar with this well-known procedure. Now, we will generalize this division method known as Euclid’s Divison Lemma.

NCERT Exemplar Solutions for Class 10 Maths Chapter 1 Real Numbers Euclid’s Division Lemma Statement

For any two given positive integers a and b, there exists unique whole numbers q and r such that

a = bq +r

Real Numbers Positive Integers A and B There Exists Unique Whole Numbers Q and R

Where 0 ≤ r< b

Here, a = dividend, b = divisor, q = quotient, r= remainder

Euclid’s Division Algorithm (To Find The HCF Of Two Positive Integers)

It is very useful to obtain the H.C.F. of two positive integers. Let c and d be two positive integers with c > d.

Now, to find the H.C.F. of c and d, follow the following steps:

Step 1: Apply Euclid’s division lemma to c and d.

We find whole numbers q and r such that

c = dq + r

where, 0 ≤ r < d,

Step 2: If r = 0 then d (recent divisor) is the H.C.F. of c and d,  and if r ≠ 0 then, apply the division lemma to d and r.

Step 3: Continue the process till the remainder is zero. The divisor at this stage will be the required H.C.F.

NCERT Exemplar Solutions for Class 10 Maths Chapter 1 Euclid’s Division Algorithm Solved Examples

Question 1. Use Euclid’s algorithm to find the H.C.F. of 4052 and 12570.
Solution:

Here, 12576 >4052

Real Numbers Euclids Division Algorithm

∴ 12576 = 4052 × 3 + 420

4052 = 420 × 9 + 272

420 =272 × 1 + 148

272 = 148 × 1 + 124

148 = 124 × 1 + 24

124 =24 × 5 + 4

24 = 4 × 6 + 0

Since remainder = 0, ⇒ recent divisor is the H.C.F.

∴ H.C.F. (12576, 4052)= 4

The H.C.F. of 4052 and 12570 = 4

Question 2. If the H.C.F. of 210 and 55 is expressible in the form 210×5 + 55x, then find the value of x.
Solution:

First, we will find the H.C.F. of 210 and 55.

Here, 210 >55

Real Numbers The HCF Of 210 And 55

210 = 55 × 3 + 45

55 =45 × 1 + 10

45 = 10 × 4 + 5

10 =5 × 2 + 0

Since remainder = 0

⇒ The recent divisor is the H.C.F.

∴ H.C.F. of (210. 55) = 5

Now, 210 × 5 + 55x = 5

55x = 5 – 210 x 5

= 5- 1050

=-1045

⇒ \(x=-\frac{1045}{55}\)

=-19

The value of x =-19

Question 3. Find die H.C.F. of 65 and 117 and express it in the form of 65x + 117y.
Solution:

Here, 117 > 65

Real Numbers The HCF Of 65 And 117

∴ 117 =65 × 1 + 52

65 = 52 × 1 + 13

52 = 13 × 4 + 0

Since remainder = 0

⇒ recent divisor is die H.C.F.

∴ H.C.F. (117, 65) = 13

Now, 13 = 65 -52 × 1

13 =65 – (117- 65 × 1)

= 65 – 117 + 65 × 1

= 65 × 2-117

= 65x + 117y

where, x = 2 and y = -1

Remark:

It follows from the above example that the H.C.F. (say h) of two positive integers o and b can be expressed as a linear combination of o and b i.e., h = xa + yb for some integers x and y. This representation is not unique. Because, h = xa + yb

⇒ \(h=x a+\underbrace{a b-a b}_{\begin{array}{c} \text { Add and } \\ \text { subtract } \end{array}}+y b=a(x+b)+b(y-a)\)

⇒ \(h=x a-\underbrace{a b+a b}_{\begin{array}{c} \text { Subtract } \\ \text { and add } \end{array}}+y b=a(x-b)+b(y+a)\)

So, the coefficients of a and b may be

  1. x and y
  2. x + band y-a
  3. x-b and y+ a

Hence, the linear combination of a and b is not unique.

Factor

It divides the number whose factor is this.

For example: 1,2,3 and 6 are the factors of 6. So, each factor divides 6 completely.

Multiple

It is divided by the number whose multiple is this.

For example : 3, 6, 9 12, 15 and 18 are the multiples of number 3. So, each multiple can be divided by the number 3.

Remember:

  1. A smaller number divides the larger while larger number is divided by smaller.
  2. Factors are smaller while multiples are larger.

H.C.F. (Highest Common Factor)

H.C.F. is the highest common factor of two or more numbers which divides each of the numbers.

For example: Two numbers are 24 and 36.

Real Numbers Highest Common Factor

Highest (maximum) common factor = 12

∴ H.C.F. = 12

So, 12 is the highest number which divides 24 and 36 completely.

L.C.M. (Least Common Multiple)

L.C.M. of two or more numbers is the least common multiple that is divided by all the numbers.

For example: Two numbers are 24 and 36.

Real Numbers Least Common Multiple

∴ The least (minimum) common multiple is 72.

So, 72 is the least number which is divided by 24 and 36 both.

Question 4. Find the largest number, which divides 246 and 1030 leaving the remainder 6 in each case.

Solution:

We have to find a number, which divides the other numbers means → H.C.F.

It is given that the required number, when divided between 246 and 1030 leaves the remainder 6 i.e.; 6 is extra in each number. It means that if these numbers are 6 less, then there is no remainder in each case.

∴ 246- 6 = 240 and 1030- 6 = 1024 are completely divisible by the required number.

Now, 1024 > 240

Real Numbers The Largest Number Divides 246 And 1030

∴ H.C.F. (1024,240)= 16

Hence, required no. = 16

Question 5. Find the largest number that will divide 400, 437, and 542 leaving the remainder 9, 12,15 respectively.

Solution:

We have to find a number, which divides the other numbers means → H.C.F.

It is given that the required number, when divided into 400, 437, and 542, leaves the remaining 9, 12, and 15 respectively. It means that if 400 is 9 less, 437 is 12 less, and 542 is 15 less, then on division, gives no remainder (extra).

∴ 400 – 9 = 391,437-12 = 425 and 542 – 15 = 527 are completely divisible by the required number.

First, we will find the H.C.F. of 391 and 425.

Real Numbers We Will Find The HCF Of 391 And 425

∴ 425 =391 × 1 +34

391 = 34 × 11 + 17

34 =17 × 2 + 0

∴ H.C.F. (391,425) = 17

Now, we will find the H.C.F. of 17 and 527.

