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

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If you start at 56 cents, the combination that you must have is 8 oranges and 2 apples.

An equation for this is [40(a) + 60(10 - a)]/10 = 56

a is the number of apples.

The next step is to solve the following:

[40(2) + 60(o)]/(o+2) = 52

Solving for o gives you 3, meaning she has to put back 5.

There are 'easier' ways to get there, but understand this method first.

An equation for this is [40(a) + 60(10 - a)]/10 = 56

a is the number of apples.

The next step is to solve the following:

[40(2) + 60(o)]/(o+2) = 52

Solving for o gives you 3, meaning she has to put back 5.

There are 'easier' ways to get there, but understand this method first.

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This is a mixture problem.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

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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Alternate approach: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 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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An alternate approach is to PLUG IN THE ANSWERS, which represent the number of oranges that must be put back.datonman 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

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!

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Mitch,GMATGuruNY wrote: 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.

Could you please post links to some problems, which can be solved using ALLIGATION method.

I want to familiarize myself with different situations in which I can use this method.

Thanks!

Manik

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See here: https://www.beatthegmat.com/seed-mixture-t21603.htmlmanik11 wrote:Mitch,GMATGuruNY wrote: 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.

Could you please post links to some problems, which can be solved using ALLIGATION method.

I want to familiarize myself with different situations in which I can use this method.

Thanks!

Manik

And here: https://www.beatthegmat.com/managers-and ... 38687.html

And here: https://gmatclub.com/forum/during-a-cert ... 44454.html

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Hi All,

We’re told that apples cost 40 cents each and oranges cost 60 cents each and that Mary selects 10 pieces of fruit (re: some apples and some oranges). The average price of those 10 pieces is 56 cents. We’re asked how many of the oranges Mary must put back so that the average price drops to 52 cents per piece. This question can be solved in a couple of different ways, including by doing just a bit of ‘brute force’ arithmetic and TESTing THE ANSWERS.

To start, we have to figure out how to get an average of 56 cents for the 10 pieces. At that average, the TOTAL PRICE of the 10 pieces would be $5.60. If there were 5 apples and 5 oranges, then the average would be exactly 50 cents per piece (and the total price would be $5.00). If we ‘trade’ an apple for an additional orange, then the total price will increase by 20 cents (since an apple is 40 cents and an orange is 60 cents), meaning that trading 3 apples for 3 additional oranges will give us the $5.60 total that we’re looking for.

This means that we start off with 2 apples and 8 oranges.

Now we have to remove enough oranges to lower the average from 56 cents to 52 cents. From the answer choices, we can see that we’re removing from 1-5 oranges, so we can simply TEST THE ANSWERS at this point. Let’s start with Answer B.

IF… we remove 2 oranges, then we’ll have 2 apples and 6 oranges. The average price of that group would be [2(40) + 6(60)]/8 = [80 + 360]/8 = 440/8 = 55 cents. This is far too high (we need the average to be 52 cents), so we clearly need to remove far MORE oranges.

IF… we remove 4 oranges, then we’ll have 2 apples and 4 oranges. The average price of that group would be [2(40) + 4(60)]/6 = [80 + 240]/6 = 320/6 = 53.333 cents. This is still too high, so we need to remove MORE oranges. There’s only one answer left…

Final Answer: E

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