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

AlgebraIf 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 AAnswer choicesTypically, 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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The only thing more dangerous than ignorance is arrogance.

-- Albert Einstein