Real Numbers We Will Find The HCF Of 17 And 527

527 = 17 × 31 +0

⇒ H.C.F. (17,527) = 17

∴ Required number = 17

Question 6. Show that one and only one out of n, (n+ 1) and (n + 2) is divisible by 3, where n is any positive integer.

Solution:

When n is divided by 3, let q and r be the quotient and remainder respectively.

∴ n = 3q + r

where, 0 ≤ r < 3 i.e.,

r = 0 or r = 1 or r = 2

1. When, r = 0, then

n = 3q,

which is divisible by 3.

n + 1 = 3q + 1

which leaves a remainder of 1 when divided by 3.

i. e., (n + 1) is not divisible by 3.

n + 2 =3q + 2

which leaves a remainder of 2 when divided by 3.

i.e., (n + 2) is not divisible by 3.

So, only n = 3q is divisible by 3 when r = 0.  …….(1)

2. When, r = 1, then

n= 3q + 1

which is not divisible by 3. (∵ it leaves a remainder 1)

n + 1 = 3q + 2

which is not divisible by 3. (∵ remainder = 2)

n + 2 = 3q + 3 = 3 (q + 1)

which is divisible by 3.

So, only n + 2 is divisible by 3 when r= 1.  ……….(2)

3. When, r = 2, then

n = 3q + 2

which is not divisible by 3. (∵ remainder = 2)

⇒ n + 1 = 3q + 3 = 3 (q + 1)

which is divisible by 3.

n + 2 =3q + 4 = 3q + 3 + 1= 3 (q + I) + 1

which is not divisible by 3 (∵ remainder = 1)

So, only n + 1 is divisible by 3 when r = 2.  ………(3)

Hence, from equations (1), (2) and (3), we can say that one and only one out of n, (n + 1) and (n + 2) is divisible by 3.

Hence Proved.

Question 7. There are 156, 208, and 260 students in groups A, B, and C respectively. Buses are to be hired to take them for a field trip. Find the minimum number of buses to be hired, if the same number of students should be accommodated in each bus, and if a separate bus for separate groups is needed.

Solution:

Given:

There are 156, 208, and 260 students in groups A, B, and C respectively. Buses are to be hired to take them for a field trip.

First of all we shall find the H.C.F. of number of students in each group. This will give tls the maximum number of students of the same group in each bus.

Given numbers are 156, 208 and 260. ,

Here, 260 > 208 > 156

Let us first find the H.C.F. of 156 and 208

By using Euclid’s division lemma for 156 and 208,

we get 208 = 156 × 1 + 52. Here remainder ≠ 0

156 = 52 × 3 + 0.

Here, remainder = 0

⇒ The recent divisor is the H.C.F.

∴ H.C.F. of 156 and 208 is 52.

Now, 260 > 52

So, we shall find the H.C.F. of 260 and 52 by Euclid’s division lemma.

∴ 260 =52 × 5 + 0.

Here, remainder = 0

⇒ The recent divisor is the H.C.F.

∴ H.C.F. of 260 and 52 is 52.

Thus, H.C.F. of 156, 208 and 260 is 52.

Hence, the minimum number ot buses = \(\frac{156}{52}+\frac{208}{52}+\frac{260}{52}\)

= 3 + 4 + 5

= 12

The minimum number ot buses = 12

Question 8. A sweetseller has 420 kaju burfis and 150 badam burfis. Mo wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. How many of these can be placed in each stack? How many stacks are formed?

Solution:

Given:

A sweetseller has 420 kaju burfis and 150 badam burfis. Mo wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray.

Maximum number of burfis in each stack = H.C.F. of 420 and 150

420 =2 × 2 × 3 × 5 × 7,

150 = 2 × 3 × 5 × 5

∴ H.C.F. =2 × 3 × 5

= 30

∴ Maximum number of burfis in each stack= 30

Also, number of stacks = \(\frac{420}{30}+\frac{150}{30}\)

= 14 = 5

= 19

Maximum number of burfis in each stack = 19

NCERT Exemplar Solutions for Class 10 Maths Chapter 1 Real Numbers Fundamental Theorem Of Arithmetic

Every composite number can be uniquely expressed as a product of primes, except for the order in which these prime factors occurs.

For example., 143 =11 × 13

24 = 2 × 2 × 2 × 3

= 23 × 3

416 = 2 × 2 × 2 × 2 × 2 × 13

= 25 × 13

Real Numbers Difference Between Composite Number And Primary Number

Remember:

  1. 1 is not a prime number as it has no two different factors.
  2. 1 is not a composite number also because at least one factor other than 1 and the number must be there.
  3. 2 is the smallest prime number and it is the only even prime number also.
  4. The smallest even composite number is 4 while the smallest odd composite number is 9.

NCERT Exemplar Solutions for Class 10 Maths Chapter 1 Real Numbers Fundamental Theorem Of Arithmetic Solved Examples

Question 1. Express each of the following as a product of prime factors:

  1. 1400
  2. 7650

Solution:

1. 1400

Real Numbers A Product Of Primary Factor 1

∴ 1400 = 2 × 2 × 2 × 5 × 5 × 7

Product of prime factors of 1400 = 2 × 2 × 2 × 5 × 5 × 7

2. 7650

Real Numbers A Product Of Primary Factor 2

∴ 7650 = 2 × 5 × 5 × 3 × 3 × 17

= 2 × 3 × 3 × 5 × 5 × 17

Product of prime factors of 7650 = 2 × 3 × 3 × 5 × 5 × 17

Question 2. Find the missing numbers in the following prime factorization.

Real Numbers The Missing Numbers In The Following Prime Factorisation

Solution:

The product of primes starts at the bottom of the factor tree and this product goes up to the top. The upper box, on 3 and 13 is filled by the product of 3 and 13, i.e., 39.

The upper next box on 2 and 39 will be filled by the product of 2 and 39, i.e., 78.

The topmost box on 2 and 78 will be filled by the product of 2 and 78, i.e., 156.

Real Numbers The Product Of Primes Factorisation

Question 3. Show that 5×11×17+17 is a composite number.

Solution:

5 × 11 × 17+ 17 = 17 × (5 × 11 + 1)

= 17 × (55 + 1)

= 17 × 56

∴ It has more than two factors including 1 and number itself.

Hence, it is a composite number.

Question 4. Find the H.C.F. and L.C.M. of 140 and 154 using the prime factorization method.

