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NCERT Solutions for Class 7th Maths Chapter 3 In-text Questions — The Greatest of All

Book page 48 Updated on2026-09-19

Q1.
How many tiles of this size should she purchase?
Answer

She should purchase 12 tiles.

Along the breadth: 12 ÷ 4 = 3 tiles
Along the length: 16 ÷ 4 = 4 tiles
Total = 3 × 4 = 12 tiles

Checking by area: 192 sq ft ÷ 16 sq ft per tile = 12 tiles ✓

Tip: counting rows × columns and counting by area must agree. If they do not, one of the two is wrong — a quick way to catch a slip.
Q2.
What if Sameeksha did not insist on the length of the tile to be a whole number of feet and the length could be a fractional number of feet? Would the answer change?
Answer

No, the answer would not change. Even with fractional sides allowed, 4 ft is still the largest tile that fits, so 12 tiles is still the smallest number.

Many fractional sides do fit. For example a side of 4⁄3 ft works:

12 ÷ 4⁄3 = 12 × 3⁄4 = 9 tiles across the breadth ✓
16 ÷ 4⁄3 = 16 × 3⁄4 = 12 tiles along the length ✓
Number of tiles = 9 × 12 = 108 — far more than 12

But no side bigger than 4 ft can fit. Here is the reason.

Let the side be s ft, with 12 ÷ s = a tiles and 16 ÷ s = b tiles, a and b whole numbers
So 12 = a × s and 16 = b × s
Subtracting: 16 − 12 = (b − a) × s
4 = (b − a) × s
Since 16 > 12, (b − a) is a whole number that is at least 1
So s = 4⁄(b − a) ≤ 4
Why it happens: the strip of floor left over when a 12 ft length is laid against a 16 ft length is exactly 4 ft wide, and that strip must itself hold a whole number of tiles. So the tile can never be wider than 4 ft — fractions or no fractions. Dropping the whole-number condition only adds smaller tiles such as 2 ft, 4⁄3 ft, 1 ft, 4⁄5 ft …, and every one of them needs more tiles.
Tip: notice what the argument really used — that 4 = 16 − 12 must also be a whole number of tiles. Taking differences like this is a very old and very quick way to hunt for the HCF.
Q3.
Lekhana purchases rice from two farms and sells it in the market. She bought 84 kg of rice from one farm and 108 kg from the other farm. She wants the rice to be packed in bags, so each bag has rice from only one farm and all bags have the same weight that is a whole number of kg. If she wants to use as few bags as possible, what should the weight of each bag be?
Answer

Each bag should hold 12 kg.

Factors of 84 = 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
Factors of 108 = 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108
Common factors = 1, 2, 3, 4, 6, 12
HCF = 12

With 12 kg bags:

From the first farm: 84 ÷ 12 = 7 bags
From the second farm: 108 ÷ 12 = 9 bags
Total = 16 bags
Why it happens: the bag weight must divide 84 exactly (no rice left loose) and must divide 108 exactly as well, since the same size bag is used for both farms. So it has to be a common factor. Heavier bags hold more, so the fewest bags come from the heaviest allowed bag — the HCF.
Q4.
She can use any of these weights to pack rice from both farms in bags of equal weight. But, she wants to minimise the number of bags. Which weight should she choose to minimise the number of bags?
Answer

She should choose the largest common factor, 12 kg.

Bag weightBags for 84 kgBags for 108 kgTotal bags
1 kg84108192
2 kg425496
3 kg283664
4 kg212748
6 kg141832
12 kg7916
Why it happens: the total rice is 84 + 108 = 192 kg whatever she does. The number of bags is 192 ÷ (bag weight), so the heaviest bag gives the fewest bags. The heaviest bag allowed is the HCF, 12 kg.
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