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MGMAT - Word Problem

by diehard_gmat » Fri Mar 18, 2011 12:30 am
Each of the 30 boxes in a certain shipment weighs either 10 pounds or 20 pounds, and 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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by Anurag@Gurome » Fri Mar 18, 2011 12:56 am
diehard_gmat wrote:Each of the 30 boxes in a certain shipment weighs either 10 pounds or 20 pounds, and 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
Before removal:
  • Say, number of 10 pound boxes is n.
    Hence, number of 20 pound boxes is (30 - n)

    Therefore, [10n + 20*(30 - n)] = 18*30 ........................... (A)
After removal:
  • Number of 10 pound boxes is n
    Say, number of 20 pound boxes removed is m
    Hence, number of 20 pound boxes is (30 - n - m)

    Therefore, [10n + 20*(30 - n - m)] = 14*(30 - m) ............ (B)
Subtracting (B) from (A)
  • 20m = 18*30 - 14*30 + 14m
    => 6m = 4*30
    => m = 4*5 = 20
The correct answer is D.
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by sanju09 » Fri Mar 18, 2011 1:05 am
diehard_gmat wrote:Each of the 30 boxes in a certain shipment weighs either 10 pounds or 20 pounds, and 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

Let there be x number of 20-pound boxes and hence (30 - x) number of 10-pound boxes in the shipment, such that the average (arithmetic mean) weight of the boxes in the shipment, which is

[20 x + 10 (30 - x)]/ 30 = 18 (given)

or, 10 x = 540 - 300, or x = 24

Now, there are 24 20-pound boxes and 6 10-pound boxes, in the shipment. Coming to the question, let's say we need to remove n number of 20-pound boxes from the shipment in order to reduce the average weight of the boxes in the shipment to 14 pounds, which means that

[20 (24 - n) + 10 × 6]/ (30 - n) = 14

or, 540 - 20 n = 420 - 14 n

or, 6 n = 120

or, n = [spoiler]20


D[/spoiler]
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by manpsingh87 » Fri Mar 18, 2011 1:19 am
diehard_gmat wrote:Each of the 30 boxes in a certain shipment weighs either 10 pounds or 20 pounds, and 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
let no. of 10 pound boxes be x and no. of 20 pound boxes be y

10x+20y=540;

also x+y=30

solving we get x=6; y=24

now as average will reduced to 14 pounds after removing m boxes of 20 pound,therefore;

10*6+20(24-m)=14*(30-m);
solving this we get m=20;

hence correct answer is D
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by kevincanspain » Fri Mar 18, 2011 3:18 am
diehard_gmat wrote:Each of the 30 boxes in a certain shipment weighs either 10 pounds or 20 pounds, and 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
The 30 boxes weigh a total of 18(30) pounds

If x 20-pound boxes are removed, the remaining ones will weigh 540 - 20x

Thus 14= (540 - 20x)/(30 - x)

14/20 = (27 - x)/(30 - x)

7/10 = (27 - x)/(30 - x)

x= 20
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by sanju09 » Fri Mar 18, 2011 3:25 am
kevincanspain wrote:
diehard_gmat wrote:Each of the 30 boxes in a certain shipment weighs either 10 pounds or 20 pounds, and 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
The 30 boxes weigh a total of 18(30) pounds

If x 20-pound boxes are removed, the remaining ones will weigh 540 - 20x

Thus 14= (540 - 20x)/(30 - x)

14/20 = (27 - x)/(30 - x)

7/10 = (27 - x)/(30 - x)

x= 20
that's the quickest one
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by GMATGuruNY » Fri Mar 18, 2011 3:26 am
diehard_gmat wrote:Each of the 30 boxes in a certain shipment weighs either 10 pounds or 20 pounds, and 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
Total current weight = 30*18=540.

An easy and efficient solution is to plug in the answer choices, which represent the number of 20-pound boxes that need to be removed.

Answer choice C: 10 of the 20-pound boxes removed
New total weight = 540 - 10*20 = 340.
New number of boxes = 30-10 = 20.
New average = 340/20 = 17.
Need to remove more boxes.
Eliminate A, B, and C.

Answer choice D: 20 of the 20-pound boxes removed
New total weight = 540 - 20*20 = 140.
New number of boxes = 30-20 = 10.
New average = 140/10 = 14. Success!

The correct answer is D.
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