Solution:

Real Numbers HCF And LCM Of 140

Real Numbers HCF And LCM Of 154

Now,

Real Numbers A Product Of Prime Factorisation Method

∴ H.C.F = 2

L.C.M. = 2 x 2 x 5 x 7 x 11

= 1540

The H.C.F. and L.C.M. of 140 and 154 are 2 and 1540.

Question 5. Find the H.C.F. and L.C.M. of 12, 18, 24 by prime factorisation method.

Solution:

Real Numbers HCF And LCM Of 12

Real Numbers HCF And LCM Of 18

Real Numbers HCF And LCM Of 24

∴ 12 = 2 × 2 × 3

18 = 2 × 3 × 3

24 = 2 × 2 × 3 × 2

Now, H.C.F. = 2 × 3 = 6

and L.C.M. = 2 × 2 × 3 × 3 × 2

= 72

Question 6. Find the L.C.M. and H.C.F. of 36 and 48 and verify that H.C.F. x L.C.M. = product of the given two numbers.

Solution:

Real Numbers The LCM And HCF Of 36

Real Numbers The LCM And HCF Of 48

36 =2×2×3×3

48 =2×2×3×2×2

H.C.F. = 2×2×3=12

L.C.M. =2×2×3×3×2×2 = 144

Now H.C.F. x L.C.M. = 12 × 144

= 1728

and product of two numbers = 36 × 48

= 1 728

Hence, H.C.F. × L.C.M. = product of two numbers

An Important Result

Product of two given numbers = Product of their H.C.F. and L.C.M.

This result is true only for two numbers.

Question 7. The H.C.F of two numbers is 23 and their L.C.M. is 1449. If one number is 207, find the other number.

Solution:

Here, H.C.F = 23

L.C.M. = 1449

Now first no. x second no. = H.C.F. × L.C.M

⇒ Second no. = \(\frac{\text { H.C.F. } \times \text { L.C.M. }}{\text { first no. }}\)

⇒ \(\frac{23 \times 1449}{207}\)

= 161

The other number is 161

Question 8. Show that 12n cannot end with the digit 0 or 5 for any natural number n.

Solution:

Real Numbers Natural Number n

∴ 12 = 2×2×3 = 22×3

⇒ 12n = (22 x 3)n = 22n × 3n

∵ it has no term containing 5.

∴ no value of n eN for which 12″ ends with digit 0 or 5.

Hence Proved.

Question 9. On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm, and 45cm respectively. What is the minimum distance each should walk, so that each can cover the same distance in complete steps?

Solution:

Given:

On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm, and 45cm respectively.

We have to find a number (distance) which is divided by each number completely, which means → L.C.M.

We have to find the L.C.M. of 40 cm, 42 cm, and 45 cm to get the required distance.

Real Numbers The LCM Of 40 Cm

Real Numbers The LCM Of 42 Cm

Real Numbers The LCM Of 45

∴ 40 =2×2×2×5

42 =2 × 3 × 7

45 = 5 × 3 ×x 3

Now, L.C.M. = 2×2×2×5×3×7×3

= 2520

∴ The minimum distance each should walk = 2520 cm

Question 10. What is the smallest number which, when divided by 35, 56, and 91 leaves a remainder of 7 in each case?

Solution:

A number is divided by 3 numbers, which means → L.C.M. of 3 numbers

35 = 5 × 7

56=2×2×2×7

91 = 13 × 7

∴ L.C.M. = 5×7×2×2×2×13

= 3640

∴ The smallest number completely divisible by 35, 36, and 91 is 3640.

Hence, the smallest numbers which when divided by 35, 56, and 91 leave a remainder of 7 in each case will be 3640 + 7 = 3647

Question 11. Find the greatest number of 5 digits exactly divisible by 35, 56, and 91.

Solution:

A number is divided by 3 numbers.

It means → L.C.M. of 3 numbers.

35 = 5×7,56 = 2×2×2×7,

91 = 13 × 7

∴ L.C.M = \(\underbrace{5 \times 7}_{\text {1 as it is }} \times \underbrace{2 \times 2 \times 2}_{\begin{array}{c}
\text { from 2 } \\
\text { not taken yet }
\end{array}} \times \underbrace{13}_{\begin{array}{c}
\text { from 3 } \\
\text { not taken yet }
\end{array}}\)

= 3640

So, 3640 is the smallest number which is divided by all the 3 numbers 35, 56, and 91 completely. But we have to find the greatest number of 5 digits.

Greatest number of 5 digits = 99999

So, required number = 99999 – remainder when 99 is divided by 3640

= 99999- 1719 = 98280.

Alternatively: You can find the multiples of 3640 (L.C.M. of 35, 56 and 91) as

3640 × 1, 3640 × 2 3640 × 10 = 36400,…

3640 × 20 = 72800, 3640 × 25 = 91000,…

3640 × 27 = 98280 (It may be the largest 5 digit no.)

3640 × 28 = 101920 (No, it is 6 digit no.)

So, 98280 is the required number. But this is the time-talking method. So, avoid it)

Rational Numbers

The number which are in the form of \(\frac{p}{q}\) where p and q are integers and q ≠ 0 are called rational numbers.

Decimal Representation of a Rational Number

Consider the following examples:

1. \(\frac{1}{4}\)

= 0.25 4

2. \(\frac{2}{3}\)

In the above example 1, the decimal representation of the rational number \(\frac{1}{4}\) is terminating while in example 2, the decimal representation of the rational number \(\frac{2}{3}\) is non-terminating.

∴ Every rational number, when expressed in decimal form is expressible either in terminating or in non-terminating repeating decimal form.

Important Observation:

  1. A rational number \(\frac{p}{q}\) is a terminating decimal only if q can be written in the form of 2m × 5n for some non-negative integers in and n.
  2. A rational number — is a non-terminating repeating decimal if q cannot be written in the form of 2n × 5n.

For example:

  1. \(\frac{1}{14}=\frac{1}{2 \times 7}\) non-terminating after decimal, as denominator consists of 7 which cannot be the part of 2m x 5n.
  2. \(\frac{1}{3600}=\frac{1}{2^4 \times 3^2 \times 5^2}\) is non-terminating after decimal as denominator consists of 3’s which cannot be the part of \(2^m \times 5^n,\)
  3. \(\frac{91}{8750}=\frac{91}{2 \times 5^4 \times 7}\)
    1. Although it seems in first view that denominator is not the form \(2^m \times 5^n,\), after simplifying \(\frac{7 \times 13}{2 \times 5^4 \times 7}=\frac{13}{2 \times 5^4}\) we see that denominator is a part of \(2^m \times 5^n,\). So, \(\frac{91}{8750}\)is a terminating decimal.
    2. So, a rational number must be written in simplest form i.e., no factor other than 1 must be common to both the numerator and the denominator.
  4. \(\frac{3}{625}=\frac{3}{5^4}\)
    1. Although,it seems in the first view that the denominator is not in the form 2m x 5″, butit is our mistake.
    2. Actually, 5 =1×5 =2×5, which is of the form 2 x 5 . So,\(\frac{3}{625}\) is a terminating decimal

NCERT Exemplar Solutions for Class 10 Maths Chapter 1 Real Numbers Irrational Numbers

The numbers, which when expressed in the decimal form are expressible as non-terminating and non-repeating decimals, are known as irrational numbers.

