Paving the Hall Floor: Slab Size, a Diagonal Claim and a Second Room
The mathematics club of a higher secondary school has taken up the paving of the school hall floor for its annual exhibition. The floor is a rectangle measuring 4.2 m by 3.6 m. The club will buy square marble slabs of one single size, laid edge to edge with no gaps, and no slab may be cut; the side of a slab must be a whole number of centimetres. The supplier stocks only 30 cm, 40 cm, 50 cm and 60 cm slabs and quotes ₹540 for a 60 cm slab. A brass strip is to be fixed along one diagonal of the floor. The mason, who measured the diagonal with a tape, declares it is "exactly 5.6 m" and wants 5.6 m of strip ordered with no wastage allowance. The treasurer is uneasy, remembering that a length obtained from Pythagoras need not be a terminating decimal. The strip is sold only in steps of 0.25 m, and the order must be placed today. A second room of 3.0 m by 2.4 m is to be paved under the same programme.
1. Which slab size should be ordered, and how many slabs will the hall floor need? Justify the choice against every other size the supplier stocks.
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Work in centimetres: the floor is 420 cm × 360 cm. A slab of side s fits with no cutting only if s divides both 420 and 360, so s must be a common divisor; the largest such s is the HCF.
- HCF(420, 360): 420 = 2²·3·5·7, 360 = 2³·3²·5, so HCF = 2²·3·5 = 60 cm.
- Check the others: 420 ÷ 50 = 8.4 (cutting needed), 420 ÷ 40 = 10.5 (cutting needed), 30 cm divides both but is not the largest.
Number of slabs = (420 ÷ 60) × (360 ÷ 60) = 7 × 6 = 42 slabs.
Cost at ₹540 each = 42 × 540 = ₹22,680. The answer must be the largest common divisor, not merely any one: 30 cm also leaves no gaps but needs 14 × 12 = 168 slabs, and 40 cm and 50 cm are disqualified by the no-cutting rule, not by cost.
2. Is the mason's figure of 5.6 m the diagonal? Decide what length of strip to order, and state whether the diagonal can ever be written as an exact terminating decimal.
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By Pythagoras, diagonal² = 4.2² + 3.6² = 17.64 + 12.96 = 30.6, so diagonal = √30.6.
Bracket it: 5.5² = 30.25 and 5.6² = 31.36, so 5.5 < √30.6 < 5.6. More precisely √30.6 ≈ 5.532 m. The mason's 5.6 m is an over-estimate of about 7 cm; it happens to be long enough, but it is not the diagonal, and it leaves too little margin once wastage is counted.
Exactness: 30.6 = 153/5. If √(153/5) = p/q in lowest terms, then 153q² = 5p². Since 5 does not divide 153, 5 | q², so 5 | q; then 25 | 5p² gives 5 | p, contradicting lowest terms. So √30.6 is irrational - no terminating or recurring decimal equals it.
Ordering: 5.532 × 1.05 ≈ 5.809 m. In steps of 0.25 m the next available length is 6.00 m; 5.75 m would leave only about 4% wastage, below the intended 5%.
3. The second room measures 3.0 m by 2.4 m. Will the 60 cm slab chosen for the hall suit it without cutting, and how many slabs in all should be ordered?
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Convert: 300 cm × 240 cm. Test divisibility rather than assuming a new size is needed:
- 300 ÷ 60 = 5 exactly, 240 ÷ 60 = 4 exactly.
- So the same 60 cm slab fits the second room with no cutting: 5 × 4 = 20 slabs.
Confirm it is also the largest possible: HCF(300, 240) = 60, so 60 cm is again the maximum square that tiles the room exactly. Buying a separate size would raise the price per square metre and create two leftover stocks of off-cuts.
Total order = 42 + 20 = 62 slabs, costing 62 × 540 = ₹33,480. The judgement here is to check the HCF of the second room before ordering, because a single slab size serving both rooms avoids waste and a second purchase.