At a certain fruit stand, the price of each apple is 40 cent

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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

OA E

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by Brent@GMATPrepNow » Sat Jan 19, 2019 5:30 am
BTGmoderatorDC wrote: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

OA E

Source: Official Guide
It turns out that the cost per apple is irrelevant. Here's why:

The average (arithmetic mean) price of the 10 pieces of fruit is 56 cents
So, (total value of all 10 pieces of fruit)/10 = 56 cents
This means, total value of all 10 pieces of fruit = 560 cents

How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is 52 cents?
Let x = the number of oranges to be removed.
Each orange costs 60 cents, so the value of the x oranges to be removed = 60x
This means 560 - 60x = the value of the REMAINING fruit
Also, if we remove x oranges, then 10 - x = the number of pieces of fruit REMAINING.

We want the REMAINING fruit to have an average value of 52 cents.
We can write: (value of REMAINING fruit)/(number of pieces of fruit REMAINING) = 52
Rewrite as: (560 - 60x)/(10 - x) = 52
Multiply both sides by (10-x) to get: 560 - 60x = 52(10 - x)
Expand right side to get: 560 - 60x = 520 - 52x
Add 60x to both sides: 560 = 520 + 8x
Subtract 520 from both sides: 40 = 8x
Solve: x = 5

Answer: E

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by ceilidh.erickson » Mon Jan 21, 2019 1:43 pm
We can also think of this as a WEIGHTED AVERAGE question.

If we buy some 40 cent apples and 60 cent oranges, the average price for all fruit will be between 40 and 60. If we buy more apples, it'll be closer to 40. If more oranges, it'll be closer to 60.
--|--------------|----|
40------------56--60
ap.-----------------or.

With weighted averages, the proportion of part:part is the same as the proportion of the differences between the individual values and the actual average. In other words, the ratio of oranges : apples = (differences btw actual average & 40) : (differences btw actual average & 60).

The difference btw 40 and 56 = 16
The difference btw 56 and 60 = 4
Thus, the ratio of oranges to apples = 16:4 = 4:1.

Since our total was 10, that must be 8 oranges and 2 apples.

We now want to put back some oranges and get an average of 52:
--|----------|--------|
40---------52-----60
ap.-----------------or.

The difference btw 40 and 52 = 12
The difference btw 52 and 60 = 8
Thus, the NEW ratio of oranges to apples = 12:8 = 3:2.

Since we're keeping the 2 apples we already had, we must end up with 3 oranges. Since we had 8 to begin with, we must put back 5.

The answer is E.
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by Scott@TargetTestPrep » Mon Feb 18, 2019 5:59 pm
BTGmoderatorDC wrote: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

OA E

Source: Official Guide
Since the average price of the 10 pieces of fruit is 56 cents, in total, she paid 560 cents.

Let n denote the number of oranges Mary must put back. Then, since each orange costs 60 cents, she will have paid 560 - 60n cents and she will have bought 10 - n pieces of fruit. Since we want the average to be 52 cents, we can create the following equation:

(560 - 60n)/(10 - n) = 52

560 - 60n = 520 - 52n

40 = 8n

n = 5

So, Mary must put back 5 oranges.

Answer: E

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