For example.,

  1. 2.101001000 is a non-terminating non-repeating decimal, so it is an irrational number.
  2. 1.767767776 is a non-terminating and non-repeating decimal, so it is an irrational number.

If x is a positive integer which is not a perfect square, then \(\sqrt{x}\) is irrational.

For example., \(\sqrt{2}, \sqrt{5}, \sqrt{7}\), etc., are irrational numbers.

Similarly \(\sqrt[3]{7}, \sqrt[4]{10}\), etc., are irrational numbers.

π is irrational and \(\frac{22}{7}\) is rational

Theorem: Let p is a prime number and V be a positive integer. If p divides a2, then show that p divides a.

Proof: We know that every positive integer can be expressed as a product of primes, not necessarily all distinct.

Let a = p1,p2,p3,……..pn

where, p1,p2,p3,……..pn are primes, not necessarily distinct.

∴ a2 = (p1,p2,p3,……..pn) (p1,p2,p3,……..pn)  = p21,p22,p23,……..p2n

Now, p divides a2

⇒ p is a prime factor of a2.

⇒ P is one of P1,p2,P3 …….. P„-

⇒ p divides a.

NCERT Exemplar Solutions for Class 10 Maths Chapter 1 Real Numbers Irrational Numbers Solved Examples

Question 1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

  1. \(\frac{12}{125}\)
  2. \(\frac{7}{1600}\)

Solution:

1. \(\frac{12}{125}=\frac{12}{5 \times 5 \times 5}=\frac{12}{5^3}=\frac{12}{2^0 \times 5^3}\)

Now, the denominator is in the form of 2m x 5”.

∴ \(\frac{12}{125}\) is a terminating decimal.

2. \(\frac{7}{1600}\)

⇒ \(\frac{7}{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5}\)

⇒ \(\frac{7}{2^6 \times 5^2}\)

Now, the denominator is in the form of 2m x 5″.

∴ \(\frac{7}{1600}\) is a terminating decimal.

Question 2. Show that :

  1. \(\frac{3}{250}\)
  2. \(\frac{11}{50}\)

are terminating decimals. Express each of them in decimal form without actual division (long division).

Solution:

1. \(\frac{3}{250}=\frac{3}{2 \times 5 \times 5 \times 5}\)

\(=\frac{3}{2^1 \times 5^3}\)

Now, the denominator is in the form of 2m × 5n

∴ \(\frac{3}{250}\) is a terminating decimal.

Agan, \(\frac{3}{250}\)

⇒ \(\frac{3}{2^1 \times 5^3}\)

⇒ \(\frac{3 \times 2^2}{2^3 \times 5^3}\)

⇒ \(\frac{12}{10^3}]\)

= 0.012

2. \(\frac{11}{50}\)

⇒ \(\frac{11}{2 \times 5 \times 5}\)

⇒ \(\frac{11}{2^1 \times 5^2}\)

Now, the denominator is in the form of 2m × 5n

∴ \(\frac{11}{50}\)

Again, \(\frac{11}{50}\)

\(\frac{11}{2^1 \times 5^2}\) \(\frac{11 \times 2}{2^2 \times 5^2}\) \(\frac{22}{10^2}\)

= 0.22

Question 3.  Show that each of the following are non-terminating repeating decimal:

  1. \(\frac{5}{12}\)
  2. \(\frac{7}{75}\)

Solution:

1. \(\frac{5}{12}\)

⇒ \(\frac{5}{2 \times 2 \times 3}\)

⇒ \(\frac{5}{2^2 \times 3}\)

the denominator 22 x 3 is not in the form of 2m x 5n

∴ it is non-terminating repeating decimal.

2. \(\frac{7}{25}\)

⇒ \(\frac{7}{3 \times 5 \times 5}\)

⇒ \(\frac{7}{3 \times 5^2}\)

the denominator 3 x 52 is not in the form of 2n x 5m

∴ it is non-terminating repeating decimal.

Question 4. The decimal expansion of the rational number \(\frac{43}{2^4 \times 5^3}\) will terminate after how many places of decimals?

Solution:

\(\frac{43}{2^4 \times 5^3}\) \(\frac{43 \times 5}{2^4 \times 5^4}\) \(\frac{215}{10^4}\)

= 0.0215

∴ it will terminate after 4 places of decimals.

Question 5. Express each of the following in the simplest form:

  1. \(0. \overline{6}\)
  2. \(3. \overline{3}\)

Solution:

1. Let x = \(0 . \overline{6}\)

⇒ x = 0.666….. (1)

⇒ 10x = 6.666….. (2)

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

9x = 6

⇒ \(x=\frac{6}{9}=\frac{2}{3}\)

⇒ \(0. \overline{6}=\frac{2}{3}\)

2. Let x = \(3 . \overline{3}\)

⇒ x = 3.333…(1)

⇒ 10x = 33.333…(2)

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

9x = 30

⇒ \(x=\frac{30}{9}=\frac{10}{3}\)

⇒ \(3. \overline{3}=\frac{10}{3}\)

Question 6. Express each of the following in the simplest form :

Let x = \(0. \overline{36}\)

⇒ \(1. \overline{046}\)

⇒ \(Let x=0. \overline{36}\)

⇒\(x=0.363636 \ldots . . .\)

⇒ \(100 x=36.363636 \ldots . . \)

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

⇒ \(99 x =36\)

⇒ \(x =\frac{36}{99}=\frac{4}{11}\)

⇒ \( 0 . \overline{36}=\frac{4}{11}\)

2. Let \(x=1. \overline{046}$

⇒ [latex] x  =1.046046046 \ldots\)

⇒ \(1000 x  =1046.046046046 \ldots \)

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

⇒ \(999 x =1045\)

⇒ \(x=\frac{1045}{999}\)

⇒ \(1. \overline{046}\)

⇒ \(\frac{1045}{999}\)

Question 7. A rational number in its recurring decimal expansion is 327.7081. What can you say about the prime factors of q, when this number is expressed in the form \(\frac{p}{q}\)? Give <l reasons.

