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# Class 10 Maths Chapter 1 Real Numbers

Class 10 Maths Chapter 1 Real Numbers – Here you can find the NCERT solutions of Class 10 Maths. All Chapter 1 Real Numbers questions are explained readibluly and beautifully by an expert maths teacher. While going through these class 10 maths NCERT solutions of chapter 1 we assure you, your interest in maths will be increased. In Chapter 1 Real Number you will learn Euclid’s division theorem and questions based on it. The purpose of introducing chapter 1 real number in NCERT maths class 10 is to aware of the properties of the integers. You can see the video for Solutions of class 10 Maths NCERT Solutions Chapter 1 updated for 2024-25,

Each question of chapter 1 real number is explained beautifully so that our readers can understand the concept of maths completely.

# Class 10 Maths Chapter 1 Real Numbers

### Exercise 1.1

Q1. Use Euclid’s division algorithm to find the HCF of :

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

Ans. (i) 135 and 225

Dividing 225 by 135, we can write 225 in the following form

225 = 135 × 1 + 90

Dividing 135 by 90, we can write the number 135 in the form

135 = 90 × 1 + 45

Dividing 90 by 45, we can write the number 90 in the following form

90 = 45 × 2 + 0

We ger r, remainder =0, therefore the HCF of the numbers (225, 135) is 45

(ii) 196 and 38220

Dividing the larger number by the smaller number, then we can write the larger number in the following form

38220 = 196 × 195 + 0

Since we have got r, remainder =0, therefore the divisor 196 is the HCF of (388220,196)

(iii) 867 and 255

Dividing 867 by 255, we can write 855 in the following form

867 = 255 × 3 + 102

255 = 102 × 2 + 51

102 = 51 × 2 + 0

So, 51 is the HCF of the given numbers

Q2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5 where q is some integers?

Ans.  According to Euclid’s division Lemma, any positive number ‘a’ can be written in the form of a= bq + r  where

0≤ r < b

Comparing bq + r with the given forms of the number, let b =6, so possible values of r are= 0,1,2,3,4,5

Therefore corresponding values of positive integers are = 6q, 6q +1, 6q +2, 6q +3, 6q + 4, 6q + 5

The numbers 6q +1, 6q +3 and 6q +5 are not divisible by 2

So, it has been proved that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5

Q3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to match the same number of columns. What is the maximum number of columns in which they can march?

Ans. Let’s find the maximum number of columns in which army band of members 616 to march behind 32 band members.

We have to find out the HCF of the (616, 32)

Using Euclid’s theorem we can write 616 in the following form

616 = 32 × 19 +  8

32 = 8 × 4 + 0

HCF of (616,32) is = 8

So, the maximum number of columns in which army members can march = 8

Q4. 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.

Ans.  According to Euclid’s theorem, any positive integer can be written in the form of a = bq + r  where 0≤ r < b

Comparing the number 3m or 3m +1, we have to prove with the standard form of the number a = bq +r

Let b = 3, then possible values of r, are 0,1,2

Corresponding values of positive integers are 3q or 3q +1 or 3q +2

Squaring the positive integers

(3q)² = 3 (3q²) = 3m, where m= 3q²,is another positive integer

(3q+1)² = (3q)² + 1²  + 6q = 9q²  + 6q +1 = 3(3q²  + 2q ) +1 = 3m +1, where m= 3q²  + 2q is another positive integer

Therefore it has been proved that the square of any positive integer can be written in the form of 3m or 3m+1

Q5. Use Euclid’s division Lemma to show that the cube of any positive integer is of the form 9m, 9m +1  or 9m +8.

Ans.  According to Euclid’s theorem, any positive integer can be written in the form of a = bq + r  where 0≤ r < b

Comparing the number 9m or 9m +1 or 9m +8, we have to prove with the standard form of the number a = bq +r

Let 9 = 3, then possible values of r, are 0,1,2,3,4,5,6,7,8

Corresponding values of positive integers are 9q or 9q +1 or 9q +2 ……9q +8

Cubing the positive integers

(9q)³ = 9(81q³) = 9m , where m = 81q³ is another positive integer

(9q +1)³ = 729q³ + 1 + 3×(9q)²×1 + 3×1²× 9q = 729q³ +243q²+27q +1 = 9(81q³+27q²+3q) + 1= 9m+1, where m=81q³+27q²+3q   is another positive integer

(9q +2)³ = (9q)³ + 2³ + 3× (9q)²× 2 + 3× 2³× 9q = 9(81q³ +27q² +216q) +8= 9m +8, where m= 81q³ +27q² +216q, is another positive integer

Hence it has been proved that cube of any positive integer can be written in the form of 9m or 9m +1 or 9m +8

### Exercise 1.2

Q1. Express each number as a product of its prime factors.

