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

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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


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[Reveal] Spoiler: OA

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Bunuel 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


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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

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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]

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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
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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.

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Bunuel 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


Kudos for a correct solution.


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

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New post 22 Mar 2016, 03:34
v12345 wrote:
Bunuel 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


Kudos for a correct solution.


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


Hi!

I used the exact same way as you did in solving this question, but when reaching the last part 2:3 I interpereted it as 3/5 oranges and 2/5 apples? Seeing as we're still closer to 60 than 40. And then we should have 6 oranges and 4 apples? Please explain!

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My bad, I see now that we were not supposed to replace them with any apples but just remove them from the total. Didn't read carefully enough :)

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Bunuel 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


Kudos for a correct solution.


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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Last edited by ScottTargetTestPrep on 04 Sep 2016, 06:29, edited 1 time in total.

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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]

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New post 04 May 2016, 15:07
let a=#apples
r=#oranges
we know that .4a+.6r=(.56)(10)=5.6➡.4a=5.6-.6r
after mary returns x oranges, then
.4a+.6(r-x)=(.52)(10-x)
substituting for .4a,
5.6-.6r+.6r-.6x=5.2-.52x
.08x=.4
8x=40
x=5 oranges put back

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At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]

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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

Last edited by Alex75PAris on 18 May 2016, 10:02, edited 1 time in total.

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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]

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New post 14 May 2016, 11:28
Using Averages and Allegations

40 60
56
4 16

A:O =1: 4==>A=2 ,O=8(as total no of apples+ oranges =10)
now we know there are 2 Apples and 8 Oranges

Using Statistics


when Mean =52

52-40=12
12*2(bcos 2 apples)
60-52=8
x= no. of remaining oranges
24-(8*x)=0
==>x=3
Hence 5 oranges must be removed.

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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!

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Although I got this right, it took me 10 minutes to solve it. Can anyone help me with a strategy to tackle this type of question on GMAT? There is no way im going to solve this question on the exam.

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New post 18 Jun 2016, 05:08
total amount = avg * number of items
old amount = 56 * 10 = 560

new amount will be 52 *10-x
10-x , you have same number of apples , only decrease in number of oranges
now new amount is equal to what
560 - 60 x = new amount
solve and x= 5

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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.

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Bunuel 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


Kudos for a correct solution.


\(\frac{Wa}{Wo} = \frac{(60-56)}{(56-40}) = \frac{1}{4}\)
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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Re: At a certain fruit stand, the price of each apple is 40 cents and the [#permalink]

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New post 15 Oct 2016, 03:01
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.

colorblind wrote:
Bunuel 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


Kudos for a correct solution.


\(\frac{Wa}{Wo} = \frac{(60-56)}{(56-40}) = \frac{1}{4}\)
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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Dev1212 wrote:
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.

Bunuel 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


Kudos for a correct solution.




Please find the solution as attached.

Answer: option E

I hope this helps!!!
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