Solution:

Given:

A rational number in its recurring decimal expansion is 327.7081.

Let x = \(327.7 \overline{081}\)

⇒ x =327.7081081081…..

10x =3277.081081081…..

10000 A- =3277081.081081081…..

On subtraction, 9990x =3273804

∴ x = \(\frac{3273804}{9990}=\frac{60626}{185}\)

which is of the form =\(\frac{p}{q}\)

Now, q= 185 = 5 × 37,

which cannot be written in the form 2m x 5n.

∴ It is a non-terminating and repeating decimal.

Question 8. Prove that \(\sqrt{2}\) is irrational.

Solution:

Let, if possible. \(\sqrt{2}\) be rational and its simplest form be \(\frac{a}{b}\)

Then a and b are integers and have no common factor other than 1 and b ≠ 0

Now, \(\sqrt{2}=\frac{a}{b}\)

⇒ \(2=\frac{a^2}{b^2}\)

⇒ \(a^2=2 b^2\)

As 2b2 is divisible by 2.

∴ a2 is divisible by 2.

⇒ a is divisible by 2.

Let a = 2c, for some integer c.

∴ From equation (1)

(2c)2 = 2b2

⇒ b2 = 2c2

But 2c2 is divisible by 2.

∴ b2 is divisible by 2.

⇒ b is divisible by 2.

Let b = 2d, for some integer d.

Thus, 2 is a common factor of a and b both.

But it contradicts the fact that a and b have no common factor other than 1.

So, our supposition is wrong.

Hence, \(\sqrt{2}\) is irrational.

Hence Proved.

Question 9. If p is a prime number, then prove that \(\sqrt{p}\) is irrational. (Treat this result as a theorem)

Solution:

Let p be a prime number and if possible let \(\sqrt{p}\) be irrational.

Let the simplest form of \(\sqrt{p} \text { be } \frac{a}{b} \text {. }\)

Then a and b are integers and having no common factors other than 1 and b ≠ 0.

Now, \( \sqrt{p}=\frac{u}{b}\)

⇒ \(p=\frac{a^2}{b^2}\)

⇒ \(a^2=p b^2\)

As pb2 is divisible by p.

∴ a2 is divisible by p.

⇒ a is divisible by p.

Let a = pc for some integer c.

From equation (1)

⇒ \((p c)^2=p b^2\)

⇒ \(b^2=p c^2\)

But pc2 is divisible by p.

∴ b2 is divisible by p.

⇒ b is divisible by p.

Let b = pd for some integer d.

Thus, p is a common factor of both a and b.

But it contradicts the fact that a and b have no common factor other than 1.

So, our supposition is wrong.

Hence, \(\sqrt{2}\) is irrational.

Question 10. Show that \((2+\sqrt{3})\) is an irrational number.

Solution:

Let, if possible \((2+\sqrt{3})\) is rational

then, \(2+\sqrt{3}=\frac{a}{b} \text { (say) }\)

where a and b are integers and b ≠ 0

⇒ \(\sqrt{3}=\frac{a}{b}-2\)

⇒ \(\sqrt{3}=\frac{a-2 b}{b}\)

a and b are integers

∴ a – 2b is also an integer.

⇒ \(\frac{a-2 b}{b}\) is rational

Now, L.H.S of equation (1) is the square root of a prime number. So, it is irrational and R.H.S is rational.

Which is a contradiction because a rational number and an irrational number can never be equal.

So, our supposition i.e., \((2+\sqrt{3})\) is rational, is wrong.

Question 11. Prove that \(5 \sqrt{7}\) is irrational

Solution:

Let if possible \(5 \sqrt{7}\) is rational.

Now, \(5 \sqrt{7}\)is rational and \(\frac{1}{5}\) is rational.

We know that the product of two rational numbers is rational.

∴ \(5 \sqrt{7} \times \frac{1}{5} \text { is rational. }\) is rational.

⇒ \(\sqrt{7}\)is rational.

But square root of a prime number is always an irrational number. This contradicts the fact because an irrational number cannot be equal to a rational number.

So, our supposition is wrong.

Hence, \(5 \sqrt{7}\) is irrational.

Question 12. A rational number in its decimal is 327.7081. What can you say about the prime factor of q. when this number is expressed in the form \(\frac{p}{q}\)? Give reasons.

Solution:

Given:

A rational number in its decimal is 327.7081.

The decimal expansion is 327.7081.

∴ It is a rational number.

Now, 327.7081 = \(\frac{3277081}{10000}\)

⇒ \(\frac{3277081}{2^4 \times 5^4}\)

⇒ \(\frac{p}{q}\)

Here q = \(2^4 \times 5^4\) which is in the form of \(2^m \times 5^n\)

∴ The prime factors of q are 2 and 5

NCERT Exemplar Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.1

Question 1. Use Euclid’s division algorithm to find the H.C.F. of :

  1. 135 and 225
  2. 196 and 38220
  3. 867 and 255

Solution:

1. Given numbers 135 and 225

Real Numbers The HCF Of 135 And 225

Here, 225 > 135

∴ 225 = 135 × 1 + 90

Remainder = 90 ≠ 0

135 = 90 × 1 + 45

Remainder = 45 ≠ 0

90 = 45 × 2 + 0

Remainder = 0

∵ The remainder is zero and the divisor is 45

∴ H.C.F. = 45

The H.C.F. of 135 and 225 = 45

2. Given numbers 196 and 38220

Real Numbers The HCF Of 196 And 38220

38220 > 196

∴ 38220 = 196 × 195 + 0

Remainder = 0

∵ The remainder is zero and the divisor is 196.

∴ H.C.F. = 196

The H.C.F. of 196 and 38220 is 196

3. Given numbers 867 and 255

Real Numbers HCF Of 255 And 869

867 > 255

∴ 867 = 255 x 3 + 102

Remainder = 102 ≠ 0

255 = 102 × 2 + 51

Remainder = 51 ≠ 0

102 = 51 × 2 + 0

Remainder = 0

∵ The remainder is zero and the divisor is 51.

∴ H.C.F. =51

The H.C.F. of 867 and 255 is 51

Question 2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Solution:

Let ‘a’ be an odd positive integer.

From Euclid’s division algorithm,

Let q be the quotient and the remainder when a is divided by 6.

