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

Question

At a certain fruit stand, the price of each apple is 40 cents and the price of each orange is 60 cents. Mary selects both apples and oranges from the fruit stand, and the average (arithmetic mean) price of the 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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E7B520F4-FAD9-4574-8D3E-E1554A4658BA.jpeg [ 783.46 KiB | Viewed 3264 times ]

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

DanTe02 · Answer

Note this is a modified question from Official Guide not exactly the same

Abhishek009 · Answer

40 ____________________ 60

__________ 56 __________

4 ____________________ 16

Or,  2 : 8 , Total 10 fruits, 2 apples and  8 oranges

40 ____________________ 60

__________ 52 __________

8 ____________________ 12

Or,  2 : 3  Total 5 fruits, 2 apples and  3 oranges

T hus, she must keep back 5 fruits, (ie 8 - 3 ) Oranges, Answer must be (E)

MBAHOUSE · Answer

At a certain fruit stand, the price of each apple is 40 cents and the price of each orange is 60 cents. Mary selects both apples and oranges from the fruit stand, and the average (arithmetic mean) price of the 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

Attachments

E7B520F4-FAD9-4574-8D3E-E1554A4658BA.jpeg [ 783.46 KiB | Viewed 3264 times ]

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

L2 · Answer

Let the number of apples is x and oranges is y. and we have to reduce n number apples. 
so (40x+60y)/(x+y)=56 
 => y =4x
 the number of oranges is 4 times more than apples
again we can write, 
{40x+60(4x-n)}/(x+4x-n)=52 
=> n=5x/2 
since the number of apple should be integer, we should assume the even value for x.
if x=2 then y= 8 
 then n=5 
 if x=4, then y=16 
 and n will be 10 
if we take the minimum value for x is 2 
then n=5 
ans: E

Posted from my mobile device

samarpan.g28 · Answer

Athough we can solve using algebra with the weighted average method assuming variables for the number of apples and oranges, alligation will be a faster approach. In allegation, the smaller quantity is written in the numerator and the larger in the denominator, after that, we have to subtract the average from each ensuring the ratio is positive. I am keeping orange on top for my ease.

Let's find the ratio of the prices of Oranges and Apples from the mean price given as 60: (56-40)/(60-56)=16/4=4/1...(i)
Later, when the mean changes to 52, the ratio will be: (52-40)/(60-52)=12/8=3/2...(ii)

For ratio (i), we can also assume it as 8/2 (multiplying 2 with both num and denom) so that it aligns with ratio (ii).
Therefore, we can say that the number of oranges changed from 8 to 3 i.e. by 8-3=5. Option (E) is correct.

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

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GMAT Problem Solving (PS) Questions

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

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