A trader has three kinds ofoils. the first kind 462 liters, the second kind 286 liters and the third kind 187 liters. What is the least numberof tins of equal volume required to fill them completely and store these oils without mixing?

As per the given question,

The first type of oil have 462 liters

2nd type of oil have 286 liters

3rd type of oil have 187 litrs of oil

Hence $$462=2\time 3\times 7 \times 11$$

$$286=2\times 11\times 13$$

$$187=11\times 17$$

Hence the required capacity= $$11 liter$$

The number of tin required $$=\dfrac{462}{11}+\dfrac{286}{11}+\dfrac{187}{11}$$

Hence, the number of tin of 11 liter capacity required $$=42+26+17=85$$