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

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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

Source: Official Guide

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by Jay@ManhattanReview » Thu Jan 10, 2019 10:43 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
Say Mary x numbers of apples and y numbers of oranges. Thus, x +y = 10 and

40x + 60y = 56*10
40x + 60y = 560

Say Mary put back p numbers of oranges, thus, we have

40x + 60y - 60 p = 52(10 - p)

560 - 60 p = 520 - 52p

p = 5.

The correct answer: E

Hope this helps!

-Jay
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by Brent@GMATPrepNow » Fri Jan 11, 2019 9:32 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
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 GMATGuruNY » Fri Jan 11, 2019 9:48 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
The original total cost of the 10 pieces of fruit = 10*56 = 560.
According to the answers, after 1, 2, 3, 4, or 5 oranges are removed -- so that 9, 8, 7, 6, or 5 pieces of fruit remain -- the average cost per piece decreases to 52.
Since the original total cost is a multiple of 10, and the price per orange is a multiple of 10, the new total cost after the oranges are removed must also be a multiple of 10.
Only answer choice E will yield a new total cost that is a multiple of 10:
5 remaining pieces of fruit with an average price of 52 cents = 5*52 = 260.

The correct answer is E.
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by fskilnik@GMATH » Sun Jan 13, 2019 5:10 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
Source: Official Guide
$$? = b\,\,\left( {{\rm{back}}} \right)$$
$$10 \to 10 - b\,\,{\rm{fruits}}\,\,\,\left\{ \matrix{
\,{\rm{apples}}\,\, = a\,\,{\rm{units}}\,\,,\,\,40\,\,{\rm{cents}}\,\,{\rm{each}} \hfill \cr
\,{\rm{oranges}}\,\,{\rm{ = }}\,\,10 - a\,\,\, \to \,\,\,{\rm{10}} - b - a\,\,{\rm{units}}\,\,,\,\,60\,\,{\rm{cents}}\,\,{\rm{each}} \hfill \cr} \right.$$
$$\sum\nolimits_{10} {\,\, = \,\,\,10 \cdot 56 = 560\,\,\,{\rm{cents}}\,\,\,\,\,\,\left( {{\rm{homogeneity}}\,\,{\rm{nature}}\,\,{\rm{of}}\,\,{\rm{the}}\,\,{\rm{average}}} \right)} $$
$$\sum\nolimits_{10 - b} {\,\, = \,\,\,\left( {10 - b} \right) \cdot 52\,\,\,{\rm{cents}}\,\,\,\,\,\,\left( {{\rm{homogeneity}}\,\,{\rm{nature}}\,\,{\rm{of}}\,\,{\rm{the}}\,\,{\rm{average}}} \right)} $$
$$\sum\nolimits_{10 - b} {\,\, + b \cdot 60\,\,\, = } \,\,\,\,\sum\nolimits_{10} {\,\,\,\,\, \Rightarrow \,\,\,\,\,\,52\left( {10 - b} \right) + 60b = 560\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\, \ldots \,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = b = 5} $$


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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by Scott@TargetTestPrep » Mon Jan 14, 2019 5:56 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
We can let the number of apples = x and the number of oranges = y. Using these variables we can create the following two equations:

1) x + y = 10

Using the formula average = sum/quantity, we have:

2) (40x + 60y)/10 = 56

Let's first simplify equation 2:

40x + 60y = 560

4x + 6y = 56

2x + 3y = 28

Isolating for y in equation 1 gives us: y = 10 - x.

Since y = 10 - x, we can substitute 10 - x for y in the equation 2x + 3y = 28. This gives us:

2x + 3(10 - x) = 28

2x + 30 - 3x = 28

-x = -2

x = 2

Since x + y = 10, then y = 8.

We thus know that Mary originally selected 2 apples and 8 oranges.

We must determine the number of oranges that Mary must put back so that the average price of the pieces of fruit that she keeps is 52¢. We can let n = the number of oranges Mary must put back.

Let's use a weighted average equation to determine the value of n.

[40(2) + 60(8-n)]/(10 - n) = 52

(80 + 480 - 60n)/(10 - n) = 52

560 - 60n = 520 - 52n

40 = 8n

5 = n

Thus, Mary must put back 5 oranges so that the average cost of the fruit she has kept would be 52 cents.

Answer: E

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