Problem Solving

Hello Everyone !!

This is the question:

At a certain food stand, the price of each apple is $0.4 and the price of each orange is $0.6. Mary selects a total of 10 apples and oranges from the food stand, and the average (arithmetic mean) price of the 10 pieces of fruit is $0.56. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is $0.52?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

OA is E

I did like this--

Let no. of apples = x and no. of oranges = y
Then x + y = 10...equation 1
(0.40x + 0.60y)/10 = 0.56 or 4x + 6y = 56 or 2x + 3y = 28....equation 2
Solving equation 1 and 2, we get x = 2, y = 8

Now let us assume that Mary puts back P oranges. So, the new average is:
{0.60(8 - P) + 0.40 * 2}/(10 - P) = 0.52
(5.6 - 0.60P) = 5.2 - 0.52P
0.4 = 0.08P
P = 5

The correct answer is [spoiler]E[/spoiler].

Is there any shortcut method for this question ? The reason why I am asking is that I think it is little bit difficult to solve this question in less than 2 min.

Thank You

Check my two posts here:

https://www.beatthegmat.com/a-certain-fo ... 16515.html

Hey Param,

I solved exactly as you did and it took me 4 minutes . Looked at Mitch's reasoning, which seems very fast. I guess its all about recognizing the short way ...

Exactly, I liked the Mitch's reasoning especially the one in which he uses the method of ALLIGATION... it was supper fast

gander123 wrote:Hey Param,

I solved exactly as you did and it took me 4 minutes . Looked at Mitch's reasoning, which seems very fast. I guess its all about recognizing the short way ...