NCERT Solutions for Class 7th Maths Chapter 3 In-text Questions — Doubling both numbers · Multiples of the same number

Book page 59–60 Updated on2026-09-19

Q1.

What happens to the HCF of two numbers if both numbers are doubled? Take some pairs of numbers and explore. Are you able to see why the HCF will also double?

Answer

The HCF also doubles.

Pair	HCF	Doubled pair	New HCF
270, 50	10	540, 100	20
12, 18	6	24, 36	12
30, 72	6	60, 144	12
7, 11	1	14, 22	2

270 = 2 × 3 × 3 × 3 × 5, 50 = 2 × 5 × 5 → HCF = 2 × 5 = 10
540 = 2 × 2 × 3 × 3 × 3 × 5, 100 = 2 × 2 × 5 × 5 → HCF = 2 × 2 × 5 = 20

Why it happens: doubling a number puts one extra 2 into its prime factorisation. Do it to both numbers and each gains one 2, so the count of 2s that they share goes up by exactly one. Every other prime is untouched, so the largest common subpart is the old HCF with one more 2 in it — that is, twice as big.

Check it yourself: the same reasoning works for tripling (HCF triples) or multiplying both by 10 (HCF becomes ten times). In general, HCF(ka, kb) = k × HCF(a, b).

Q2.

Consider the following two multiples of 14 — 14 × 6, 14 × 9. What is their HCF?

Answer

HCF = 42, not 14.

14 × 6 = 84 = 2 × 7 × 2 × 3
14 × 9 = 126 = 2 × 7 × 3 × 3
Shared primes: one 2, one 3, one 7
HCF = 2 × 3 × 7 = 14 × 3 = 42
Check: 84 ÷ 42 = 2 ✓ and 126 ÷ 42 = 3 ✓

Why it happens: 14 is certainly a common factor, but it is not the highest one. The multipliers 6 and 9 are themselves not co-prime — both contain a 3. That extra shared 3 joins the 14, giving 14 × 3 = 42. In general, HCF(14a, 14b) = 14 × HCF(a, b), and here HCF(6, 9) = 3.

Q3.

Here are some more numbers where both numbers are multiples of the same number. Find their HCF: (a) 18 × 10, 18 × 15 (b) 10 × 38, 10 × 21 (c) 5 × 13, 5 × 20 (d) 12 × 16, 12 × 20

Answer

Use HCF(ka, kb) = k × HCF(a, b) — the common multiplier comes out, and whatever the multipliers still share joins it.

	Numbers	Multiplier k	HCF of the multipliers	HCF
(a)	180, 270	18	HCF(10, 15) = 5	18 × 5 = 90
(b)	380, 210	10	HCF(38, 21) = 1	10 × 1 = 10
(c)	65, 100	5	HCF(13, 20) = 1	5 × 1 = 5
(d)	192, 240	12	HCF(16, 20) = 4	12 × 4 = 48

Checks by prime factorisation:
(a) 180 = 2 × 2 × 3 × 3 × 5, 270 = 2 × 3 × 3 × 3 × 5 → 2 × 3 × 3 × 5 = 90 ✓
(b) 380 = 2 × 2 × 5 × 19, 210 = 2 × 3 × 5 × 7 → 2 × 5 = 10 ✓
(c) 65 = 5 × 13, 100 = 2 × 2 × 5 × 5 → 5 ✓
(d) 192 = 2⁶ × 3, 240 = 2⁴ × 3 × 5 → 2 × 2 × 2 × 2 × 3 = 48 ✓

Q4.

In which of these cases is the HCF the same as the common multiplier, like problem (b) where the HCF is 10? Explore a few more examples of this type to understand when this happens.

Answer

In (b) and (c) only.

	Multipliers	Are they co-prime?	HCF = multiplier?
(a)	10 and 15 (share 5)	no	no — HCF 90, multiplier 18
(b)	38 and 21	yes	yes — both 10
(c)	13 and 20	yes	yes — both 5
(d)	16 and 20 (share 4)	no	no — HCF 48, multiplier 12

The rule: HCF(ka, kb) equals k exactly when a and b are co-prime.

More examples:
9 × 4, 9 × 25 → 4 and 25 co-prime → HCF = 9 ✓
9 × 4, 9 × 26 → 4 and 26 share 2 → HCF = 9 × 2 = 18 ✗
7 × 6, 7 × 35 → 6 and 35 co-prime → HCF = 7 ✓

Why it happens: HCF(ka, kb) = k × HCF(a, b). The answer is k itself only when the second piece, HCF(a, b), equals 1 — that is, when the multipliers share no prime factor at all. If they do share something, that extra piece rides along and makes the HCF bigger than k.

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