Question

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

Hint:

In problems asking for "least number of items" based on a capacity/size, you are looking for the Highest Common Factor (HCF) of the given quantities. Dividing the total amount by this HCF gives the minimal count.

Answer 1

Step 1: Find HCF of 870, 812 and 638
Using prime factorisation:

870 = 2 × 3 × 5 × 29
812 = 2² × 7 × 29
638 = 2 × 11 × 29

Common prime factors:
2 and 29

HCF = 2 × 29
= 58 litres

Step 2: Find Number of Containers
For 870 litres:
870 ÷ 58 = 15

For 812 litres:
812 ÷ 58 = 14

For 638 litres:
638 ÷ 58 = 11

Step 3: Total Containers
Total = 15 + 14 + 11
= 40

Final Answer:
Capacity of each container = 58 litres
Least number of containers required = 40