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
_________________

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

Cheers,
Brent
_________________

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 correct answer is (E). People who are getting (B) need to think about this: The second time when the average is 52, does Mary still have 10 fruits? She ONLY needs to put back oranges. Not replace them with apples. So when you use alligation again and get the ratio as 2:3, you don't get 6 oranges since total number of fruits are not known. Let me solve it step by step.

When avg = 56

wa/wo = (60 - 56)/(56 - 40) = 1/4
So number of apples = 2, number of oranges = 8

When avg - 52
wa/wo = (60 - 52)/(52 - 40) = 2/3
What is the total number of fruit now? We don't know. We know Mary put back some oranges. We don't know how many. What we do know is that then number of apples she had stayed the same. She had 2 apples before so she still has 2 apples. If number of apples now is 2, number of oranges must be 3. So she must have put back 8 - 3 = 5 oranges.

For a discussion on this question, check out: http://www.veritasprep.com/blog/2013/12 ... -question/
_________________

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
_________________

Video solution from Quant Reasoning starts at 0:27
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1

_________________

Let the number of apples be 'x' and oranges be 'y'

Apple: 40 cents; Orange: 60 cents.

10 (Apples+Oranges) was selected

Average (Arithemtic Mean) = \(\frac{Sum }{ Total}\)

=> Average (Arithemtic Mean) 56 = \(\frac{40x + 60y }{ 10 }\)

=> 40x + 60y = 560 -----(1)

Let 'p' oranges are put back. Then,

Average (Arithemtic Mean) = \(\frac{Sum }{ Total}\)

=> Average (Arithemtic Mean) 52 =\( \frac{40x + (y - p) 60 }{ (10 - p)}\)

=> 40x + 60y - 60p = 520 - 52 p-----(2)


(1) - (2) gives,

=> 60 p = 40 + 52p

=> 8p = 40

=> p = 5

Answer E
_________________

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:

GMAT Assassins aren’t born, they’re made,
Rich
_________________

ALWAYS KEEP YOUR EYE ON THE ANSWER CHOICES.

The average price of the pieces of fruit that she keeps is 52 cents.
Since the cost of each piece of fruit is either 40 or 60, the total cost must be a MULTIPLE OF 10.
Given an average price of 52, the total cost will be a multiple of 10 only if the quantity of fruit is an integer that ends in 5 or 0.

How many oranges must Mary put back?
Since the original quantity of fruit = 10, Mary must put back option E -- 5 pieces of fruit -- to yield a new quantity that ends in 5 or 0:
10-5 = 5


_________________

here price of apple = 40 cents and price of orange = 60 cents

average price = 56 cents

so, the apples and oranges are in ratio

apple : orange = 60 - 56 : 56 - 40
=> apple : orange = 4 : 16
=> apple : orange = 1 : 4

total no =10, therefore apples = 2 nos. and oranges = 8 nos.


now we have to make average price = 52 cents
so, the apples and oranges will be in ratio

apple : orange = 60 - 52 : 52 - 40
=> apple : orange = 8 : 12
=> apple : orange = 2 : 3

now we have apples = 2 nos. from above, therefore oranges must be 3 in number

oranges to be put back = 8 - 3 = 5 nos

Answer choice E

kudos if you like the explanation

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

Let number of Apples = A
number of oranges = B
A+B=10 --- 1
.56 =(.4A + .6 B)/10
=> 56 = 4A + 6B ----2
Solving 1 and 2, we get
A= 2
B= 8

Let the number of oranges put back = C
52*(10-c) = 40*2 + 60(8-C)
=> C= 5
Answer E

Alternatively , we can use Scale method
Since average is .56 , distance of the average from price of apple =.56-.4=.16
and distance of the average from price of orange = .6 - .56 = .04
Ratio of distances is .16:.04 = 4:1

Therefore , number of apples and oranges will be in inverse proportion =1:4
Number of apples = 2
Number of oranges = 8

After removing a certain number of oranges C , the new average will be .52
Ratio of distances is .08:.12 = 2:3

Number of oranges will be 3
Therefore Mary must put 5 oranges back .

Answer E
_________________

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.

A = number of apples
B = number of oranges

We have :

\(\frac{40A + 60 B}{10}\) = 56


and \(\frac{40A + 60 (B - x)}{10 - x}\) = 52

\(\frac{40A + 60 B - 60 x}{10 - x}\) = 52

\(\frac{560 - 60 x}{10 - x}\) = 52

560 - 60 x = 520 - 52 x

40 = 8 x

x = 5

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!

The current total price is divisible by 10, and so is the price of each orange. The adjusted total price therefore will have to be divisible by 10, and at the same time divisible by 52. The only possible value for the adjusted total price is 260. Hence 300 cents' worth of oranges, or 5 oranges, will have to be removed.

\(\frac{Wa}{Wo} = \frac{(60-56)}{(56-40}) = \frac{1}{4}\)
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
_________________

Not convinced mate... I am able to get to fractions 1/4 and 2/3.... but given the total number of items is 10, why would make the ratio of 1/4 to 2/8 and keep the ratio 2/3 as it is ?

If someone can breakdown this using the same formula that would be helpful.

Please find the solution as attached.

Answer: option E

I hope this helps!!!

_________________

The original ratio of number of apples to number of oranges we found is 1:4. So there is 1 apple for every 4 oranges. Total number of fruits on the ratio scale is 1+4 = 5. Total number of fruits is actually 10. So actually, there must have been 2 apples and correspondingly, 8 oranges.

With the new average of 52, the ratio of apples to oranges is 2:3. So for every 2 apples, there are 3 oranges. We don't know the total number of fruits after putting oranges back.

Now note that number of apples doesn't change. Only oranges are put back. In original case, we found that there were 2 apples. So finally also there will be 2 apples only. Since the ratio of apples to oranges is 2:3 and number of apples actually is 2, number of oranges actually must be now 3.

From 8, number of oranges has gone down to 3. So 5 oranges have been put back.
_________________