NCERT Exemplar Class 10 Maths Chapter 1 Real Numbers

Real Numbers Chapter 1: NCERT Exemplar for Class 10

The NCERT Exemplar Class 10 Maths Chapter 1 Real Numbers is provided here for students to practise and prepare for the CBSE first- and second-term exams. Simplyacad provides free pdf access to ncert exemplar pdf created by our experts as per the CBSE guidelines (2024-2025). This material will help students to revise the syllabus of the Real Numbers chapter and score numbers with flying colors in their board exams.

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NCERT Solutions of Class 10 Chapter 1 – Real Numbers Exercise 1.1

Question 1. For some integer x, every even integer is of the form.

(a) x

(b) x + 1

(c) 2x

(d) 2x + 1

Answer 1: (C) 2x

Explanation: An integer is called even if it is divided by 2. Let m be an integer.

We get, x = …,-6, −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, 6 …

so x can be even or odd.

Therefore, ‘x’ cannot always be even.

Now, x + 1 = …, -5,−4, −3, −2, −1, 0, 1, 2, 3, 4, 7 …

We observe that ‘x+1’ cannot always be even.

Now consider 2x =…, -12,−10, −8, −6, −4, −2, 0, 2, 4, 6, 8, 10, …

We observe that ‘2x’ is always even.

Consider 2x + 1 = …, -11, −9, −7, −5, −3, −1, 1, 3, 5, 7, 9, 11, ….

We observe that ‘2x+1’ is always an odd integer.

So, ‘2x’ is the answer.

Therefore, the correct answer is option C.

Question 2. The decimal form of 129/225775 is

(a) terminating

(b) non-terminating

(c) non-terminating non-repeating

(d) none of the above

Answer 2 : (c) non-terminating non-repeating

Question 3. The decimal expansion of 22/7 is

(a) Terminating

(b) Non-terminating and repeating

(c) Non-terminating and Non-repeating

(d) None of the above

Answer 3 : (b) Non-terminating and repeating

Explanation: 22/7 = 3.14285714286.

Question 4. For some integer m, every odd integer is of the form

(a) m

(b) m + 1

(c) 2m

(d) 2m + 1

Answer 4 : (D) 2m + 1

Explanation: An integer is called odd if it is not divided by 2. Let m be an integer i.e m = …, −2, −1, 0, 1, 2, …

Multiplication of both sides by 2 gives, 2m = …,-6, −4, −2, 0, 2, 4, 6 …

Adding 1 on both sides, 2m + 1 = …,-5, −3, −1, 1, 3, 5, 7 …

Hence, for any integer m, (2m +1) is always odd.

Therefore, the correct answer is option D.

 

Question 5. For some integer a, the odd integer is represented in the form of:

(a) a

(b) a+ 1

(c) 2a + 1

(d) 2a

Answer 5 : (c) 2a + 1

Explanation: Since 2a represents even numbers, hence 2a + 1 will always represent odd numbers. Suppose if n = 2, then 2a = 4 and 2a + 1 = 5.

Question 6. n 2 – 1 is divisible by 8, if n is

(a) an integer

(b) a natural number

(c) an odd integer

(d) an even integer

Answer 6 : (C) an odd integer

Explanation: Let a = n 2 − 1. Here, n can be odd or even.

Case 1: When n = even. Let n = 4k, where k is an integer.

⇒a=(4k) 2-1

⇒a=16k 2-1

which is not divisible by 8.

Case 2: When n = odd. Let n = 4k + 1, where k is an integer.

⇒a=(4k + 1) 2-1

⇒a=16k 2+1+8k-1⇒a=16k 2+8k

⇒a=8(2k 2+k)

which is always divisible by 8. Thus, if n is odd, then n 2-1 is divisible by 8.

Hence, the correct answer is option C.

 

Question 7. HCF of 26 and 91 are:

(a) 15

(b) 13

(c) 19

(d) 11

Answer 7 : (b) 13

Explanation:

Let us do the prime factorisation of 26 and 91;

26 = 2 x 13

91 = 7 x 13

Hence, HCF (26, 91) = 13

Question 8. If HCF of 65 and 117 is expressible in form 65m– 117, then the value of m is

(a) 4

(b) 2

(c) 1

(d) 3

Answer 8 : (b) 2

Explanation: By Euclid’s division lemma, b = aq + r, here 0 ≤r<a

117=65×1+52

65=52×1+13

52=13×4+0

∴ HCF(65,117) = 13 …..(1)

We observe that HCF (65,117) = 65m − 117 …..(2)

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

65m−117=13

65m=130

m=2

Therefore, the right answer is option (b).

