NCERT Maths Class 10th Chapter 1st Real Numbers Solution

In this section, we shall study the NCERT Maths Class 10^th solution in English. In Maths class 10^th, there are 15 chapters. The solution is given chapter-wise in PDF format. In each PDF, every question of each exercise is solved in detail.

Chapter 1^st Real Numbers

NCERT maths class 10th solution of chapter 1st is given here. In Chapter 1st Real Numbers, in exercise 1.1, questions based on Euclid’s division lemma and Euclid’s division algorithm are solved with the help of an easy explanation. In exercise 1.2, questions based on the fundamental theorem of arithmetic, LCM, and HCF are solved. In exercise 1.3, 3 questions are based on the irrationality theorem. We use the technique called ‘proof by contradiction’ to solve these questions. In exercise 1.4, there are 3 questions based on rational numbers and their decimal expansions, and their theorems. All the questions are solved in detail.

Chapter 1^st Real Numbers

Solution

Exercise 1.1

Q (1) Use Euclid’s division algorithm to find the HCF of:

(i) 135 and 225     (ii) 196 and 38220     (iii) 867 and 255

Solution – (i) 135 and 225

Using Euclid’s division algorithm        

Step 1 – Here, two positive integers are 135 and 225 and 225 > 135 so, by Euclid’s Division lemma.

225 = 135×1 + 90

Step 2 – Since the Remainder is not zero, 90 ≠ 0 so, we shall apply Euclid’s Division Lemma to Divisor 135 and Remainder 90.

135 = 90×1 + 45

Step 3 – Again, the Remainder is not zero 45 ≠ 0 so, we shall apply Euclid’s Division Lemma to Divisor 90 and Remainder 45.

90 = 45×2 + 0

Since the remainder is zero in step 3 and the divisor in this step is 45.

Therefore, the required HCF of 135 and 225 is 45.                Ans.

We can also understand it by the division process.

(ii) 196 and 38220

Using Euclid’s division algorithm

Step 1 – Here, 38220 > 196 so, by Euclid’s Division lemma.

38220 = 196×195 + 0

Since the remainder is zero in step 1 therefore, the HCF of 196 and 38220 is 196. We can also understand it by the division process.

(iii) 867 and 255

Using Euclid’s division algorithm

Step 1 – Here, 867 > 255 so, by Euclid’s Division lemma.

867 = 255×3 + 102

Step 2 – Since the Remainder is not zero 102 ≠ 0 so, we shall apply Euclid’s Division Lemma to Divisor 255 and Remainder 102.

255 = 102×2 + 51

Step 3 – Again, the Remainder is not zero 51 ≠ 0 so, we shall apply Euclid’s Division Lemma to Divisor 102 and Remainder 51.

102 = 51×2 + 0

Since the remainder is zero in step 3 and the divisor in this step is 51. Therefore, the required HCF of 867 and 255 is 51.                Ans.

We can also understand it by the division process.

Q (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 any positive odd integer and b be any positive integer. According to the question, we are taking b = 6.

By Euclid’s Division lemma,                  a = bq + r                where,  0  r < b

                          Here,      b = 6          so, 0 ≤ r < 6    i.e. r = 0, 1, 2, 3, 4, 5

Putting the values, b = 6 and r = 0 then      a = 6q + 0; a = 6q                                 

                                   b = 6 and r = 1 then     a = 6q + 1

                                   b = 6 and r = 2 then     a = 6q + 2

                                   b = 6 and r = 3 then     a = 6q + 3

                                   b = 6 and r = 4 then     a = 6q + 4

                                   b = 6 and r = 5 then     a = 6q + 5

Now, since a is a positive odd integer so, it cannot be of the form 6q, or 6q + 2, or 6q + 4 because all these are even positive integers and divisible by 2. 

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

Q (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 – In this question, there are two groups of 616 members and 32 members.

To find the maximum number of columns, we’ll find out the HCF of 616 and 32 by Euclid’s division lemma.

Using Euclid’s division algorithm,

Step 1 – Here, 616 > 32 so, by Euclid’s Division lemma.

616 = 32×19 + 8

Step 2 – Since the Remainder is not zero 8 ≠ 0 so, we shall apply Euclid’s Division Lemma to Divisor 32 and Remainder 8.