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Ans. (i) 140= 2 × 70 = 2 ×2 × 35 = 2×2×5× 7 = 2² × 5 × 7

(ii) 156 = 2 × 78 = 2× 2× 39 = 2×2×3 ×13 = 2²× 3 × 13

(iii)3825 = 5 ×765 = 5×5  ×153 = 5 × 5 × 3 ×51 = 5 × 5 × 3 × 3× 317 = 5² × 3² × 17

(iv) 5005 = 5 × 1001 = 5 × 7 × 143 = 5 × 7 × 11 × 13

(v) 7429 = 17 × 19 × 23

Q2. Find the LCM and HCF of the following pairs of integers and verify that LCM×HCF=Product of two numbers.

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

Ans. (i) 26 and 91

26 =  2 × 13

91 = 7 × 13

HCF = 13

LCM = 2 × 13 × 7 = 182

LCM×HCF=Product of two numbers

182 × 13 = 26 × 91

2366 = 2366

Verified

(ii) 510 and 92

510 = 3 × 170 = 3 × 2 × 85 = 3 × 2 × 5 × 17

92 = 2 × 46 = 2² × 23

HCF(510 , 92) = 2

LCM(510 , 92) = 3 × 2² × 5 × 17 × 23  = 23460

HCF × LCM = Product of numbers

2 × 23460 = 510 × 92

46920 = 46920

LHS = RHS

Verified

(iii) 336 and 54

336 = 2 × 2 × 2 × 2× 3 × 7 =  2 ×3 × 7

54 = 2 × 27 = 2 × 3 × 3 × 3 = 2 × 3³

HCF (336,54) = 2 ×  3 = 6

LCM(336,54) = 24× 3³ × 7 = 16 × 27 × 7 = 3024

HCF × LCM =  Product of the numbers

6 × 3024 =  336 × 54

18144 = 18144

LHS = RHS

Verified

Q3. Find the LCM and HCF of the following integers by applying the prime factorization method.

(i) 12,15 and 21

(ii) 17,23 and 29

(iii) 8,9 and 25

Ans. (i) 12,15 and 21

12 = 2 × 2 × 3 = 2² ×  3

15 = 3 ×  5

21 = 3 ×  7

HCF(12,15 , 21) = 3

LCM(12,15 , 21) = 2² × 3 × 5 × 7 =420

(ii) 17,23 and 29

17 = 1 × 17

23 = 1 × 23

29 = 29 × 1

HCF(17,23,29) = 1

LCM(17,23,29) = 17 × 23 × 29 = 11339

(iii) 8, 9 and 25

8 = 2³ × 1

9 = 3² × 1

25 = 5² × 1

HCF(8,9 , 25) = 1

LCM(8,9 , 25) = 2³ × 3² ×  5² = 1800

Q4. Given that HCF(306,657)=9,find LCM(306,657)?

Ans. We are given that HCF(306,657) = 9

The relationship between the numbers, their HCF and LCM is as follows

HCF (306,657 ×  LCM  (306,657)  = Product of the numbers(306,657)

9 ×  LCM  (306,657) = 306 ×  657

Q5. Check whether 6n can end with the digit 0 for any natural number n?

Ans. Any number can end with the digits 0 if its prime factors are in the form of 2n × 5n

Since  6n = (2× 3)n = 2n × 3n

6n = (2× 3)n = 2n × 3n, the factors are not in the form of 2n × 5n

Therefore 6n can not end with the digit 0 for any natural number

Q6. Explain why 7×11×13+13 and 7×6×5×4×3×2×1+5 are composite numbers?

Ans. 7×11×13+13 = 13 (7 × 11 + 1) = 7 × 78 = 7 × 2 × 3 × 13

Since the number which has more than two factors or the number divisible by other numbers in addition to 1 and the number itself is known as a composite number

Therefore 7×11×13+13 is a composite number

7×6×5×4×3×2×1+5 = 5( 7 × 6 × 4 × 3 × 2 × 1 +1) = 7× 1009

It has the factors 7 and 1009, so 7×6×5×4×3×2×1+5 is a composite number

Q7. There is a circular path around the sports field. Sonia takes 18minutes to drive one round of the field, while Ravi takes 12 minutes for 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?

