Apple (A) = 40 cemts
Orange (B) = 60 cents
Average of 10 fruits = 56 cents
u.e. Total Price of 10 Fruits = 56*10 = 560 cents

i.e. 40A + 60 B = 560
and A + B = 10

i.e. 40A + 60 (10-A) = 560
i.e. -20A + 600 = 560
i.e. A = 2
i.e. B = 8

Average to be brought at = 52 cents
let oranges to be put back = C
i.e. Total Fruits = 10-C
where total Oranges left = 8-C
i.e. Total Price of New lot of fruits = 52*(10-C)

i.e. 40*2 + 60*(8-C) = 52*(10-C)
i.e. 80 + 480 - 60C = 520 - 52C
i.e. 8 C = 40
i.e. C = 5

Answer: Option E
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Ans E

Sum of Apples and Oranges is 10

A + O = 10

Apple cost 40 cent and Orange cost 60 cent and there mean is 56 , therefore

(40*A + 60*O)/10 = 56 , which gives us

4A + 6O = 56

Solving the above 2 equation we get , Apples = 2 , Oranges = 8.

Now as she has remove some oranges ( x ) such that the new mean of the remaining fruits become 52

note that the number of Apples remain the same , therefore new equation will become

( 2*40 + (8-x)60 ) / 10 -x = 52

As the remaining fruits will be 10 - x

On solving for x we will get our ans 5

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

Using the formula : Sum = Average * Number of terms
Sum = 56 * 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 :
(560-60x) / (10-x) = 52
560 - 60x = (10-x) * 52
560 - 520 = -52x + 60x
40 = 8x
x=5

The answer is E.

Solution:

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 one 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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i did this one with a scale method

Apples : Price : Oranges
40 cents : 56 average : 60 cents
= 16 cents from average : average : 4 cents from average
= 1 apple for every 4 oranges
= 2 apples and 8 oranges = 10 fruits = 2*.4 + 8*.6 = 5.60 / 10 = .56

40 cents : 52 average : 60 cents
= 12 cents from average : average : 8 cents from average
= 2 apples for every 3 oranges

8 oranges - 3 oranges = 5 oranges must be put back

Hope this helps!

\(\frac{Wa}{Wo} = \frac{(60-56)}{(56-40}) = \frac{1}{4}\)
Apple = 2 & Orange = 8
New Avg
\(\frac{Wa}{Wo} = \frac{(60-52)}{(52-40)} = \frac{2}{3}\)
\(\frac{2}{Wo} = \frac{2}{3}\)
Orange = 3

Hence, Oranges have to reduce from 8 to 3 i.e. 5
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Please find the solution as attached.

Answer: option E

I hope this helps!!!

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1. Total cost is 560 cents
2. If she puts 1 orange back , average cost is 500/9. Similarly for 2,3,4, and 5 oranges back it is 440/8, 380/7, 320/6, 260/5
We see 260/5 gives 52 cents average cost.
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The question tells you that initial quantity is not maintained. It doesn't talk about picking more fruits/replacing fruits etc.
If you want to maintain the number of fruits to 10 and achieve 52 cents as the average price, you know that

w1/w2 = (60 - 52)/(52 - 40) = 8/12 = 2/3

So there should be 2 apples for every 3 oranges. This means there should be 4 apples and 6 oranges.
But actually she has 2 apples and 8 oranges. So she needs to put back 2 oranges and pick 2 more apples.
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For those who are thinking about how to approach this question efficiently, here are my GMAT Timing Tips (the links below have growing lists of questions that you can use to practice these timing tips):

Weighted average mapping strategy: The weighted average mapping strategy is the same as the "teeter-totter" approach shown in the great solution by FANewJersey. This really is a time-saving approach. The key thing that you get from this approach is that the ratio of oranges to apples is 4 to 1 before (so 10 total fruits gives us 8 oranges and 2 apples) and 3 to 2 after putting back the apples.

Track how the parts of a whole change: The key thing to notice is that there are the same number of apples (2) before and after putting back the oranges. So, if the ratio is 3 oranges to 2 apples after putting back the oranges, then we end up with 2 apples and 3 oranges. This means that Mary put back 8 – 3 = 5 oranges.

Please let me know if you have any questions, and if you would like me to post a video solution!
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