∴ a = 6q +r

where, r = 0,1,2,3,4,5

Now, a = 6q+ 0 or a = 6q + 1 or a = 6q + 2 or a = 6q + 3 or a = 6q + 4 or a = 6q + 5 but 6q + 0, 6q + 2, 6q + 4 are even numbers.

∴ a = 6q + 1 or 6q + 3 or 6q + 5

Therefore, any positive odd integer is of the form 6q+1 or 6q + 3 or 6q + 5 where q is some integer.

Question 3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Solution :

Given:

An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns.

The maximum number of columns will be the H.C.F. of the number of army contingent of 616 members and the number of army band of 32 members.

Here, 616> 32

Real Numbers The HCF Of 32 And 616

Now, 616 = 32 × 19 + 8

Remainder = 8≠0

32 = 8 × 4 + 0

Remainder = 0

∵ The remainder is zero and the divisor is 8.

∴ H.C.F. = 8

Therefore, an army can march in 8 columns.

Question 4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Solution :

Let ‘a’ be a positive integer.

From Euclid’s division algorithm,

Let q be the quotient and r the remainder when a is divided by 3,

∴ a = 3q + r where r = 0,1,2

⇒ a = 3q + 0 or a = 3q + 1 or a = 3q + 2

Now a = 3q

⇒ a2 = (3q)2

= 9q2 = 3(3q2)

= 3m

where, m = 3q2 is an integer.

Again, a = 3q + 1

⇒ a2 = (3 q+1)2 = 9q2+ 6q + 1

= 3(3q2+2q) +1 = 3m + 1

where, m = 3q2 + 2q is an integer and

a = 3q +2

⇒  a2 = (3 q + 2)2

= 9q2 + 12q + 4

= 3(3q2 + 4q + 1) + 1

= 3m + 1

where, m = 3q2 + 4q + 1 is an integer.

So, a2 = 3m or 3m + 1.

The square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Hence Proved.

Question 5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

Solution:

Let ‘a’ be a positive integer,

From Euclid’s division algorithm,

Let q be the quotient and r, be the remainder when a is divided by 3.

∴ a = 3q + r

where r = 0, 1, 2

⇒ a = 3q + 0 or a = 3q + 1 or a = 3q + 2

Now, a = 3

⇒ a3 = (3q)3 = 27q3 = 9(3q3) = 9m.

where = 3q3 is an integer.

Again, a = 3q + 1

⇒ \( a=3 q+1\quad a^3=(3 q+1)^3\)

⇒ \(27 q^3+3(3 q)(1)(3 q+1)+1\)

⇒ \(27 q^3+9 q(3 q+1)+1\)

⇒ \(9\left[3 q^3+3 q^2+q\right]+1=9 m+1\)

where, m = 3q3 + 3q2+q is an integer.

Again, a = 3q + 2

⇒ \( a^3=(3 q+2)^3\)

⇒ \(27 q^3+3(3 q)(2)(3 q+2)+8\)

⇒ \(9\left[3 q^3+6 q^2+4 q\right]+8\)

= 9m+8

Where \(m=3 q^3+6 q^2+4 q \text { is an integer. }\)

∴ a3 = 9m or 9m + 1 or 9m + 8

The cube of any positive integer is of the form 9m or 9m + 1 or 9m + 8.

NCERT Exemplar Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.2

Question 1. Express each number as a product of its prime factors :

  1. 140
  2. 156
  3. 3825
  4. 5005
  5. 7429

Solution:

1. 140 = 2 × 2 × 5 × 7

= 22 × 5 × 7

Real Numbers Product Of Prime Factor 1

2.  156 = 2 × 2 ×3 × 13

= 22 × 3 × 13

Real Numbers Product Of Prime Factor 2

3. 3825 = 3 × 3 × 5 × 5 × 17

= 32 × 52 × 17

Real Numbers Product Of Prime Factor 3

4. 5005 = 5 × 7 × 11 × 13

Real Numbers Product Of Prime Factor 4

5. 7429 = 17 × 19 × 23

Real Numbers Product Of Prime Factor 5.

Question 2. Find the L.C.M. and H.C.F. of the following pairs of integers and verify that L.C.M. x H.C.F. = product of the two numbers.

  1. 26 and 91
  2. 510 and 92
  3. 336 and 54

Solution:

1. 26 = 2 × 13

91 = 13 × 7

H.C.F. = 13

L.C.M. = 2 × 13 × 7

= 182

Product of numbers = 26 x 91 = 2366

H.C.F. x L.C.M. = 13 x 182 = 2366

∴ Product of numbers = H.C.F. × L.C.M.

2. 510 = 2 × 3 × 5 × 17

92 = 2 × 2 × 23

H.C.F. = 2

L.C.M. =2 × 3 × 5 × 17 × 2 × 23

= 23460

Now, product of numbers = 510 × 92

= 46920

H.C.F. x L.C.M. = 2 x 23460 = 46920

∴ Product of numbers = H.C.F. x L.C.M.

336 = 2x2x2x2x3x7

54 = 2x3x3x3

H.C.F. =2×3 = 6

L.C.M. = 2x2x2x2x3x7x3x3

= 3024

Now, product of numbers = 336 x 54 = 18144

H.C.F. x L.C.M. = 6 x 3024 = 18144

∴ Product of numbers

H.C.F. x L.C.M

Question 3. Find the L.C.M and H.C.F of the following integers by applying the prime factorization method:

  1. 12. 15 and 21
  2. 17, 23 and 29
  3. 8. 9 and 25

Solution:

1. 12 = 2 × 2 × 3

15 =3 × 5

21 =3 × 7

∴ H.C.F. = 3

L.C.M. = 2×2×3×5×7

= 420

The L.C.M and H.C.F of 12. 15 and 21 is 420 and 3.

2. 17 = 17 × 1

23 = 1 × 23

29 = 1 × 29

∴ H.C.F. = 1

L.C.M. = 1 7 × 23 × 29 = 11339

The L.C.M and H.C.F of 17, 23 and 29 is 11339 and 1.

3. 8 = 2 × 2 × 2

9 = 3 × 3

25 =5 × 5

∴ H.C.F. = 1

L.C.M. = 2×2×2×3×3×5×5

= 1800

The L.C.M and H.C.F of 8. 9 and 25 is 1800 and 1.

Question 4. Given that H.C.F. (306, 657) = 9, find L.C.M. (306, 657).