32 = 8×4 + 0

Since the remainder is zero in step 2 and the divisor in this step is 8.

Therefore, the HCF of 616 and 32 is 8. Which means the maximum number of columns in which they can march is 8.                 Ans.

Q (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.

[Hint: Let x be any positive integer, then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]

Solution – Let x be any positive integer, then it can be of the form 3q, 3q + 1 or 3q + 2.

Now, if x = 3q

As given in the question, squaring both sides.

x² = (3q)²

x² = 9q²

x² = 3(3q²)

x² = 3m       …………….(1)

Where, m = 3q² and m is also an integer.

Again, if x = 3q + 1

Squaring both sides,

x² = (3q + 1)²

Using identity, (a + b)² = a² + 2ab + b²

x² = 9q² + 6q + 1

x² = 3(3q² + 2q) + 1

x² = 3m + 1       ……………..(2)

Where, m = 3q² + 2q and m is also an integer.

Again, if x = 3q + 2

Squaring both sides,

x² = (3q + 2)²

Again, using identity, (a + b)² = a² + 2ab + b²

x² = 9q² + 12q + 4

We can also write, x² = 9q² + 12q + 3 + 1

x² = 3(3q² + 4q + 1) + 1

x² = 3m + 1       ……………..(3)

Where, m = 3q² + 4q + 1 and m is also an integer.

From equations (1), (2), and (3)

x² = 3m or 3m + 1

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

Q (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 any positive integer.

By Euclid’s division lemma, a = bq + r          where, q = quotient and r = remainder

Let b = 3

So, 0 £ r < 3       i.e. r = 0, 1, 2 (As required in the question)

Now, putting the values,

b = 3 and r = 0 then      a = 3q + 0; a = 3q

b = 3 and r = 1 then      a = 3q + 1

b = 3 and r = 2 then      a = 3q + 2

It means a is of the form 3q, 3q + 1 or 3q + 2.

Now, if a = 3q

Cubing both sides,

a³ = (3q)³

a³ = 27q³

a³ = 9(3q³)

a³ = 9m       …………..(1)

Where, m = 3q³ and m is also an integer.

Again, if a = 3q + 1

Cubing both sides,

a³ = (3q + 1)³

Using identity, (a + b)³ = a³ + 3a²b + 3ab² + b³

a³ = 27q³ + 27q² + 9q + 1

a³ = 9(3q³ + 3q² + q) + 1

a³ = 9m + 1      ……………(2)

Where, m = 3q³ + 3q² + q and m is also an integer.

Again, if a = 3q + 2

Cubing both sides,

a³ = (3q + 2)³

Again, using identity, (a + b)³ = a³ + 3a²b + 3ab² + b³

a³ = 27q³ + 54q² + 36q + 8

a³ = 9(3q³ + 6q² + 4q) + 8

a³ = 9m + 8      ……………(3)

Where, m = 3q³ + 6q² + 4q and m is also an integer.

From equations (1), (2), and (3)

a³ = 9m, 9m + 1 or 9m + 8

Therefore, the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.            Ans.

Exercise 1.2

Q (1) Express each number as a product of its prime factors:

(i) 140     (ii) 156     (iii) 3825     (iv) 5005     (v) 7429

Solution – (i) 140

Prime factors of 140 = 2×2×5×7 = 2²×5×7

(ii) 156

Prime factors of 156 = 2×2×3×13 = 2²×3×13

(iii) 3825

Prime factors of 3825 = 3×3×5×5×17 = 3²×5²×17

(iv) 5005

Prime factors of 5005 = 5×7×11×13

(v) 7429

Prime factors of 7429 = 17×19×23

Q (2) Find the LCM and HCF of the following pairs of integers and verify that LCM×HCF = product of the two numbers.