 

Question 9. Which of the following is not irrational?

(a) (3 + √7)

(b) (3 – √7)

(c) (3 + √7) (3 – √7)

(d) 3√7

Answer 9 : (c) (3 + √7) (3 – √7)

Explanation: When we solve (3 + √7) (3 – √7), we would get;

(3 + √7) (3 – √7) = 3*3 – (√7)*(√7) = 9 – 7 = 2 [using a2 – b2 = (a – b) (a + b)]

 

Question 10: The largest number, which divides 70 and 125, leaving remainders 5 and 8 respectively are

(a) 13

(b) 65

(c) 875

(d) 1750

Answer 10 : (a) 13

Explanation: Firstly, we subtract the remainder 5 and 8, respectively, from corresponding numbers and then get the HCF of the resulting numbers using Euclid’s division lemma to get the associated number. After removing these remainders from the numbers, we have:

(70 − 5) = 65

(125 − 8) = 117

Now, the required number is HCF(65,117).

Using Euclid’s division algorithm,

117=65×1+52

65=52×1+13

52=13×4+0

∴ HCF = 13. Thus, 13 is the largest number which is divisible by 70 and 125 leaving remainder 5 and 8.

Hence, the correct answer is option (a) .

Question 11. The addition of a rational number and an irrational number is equal to:

(a) a rational number

(b) Irrational number

(c) Both

(d) None of the above

Answer 11 : (b) Irrational number

Explanation: The addition of a rational number and an irrational number equals an irrational number.

Question 12. If two positive integers ‘a’ as well as ‘b’ are written as a = x3y2 and b = xy3; x, y are prime numbers, then HCF (a, b) is

(a) xy

(b) xy2

(c) x3y3

(d) x2y2

Answer 12 : (b) xy2

Explanation: Given: a= x3y2 = x×x×x×y×y b= xy3= x×y×y×y

HCF has full-form as the highest common factor of two or more related numbers.

Here, HCF(a, b) = HCF(x3y3, xy3) = x×y×y = xy2

Hence, the correct answer is option B.

Question 13. The multiplication of two irrational numbers is:

(a) irrational number

(b) rational number

(c) Maybe rational or irrational

(d) None

Answer 13 : (c) Maybe rational or irrational

Explanation: The multiplication of two irrational numbers is maybe rational or irrational.

Question 14. If two positive integers ‘p’ as well as ‘q’ can be expressed as p = ab2 and q = a3b; a, b being prime numbers, then LCM (p, q) is

(a) ab

(b) a2b3

(c) a3b2

(d) a3b3

Answer 14 : (c) a3b2

Explanation: Given that,

p=ab2=a×b×b

q=a3b=a×a×a×b

The LCM has the full-form as the smallest common multiple that can be divided by both the numbers.

Thus,

LCM(p, q)

= LCM (ab2, a3b)

Hence,

=a×b×b×a×a =a3b2

Therefore, the right answer is option (c)

 

Question 15. If the set A = {1, 2, 3, 4, 5, 6, 7…} is given, then it represents:

(a) Whole numbers

(b) Rational Numbers

(c) Natural numbers

(d) Complex numbers

Answer 15 : (c) Natural numbers

Explanation: If set A = {1, 2, 3, 4, 5, 6, 7…} is given, then it represents natural numbers.

Question 16. The product of an irrational number and a non-zero rational is

(a) it’s always irrational

(b) it’s always rational

(c) rational or irrational

(d) one

Answer 16 : (a) always irrational

Explanation: Consider the rational number as 7/2 and the irrational number as √5/2

Their product is given as:

7/2×√5/2=7√5/4, which is irrational.

Hence, the correct answer is option A.

Question 17. If a and b are integers and is represented in the form of a/b, then it is a:

(a) Whole number

(b) Rational number

(c) Natural number

(d) Even number

Answer 17 : (b) Rational number

Explanation: If a and b are integers and are represented in the form of a/b, then it is a rational number.

 

Question 18. The least number which is divisible by all the numbers from 1 to 10 (both 1 and 10 inclusive) is

(a) 10

(b) 100

(c) 504

(d) 2520

Answer 18 : (d) 2520

Explanation: The LCM is the smallest positive integer that is divisible by the given numbers.