Solution:

H.C.F. (306, 657) = 9

H.C.F. × L.C.M. = Product of numbers

9 × L.C.M. = 306 x 657

306 × 657

⇒ L.C.M. = \(\frac{306 \times 657}{9}\)

= 34 × 657

= 22338

L.C.M. of (306, 657) = 22338

Question 5. Check whether 6″ can end with the digit 0 for any natural number n.

Solution:

6 = 2 × 3

6n = (2 x 3)n

5 is not a factor of 6.

NCERT Exemplar Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.3

Question 1. Prove that \(\sqrt{5}\) is irrational.

Solution:

Let \(\sqrt{5}\) is a rational number.

Let \(\sqrt{5}\) = where\(\frac{a}{b}\) and a andb are integers,

which have no common prime factors other than 1.

Now, \(\) = \(\frac{a}{b}\) ….(1)

⇒ a2 is divisible by 5.

⇒ a is divisible by 5

Let a = 5c

⇒ a2 = 25c2

⇒ 5b2 = 25c2 (From equation (1) )

⇒ b2 = 5c2

Now, 5c2 is divisible by 5.

⇒ b2 is divisible by 5.

⇒ b is divisible by 5

∴ 5 is a common factor of a and b.

It is opposite to our assumption.

∴ Our assumption is wrong.

i.e., \(\sqrt{5}\) is an irrational number.

Hence proved.

Question 2. Prove that \(3+2 \sqrt{5}\) is irrational:

Solution:

Let \(3+2 \sqrt{5}\) is a rational number

Let \(\) = \(\) where q* 0 and p and q are positive integers.

⇒ \(2 \sqrt{5}=\frac{p}{q}-3\)

⇒ \(\frac{p-3 q}{q}\)

⇒ \(\sqrt{5}=\frac{p-3 q}{2 q}\)

∵ p and q are integers and q≠0.

∴ \(\frac{p-3 q}{2 q}\) is a rational number.

From question 1,\(\sqrt{5}\) is an irrational.

Now, an irrational number and a rational number are equal.

Which is not possible.

So, it is opposite to our assumption.

∴ Our assumption is wrong.

i.e., \(3+2 \sqrt{5}\) is an irrational number.

Question 3. Prove that the following are irrationals :

  1. \(\frac{1}{\sqrt{2}}\)
  2. \(\frac{7}{\sqrt{5}}\)
  3. \(6+\sqrt{2}\)

Solution:

1. \(\frac{1}{\sqrt{2}}\)

Let \(\frac{1}{\sqrt{2}}\) is a rational number

Let \(\frac{1}{\sqrt{2}}=\frac{p}{q}\)= where q≠0 andp and q are integers.

⇒ \(q=\sqrt{2} p \Rightarrow q^2=2 p^2\) ……(1)

Now, 2p2 is divisible by 2.

⇒ q2 is divisible by 2.

⇒ q is divisible by 2.

Let q = 2r

⇒ q2 = 4r2

⇒ 2p2 = 4r2    (from equation 1)

⇒ p2 = 2r2

Now, 2r2 is divisible by 2.

⇒ p2 is divisible by 2.

⇒ p is divisible by 2.

∴ 2 is a common factor of p and q, which is opposite to our assumption.

So, our assumption is wrong i.e., \(\frac{1}{\sqrt{2}}\) is an So, our assumption is wrong i.e., irrational number.

2.  Let \(7 \sqrt{5}\) is a rational number.

Let \(7 \sqrt{5}=\frac{p}{q}\) where q*0 andp andq are integers.

⇒ \(\sqrt{5}=\frac{p}{7 q}\)

∵ p and are integers and 4 * 0

∴ \(\frac{p}{7 q}\) is a rational number

From question 1,\(\sqrt{5}\) is an irrational number. Now, an irrational number and a rational number are equal, which is impossible. So, it is opposite to our assumption:

∴ Our assumption is wrong, i.e., \(7 \sqrt{5}\) is an irrational number.

3.  Let \(6+\sqrt{2}\) is a rational number.

Let \(6+\sqrt{2}=\frac{p}{q}\) where q * 0 and p and q are integers.

⇒ \(\sqrt{2}=\frac{p}{q}-6\)

Now,  \(\frac{p}{q}-6\)  is rational number and \(\sqrt{2}\) is an irrational number.

Now, an irrational number and a rational number are equal.

Which is not possible.

So, it is opposite to our assumption.

∴ Our assumption is wrong.

Therefore, \(6+\sqrt{2}\) is an irrational number.

NCERT Exemplar Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.4

Question 1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion :

  1. \(\frac{13}{3125}\)
  2. \(\frac{17}{8}\)
  3. \(\frac{64}{455}\)
  4. \(\frac{15}{1600}\)
  5. \(\frac{29}{343}\)
  6. \(\frac{23}{2^3 5^2}\)
  7. \(\frac{129}{2^2 5^7 7^5}\)
  8. \(\frac{6}{15}\)
  9. \(\frac{35}{50}\)
  10. \(\frac{77}{210}\)

Solution:

1. \(\frac{13}{3125}\)

⇒ \(\frac{13}{3125}\)

⇒ \(=\frac{13}{5 \times 5 \times 5 \times 5 \times 5}\)

⇒ \(=\frac{13}{5^5}\)

⇒ \(\frac{13}{2^0 \times 5^5}\)

Its denominator is 2° × 55 whose prime factors are 2 and 5 only.

Therefore, the decimal expansion of \(\frac{13}{3125}\) is terminating.

2. \(\frac{17}{8}\)

⇒ \(\frac{17}{2 \times 2 \times 2}\)

⇒ \(=\frac{17}{2^3 \times 5^0}\)

Its denominator is 23 × 5° whose prime factors are 2 and 5 only.

Therefore, the decimal expansion of \(\) is terminating

3. \(\frac{64}{455}\)

⇒ \(\frac{64}{5 \times 7 \times 13}\)

Its denominator has the prime factors 7 and 13 other than 5, which is not in form 2m × 5n. So the decimal expansion of \(\frac{64}{455}\) is non-terminating and repeating.

4. \(\frac{15}{1600}\)

⇒ \(=\frac{3 \times 5}{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5}\)

⇒ \(\frac{3}{2^6 \times 5^1}\)

Its denominator is 26 × 51 whose prime factors are 2 and 5 only.

Therefore, the decimal expansion of \(\frac{15}{1600}\) is terminating.

5. \(\frac{29}{343}\)

⇒ \(\frac{29}{7 \times 7 \times 7}\)

⇒ \(\frac{29}{7^3}\)

7 is a prime factor of its denominator which is not in the form 2m x 5n.

