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GMAT Prep / Arithmetic

Expert replies
by Blast » Wed Dec 03, 2008 1:09 pm
Anyone?
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Source: — Problem Solving |

by Zipper » Wed Dec 03, 2008 1:50 pm
We have:

x=lightest box
y=median box
z=3rd box
y=9

x+y+z/3=7 => x+y+z=21 => x+z=12

We replace z with the lowest possible weight (in order to find the maximum possible wight for x), so z=9 (they CAN be equal with y, 9 is still the median value). So x=3.

hence, C
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by lunarpower » Thu Dec 04, 2008 12:19 am
a good, and surprisingly simple, way to impose structure on problems involving the median of a set is to write a series of BLANKS that represent the numbers in the set IN ORDER, and then fill them in as appropriate.

start with 3 blanks:
_ _ _

in this set, we're given that the median is 9. this means that the middle one of the 3 blanks is 9:
_ 9 _

we want to maximize the left hand blank.
general takeaway: to MAXIMIZE a quantity, MINIMIZE all OTHER QUANTITIES over which you have discretion.
in this problem, that just means minimize the right-hand blank.
the right-hand blank must be > 9; otherwise it wouldn't be the right-hand blank anymore (remember, the blanks are in increasing order).
so, to minimize it, put a 9 in it.
_ 9 9

finally, the average is 7, meaning that the sum is 7 x 3 = 21.
21 - 9 - 9 = 3.
3 9 9

done.
Ron has been teaching various standardized tests for 20 years.

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by jnellaz » Thu Dec 04, 2008 7:36 am
Great approach Lunarpower. Thanks!
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why not 1, 9, 11?

by mayuran23 » Tue Apr 14, 2009 3:52 pm
never mind... my biggest mistake is in not reading the question thoroughly.

thanks very much for the explanation
Last edited by mayuran23 on Tue Apr 14, 2009 7:18 pm, edited 1 time in total.
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by mbadrew » Tue Apr 14, 2009 5:44 pm
Ron,

thanks for the explanation.

Andrew
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