(i) 26 and 91          (ii) 510 and 92          (iii) 336 and 54

Solution – (i) 26 and 91

By prime factorisation method,

LCM of 26 and 91,

Prime factors of 26 = 2×13

Prime factors of 91 = 7×13

Therefore, LCM of 26 and 91 = 2×7×13 = 182

HCF of 26 and 91,                                                                                                                                             

Prime factors of 26 = 2×13

Prime factors of 91 = 7×13

Therefore, HCF of 26 and 91 = 13

Now, verification of LCM×HCF = product of the two numbers

By putting the values from above,

182×13 = 26×91

2366 = 2366

LHS = RHS

(ii) 510 and 92

By prime factorisation method,

LCM of 510 and 92,

Prime factors of 510 = 2×3×5×17

Prime factors of 92 = 2×2×23 = 2²×23

Therefore, LCM of 510 and 92 = 3×5×17×2²×23 = 23460

HCF of 510 and 92,

Prime factors of 510 = 2×3×5×17

Prime factors of 92 = 2×2×23

Therefore, HCF of 510 and 92 = 2

Now, verification of LCM×HCF = product of the two numbers

By putting the values from above,

23460×2 = 510×92

46920 = 46920

LHS = RHS

(iii) 336 and 54

By prime factorisation method,

LCM of 336 and 54,

Prime factors of 336 = 2×2×2×2×3×7 = 2⁴×3×7

Prime factors of 54 = 2×3×3×3 = 2×3³

Therefore, LCM of 336 and 54 = 2⁴×7×3³ = 3024

HCF of 336 and 54,

Prime factors of 336 = 2×2×2×2×3×7

Prime factors of 54 = 2×3×3×3

Therefore, HCF of 336 and 54 = 2×3 = 6

Now, verification of LCM×HCF = product of the two numbers

By putting the values from above,

3024×6 = 336×54

18144 = 18144

LHS = RHS

Q (3) Find the LCM and HCF of the following integers by applying the prime factorisation method.

(i) 12, 15 and 21          (ii) 17, 23 and 29          (iii) 8, 9 and 25

Solution – (i) 12, 15 and 21

By prime factorisation method,

Prime factors of 12 = 2×2×3

Prime factors of 15 = 3×5

Prime factors of 21 = 3×7

LCM of 12, 15 and 21 = 2×2×3×5×7 = 420

HCF of 12, 15 and 21 = 3                                  Ans.

(ii) 17, 23 and 29

By prime factorisation method,

Prime factors of 17 = 17×1

Prime factors of 23 = 23×1

Prime factors of 29 = 29×1

LCM of 17, 23 and 29 = 17×23×29 = 11339

HCF of 17, 23 and 29 = 1                                     Ans.

(iii) 8, 9 and 25

By prime factorisation method,

Prime factors of 8 = 2×2×2

Prime factors of 9 = 3×3

Prime factors of 25 = 5×5

LCM of 8, 9 and 25 = 2×2×2×3×3×5×5 = 1800

HCF of 8, 9 and 25 = 1                                              Ans.

Q (4) Given that HCF (306, 657) = 9, find LCM (306, 657).

Solution – As given in the question, two numbers are 306 and 657.

We know that, LCM×HCF = product of the two numbers

By putting the values from the question,

LCM×9 = 306×657

LCM = 306×657 / 9

LCM = 306×73

LCM = 22338

Therefore, LCM (306, 657) = 22338                Ans.

Q (5) Check whether 6ⁿ can end with the digit 0 for any natural number n.

Solution – First, we suppose that for any natural number n (n ϵ N), 6ⁿ ends with the digit 0. It means 6ⁿ will be divisible by 5 and 10.

But prime factors of 6 = 2×3

So, prime factors of 6ⁿ will be, 6ⁿ = (2×3)ⁿ

In the prime factors of 6ⁿ, we can see that the number 5 is not present.

By the fundamental theorem of arithmetic, we know that every composite number can be factorised as a product of unique prime numbers.

The above explanation shows that our assumption is wrong in the beginning.

Hence, there is not any natural number n for which 6ⁿ ends with the digit 0.             Ans.

Q (6) Explain why 7×11×13 + 13 and 7×6×5×4×3×2×1 + 5 are composite numbers.

Solution – According to the question,

The first number is 7×11×13 + 13

By taking 13 as a common factor,

13 (7×11 + 1)

13 (77 + 1)

13 (78)

13×78                                                                                                                                             

Since 13 is a factor of this number.

Therefore, it is a composite number.

The second number is 7×6×5×4×3×2×1 + 5

By taking 5 as a common factor,

5 (7×6×4×3×2×1 + 1)

5 (1008 + 1)

5 (1009)

5×1009

Since 5 is a factor of this number.

