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apple is 40 cents and the price of each orange is 60 cents.

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apple is 40 cents and the price of each orange is 60 cents.

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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

OAE

Please explain.

Thanks...

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Quote:
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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kamalj 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

OAE

Please explain.

Thanks...
average price of 10 apples+oranges =56
therefore cost of 10 apples and oranges= 560
if she puts back say x oranges and retains (10- x) apples and oranges
then the cost of apples and oranges she retains = 52(10-x)
and the cost of (x)oranges which she puts back=60*x
therefore 560= 60x + 52*(10-x)
=60x+520-52x
or 8x=40
or x=5
hence option E

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kamalj 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

OAE

Please explain.

Thanks...
Alternate approach:
Because the average cost ($.56)is much closer to oranges than to apples, we know there are initially more oranges among the 10 purchased pieces of fruit. Plotting on a number line

40------------------------------56---------60
Gap--------------16------------------4------
So there's a ratio of 16:4 or 4:1 in favor of oranges, meaning there must be 8 oranges and 2 apples initially.

If we want an average of $.52, our number line would look like this
40----------------------52---------60
Gap--------------12------------------8------
So we'd need a new orange:apple ratio of 12:8, or 3:2. If we have 8 oranges and 2 apples, and we want a ratio of 3 oranges: 2 apples, we'd need to put back 5 oranges. The answer is E

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Quote:
At a certain food 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 food stand, and the average (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?
1
2
3
4
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.

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Mitch Hunt
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GMATGuruNY@gmail.com

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Post
Quote:
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
An alternate approach is to PLUG IN THE ANSWERS, which represent the number of oranges that must be put back.
Before any oranges are put back, the total cost of the 10 pieces of fruit = 10*56 = 560.
When the correct answer choice is plugged in, the average cost will decrease to 52 cents.

Answer choice D: 4 oranges put back
Resulting total cost after 4 60-cent oranges are put back = 560 - (4*60) = 320.
Average price of the remaining 6 pieces of fruit = 320/6 = 53+.
The average price is too high.
Eliminate D.

Since oranges are more expensive than apples, the average price will decrease to 52 cents only if MORE oranges are put back.

The correct answer is E.

Answer choice E: 5 oranges put back
Resulting total cost after 5 60-cent oranges are put back = 560 - (5*60) = 260.
Average price of the remaining 5 pieces of fruit = 260/5 = 52.
Success!

_________________
Mitch Hunt
Private Tutor for the GMAT and GRE
GMATGuruNY@gmail.com

If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon.

Available for tutoring in NYC and long-distance.
For more information, please email me at GMATGuruNY@gmail.com.
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