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Class 10 Maths Chapter 1: Real Numbers worksheet with answer key

A 20-mark worksheet on Real Numbers for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.

Download the worksheet (PDF)Answer key (PDF)
Question 1
The HCF of 23 × 52 × 7 and 22 × 53 × 11 is
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Answer: (d) 100

Common primes are 2 and 5. Smallest powers: 22 and 52. HCF = 4 × 25 = 100. 7 and 11 are not common, so they do not enter.

Question 2
Which of the following is irrational?
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Answer: (c) √50

√196 = 14 and 2.5 = 5/2 are rational, and 22/7 is a ratio of integers. √50 = 5√2, and 5√2 cannot be written as p/q, so it is irrational.

Question 3
In the prime factorisation of 1764, the power of 7 is
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Answer: (c) 2

1764 = 2 × 882 = 2 × 2 × 441 = 22 × 32 × 49 = 22 × 32 × 72. The power of 7 is 2.

Question 4
The LCM of 11, 13 and 19 is
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Answer: (a) 2717

No two of 11, 13, 19 share a prime, so the LCM is the product 11 × 13 × 19 = 2717. 1 is their HCF.

Question 5
If a and b are two different primes, the LCM of a2b3 and a3b is
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Answer: (a) a3b3

LCM takes the greatest power of each prime involved: a3 and b3. a2b is the HCF; a5b4 is the product.

Question 6 · 2 marks
Find the HCF and LCM of 84 and 126 by prime factorisation.
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84 = 22 × 3 × 7, 126 = 2 × 32 × 7. HCF = 2 × 3 × 7 = 42. LCM = 22 × 32 × 7 = 252. Check: 42 × 252 = 10584 = 84 × 126.

  • both prime factorisations (1)
  • HCF 42 and LCM 252 (1)
Question 7 · 2 marks
Can 14n end with the digit 5 for any natural number n? Give a reason.
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A number ending in 5 is divisible by 5, so 5 must be in its prime factorisation. 14n = 2n × 7n. By the uniqueness of prime factorisation, its only primes are 2 and 7. So 14n never ends in 5.

  • ending in 5 needs the prime 5 (1)
  • 14n = 2n × 7n; uniqueness rules out 5, so no (1)
Question 8 · 2 marks
Find the least natural number that is divisible by every number from 1 to 10.
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The number must contain each prime up to 10 at its greatest power among 1 to 10: 23 (from 8), 32 (from 9), 5 and 7. So the least such number is 23 × 32 × 5 × 7 = 2520.

  • greatest prime powers 23, 32, 5, 7 (1)
  • 2520 (1)
Question 9 · 3 marks
Two numbers are in the ratio 3 : 7 and their LCM is 357. Find the numbers and their HCF.
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Let the numbers be 3k and 7k; 3 and 7 are coprime, so HCF = k and LCM = 21k. 21k = 357 gives k = 17. Numbers 51 and 119, HCF 17. Check: 17 × 357 = 6069 = 51 × 119.

  • writes 3k, 7k with LCM 21k (1)
  • k = 17, numbers 51 and 119 (1)
  • HCF 17 (1)
Question 10 · 3 marks
Given that √7 is irrational, prove that 3 − 2√75 is irrational.
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Suppose (3 − 2√7)/5 = a/b, with a, b integers, b ≠ 0. Then 3 − 2√7 = 5a/b, so 2√7 = 3 − 5a/b = (3b − 5a)/b and √7 = (3b − 5a)/(2b). The right side is a ratio of integers, so √7 would be rational. This contradicts the given fact. So (3 − 2√7)/5 is irrational.

  • assumes the number equals a/b (1)
  • isolates √7 = (3b − 5a)/(2b) (1)
  • notes the right side is rational, states the contradiction (1)
Question 11 · 3 marks
Find the HCF and LCM of 28, 42 and 70 by prime factorisation, and show that their product is not the product of the three numbers.
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28 = 22 × 7, 42 = 2 × 3 × 7, 70 = 2 × 5 × 7. HCF = 2 × 7 = 14. LCM = 22 × 3 × 5 × 7 = 420. HCF × LCM = 5880, but 28 × 42 × 70 = 82320. They differ: the product rule holds for two numbers only.

  • the three prime factorisations (1)
  • HCF 14 and LCM 420 (1)
  • 5880 ≠ 82320 (1)
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