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OG12 Problem... What am I doing wrong?

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OG12 Problem... What am I doing wrong?

by thp510 » Mon Feb 07, 2011 7:35 am
If 3 liters of water are added to Maria's bucket when it is half full of water, the amount of water in the bucket will increase by (1/3). Marcia's bucket can hold a maximum of how many liters of water?


18 liters of water

I initially approached it using the following equation. What is my incorrect equation saying???

B=Marcia's bucket at full capacity.

[(1/2)B]+3 = 1/3B+[(1/2)B]

(B/2)+(6/2) = (2B/6) (3B/6)

(B+6)/2 = 5B/6

Solve for B, [spoiler]B=8[/spoiler]
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by Geva@EconomistGMAT » Mon Feb 07, 2011 8:09 am
thp510 wrote:If 3 liters of water are added to Maria's bucket when it is half full of water, the amount of water in the bucket will increase by (1/3). Marcia's bucket can hold a maximum of how many liters of water?


18 liters of water

I initially approached it using the following equation. What is my incorrect equation saying???

B=Marcia's bucket at full capacity.

[(1/2)B]+3 = 1/3B+[(1/2)B]

(B/2)+(6/2) = (2B/6) (3B/6)

(B+6)/2 = 5B/6

you're taking the additional 1/3 out of the B, instead of the current amount of water (half of B)...which is exactly what whoever wrote this question wanted you to do to get it wrong.

the correct formula would be
[(1/2)B]+3 = [(1/2)B] + 1/3*[(1/2)B, or 4/3([(1/2)B]

In order to avoid the problem altogether, use real numbers instead: if 3 liters are 1/3 of the amount in the bucket, then the bucket currently holds 9 liters, which is half of 18.

Solve for B, [spoiler]B=8[/spoiler]
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by thp510 » Mon Feb 07, 2011 8:17 am
Geva@MasterGMAT wrote:
thp510 wrote:If 3 liters of water are added to Maria's bucket when it is half full of water, the amount of water in the bucket will increase by (1/3). Marcia's bucket can hold a maximum of how many liters of water?


18 liters of water

I initially approached it using the following equation. What is my incorrect equation saying???

B=Marcia's bucket at full capacity.

[(1/2)B]+3 = 1/3B+[(1/2)B]

(B/2)+(6/2) = (2B/6) (3B/6)

(B+6)/2 = 5B/6

you're taking the additional 1/3 out of the B, instead of the current amount of water (half of B)...which is exactly what whoever wrote this question wanted you to do to get it wrong.

the correct formula would be
[(1/2)B]+3 = [(1/2)B] + 1/3*[(1/2)B, or 4/3([(1/2)B]

In order to avoid the problem altogether, use real numbers instead: if 3 liters are 1/3 of the amount in the bucket, then the bucket currently holds 9 liters, which is half of 18.

Solve for B, [spoiler]B=8[/spoiler]

Awesome. Thanks!
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by GMATGuruNY » Mon Feb 07, 2011 9:05 am
thp510 wrote:]Marcia's bucket can hold a maximum of how many liters of water?

(1) The bucket currently contains 9 liters of water.
(2) If 3 liters of water are added to the bucket when it is half full of water, the amount of water in the bucket will increase by 1/3 .
Since this is a DS question, we shouldn't be solving for B. We should determine only whether there is sufficient information to solve for B.

Statement 1: Bucket currently contains 9 liters.
No way to determine the maximum amount that the bucket can hold.
Insufficient.

Statement 2: If 3 liters of water are added to the bucket when it is half full of water, the amount of water in the bucket will increase by 1/3.
Since the bucket is half full (1/2*B), and 3 liters increases the amount in the bucket by 1/3, we get:
3 = 1/3 * 1/2 * B
Since we can solve for B, sufficient.

The correct answer is B.
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