Here, we need to find the LCM that is the lowest common multiple of the numbers starting from 1 to 10 (both inclusive). Factors of 1 to 10 numbers:

1=1

2=1×2

3=1×3

4=1×2×2

5=1×5

6=1×2×3

7=1×7

8=1×2×2×2

9=1×3×3

10=1×2×5

Therefore, LCM of numbers from 1 to 10 = LCM (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

Hence, LCM = 1×2×2×2×3×3×5×7=2520

Therefore, the right answer is option (d)

Question 19. The largest number that divides 70 and 125, and which leaves the remainders 5 and 8, is:

(a) 65

(b) 15

(c) 13

(d) 25

Answer 19 : (c) 13

Explanation:

Step 1: 70 – 5 = 65 and 125 – 8 = 117

Step 2: HCF (65, 117) is the largest number that divides 70 and 125 and leaves the remainder 5 and 8.

HCF (65, 117) = 13

Question 20. The decimal expansion of rational number 14587/1250 will terminate after:

(a) one decimal place

(b) two decimal places

(c) three decimal places

(d) four decimal places

Answer 20 : (d) four decimal places

Explanation: For terminating decimal expansion of a rational number, the denominator is of the form 2m×5n.

Here, 14587/1250=14587/21×54

⇒[14587/10×53]×[23/23\ = [14587×8]/[10×1000] = 116696/10000=11.6696

Hence, the decimal expansion of the given rational number terminates after four decimal places.

Thus, the right answer is option (d).

Question 21. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = the product of the two numbers.

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

Answer 21:

(i) 26 and 91

Explanation:

Expressing 26 and 91 as products of the prime factors, we observe,

26 = 2 × 13 × 1

91 = 7 × 13 × 1

Hence, LCM (26, 91) = 2 × 7 × 13 × 1 = 182

And HCF (26, 91) = 13

Verification

Now, multiplication of 26 and 91 = 26 × 91 = 2366

And multiplication of LCM and HCF = 182 × 13 = 2366

Therefore, LCM × HCF = product of the 26 and 91.

(ii) 510 and 92

Expressing 510 and 92 as products of the prime factors, we observe,

510 = 2 × 3 × 17 × 5 × 1

92 = 2 × 2 × 23 × 1

Hence, LCM(510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460

And HCF (510, 92) = 2

Verification

Now, multiplication of 510 and 92 = 510 × 92 = 46920

And multiplication of LCM and HCF = 23460 × 2 = 46920

Therefore, LCM × HCF = product of the 510 and 92.

(iii) 336 and 54

Expressing 336 and 54 as products of the prime factors,

we observe, 336 = 2 × 2 × 2 × 2 × 7 × 3 × 1 and, 54 = 2 × 3 × 3 × 3 × 1

Hence, LCM(336, 54) = 3024

And HCF (336, 54) = 2×3 = 6

Verification

Now, multiplication of 336 and 54 = 336 × 54 = 18,144

And multiplication of LCM and HCF = 3024 × 6 = 18,144

Therefore, LCM × HCF = product of 336 and 54.

Question 22: Express each number as a product of its prime factors:

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Answer 22:

(i) 140

Explanation:

With the help of division of the number by the prime numbers method, we observe the multiple of prime factors of 140

hence,

140 = 2 × 2 × 5 × 7 × 1 = 22 × 5 × 7

(ii) 156

With the help of division of the number by the prime numbers method, we observe the multiple of prime factors of 156

here, 156 = 2 × 2 × 13 × 3 = 22 × 13 × 3

(iii) 3825

With the help of division of the number by the prime numbers method, we observe the multiple of prime factors of 3825

here, 3825 = 3 × 3 × 5 × 5 × 17 = 32 × 52 × 17

(iv) 5005

With the help of division of the number by the prime numbers method, we observe the multiple of prime factors of 5005

here, 5005 = 5 × 7 × 11 × 13 = 5 × 7 × 11 × 13

(v) 7429

With the help of division of the number by the prime numbers method, we observe the multiple of prime factors of 7429

here, 7429 = 17 × 19 × 23 = 17 × 19 × 23

Question 23. 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

Answer 23:

Note: If the denominator of a number has only factors of 2 and 5 in the form of 2m ×5n, then it shows terminating decimal expansion.

