A milk vendor has 75 litres of cow milk, 45 litres of toned milk and 30 litres of full cream milk. These should be placed in
different trays such that each tray contains the same quantity of milk and two kinds of milk should not be mixed in a tray.
What is the least number of trays required to place all the milk?

Correct option is D

Given:
Cow milk = 75 litres
Toned milk = 45 litres
Full cream milk = 30 litres
Formula Used:
The least number of trays = $$\frac{\text{Total milk quantity}}{\text{HCF of milk quantities}}$$

HCF = Highest  Common Factor 
Solution:
Find the HCF of 75, 45, and 30:
Prime factorization of 75: $$75 = 3 \times 5^2$$​​
Prime factorization of 45: $$45 = 3^2 \times 5$$​​
Prime factorization of 30: $$30 = 2 \times 3 \times 5$$​

The HCF is $$3 \times 5$$​= 15.

The total milk quantity = 75 + 45 + 30 = 150 litres.

Number of trays required: $$\frac{150}{15} = 10 \,$$​ trays

Thus, the correct option is (d) 10