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A house wife, with a given amount can buy either 10 apples or 15 oranges or 2 watermelons. Find the maximum number of oranges which she can buy with six times the initial amount such that she gets each of the three varieties of fruits.

A.	75	B.	81
C.	60	D.	72
E.	Cannot be determined

Answer and Explanation

Answer:81

Explanation:
Let the initial amount be a
Cost of 1 apple = a/10
Cost of 1 orange = a/15
Cost of 1 watermelon = a/2
As all the 3 types of fruits are bought, the minimum shall be 1
As oranges are the max, others are one each.
Amount spent = 6a
Hence the number of oranges
= [6a-{a/10 + a/2}] / a/15
= [6a – 6a/10] / a/15 = (54a/10) / (a/15) = 81

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