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by dunkin77 » Thu Jun 14, 2007 11:52 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


Hi,

For some reason my answer is B...which is not correct. Can anyone help?
Source: — Problem Solving |

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by drhomler » Thu Jun 14, 2007 12:13 pm
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?


30*18=540 total weight
30*14=420 total weight desired
120 is the differnce which if you divide 120 by 20 you get 6

Im getting B too-not sure if Im missing something here but I dont see another way to do it. What is the OA? Jay(or anyone), assistance s'il vous plait

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by dunkin77 » Thu Jun 14, 2007 12:22 pm
OA is D :? - Jay pls help....

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by jayhawk2001 » Thu Jun 14, 2007 2:18 pm
Let x be the number of 10 lb boxes. We have

10x + 20(30-x) = 18*30

x = 6. So, there are 6 10lb boxes and 24 20lb boxes.

Let y be the number of 20lb boxes you want to remove

10*6 + 20*(24-y) = 14 * (6+24-y)
60 + 480 - 20y = 420 - 14y
6y = 120
y = 20

Hence D

Hope this helps, guys.

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by jayhawk2001 » Thu Jun 14, 2007 2:22 pm
drhomler wrote: 30*18=540 total weight
30*14=420 total weight desired
Just wanted to mention that you cannot do 30*14, as the average is
computed over total number of boxes i.e. excluding the boxes removed.

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by drhomler » Thu Jun 14, 2007 2:46 pm
Thanks-I had a feeling like it was something to that affect but I wasnt quite grasping the two seperate equations and I ended up with two variables and thought that couldnt be the way. What difficulty level would you peg this one at?

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by dunkin77 » Thu Jun 14, 2007 3:16 pm
I made mistake that you pointed out - thank you for your help!