After a fisherman sold 1/4 of the fish he had caught and gave away 2/3 of the remaining fish, he had 4 fish left. What was the total number of fish he had caught?
(A) 8
(B) 16
(C) 20
(D) 32
(E) 40
OA: B
After a fisherman sold 1/4 of the fish he had
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Let the number of fishes caught \(= x\)
Number of fishes sold \(= \frac{1}{4}x\)
Number of fishes UNsold \(= xâˆ’\frac{1}{4}x=\frac{3}{4}x\)
Gave away \(\frac{2}{3}\) of the remaining fish \(=\frac{2}{3}\cdot \frac{3}{4}x = \frac{1}{2}x\)
After all the above he had 4 fish left. Therefore, __B__ is the correct option.
Number of fishes sold \(= \frac{1}{4}x\)
Number of fishes UNsold \(= xâˆ’\frac{1}{4}x=\frac{3}{4}x\)
Gave away \(\frac{2}{3}\) of the remaining fish \(=\frac{2}{3}\cdot \frac{3}{4}x = \frac{1}{2}x\)
After all the above he had 4 fish left. Therefore, __B__ is the correct option.

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Let the No. of fishes be X
After selling 1/4x fishes, fishes left = 3/4x.
He gave away 2/3rd of fishes left. Thus, he now has 1/3rd of fishes left = 1/3 * 3/4x = 1/4x
This 1/4x is equal to 4. Thus, x = 16
Answer is B
After selling 1/4x fishes, fishes left = 3/4x.
He gave away 2/3rd of fishes left. Thus, he now has 1/3rd of fishes left = 1/3 * 3/4x = 1/4x
This 1/4x is equal to 4. Thus, x = 16
Answer is B
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We can let n = the number of fish the fisherman caught and create the equation:NandishSS wrote:After a fisherman sold 1/4 of the fish he had caught and gave away 2/3 of the remaining fish, he had 4 fish left. What was the total number of fish he had caught?
(A) 8
(B) 16
(C) 20
(D) 32
(E) 40
OA: B
n  (1/4)n  (2/3)(3/4)n = 4
n  (1/4)n  (1/2)n = 4
Multiplying the equation by 4, we have:
4n  n  2n = 16
n = 16
Answer: B
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