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NCERT Solutions · Class 10 · Chapter 1: Real Numbers

NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1

Exercise 1.1: Prime factorisation, HCF and LCM. Fundamental Theorem of Arithmetic: every composite number is a product of primes in exactly one way. HCF = product of the smallest powers of common primes; LCM = product of the greatest powers of all primes.

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Exercise 1.1 questions and solutions

Exercise 1.1, Question 1

Write each number as a product of its prime factors.
(i) \(140\)
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  1. Divide by the smallest prime each time: \(140 \div 2 = 70\), \(70 \div 2 = 35\), \(35 \div 5 = 7\), and \(7\) is prime.
Answer: \(140 = 2^2 \times 5 \times 7\)
(ii) \(156\)
Show solution
  1. \(156 \div 2 = 78\), \(78 \div 2 = 39\), \(39 \div 3 = 13\), and \(13\) is prime.
Answer: \(156 = 2^2 \times 3 \times 13\)
(iii) \(3825\)
Show solution
  1. The digit sum is \(18\), so \(3825\) is divisible by \(9\): \(3825 \div 3 = 1275\), \(1275 \div 3 = 425\).
  2. It ends in 5: \(425 \div 5 = 85\), \(85 \div 5 = 17\), and \(17\) is prime.
Answer: \(3825 = 3^2 \times 5^2 \times 17\)
(iv) \(5005\)
Show solution
  1. \(5005 \div 5 = 1001\).
  2. \(1001 = 7 \times 143\) and \(143 = 11 \times 13\).
Answer: \(5005 = 5 \times 7 \times 11 \times 13\)
(v) \(7429\)
Show solution
  1. No factor 2, 3, 5, 7, 11 or 13 works. Try \(17\): \(7429 \div 17 = 437\).
  2. \(437 \div 19 = 23\), and \(23\) is prime.
Answer: \(7429 = 17 \times 19 \times 23\)

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Exercise 1.1, Question 2

Find the LCM and HCF of each pair, and check that LCM × HCF equals the product of the two numbers.
(i) \(26\) and \(91\)
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  1. \(26 = 2 \times 13\) and \(91 = 7 \times 13\).
  2. HCF: the only common prime is \(13\), so HCF \(= 13\).
  3. LCM: every prime to its highest power, \(2 \times 7 \times 13 = 182\).
  4. Check: \(13 \times 182 = 2366\) and \(26 \times 91 = 2366\). ✓
Answer: HCF \(= 13\), LCM \(= 182\); \(13 \times 182 = 26 \times 91 = 2366\)
(ii) \(510\) and \(92\)
Show solution
  1. \(510 = 2 \times 3 \times 5 \times 17\) and \(92 = 2^2 \times 23\).
  2. HCF: common prime \(2\) to the smaller power, so HCF \(= 2\).
  3. LCM \[= 2^2 \times 3 \times 5 \times 17 \times 23 = 23460\]
  4. Check: \(2 \times 23460 = 46920\) and \(510 \times 92 = 46920\). ✓
Answer: HCF \(= 2\), LCM \(= 23460\); product \(= 46920\) both ways
(iii) \(336\) and \(54\)
Show solution
  1. \(336 = 2^4 \times 3 \times 7\) and \(54 = 2 \times 3^3\).
  2. HCF \(= 2^1 \times 3^1 = 6\) (smallest powers of the common primes).
  3. LCM \(= 2^4 \times 3^3 \times 7 = 3024\) (greatest powers of all primes).
  4. Check: \(6 \times 3024 = 18144\) and \(336 \times 54 = 18144\). ✓
Answer: HCF \(= 6\), LCM \(= 3024\); product \(= 18144\) both ways

Where marks slip: The product rule works for two numbers only. Do not use HCF × LCM = product for three numbers (see Q3).

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Exercise 1.1, Question 3

Find the HCF and LCM by prime factorisation.
(i) \(12, 15, 21\)
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  1. \(12 = 2^2 \times 3\), \(15 = 3 \times 5\), \(21 = 3 \times 7\).
  2. The only prime in all three is \(3\): HCF \(= 3\).
  3. LCM \(= 2^2 \times 3 \times 5 \times 7 = 420\).
Answer: HCF \(= 3\), LCM \(= 420\)
(ii) \(17, 23, 29\)
Show solution
  1. All three are prime, so they share no prime factor: HCF \(= 1\).
  2. LCM \(= 17 \times 23 \times 29 = 11339\).
Answer: HCF \(= 1\), LCM \(= 11339\)
(iii) \(8, 9, 25\)
Show solution
  1. \(8 = 2^3\), \(9 = 3^2\), \(25 = 5^2\): no common prime, so HCF \(= 1\).
  2. LCM \(= 2^3 \times 3^2 \times 5^2 = 1800\).
Answer: HCF \(= 1\), LCM \(= 1800\)

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Exercise 1.1, Question 4

Given HCF\((306, 657) = 9\), find LCM\((306, 657)\).
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  1. For two numbers, HCF × LCM = product of the numbers.
  2. \[\begin{aligned}\text{LCM} &= \dfrac{306 \times 657}{9} \\ &= \dfrac{201042}{9} \\ &= 22338\end{aligned}\]
Answer: LCM \(= 22338\)

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Exercise 1.1, Question 5

Can \(6^n\) end in the digit \(0\) for a natural number \(n\)?
Show solution
  1. A number ends in \(0\) only if it is divisible by \(10 = 2 \times 5\), so its prime factorisation must contain \(5\).
  2. \(6^n = (2 \times 3)^n = 2^n \times 3^n\). By the Fundamental Theorem of Arithmetic this factorisation is unique, and it has no factor \(5\).
  3. So \(6^n\) is never divisible by \(5\).
Answer: No: \(6^n = 2^n \times 3^n\) has no prime factor \(5\), so it never ends in \(0\).

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Exercise 1.1, Question 6

Explain why each number is composite.
(i) \(7 \times 11 \times 13 + 13\)
Show solution
  1. Take out the common factor \(13\): \(13(7 \times 11 + 1) = 13 \times 78\).
  2. It has factors other than 1 and itself (for example \(13\)), so it is composite. (Its value is \(1014\).)
Answer: \(13 \times 78 = 1014\), which has factor \(13\): composite
(ii) \(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5\)
Show solution
  1. Take out the common factor \(5\): \[5(7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) = 5 \times 1009\]
  2. So it has the factor \(5\) and is composite. (Its value is \(5045\).)
Answer: \(5 \times 1009 = 5045\): composite

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Exercise 1.1, Question 7

Two people start together at the same point of a circular track and walk the same way; one takes 18 minutes per round and the other 12. After how many minutes are they next together at the start?
Show solution
  1. They are both at the start at times that are multiples of \(18\) and of \(12\), so the first such time is LCM\((18, 12)\).
  2. \(18 = 2 \times 3^2\), \(12 = 2^2 \times 3\), so LCM \(= 2^2 \times 3^2 = 36\).
Answer: After \(36\) minutes

Where marks slip: "Next time together" questions use the LCM; "largest equal groups" questions use the HCF.

Practise this: Step 3, Full marks on long answers →

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