Bunuel wrote:

Ed has some fish, and the combined weight of all the fish is 100 pounds. The weights of the fish are all different. Ed gives the 3 heaviest fish to Ann, the 2 lightest fish to Bob, and the remaining fish to Carol. If the combined weight of the 3 heaviest fish is 50 pounds, how many fish does Carol receive?

(1) The combined weight of the fish Carol receives is 30 pounds.

(2) The combined weight of the fish Bob receives is 20 pounds.

2 lightest fish are \(A < B\)

3 heaviest fish rea \(X< Y < Z\).

The combined weight of fish that Carol received is R.

We have \(A+B+R+X+Y+Z=100\)

\(X + Y + Z = 50 \implies A+B+R=50\)

(1) We have \(R=30 \implies A+B=20\).

Note that \(A+B=20 \implies A<10<B\)

\(X+Y+Z=50 \implies X < \frac{50}{3}(\approx 16.77) < Z\)

R cant be a fish since \(R=30 > X\)

If R contains 2 fish. Let \(R=U+V=30\) so \(U<15<V \). Let find a root for this case.

Let \(V= 15.1 \implies U=14.9\)

Let \(X=16\), \(Y=16.5 \implies Z=17.5\).

Let \(B=12 \implies A=8\).

The weight of fish Ed has are \(\{8, 12, 14.9, 15.1, 16, 16.5, 17.5\}\). Carol received 2 fish.

If R contains more than 2 fish, then one of them must weigh lighter than 10. Insufficient.

Hence (1) is sufficient.

(2) We have \(A+B=20 \implies R=30\). The same as (1). Sufficient.

The answer is D.

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