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# Ed has some fish, and the combined weight of all the fish is 100 pound

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Ed has some fish, and the combined weight of all the fish is 100 pound  [#permalink]

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21 May 2017, 06:12
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49% (02:48) correct 51% (02:14) wrong based on 99 sessions

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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.

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Ed has some fish, and the combined weight of all the fish is 100 pound  [#permalink]

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21 May 2017, 07:26
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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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Re: Ed has some fish, and the combined weight of all the fish is 100 pound  [#permalink]

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24 Aug 2017, 17:35
yeah, test takers should bear in their mind that the weights need not be integer numbers.
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Re: Ed has some fish, and the combined weight of all the fish is 100 pound  [#permalink]

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25 Aug 2017, 06:12
Hi,

Given: Ann = a1 + a2 + a3 = 50
Bob = b1 + b2 = ?
Carol = c1+c2.. cn = ?

a1 + a2 + a3 + b1 + b2 + c1 + c2 + ... cn = 100
b1 + b2 + c1 + c2 + ... cn = 50

a1 , a2, a3 lie between 15-20 (maximizing)

Question: n=?

Statement1 = b1+ b2 = 20 , then, c1 + c2 + .... cn = 30
minimizing: b1, b2 = 10, then c1=c2=b1=b2. Therefore, Carol has three fish. Hence, sufficient.

Statement2 = b1 + b2 = 20, same as statement 1. Hence, sufficient.

Therefore, D.

Alternatively, you can clearly see both the statements are the same, therefore, the answer must be either D or E. Once you delve in further by looking for maximums and minimums (optimisations) you get the correct answer.
Re: Ed has some fish, and the combined weight of all the fish is 100 pound   [#permalink] 25 Aug 2017, 06:12
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# Ed has some fish, and the combined weight of all the fish is 100 pound

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