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A and B are two partially filled buckets of water. If 5 liters are tra

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A and B are two partially filled buckets of water. If 5 liters are tra [#permalink]

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A and B are two partially filled buckets of water. If 5 liters are transferred from A to B, then A would contain one-third of the amount of water in B. Alternatively, if 5 liters are transferred from B to A, B would contain one-half of the amount of water in A. Bucket A contains how many liters of water?

A. 11
B. 13
C. 17
D. 21
E. 23
[Reveal] Spoiler: OA

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Re: A and B are two partially filled buckets of water. If 5 liters are tra [#permalink]

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New post 11 Sep 2016, 04:44
Bunuel wrote:
A and B are two partially filled buckets of water. If 5 liters are transferred from A to B, then A would contain one-third of the amount of water in B. Alternatively, if 5 liters are transferred from B to A, B would contain one-half of the amount of water in A. Bucket A contains how many liters of water?

A. 11
B. 13
C. 17
D. 21
E. 23



Let A contains A, B contains B liters

so, (A-5)/(B+5)= 1/3.......(1)
again, (B-5)/(A+5)= 1/2..........(2)

from (1) & (2) we find A= 11

ans: (A)
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Re: A and B are two partially filled buckets of water. If 5 liters are tra [#permalink]

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New post 11 Sep 2016, 04:46
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Let Bucket A be A and Bucket B be B

Scenario 1
A-5 = 1/3 (B+5) ----> 3A-15 =B+5

Scenario 2
B-5 = 1/2 (A+5) -----> 2B-10 = A+5

From Scenario 1, B = 3A-20

Substitute B with this information in Stmt 2
2 (3A-20) -10 = A+5 ------> 6A -40-10 = A+5 ------> 6A- A = 50+5 ---> 5A= 55

A= 11, Answer choice A

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Re: A and B are two partially filled buckets of water. If 5 liters are tra [#permalink]

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New post 26 Dec 2017, 17:00
Bunuel wrote:
A and B are two partially filled buckets of water. If 5 liters are transferred from A to B, then A would contain one-third of the amount of water in B. Alternatively, if 5 liters are transferred from B to A, B would contain one-half of the amount of water in A. Bucket A contains how many liters of water?

A. 11
B. 13
C. 17
D. 21
E. 23

Algebra
If 5 liters are transferred from A to B, then A would contain one-third of the amount of water in B.

Either \(\frac{A-5}{B+5}=\frac{1}{3}\) OR

\(A - 5 = \frac{1}{3}(B + 5)\)
\(3A - 15 = B + 5\)
\(3A - 20 = B\)


Alternatively, if 5 liters are transferred from B to A, B would contain one-half of the amount of water in A.

\(B - 5 =\frac{1}{2}(A +
5)\)
\(2B - 10 = A + 5\)
\(2B - 15 = A\)


Rearrange so that B term is isolated:
\(A + 15 = 2B\)

Equations:
\(3A - 20 = B\)
\(A + 15 = 2B\)


B needs to be eliminated.
Multiply top equation by 2, bottom by (-1), then add

\(6A - 40 = 2B\)
\(- A - 15 = -2B\)
---------------------------------

\(5A - 55 = 0\)
\(5A = 55\)
\(A = 11\)


Answer A

Answer choices
Typically, start with Answer C because it yields a benchmark.

C) 17 liters of water in A?

5 liters from A to B, where A = 17:
A has: (17 - 5) = 12 left in A
12 is \(\frac{1}{3}\) of B. B is (3 * A)
B has: (3 * 12) = 36 after +5
Before +5, B was (36 - 5) = 31

5 liters from B to A:
B has: (31 - 5) = 26 left in B
26 is \(\frac{1}{2}\) of A. A is (2 * B)
A has: (2 * 26) = 52 after +5
Before +5, A was (52 - 5) = 47
\(47 \neq{17}\)

17 is much too great.
It sets off arithmetic where A and B are:
-- too great compared to 5 and
-- too far apart from each other
Drop down a lot.

Answer A) 11 liters of water in A

5 liters from A to B:
A has: (11 - 5) = 6 left in A
B has: B is (3 * A), so B = 18 after +5
Before +5, B is (18 - 5) = 13

5 liters from B to A
B has: (13 - 5) = 8 left in B
A has: A is (2 * B), so A = 16 after +5
A before + 5: (16 - 5) = 11

That's a match.

Answer A
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Re: A and B are two partially filled buckets of water. If 5 liters are tra   [#permalink] 26 Dec 2017, 17:00
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