Each of the 30 boxes in a certain shipment with either 10 : PS Archive
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# Each of the 30 boxes in a certain shipment with either 10

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Each of the 30 boxes in a certain shipment with either 10 [#permalink]

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12 Nov 2006, 08:50
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Each of the 30 boxes in a certain shipment with either 10 pounds or 20 pounds, and the average (arithmetic mean) weight of the boxes in the shipment is 18 pounds. If the average weight of the boxes in the shipment is to be reduced to 14 pounds by removing some of the 20-pound boxes, how many 20-pound boxes must be removed?

A. 4
B. 6
C. 10
D. 20
E. 24
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12 Nov 2006, 17:09
Answer D. Here is my approach:

Average = Sum/no of items. Since the average before loading is 18 and there are 30 items. SUM = 18*30.
Later on, we want the average to be 14. The no of items is now 30-x with x being the number of 20 lbs item we need to unload. Sum = 14 (30-x).

The diff between the sum before and after unloading is the 20*x since we take away x item of the 20 lbs. So, the equation is
18*30 - 14(30-x) = 20x <=> 18*30 - 14*30 = 20x - 14x
4*30 = 6x so x=20
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12 Nov 2006, 21:36
# of 20 pound boxes = n
# of 10 pound boxes = 30-n

Average weight = [20n + 10(30-n)]/30 = 18
20n + 300 - 10n = 540
10n = 240
n = 24

# of 20 pound boxes = 24
# of 10 pound boxes = 6

Average weight = 14

Assuming x number of 20 pound boxes removed, then average weight is:

[(24-x)(20) + 6(10)]/(30-x) = 14

480-20x+60 = 420-14x
6x = 120
x = 20
12 Nov 2006, 21:36
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