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

Question

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.

(1) The combined weight of the fish Carol receives is 30 pounds. 
(2) The combined weight of the fish Bob receives is 20 pounds.

broall · Answer

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.

chesstitans · Answer

yeah, test takers should bear in their mind that the weights need not be integer numbers.

das.sushmit · Answer

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.

Vivek1707 · Answer

The only thing you need to bear in mind is that ---> the values need not be integers

You ll have to test values here
Lets say Ann got a , b , c i.e 3 heaviest fish and Bob got x , y i.e two lightest fish. 
Since the sum is 100 = a + b + c + Fish for carol + x , y | Note we dont know how many fish Carol received

Start by placing values a = b = c = 16 and x = y = 10 ---> but since all values are different try decimals i.e try and maximize the values for a b c and minimize the value for x , y so you get to fit in as many fish for Carol as possible

a = 16.8 
b = 16.7
c = 16.5

x = 10.1
y = 9.9

At this point notice that the weight of fish for carol would be between abc and xy
Now since the sum total of the fish for Carol is 30, it can only have 2 fish

Check what happens when you put 3 fish for carol ---> values to start with would be 10 , 10 , 10 but since numbers are different you will have to pick decimals so try 10.1 and 10.15 but notice that if these are the two values then the third value will have to be less than 9.9 which is not possible since that fish would fall in the 'Bob' bucket. Similarly, Carol cannot have 1 fish since that would weigh 30 pounts and it would fall in the 'Ann' bucket.

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

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