Real Numbers Exercise 1.1 Solutions: Class 10
Original worked solutions for all seven question groups in NCERT Class 10 Maths Exercise 1.1, checked against the 2026–27 reprint.
Open the official chapter alongside this answer companion. The question numbers below refer to Exercise 1.1 on pages 5–6 of the NCERT 2026–27 reprint. Older editions may number the exercises differently. Udayy provides its own working; the full question wording remains in the linked textbook.
Open NCERT Real Numbers: Exercise 1.1
Question 1: prime factorisation
| Part | Factorisation |
|---|---|
| (i) | 140 = 2² × 5 × 7 |
| (ii) | 156 = 2² × 3 × 13 |
| (iii) | 3825 = 3² × 5² × 17 |
| (iv) | 5005 = 5 × 7 × 11 × 13 |
| (v) | 7429 = 17 × 19 × 23 |
Divide by a small prime repeatedly while it divides exactly, then move to the next possible prime. For example, 3825 ÷ 5 = 765, ÷ 5 = 153, ÷ 3 = 51, ÷ 3 = 17. Stop only when the remaining factor is prime. Multiply your factors back to verify the original number.
Question 2: HCF and LCM of pairs
| Pair | Prime factors | HCF; LCM; product check |
|---|---|---|
| 26, 91 | 26 = 2 × 13; 91 = 7 × 13 | 13; 182; 2366 |
| 510, 92 | 510 = 2 × 3 × 5 × 17; 92 = 2² × 23 | 2; 23460; 46920 |
| 336, 54 | 336 = 2⁴ × 3 × 7; 54 = 2 × 3³ | 6; 3024; 18144 |
For each row, the last number equals both HCF × LCM and the product of the pair. A factor absent from one number cannot appear in the HCF; it still belongs in the LCM.
Question 3: HCF and LCM of triples
| Numbers | HCF | LCM |
|---|---|---|
| 12, 15, 21 | 3 | 420 |
| 17, 23, 29 | 1 | 11339 |
| 8, 9, 25 | 1 | 1800 |
Use the smallest exponent shared by all three for HCF and the largest exponent present in any number for LCM. In the second row, the three different primes have no shared factor other than 1. Their LCM is their product. Do not apply the two-number product shortcut to triples.
Questions 4–7: applications
4. Recover the LCM
- 1LCM = (306 × 657) ÷ 9 = 22338.
- 2Check: 306 × 657 = 201042 and 9 × 22338 = 201042.
5. Powers of 6
- 16ⁿ = 2ⁿ × 3ⁿ for a positive integer n. No factor 5 appears.
- 2A final zero would require divisibility by 10 = 2 × 5, which is impossible.
6. Show the expressions are composite
- 17 × 11 × 13 + 13 = 13 × (77 + 1) = 13 × 78.
- 27 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × (1008 + 1) = 5 × 1009.
- 3Each expression is a product of two integers greater than 1, which proves it is composite.
7. Return to the starting point
- 1Shared return times are common multiples of 18 and 12 minutes.
- 218 = 2 × 3² and 12 = 2² × 3, so LCM = 2² × 3² = 36 minutes.
- 3In that time the two people finish 2 and 3 rounds respectively.
Try the method independently
Change the last question to lap times of 15 and 20 minutes. List multiples before factorising: the first common return is 60 minutes. Then explain why the answer is a time rather than a distance.