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balls

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balls

by ketkoag » Mon May 04, 2009 4:56 am
Each of four identical containers contains m balls. By moving some balls from the first to other three containers, the ratio of the number of the balls in them changed to 1:6:5:4. How many balls were moved from the first container?
A. 0 B. m/2 C. 3m/4 D 2m/3 E 5m/6
Answer: C
please lemme know how to solve it?
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by g2000 » Mon May 04, 2009 6:02 am
We have container A, B, C and D.
Before, they have m balls.

Let x be the # balls taken:
y be the total # of balls
b4 after
A m y/16
B m 6y/16
C m 5y/16
D m 4y/16

For container A,
m-x = y/16
y= total of all balls = 4m

m-x = 4m/16
x = m-m/4 = 3/4m
answer is C
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Re: balls

by Ian Stewart » Mon May 04, 2009 6:29 am
ketkoag wrote:Each of four identical containers contains m balls. By moving some balls from the first to other three containers, the ratio of the number of the balls in them changed to 1:6:5:4. How many balls were moved from the first container?
A. 0 B. m/2 C. 3m/4 D 2m/3 E 5m/6
Answer: C
please lemme know how to solve it?
In the end, the balls are in a 1:6:5:4 ratio. That is, the first container contains 1/(1+6+5+4) = 1/16 of the balls. If all the containers initially had the same number of balls, they each must have contained 4/16 of the total. So the number in our first container dropped from 4/16 of the total to 1/16 of the total, and we must have removed 3 of every 4 balls from the first container. Since it contained m balls to begin with, we removed 3m/4 balls.

Of course, there's no need to do this abstractly. If you recognize that, in the end, the first container contains 1/16 of the total, there's no harm in pretending that there are 16 balls in total, and that m=4, and solving the problem using that value.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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