Therefore, it is a composite number.                Ans.

Q (7) There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes of the same. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet again at the starting point?

Solution – According to the question,

Time taken by Sonia to drive one round of the field = 18 minutes

Time taken by Ravi to drive one round of the same field = 12 minutes

Now, we have two numbers, 18 and 12. So, we’ll find the LCM of these numbers to find after how many minutes will they meet again at the starting point.

By prime factorisation method,

Prime factors of 18 = 2×3×3

Prime factors of 12 = 2×2×3

LCM of 18 and 12 = 2×2×3×3 = 36

Therefore, Sonia and Ravi will meet again at the starting point after 36 minutes.               Ans.

Exercise 1.3

Q (1) Prove that √5 is irrational.

Solution – Let √5 be a rational number.

So, for integers a and b we can write,     

√5 = a/b    where b ≠ 0 and a, b = coprime numbers [not any common factor other than 1]

√5b = a

By squaring both sides

(√5b)² = a²                                                                                                                                                                                                                             

5b² = a²   

Or    b² = a²/5 

it means 5 divides a² therefore, 5 also divides a [by theorem] ………………..(1)

So, we can write for any integer c, a = 5c    …………………….(2)

Putting the value of a from equation (2) to (1)

We get,     b² = (5c)²/5

b² = 25c²/5

b² = 5c²

Or   b²/5 = c² 

it means 5 divides b² therefore, 5 also divides b [by theorem] ………………(3)

From equations (1) and (3), 5 divides a and 5 divides b this means a and b have at least 5 as a common factor.

But this contradicts (opposes) the fact that a and b have no common factor other than 1.

It shows that our assumption that √5 is a rational number is wrong.

So, we conclude that √5 is an irrational number.            Hence Proved.

Q (2) Prove that 3 + 2√5 is irrational.

Solution – Let 3 + 2√5 be a rational number.

So, for integers a and b,      

3 + 2√5 = a/b where b ≠ 0 and a, b = coprime numbers [not any common factor other than 1]

2√5 = a/b – 3

2√5 = (a – 3b)/b

√5 = (a – 3b)/2b …………………….(1)

Since a, b, 3, and 2 are integers, so (a – 3b)/2b is a rational number.

So, from equation (1), √5 will be a rational number.

But this contradicts the fact that √5 is an irrational number.

It shows that our assumption that 3 + 2√5 is a rational number is wrong.

So, we conclude that 3 + 2√5 is irrational.             Hence Proved.

Q (3) Prove that the following are irrationals:

(i) 1/√2          (ii) 7√5          (iii) 6 + √2

Solution – (i) 1/√2                                                                                                                                             

Let 1/√2 be a rational number.

We can also write, 1/√2 = 1/√2 × √2/√2 = √2/2

So, for integers a and b,

√2/2 = a/b where b ≠ 0 and a, b = coprime numbers [not any common factor other than 1]

√2 = 2a/b ………………………(1)

Since a, b and 2 are integers, so 2a/b is a rational number.

So, from equation (1), √2 will be a rational number.

But this contradicts the fact that √2 is an irrational number.

It shows that our assumption that 1/√2 is a rational number is wrong.

So, we conclude that 1/√2 is irrational.             Hence Proved.

(ii) 7√5

Let 7√5 be a rational number.

So, for integers a and b,      

7√5 = a/b where b ≠ 0 and a, b = coprime numbers

√5 = a/7b …………………….(1)

Since a, b, and 7 are integers, so a/7b is a rational number.

So, from equation (1), √5 will be a rational number.

But this contradicts the fact that √5 is an irrational number.

It shows that our assumption that 7√5 is a rational number is wrong.

So, we conclude that 7√5 is irrational.               Hence Proved.

(iii) 6 + √2

Let 6 + √2 be a rational number.

So, for integers a and b,      

6 + √2 = a/b where b ≠ 0 and a, b = coprime numbers [not any common factor other than 1]

√2 = a/b – 6

√2 = (a – 6b)/b …………………….(1)

Since a, b, and 6 are integers, so (a – 6b)/b is a rational number.

So, from equation (1), √2 will be a rational number.

But this contradicts the fact that √2 is an irrational number.

It shows that our assumption that 6 + √2 is a rational number is wrong.

