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Food Stand [#permalink] New post 07 Jun 2010, 02:31
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Q3:
At a certain food stand, the price of each apple is ¢ 40 and the price of each orange is ¢ 60. 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 ¢ 56. How many oranges must Mary put back so that the average
price of the pieces of fruit that she keeps is ¢ 52?
A. 1
B. 2
C. 3
D. 4
E. 5

[Reveal] Spoiler:
Answer:E. I solved the question and got ans as 2 :(

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Re: Food Stand [#permalink] New post 07 Jun 2010, 04:18
A + O = 10 .....1

\frac{40A + 60O}{10} = 56

=> 2A + 3O = 28

Substitute equation 1:

=> 2A + 3*(10-A) = 28

=> 2A + 30 - 3A = 28

=> A = 2 From 1: O = 8

Let's say she put back "r" oranges.

=> \frac{40*2 + 60*(8 - r)}{10 - r} = 52

=> 80 + 480 - 60r = 520 - 52r

=> 8r = 40

r = 5

Pick E
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Re: Food Stand [#permalink] New post 02 Aug 2010, 15:51
This is a funny question.
I also solved and got the answer as 2, and after checking AbhayPrasanna's method the answer can also be 5 as demonstrated and if u substitue for 6 Oranges and 4 Apples or 3 Oranges and 7 Apples, the average turns out to be 52cents.

So is the question wrong or am I missing something here ?
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Re: Food Stand [#permalink] New post 02 Aug 2010, 19:56
devashish wrote:
This is a funny question.
I also solved and got the answer as 2, and after checking AbhayPrasanna's method the answer can also be 5 as demonstrated and if u substitue for 6 Oranges and 4 Apples or 3 Oranges and 7 Apples, the average turns out to be 52cents.

So is the question wrong or am I missing something here ?


As per the first two statements in the question(A+0=10 and (.4A+.6O)/10=5.6) no of apples is 2.
The last statement specified that she must put back only oranges to make the average 52cents. The no of apples should still remain 2 and can't change.
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Re: Food Stand [#permalink] New post 02 Aug 2010, 20:28
Here is my work (after I realized I could only put back oranges and not switch an orange for an apple)

a=apples
o=oranges
t=total

10 = a + o
5.60 = .4a-.6o
Substitute
5.60 = .4a-.6(10-a)
solution is a=2
10 = 2 + o
o=8

So we have 8 oranges and 2 apples

5.20(t-2)=.4(2)-.6(t-2)
solution is t=5

Total - Apples = Oranges
5 - 2 = 3

In the beginning there were 8 oranges, now there are 3 so we must have given back 5. E!
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Re: Food Stand [#permalink] New post 02 Aug 2010, 21:08
My answer is B = 2

With the help of the alligation Rule we can determine the original ratio as being 2:8

Alligation Rule (Cost of the Highest item less the Mean Cost) : (Mean Cost minus Lowest Cost Item)

(60 - 56) : (56 - 40 )

(4) : (16) = 2 : 8

So, utilizing the same rule for the new Mean 52

We get

(60 - 52) : (52 - 40)
8 : 12
4 : 6
Therefore the answer is 2... B
Regards from Mexico City!
Re: Food Stand   [#permalink] 02 Aug 2010, 21:08
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