Ans. Sonia takes 18 minutes to complete one round of the field and Ravi takes 12 minutes to complete round of the field. Sonia and Ravi starts from the same point and goes in the same direction so they will meet again in the least time duration = LCM(18, 12)

18 = 2 × 3²

12 = 2²  × 3

LCM(18,12) = 2² × 3²  = 36

Therefore Sonia and Ravi will meet again after 36 minutes.

### Exercise 1.3

Q1. Prove that √5 is an irrational number?

Ans. Let √5  is a rational number

Where a and b are co-prime number ( study Number system)

b√5 = a

Squaring both sides

5 b² = a²………..(i)

5 is one of the factors of  a²

∴ 5 will also be one of the factors of a

Therefore a can be written as a multiple of 5

a = 5c ( where c is another positive integer)

Putting the value of ‘a’ in equation number (i)

5 b² = (5c)²  = 25c²

b² = 5c²…………(ii)

5 is one of the factors of b²

∴ 5 will also be one of the factors of b

It is evident from eq.(i) and (ii) that 5 is a common factor in ‘a’ and ‘b’, so  it is contrary to the fact we have supposed that √5  is a rational number

Therefore √5 is an irrational number.

Q2. Prove that 3 + 2√5 is an irrational number?

Ans.  Let 3 + 2√5  is a rational number

Irrational number = Rational number

This is impossible,  the fact we have assumed 3 + 2√5 a rational number is wrong, so 3 + 2√5 will be an irrational number.

Q3. Prove that following are irrational?

Where a and b are co-prime number

b = a√2

Irrational = Rational

It is impossible, so the given number 1/√2 is a rational number

### Exercise 1.4

Q1. Without actually performing the long division, state whether the following rational number will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

Hint: The decimal expansion of a/b is a terminating decimal if prime factors of b is in the form of  2m× 5

Where a and b are co-prime number

.

3125 = 5 × 625 = 5 × 5 × 5 × 5 = 20 × 54

The factor of the denominator is in the form of  2m× 5n

So, the given fraction is a finite decimal

8 = 2³ = 23 × 50

The factor of the denominator is in the form of 2m× 5n

So, the expansion of the given fraction is terminating decimal

455 = 5 × 91 = 5 ×  7 ×  13

The factors of the denominator are not in the form of 2m× 5 because 7 and 13 are one of the factors of the denominator   So, the expansion of the given fraction is a non-terminating decimal

3200 = 2 ×2×2 ×2 ×2 ×2 ×2 ×  5× 5 = 27× 5²

The factor of the denominator is in the form of 2m× 5n

Therefore the expansion of the given fraction is a finite decimal

343 = 7 × 49 = 7 ³

The factor of the denominator is not in the form of 2m× 5n

So, the expansion of the given fraction is the non-terminating decimal

The factors of the denominator are  in the form of  2m× 5n

So, its expansion is a terminating decimal

The factors of the denominator is not in the form 2m× 5n

Since 7 is also one of the factor of the denominator, therefore, the expansion of the given fractions is a non-terminating repeating decimal

5 = 20× 51

The factors of the denominator is  in the form 2m× 5n

So the decimal expansion of the given fraction is a terminating decimal expansion

The factors of the denominator is  in the form 2m× 5n

Therefore the decimal expansion of the given fraction is a terminating decimal.

210 = 3 ×  7 ×  2×  5

The factors of the denominator is not  in the form 2m× 5n

Since 3 and 7 are two of the factor of the denominator so the decimal expansion of the given fraction is a non-terminating decimal expansion

Q2. Write down the decimal expansion of those rational numbers in question 1, which have terminating decimal expansion.

Ans.

Q3.The following real numbers have decimal expansions as given below. In each case whether they are rational or not, if they are rational and of the form p/q, what can you say about the prime factor of q?

(i) 43.123456789

(ii) 0.1201200120001120000….

Ans. (i) The number 43.123456789 is a terminating decimal so it is a rational number. It can be written in the form of p/q as follows

The factor of the denominator,(1000000000) are = 109 = (2×5)9 = 29 × 59

The factors of the denominator(q) are in the form 0f 2m× 5n

(ii)  The number 0.1201200120001120000…. is  non-terminating non-repeating decimal so it is an irrational number

The factors of the denominator 33333333 can not be written in the form of 2m× 5n