So, we conclude that 6 + √2 is irrational.              Hence Proved.

Exercise 1.4

Q (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:

(i) 13/3125               (ii) 17/8               (iii) 64/455               (iv) 15/1600

(v) 29/343                (vi) 23/2³5²         (vii) 129/2²5⁷7⁵        (viii) 6/15

(ix) 35/50                 (x) 77/210

Solution – (i) 13/3125

To find the decimal expansion, first, we find the prime factors of denominator 3125.

Prime factors of 3125 = 5×5×5×5×5

Or we can write, 3125 = 5⁵×2⁰

We can see that the prime factors of the denominator 3125 are of the form 2ⁿ×5^m, where n = 0 and m = 5 are non-negative integers.

Therefore, the decimal expansion of 13/3125 is terminating.             Ans.

(ii) 17/8

Prime factors of the denominator 8 = 2×2×2

Or we can write, 8 = 2³×5⁰

The prime factors of the denominator 8 are of the form 2ⁿ×5^m, where n = 3 and m = 0 are non-negative integers.

Therefore, the decimal expansion of 17/8 is terminating.             Ans.

(iii) 64/455

Prime factors of the denominator 455 = 5×7×13

The prime factors of the denominator 455 are not of the form 2ⁿ×5^m because it contains prime factors 7 and 13 also.

Therefore, the decimal expansion of 64/455 is non-terminating repeating.             Ans.

(iv) 15/1600

Prime factors of the denominator 1600 = 2×2×2×2×2×2×5×5

Or we can write, 1600 = 2⁶×5²

The prime factors of the denominator 1600 are of the form 2ⁿ×5^m, where n = 6 and m = 2 are non-negative integers.

Therefore, the decimal expansion of 15/1600 is terminating.             Ans.

(v) 29/343

Prime factors of the denominator 343 = 7×7×7

The prime factors of the denominator 343 are not of the form 2ⁿ×5^m because it contains only prime factor 7.

Therefore, the decimal expansion of 29/343 is non-terminating repeating.             Ans.

(vi) 23/2³5²

Prime factors of the denominator 2³5² = 2³×5²

The prime factors of the denominator 2³5² are already of the form 2ⁿ×5^m, where n = 3 and m = 2 are non-negative integers.

Therefore, the decimal expansion of 23/2³5² is terminating.             Ans.

(vii) 129/2²5⁷7⁵

Prime factors of the denominator 2²5⁷7⁵ = 2²×5⁷×7⁵

The prime factors of the denominator 2²5⁷7⁵ are not of the form 2ⁿ×5^m because it contains prime factor 7 also.

Therefore, the decimal expansion of 129/2²5⁷7⁵ is non-terminating repeating.             Ans.

(viii) 6/15

We can also write, 6/15 = 3/5

Prime factors of the denominator 5 = 2⁰×5¹

The prime factors of the denominator 5 are of the form 2ⁿ×5^m, where n = 0 and m = 1 are non-negative integers.

Therefore, the decimal expansion of 6/15 is terminating.             Ans.

(ix) 35/50

We can also write, 35/50 = 7/10

Prime factors of the denominator 10 = 2×5

Or we can write, 10 = 2¹×5¹

The prime factors of the denominator 10 are of the form 2ⁿ×5^m, where n = 1 and m = 1 are non-negative integers.

Therefore, the decimal expansion of 35/50 is terminating.             Ans.

(x) 77/210

We can also write, 77/210 = 11/30

Prime factors of the denominator 30 = 2×3×5

The prime factors of the denominator 30 are not of the form 2ⁿ×5^m

because it contains prime factor 3 also.

Therefore, the decimal expansion of 77/210 is non-terminating repeating.             Ans.

Q (2) Write down the decimal expansions of those rational numbers in question (1) above which have terminating decimal expansions.

Solution – (i) 13/3125

According to question (1), we know that 13/3125 = 13/5⁵×2⁰ ………….(1)

Multiplying and dividing equation (1) by 2⁵ to make the denominator a power of 10.

13/5⁵×2⁰ × 2⁵/2⁵

13×2⁵ / 5⁵×2⁵       [⸪ 2⁰ = 1]

13×32 / (5×2)⁵

416 / 10⁵

416 / 100000

0.00416

Therefore, the decimal expansion of 13/3125 is 0.00416.             Ans.

