At a certain fruit stand, the price of each apple is 40 cents and the price of each orange is 60 cents. Mary selects a total of 10 apples and oranges from the fruit stand, and the average (arithmetic mean) price of the 10 pieces of fruit is 56 cents. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is 52 cents?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
This is a mixture problem.
An average of 40 and an average of 60 are being mixed to yield an average of 56 in the first case and of 52 in the second case.
To determine the ratio of apples to oranges in each case, use ALLIGATION.
Case 1: average cost = 56.
Step 1: Plot the 3 averages on a number line, with the average apple cost (40) and the average orange cost (60) on the ends and the average cost of all the fruit (56) in the middle.
A(40)--------------------56-------O(60)
Step 2: Calculate the distances between the averages.
A(40)----------
16----------56---
4----O(60)
Step 3: Determine the ratio of apple to oranges.
The ratio of A to O is the RECIPROCAL of the distances in red.
A : O = 4:16 = 1:4.
Since A : O = 1:4 = 2:8, we know that A=2 and O=8, for a total of 10 pieces of fruit.
Case 2: average cost = 52.
A(40)----------
12----------52---
8----O(60)
A : O = 8:12 = 2:3.
Since the number of apples isn't changing, A=2 (same as above) and new O=3.
Thus:
Old O - new O = 8-3 = 5.
The correct answer is
E.
Another approach:
The original total cost of the 10 pieces of fruit = 10*56 = 560.
According to the answers, after 1, 2, 3, 4, or
5 pieces are removed -- so that 9, 8, 7, 6, or
5 pieces remain -- the average cost decreases to 52.
Since the prices are each a multiple of 10, the new total cost after the oranges are removed must also be a multiple of 10.
Only the option in red -- implying 5 remaining pieces of fruit -- will yield a new total cost that is a multiple of 10:
5*52 = 260.
The correct answer is
E.
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