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

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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

Answer: E

Source: Official Guide

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Vincen wrote:
Wed Jul 21, 2021 12:31 pm
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

Answer: 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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Brent
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Vincen wrote:
Wed Jul 21, 2021 12:31 pm
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

Answer: E

Source: Official Guide
Step 1: We start with the first average to find the total price of the apples and oranges.

Using the formula: Sum \(=\) Average \(\ast \) Number of terms
Sum \(= 56 \ast 10 = 560\)

Step 2: Set up the new average based on the first one.
\(x:\) number of oranges we need to take away.
We will have the previous sum (i.e. \(560\)) minus the price of all oranges we will take away (i.e. \(60x\)) and the number of elements will be \(10\) minus \(x.\)

So:
\(\dfrac{560-60x}{10-x} = 52\)
\(560 - 60x = (10-x)\ast 52\)
\(560 - 520 = -52x + 60x\)
\(40 = 8x\)
\(x=5\)

Therefore, E