Therefore, the decimal expansion of is \(\frac{29}{343}\) non-terminating and repeating.

6. \(\frac{23}{2^3 \cdot 5^2}\)

Its denominator is 23. 52 whose prime factors are 2 and 5 only.

Therefore, the decimal expansion of \(\frac{23}{2^3 \cdot 5^2}\) is terminating.

7. \(\frac{129}{2^2 \cdot 5^7 \cdot 7^5}\)

Its denominator is 2n.57. 75 which is not in the form 2m × 5n.

Therefore, the decimal expansion of \(\frac{129}{2^2 \cdot 5^7 \cdot 7^5}\) is non-terminating and repeating.

8. \(\frac{6}{15}\)

⇒ \(\frac{3 \times 2}{3 \times 5}\)

⇒ \(\frac{2}{5}\)

Its denominator is 5 whose prime factor is 5 only.

Therefore, the decimal expansion of \(=\frac{6}{15}\) is terminating.

9. \(\frac{35}{50}\)

⇒ \(\frac{5 \times 7}{2 \times 5 \times 5}\)

⇒ \(\frac{7}{2 \times 5}\)

Its denominator is 2 x 5 whose prime factors are 2 and 5 only.

Therefore, the decimal expansion of \(\frac{35}{50}\) is terminating.

10. \(\frac{77}{210}\)

⇒ \(\frac{7 \times 11}{2 \times 5 \times 3 \times 7}\)

⇒ \(\frac{11}{2 \times 5 \times 3}\)

Its denominator is 2×5×3 which is not in the form 2nx5n

Therefore, the decimal expansion of \(\frac{77}{210}\) is non-terminating and repeating.

Question 2. Write down the decimal expansions of those rational numbers above which have terminating decimal expansions.

Solution:

1. \(\frac{13}{3125}\)

⇒ \(\frac{13}{5^5}\)

⇒ \(\frac{13 \times 2^5}{5^5 \times 2^5}\)

⇒ \(\frac{13 \times 32}{(5 \times 2)^5}\)

⇒ \(\frac{416}{10^5}\)

=0.00416

\(\frac{13}{3125}\) =0.00416

2. \(\frac{17}{8}\)

⇒ \(\frac{17}{2^3}\)

⇒ \(\frac{17 \times 5^3}{2^3 \times 5^3}\)

⇒ \(\frac{17 \times 125}{(2 \times 5)^3}\)

\(\frac{17}{8}\) ⇒ \(\frac{17 \times 125}{(2 \times 5)^3}\)

3. \(\frac{15}{1600}\)

⇒ \(\frac{3 \times 5}{2^6 \times 5^2}\)

⇒ \(\frac{3 \times 5 \times 5^4}{2^6 \times 5^6}\)

⇒ \(\frac{15 \times 625}{(2 \times 5)^6}\)

⇒ \(\frac{9375}{10^6}\)

=0.009375

\(\frac{15}{1600}\) =0.009375

4. \(\frac{23}{2^3 5^2}\)

⇒ \(\frac{23 \times 5}{2^3 \times 5^3}\)

⇒ \(\frac{115}{(2 \times 5)^3}\)

⇒ \(\frac{115}{10^3}\)

=0.115

\(\frac{23}{2^3 5^2}\) =0.115

5. \(\frac{6}{15}\)

⇒ \(\frac{2 \times 3}{3 \times 5}\)

⇒ \(\frac{2}{5}\)

⇒ \(\frac{2 \times 2}{5 \times 2}\)

⇒ \(\frac{4}{10}=0.4\)

\(\frac{6}{15}\) ⇒ \(\frac{4}{10}=0.4\)

6. \(\frac{35}{50}\)

⇒ \(\frac{5 \times 7}{5 \times 5 \times 2}\)

⇒ \(\frac{5 \times 7 \times 2}{5^2 \times 2^2}\)

⇒ \(\frac{70}{(10)^2}\)

=0.7

\(\frac{35}{50}\) =0.7

Question 3. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they rational’ and of the form \(\frac{p}{q}\) what can you say about the prime factors of q?

  1. 43.123456789
  2. 0.120120012000120000…
  3. \(43. \overline{123456789}\)

Solution:

1. 43.123456789 = \(\frac{43123456789}{1000000000}\)

which is in the form \(\frac{p}{q}\) So, it is a prime number

1000000000 has prime factors 2 and 5.

2. 0.120120012000120000…

Its decimal expansion is non-terminating and non-repeating. It cannot be expressed in the form \(\frac{p}{q}\)

So, it is an irrational number.

3. \(43. \overline{123456789}\)

Its decimal expansion is non-terminating and repeating.

So, it is a rational number.

The given number can be expressed in the form \(\frac{p}{q}\)

It has some other prime factor except 2 or 5.

NCERT Exemplar Solutions for Class 10 Maths Chapter 1 Real Numbers Multiple Choice Questions And Answers

Question 1. The decimal expansion of the rational number \(\frac{17}{2^2 \cdot 5}\)ends after the following decimal places :

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

Answer: 2. 2

Question 2. For some integer m, every odd integer is of the following form :

  1. m
  2. 2m
  3. m+1
  4. 2m+1

Answer: 4. 2m+1

Question 3. For some integer m, every even integer is of the following form :

  1. m
  2. 2m
  3. m+1
  4. 2m+1

Answer: 2. 2m

Question 4. 4-. The largest number from which 57 and 67 divided, leaving the remainders 5 and 7 respectively, is:

  1. 5
  2. 8
  3. 10
  4. 11

Answer: 3. 10

Question 5. The sum of a non-zero rational number and an irrational number is:

  1. Rational
  2. Irrational
  3. Zero
  4. None of these

Answer: 2. Irrational

Question 6. If a = x2y and b =xy2 then HCF (a, b) is

  1. x
  2. y
  3. xy
  4. x2y2

Answer: 3. xy

Question 7. The decimal expansion of \(\frac{3721}{625}\) ends after the following decimal places:

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

Answer: 4. 4

Question 8.  Given that LCM (132, 288) = 3168 then HCF (132, 288) is :

  1. 288
  2. 132
  3. 48
  4. 12

Answer: 4. 12

Question 9. The sum of exponents of prime factors in the prime factorization of the number 144 is :

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

Answer: 4. 6

Question 10. The prime number is:

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

Answer: 3. 2

Question 11. If the L.C.M. of 26, 156 is 156, then the value of HCF is:

  1. 156
  2. 26
  3. 13
  4. 6

Answer: 2. 26

Leave a Comment