If the denominator of a number has factors not 2 and 5 then it shows a non-terminating decimal expansion.

(i) 13/3125

When we factorise the denominator, we observe,

3125 = 5 × 5 × 5 × 5 × 5 = 55

Then, the denominator has only 5 as its factor, 13/3125 has a terminating decimal expansion.

(ii) 17/8

When we factorise the denominator, we observe,

8 = 2×2×2 = 23

then, the denominator has only 2 as its factor, 17/8 has a terminating decimal expansion.

(iii) 64/455

When we factorise the denominator, we observe,

455 = 5×7×13

then, the denominator is not in the form of 2m × 5n, thus 64/455 has a non-terminating decimal expansion.

(iv) 15/ 1600

When we factorise the denominator, we observe,

1600 = 26×52

then, the denominator is in the form of 2m × 5n, hence 15/1600 shows a terminating decimal expansion.

(v) 29/343

When we factorise the denominator, we observe,

343 = 7×7×7 = 73

then, the denominator is not in the form of 2m × 5n thus 29/343 has a non-terminating decimal expansion.

(vi)23/(2352)

It is observed that the denominator is in the form of 2m × 5n.

So, 23/ (2352) shows a terminating decimal expansion.

(vii) 129/(225775)

As you can observe, the denominator is not in the form of 2m × 5n.

Therefore, 129/ (225775) shows a non-terminating decimal expansion.

(viii) 6/15

6/15 = ⅖

Then, the denominator has only 5 as its factor, thus, 6/15 shows a terminating decimal expansion.

(ix) 35/50

35/50 = 7/10

When we factorise the denominator, we observe

10 = 2 × 5

Then, the denominator is in the form of 2m × 5n thus, 35/50 shows a terminating decimal expansion.

(x) 77/210

77/210

Then,

= (7× 11)/ (30 × 7)

= 11/30

When we factorise the denominator, we observe,

30 = 2 × 3 × 5

As you can observe, the denominator is not in the form of 2m × 5n. Therefore, 77/210 has a non-terminating decimal expansion.

Question 24. Given that HCF (306, 657) = 9, find LCM (306, 657).

Answer 24:

You are aware about,

HCF × LCM = Multiplication of the two given numbers

So,

9 × LCM = 306 × 657

LCM = (306 × 657)/9 = 22338

Hence, LCM (306,657) = 22338

Question 25. Prove that the following are irrationals:

(i) 1/√2

(ii) 7√5

(iii) 6 + √2

Answer 25:

(i) 1/√2

Consider, 1/√2 is rational.

Then we can get co-prime x and y (y ≠ 0) such that 1/√2 = x/y

Rearranging, we observe,

√2 = y/x

Then, x and y are integers, so √2 is a rational number, which tells the fact that √2 is irrational.

Therefore, we can conclude that 1/√2 is irrational.

(ii) 7√5

Consider 7√5 is a rational number.

Then we can get co-prime a and b (b ≠ 0) such that 7√5 = x/y

Rearranging, we observe,

√5 = x/7y

Then, x and y are integers, thus, √5 is a rational number, which tells the fact that √5 is irrational.

Therefore, we can conclude that 7√5 is irrational.

(iii) 6 +√2

Consider 6 +√2 is a rational number.

Then we can get co-primes x and y (y ≠ 0) such that 6 +√2 = x/y⋅

Rearranging, we observe,

√2 = (x/y) – 6

Then, x and y are integers, thus (x/y) – 6 is a rational number and therefore, √2 is rational. This tells the fact that √2 is an irrational number.

Therefore, we can conclude that 6 +√2 is irrational

Question 26. Show that any positive odd integer is in the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Answer 26:

Consider any positive integer and b = 6. Hence, by Euclid’s algorithm, a = 6q + r, if some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5, due to 0≤r<6.

Now substituting the value of r, we observe,

If r = 0, then a = 6q

if, for r= 1, 2, 3, 4 and 5, we consider the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively.

for a = 6q, 6q+2, 6q+4, then a is even number and divided by 2. A positive integer can be even or not odd, hence, any positive odd integer is of the form of 6q+1, 6q+3 and 6q+5, where q is some integer.

Question 27. “The product of three consecutive positive integers is divisible by 6”. Is this statement true or false? Justify your answer.

Answer 27:

Yes, the statement is true “the product of three consecutive positive integers is divisible by 6”.

