Exercise 1.1-Class 10th NCERT Chapter-1 – Real Numbers

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

(i) 140 = 2 × 70 = 2 × 2 × 35 = 2² × 5 × 7
(ii) 156 = 2 × 78 = 2 × 2 × 39 = 2² × 3 × 13
(iii) 3825 = 5 × 765 = 5 × 5 × 153 = 5² × 3 × 51 = 3² × 5² × 17
(iv) 5005 = 5 × 1001 = 5 × 7 × 143 = 5 × 7 × 11 × 13
(v) 7429 = 17 × 437 = 17 × 19 × 23


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

(i) 26 and 91
26 = 2 × 13; 91 = 7 × 13
HCF = 13, LCM = 182
Check: 13 × 182 = 2366 = 26 × 91 ✔

(ii) 510 and 92
510 = 2 × 3 × 5 × 17; 92 = 2² × 23
HCF = 2, LCM = 23,460
Check: 2 × 23,460 = 46,920 = 510 × 92 ✔

(iii) 336 and 54
336 = 2⁴ × 3 × 7; 54 = 2 × 3³
HCF = 6, LCM = 3024
Check: 6 × 3024 = 18,144 = 336 × 54 ✔


Q3. Find LCM and HCF by prime factorisation method

(i) 12, 15, 21
12 = 2² × 3; 15 = 3 × 5; 21 = 3 × 7
HCF = 3, LCM = 420

(ii) 17, 23, 29
All are primes, no common factor
HCF = 1, LCM = 17 × 23 × 29 = 11,339

(iii) 8, 9, 25
8 = 2³, 9 = 3², 25 = 5²
HCF = 1, LCM = 1800


Q4. If HCF (306, 657) = 9, find LCM (306, 657).

LCM = (306 × 657) ÷ 9 = 201,042 ÷ 9 = 22,338


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

6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ
For a number to end in 0, it must have factor 5. Since 6ⁿ never has factor 5, it can never end with digit 0.


Q6. Explain why these numbers are composite.

(a) 7 × 11 × 13 + 13
= 13 × (7 × 11 + 1) = 13 × 78
Divisible by 13 → composite.

(b) 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
The product includes factor 5, so the product is divisible by 5. Adding 5 makes it also divisible by 5, and since it is greater than 5, it is composite.


Q7. There are two bells — one rings in 18 minutes, another in 12 minutes. In how many minutes will they ring together again?

Required time = LCM (18, 12) = 36 minutes.

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