A merchant has 120 litres of oil of one kind, 180 liters of another kind and 240 litres of third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such a tin?

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Solution

The maximum capacity of the required tin is the HCF of the three quantities of oil.

Prime factorisation of 120 = 2 × 2 × 2 × 3 × 5
Prime factorisation of 180 = 2 × 2 × 3 × 3 × 5
Prime factorisation of 240 = 2 × 2 × 2 × 2 × 3 × 5
∴ HCF of 120, 180, and 240 = 2 × 2 × 3 × 5 = 60
Hence, the required greatest capacity of the tin must be 60 litres.

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