A person has to put each of three liquids, 420 L of diesel, 475 L of petrol and 495 L of oil in bottles of equal size without mixing any of the above three types of liquids such that each bottle is completely filled. What is the least possible number of bottles required?

Correct option is D

Given:

The person has 420 L of diesel, 475 L of petrol, and 495 L of oil.

Formula Used:
To minimize the number of bottles, we need to use Highest Common Factor  of the three amounts of liquid.
Solution:
The HCF of 420, 475, and 495 is 5 L.

Therefore, the size of each bottle must be 5 L
Number of bottles for diesel = $$\frac{420}{5}$$​ = 84
Number of bottles for petrol = $$\frac{475}{5}$$​ = 95
Number of bottles for oil = $$\frac{495}{5}$$​ = 99
The total number of bottles required is:
84 + 95 + 99 = 278
Thus, the least possible number of bottles required is 278.