Question

A trader has three different types of oils of volume 870 `l`, 812 `l` and 638 `l`. Find the least number of containers of equal size required to store all the oil without getting mixed.

Answer 1

Prime factorization of each number:

870 = 2 × 3 × 5 × 29

812 = 2² × 7 × 29

638 = 2 × 11 × 29

The common factors are 2 and 29.

HCF (870, 812, 638) = 2 × 29 = 58 litres

This is the capacity of one container.

The number of containers for each type of oil:

For `870  l: 870/58 = 15`

For `812  l: 812/58 = 14`

For `638  l: 638/58 = 11`

Total number of containers = 15 + 14 + 11 = 40.

The least number of containers required is 40.