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Apples and Oranges!

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Apples and Oranges!

by zagcollins » Mon Jul 14, 2008 8:31 am
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
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by AleksandrM » Mon Jul 14, 2008 9:32 am
E.

40A + 60Or/10 = 56 or 40A + 60Or = 560

and

40A + 60(Or - x)/10 - x = 52 or 40A + 60Or - 60x = 520

Subtract the first equation from the secon and you end up with

8x = 40

x = 5

So you need to put back 5 oranges.
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by sudhir3127 » Mon Jul 14, 2008 9:46 am
My answer is 5.

Assume no of apples = X and oranges be Y

equation one is 40X+60 Y = 560

eqn 2 is X+Y=10

solve this equation and u get X =2 and Y =8.

to get the number of oranges to be reduced to get the average @52

40(2)+60(8-x)/(10-x)=52

there fore X =5
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by egybs » Mon Jul 14, 2008 4:52 pm
Here's a little trick to make things a bit easier and (at least for me) less error prone.

Instead of using 60c and 40c as the prices or oranges and apples, use 20c and 0c, respectively. You simply subtract the same number from both values so that one becomes a 0. Then, with the averages we have and are looking for, subtract that same number. So we currently have an average of 16c, and we're looking for an average of 12c.

Then the algebra becomes simpler since we don't have to worry about the apples (as they are now free).

We get:
x: number of oranges; y: number of oranges removed

20x/10 = 16
or:
20x = 160


20(x-y)/(10-y) = 12
or:
20x-20y=120-12y

Then substitute 20x for 160:
160-20y =120-12y
40=8y
y=5
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