Justification:

we can consider the 3 consecutive numbers 2, 3, 4

(2 × 3 × 4)/6 = 24/6 = 4

Now, we can consider another 3 consecutive numbers 4, 5, 6

(4 × 5 × 6)/6 = 120/6 = 20

Now, we can consider another 3 consecutive numbers 7, 8, 9

(7 × 8 × 9)/6 = 504/6 = 84

Therefore, the statement “product of three consecutive positive integers is divisible by 6” is true.

Question 28. Check whether 6n can end with the digit 0 for any natural number n.

Answer 28:

If any number 6n ends with the digit zero (0), it will be divisible by 5, as we know any number with a unit place as 0 or 5 can be divided by 5.

Prime factorization of 6n = (2 × 3)n

Hence, the prime factorization of 6n doesn’t include the prime number 5.

Therefore, it is clear that for any natural number n, 6n is not divisible by 5 and thus it proves that 6n cannot end with the digit 0 for any natural number n.

 

Question 29. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Answer 29:

As per the definition of a composite number, we know that if the number is composite, it means it has factors one and itself. hence, for the expression;

7 × 11 × 13 + 13

Take 13 we observe as a common factor, we get,

=13(7×11×1+1) = 13(77+1) = 13×78 = 13×3×2×13

Hence, 7 × 11 × 13 + 13 is a composite number.

Now let’s take another number,

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

Take 5 we observe as a common factor, we get,

=5(7×6×4×3×2×1+1) = 5(1008+1) = 5×1009

so, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number.

 

Question 30. Write whether every positive integer can be of the form 4q + 2, where q is an integer. Justify your answer.

Answer 30:

No, each positive integer cannot take the form 4q + 2, where q is the integer.

clarification:

All the related numbers of the form 4q + 2, where ‘q’ is the integer, are even numbers which we can not divide by ‘4’.

As given example,

if q=1,

4q+2 = 4(1) + 2= 6.

if q=2,

4q+2 = 4(2) + 2= 10

if q=0,

4q+2 = 4(0) + 2= 2 etc.

Therefore, any number which has the form 4q+2 will give only even numbers which can not be multiplied by 4.

Therefore, every positive integer cannot be written in the form 4q+2

Question 31. Find HCF and LCM of 96 and 404 and then verify that HCF × LCM = Product of the two given numbers.

Answer 31:

Prime factors of 96 = 2 × 2 × 2 × 2 × 2 × 3 = 25 × 3

Prime factors of 404 = 2 × 2 × 101

HCF = 2 × 2 = 4

LCM = 25 × 3 × 101 = 9696

HCF × LCM = 4 × 9696 = 38784

Multiplication of the given two numbers = 404 × 96 = 38784

Therefore, verified that LCM × HCF = Product of the given two numbers.

 

Question 32. Prove that √5 is irrational.

Answer 32:

Consider that √5 is a rational number.

i.e. √5 = x/y (as well as, x and y are co-primes)

y√5= x

Squaring the both the sides, we observe,

(y√5)2 = x2

⇒5y2 = x2……………………………….. (1)

Therefore, x2 is divided by 5, so x is also divided by 5.

Consider, x = 5k, for some value of k and putting the value of x in equation (1), we observe,

5y2 = (5k)2

⇒y2 = 5k2

is divisible by 5 it means y is divisible by 5.

Clearly, x and y are not co-primes. Thus, our assumption about √5 is rational is incorrect.

Hence, √5 is an irrational number.

Question 33. A positive integer is of the form 3q + 1, q being a natural number. Can you write its square in any form other than 3m + 1, i.e., 3m or 3m + 2 for some integer m? Justify your answer.

Answer 33:

Answer is No.

clarification:

Let the positive integer 3q + 1, where q is the natural number.

(3q + 1)2 = 9q2 + 6q + 1

= 3(3q2 + 2q) + 1

= 3m + 1, (here, m is an integer which is equal to 3q2 + 2q.

hence (3q + 1)2 cannot be expressed in any other form apart from 3m + 1.

Question 34. Prove that 3 + 2√5 is irrational.

Answer 34:

Let us assume 3 + 2√5 to be a rational number.

if the co-primes x and y of the required rational number where (y ≠ 0) is such that:

3 + 2√5 = x/y

By rearranging, we get,

2√5 = (x/y) – 3

√5 = 1/2[(x/y) – 3]

Now x and y are integers, thus, 1/2[(x/y) – 3] is a rational number.