Three jars contain the same numbers of balls. 2 balls from t

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[GMAT math practice question]

Three jars contain the same numbers of balls. 2 balls from the first jar are moved to the second jar and 4 balls from the first jar are moved to the third jar. If the ratio between the numbers of balls in the second and third jars after moving the balls is 13 to 14, how many balls were originally in each jar?

A. 12
B. 15
C. 21
D. 24
E. 30

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by swerve » Fri Jun 22, 2018 2:34 pm
Three jars with the same number of balls. Let number be X.

So total = 3X

As per information, two balls removed from 1st jar and put into 2nd jar. So balls in 2nd jar = X + 2.

4 balls removed from 1st jar and put into 3rd jar. So, number os balls in 3rd jar = X + 4.

Also, balls ratio is given for 2nd jar / 3rd jar = 13/14.
Equation becomes, X + 2 / X + 4 = 13 / 14.
Solving we get X = 24.

Hence, D is the correct answer. Regards!

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by Max@Math Revolution » Sun Jun 24, 2018 6:05 pm
=>
Suppose each jar originally contained n balls.
After moving balls from the first jar to the second and third jars, the second and the third jars contain n + 2 and n + 4 balls, respectively.
So,
n+2 : n+4 = 13 : 14
=> 13(n+4) = 14(n+2)
=> 13n + 52 = 14n + 28
=> n = 24
Therefore, the answer is D.
Answer: D

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by Scott@TargetTestPrep » Mon Jun 25, 2018 11:10 am
Max@Math Revolution wrote:
Three jars contain the same numbers of balls. 2 balls from the first jar are moved to the second jar and 4 balls from the first jar are moved to the third jar. If the ratio between the numbers of balls in the second and third jars after moving the balls is 13 to 14, how many balls were originally in each jar?

A. 12
B. 15
C. 21
D. 24
E. 30
Let x = the number of balls in each jar originally. After all balls were moved, we see that the second jar had (x + 2) balls, and the third jar had (x + 4) balls, resulting in a new ratio between the balls in those two jars of 13 to 14. Thus, we have:

(x + 2)/(x + 4) = 13/14

14(x + 2) = 13(x + 4)

14x + 28 = 13x + 52

x = 24

Answer: D

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