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by bobdylan » Tue Jun 12, 2012 4:34 am
Each of the 30 boxes in a certain shipment weighs either 10 pound or 20 pounds, and the arithmetic mean of the boxes in the shipment is 18 pounds. If the average weight of the boxes 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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by raunekk » Tue Jun 12, 2012 5:13 am
IMO : D = 20

a = 10 pound boxes
b = 20 pound boxes

a+b = 30 --- (1)
10(a) +20 (b) = 30* 18 ( given)

10a + 20 b = 540 --(2)

Solving 1 & 2, a= 6, b = 24

To reduce average to 14 by removing some 20 pound boxes

We use equation (2) to form new equation for average 14

10(6) + 20 (24-x) = 14 (30-x) ( Where x is the number of 20 pound boxes to be removed)

Solving, we get x = 20 (D)

Hope this helps!
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by GMATGuruNY » Tue Jun 12, 2012 6:36 am
bobdylan wrote:Each of the 30 boxes in a certain shipment weighs either 10 pound or 20 pounds, and the arithmetic mean of the boxes in the shipment is 18 pounds. If the average weight of the boxes 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
Total current weight = number*average = 30*18 = 540.
We can plug in the answers, which represent the number of 20-pound boxes that must be removed to reduce the average weight to 14.

Answer choice C: 10 of the 20-pound boxes
New total weight = 540 - 10*20 = 340.
New average for the 20 remaining boxes = 340/20 = 17.
To reduce the average weight further, more 20-pound boxes must be removed.
Eliminate A, B and C.

Answer choice D: 20 of the 20-pound boxes
New total weight = 540 - 20*20 = 140.
New average for the 10 remaining boxes = 140/10 = 14.
Success!

The correct answer is D.
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