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At a certain food stand the price of each apple is 40 cents and the pr

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

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New post 03 Apr 2011, 02:35
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A
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E

Difficulty:

  55% (hard)

Question Stats:

57% (01:49) correct 43% (01:10) wrong based on 7 sessions

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

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New post 03 Apr 2011, 02:56
Acer86 wrote:
At a certain food 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 food 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


Let the number of Apples purchased be "A"
Let the number of Oranges purchased be "R"

Mary selects a total of 10 apples and oranges i.e.
\(A+R=10\) -----1

the average (arithmetic mean) price of the 10 pieces of fruit is 56 cents:
\(\frac{40A+60R}{10}=56\)
\(4A+6R=56\)
\(2A+3R=28\)-----2

Solving 1 and 2:
\(A=2\)

\(R=8\)

Let's assume that she takes "x" oranges out;
Average price of the pieces of fruit that she keeps is 52 cents

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

Substituting A and R as 2 and 8 respectively

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

\(80+480-60x = 520- 52x\)

\(-60x + 52x= -80-480+520\)

\(-8x= -40\)

\(x= 5\)

Ans: "E"
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Re: At a certain food stand the price of each apple is 40 cents and the pr  [#permalink]

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New post 03 Apr 2011, 03:04
apples cost 40 cents
oranges cost 60 cents
40 < 56 < 60. 56 cents is the average price when she bought 10 fruits.

Given Apples + Oranges = 10
Their numbers are in ratio
apples : oranges = (60 - 56) : (56 - 40) = 4 : 16 = 1:4

Hence apples = 1/5 * 10 = 2
Oranges = 4/5 * 10 = 8

New average price of the fruits is 52 cents. Their numbers are in ratio
apples : oranges = (60 - 52) : (52 - 40 ) = 8/12 = 2:3

Since apples are 2. Hence oranges are 3. She must put back (8 - 3) = 5 oranges
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Re: At a certain food stand the price of each apple is 40 cents and the pr  [#permalink]

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New post 03 Apr 2011, 05:24
Thanks guys. Any idea as to what difficulty level this type of question would be on an actual GMAT exam?
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Re: At a certain food stand the price of each apple is 40 cents and the pr  [#permalink]

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New post 03 Apr 2011, 23:21
fluke wrote:
Acer86 wrote:
At a certain food 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 food 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


Let the number of Apples purchased be "A"
Let the number of Oranges purchased be "R"

Mary selects a total of 10 apples and oranges i.e.
\(A+R=10\) -----1

the average (arithmetic mean) price of the 10 pieces of fruit is 56 cents:
\(\frac{40A+60R}{10}=56\)
\(4A+6R=56\)
\(2A+3R=28\)-----2

Solving 1 and 2:
\(A=2\)

\(R=8\)

Let's assume that she takes "x" oranges out;
Average price of the pieces of fruit that she keeps is 52 cents

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

Substituting A and R as 2 and 8 respectively

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

\(80+480-60x = 520- 52x\)

\(-60x + 52x= -80-480+520\)

\(-8x= -40\)

\(x= 5\)

Ans: "E"


Hi - Thanks for the detailed answer... I think algebraic method to solve these questions should be avoided in the exam. Using alligation/mixtures method in the exam will save time. No offence... Just an opinion...
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Re: At a certain food stand the price of each apple is 40 cents and the pr  [#permalink]

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New post 03 Apr 2011, 23:41
x + y = 10


0.40x + 0.60y = 0.56

=> 40x + 60y = 560

=> 20y = 160

=> y = 8, x = 2

(8-p)*0.60 + 2 * 0.40 = 0.52 *(10 - p)

4.8 - 0.60p + 0.80 = 5.2 - 0.52p

=> 5.60 - 5.20 = 0.08p

=> 0.4 = 0.08p

=> p = 5

Answer - E
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Re: At a certain food stand the price of each apple is 40 cents and the pr  [#permalink]

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New post 04 Apr 2011, 00:43
gmatprep2011 wrote:
Hi - Thanks for the detailed answer... I think algebraic method to solve these questions should be avoided in the exam. Using alligation/mixtures method in the exam will save time. No offence... Just an opinion...


Point taken. Thanks.
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Re: At a certain food stand the price of each apple is 40 cents and the pr  [#permalink]

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New post 18 May 2011, 22:47
allegation formula

a/ or = 60-56/56-40 = 1/4 means a=2 and or = 8

a/or = 60-52/52-40 = 2/3 mean a = 2 and or = 3

hence 8-5 = 3

5

--== Message from the GMAT Club Team ==--

THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.


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To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

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Re: At a certain food stand the price of each apple is 40 cents and the pr &nbs [#permalink] 18 May 2011, 22:47
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