A drink vendor has 80 liters of M 144 liters of P and 368 liters of S drinks He wants to pack them i...........

Math and Mental Ability LCM, HCF

A drink vendor has 80 liters of M, 144 liters of P, and 368 liters of S drinks. He wants to pack them in cans, so that each can contains the same number of liters of a drink, and doesn't want to mix any two drinks in a can. What is the least number of cans required ?

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Explanation:

Explanation:
If we want to pack the drinks in the least number of cans possible, then each can should contain the maximum numbers of liters possible.As each can contains the same number liters of a drink, the number of liters in each can is a comman factor for 80,144 and 368; and it is also the highest such factor, as we need to store the maximum number of liters in each can.
So, the number of liters in each can = HCF of 80,144 and 368 = 16 liters.
Now, number of cans of M = 80/16 = 5
Number of cans of P= 144/16 = 9
Number of cans of S = 368/16 = 23
Thus, the total number of cans required = 5 + 9 + 23 = 37