(ii) 17/8

According to question (1), we know that 17/8 = 17/5⁰×2³ ………….(1)

Multiplying and dividing equation (1) by 5³ to make the denominator a power of 10.

17/5⁰×2³ × 5³/5³

17×5³ / 2³×5³       [⸪ 5⁰ = 1]

17×125 / (2×5)³

2125 / 10³

2125 / 1000

2.125

Therefore, the decimal expansion of 17/8 is 2.125.             Ans.

(iv) 15/1600

According to question (1), we know that 15/1600 = 15/2⁶×5² ………….(1)

Multiplying and dividing equation (1) by 5⁴ to make the denominator a power of 10.

15/2⁶×5² × 5⁴/5⁴

15×5⁴ / 2⁶×5²×5⁴

15×625 / 2⁶×5⁶      [⸪ a^m×aⁿ = a^m+n]

9375 / (2×5)⁶                                                                                                                                                                                                  

9375 / 10⁶

9375 / 1000000

0.009375

Therefore, the decimal expansion of 15/1600 is 0.009375.             Ans.

(vi) 23/2³5²

According to question (1), we know that 23/2³5² = 23/2³×5² ………….(1)

Multiplying and dividing equation (1) by 5 to make the denominator a power of 10.

23/2³5² × 5/5

23×5 / 2³×5²×5¹

115 / 2³×5³      [⸪ a^m×aⁿ = a^m+n]

115 / (2×5)³

115 / 10³

115 / 1000

0.115

Therefore, the decimal expansion of 23/2³5² is 0.115.             Ans.

(viii) 6/15

According to question (1), we know that 6/15 = 2/5 = 2/5×2⁰ ………….(1)

Multiplying and dividing equation (1) by 2 to make the denominator a power of 10.

2/5×2⁰ × 2/2

2×2 / 5×2       [⸪ 2⁰ = 1]

4 / 10

0.4

Therefore, the decimal expansion of 6/15 is 0.4.             Ans.

(ix) 35/50

According to question (1), we know that 35/50 = 7/10 = 7/2×5 ………….(1)

The denominator of equation (1) is already a power of 10.

7/2×5

7 / 10

0.7

Therefore, the decimal expansion of 35/50 is 0.7.             Ans.

Q (3) The following real numbers have decimal expansions as given below.                                                                                                                                             

In each case, decide whether they are rational or not. If they are rational,

and of the form p/q, what can you say about the prime factors of q?

(i) 43.123456789               (ii) 0.120120012000120000…               (iii) 43.1̅2̅3̅4̅5̅6̅7̅8̅9̅

Solution – (i) 43.123456789

Let x = 43.123456789

x = 43123456789 / 1000000000

x = 43123456789 / 10⁹

x = 43123456789 / (2×5)⁹

x = 43123456789 / 2⁹×5⁹

The given number is in the form p/q, so it is a rational number. The prime factors of the denominator q will be in the form of 2ⁿ×5^m.               Ans.

(ii) 0.120120012000120000…

As we can see that the decimal expansion of the given number is non-terminating non-recurring so we cannot write this number in the form of p/q.

Therefore, this number is not a rational number.

This number is an irrational number.                 Ans.

(iii) 43.1̅2̅3̅4̅5̅6̅7̅8̅9̅

Let x = 43.1̅2̅3̅4̅5̅6̅7̅8̅9̅

Or we can also write, x = 43.123456789123456789…   …………(1)

Multiplying both sides of the equation (1) by 10⁹ because 9 digits are repeating.

10⁹x = 43.123456789123456789…×10⁹

10⁹x = 43123456789.123456789…   …………(2)

Subtracting equation (1) from equation (2)

10⁹x – x = 43123456789.123456789… – 43.123456789…

(10⁹ – 1)x = 43123456746

(1000000000 – 1)x = 43123456746

999999999x = 43123456746

x = 43123456746 / 999999999

dividing by 9

x = 4791495194 / 111111111

The given number is in the form of p/q, so it is a rational number.                                                                                                                                             

The prime factors of the denominator q will not be in the form of 2ⁿ×5^m.       Ans.

NCERT Maths Class 10th Chapter 1st Solution in Hindi

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