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Re: An aquarium has 7 fishes inside. The four heaviest fishes ha [#permalink]
It is D) Our teacher gave us this question in class. By the way, I also have a tank at home and not to clean it too often I decided to collect some live rocks from the sea. And I don't know how but I have observed a few small bobbit worms is the sand of my aquarium. First, they were eating just algae, and the tank was clean a longer time. But soon they started eating small fish and I had to take then out.

Originally posted by Sebastrue on 10 Apr 2020, 08:07.
Last edited by Sebastrue on 02 May 2020, 05:32, edited 1 time in total.
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Re: An aquarium has 7 fishes inside. The four heaviest fishes ha [#permalink]
Why is the answer only D and not C, D and E as Chetan has explained?
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Re: An aquarium has 7 fishes inside. The four heaviest fishes ha [#permalink]
Expert Reply
The chetan explanation is wrong.

This is a single answer choice to pick among and NOT multiple answer choices.
The answer is D
Refer to the explanation provided by Brent above

Regards
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Re: An aquarium has 7 fishes inside. The four heaviest fishes ha [#permalink]
Expert Reply
Carcass wrote:
The chetan explanation is wrong.

This is a single answer choice to pick among and NOT multiple answer choices.
The answer is D
Refer to the explanation provided by Brent above

Regards



Hi Carcass,
There is nothing wrong in the explanation give by me above. Not only I have said that C, D and E can be the answer, but also I have given the set of weights when those values can be taken.
How can a solution that gives you the set of numbers fitting in can be wrong? I really would like to know what is wrong. You could check from Brent too.
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Re: An aquarium has 7 fishes inside. The four heaviest fishes ha [#permalink]
Expert Reply
Sorry.

I badly wording myself.

This is a single answer choice to pick among and NOT multiple answer choices.

In the sense of possibly more than one answer choice.

One is the answer. The overall explanation is perfectly fit. :wink:

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Re: An aquarium has 7 fishes inside. The four heaviest fishes ha [#permalink]
GreenlightTestPrep wrote:
Zamala wrote:
An aquarium has 7 fishes inside. The four heaviest fishes have a total weight of 117 pounds. The four lightest fishes have an average weight of 20.5 pounds.
What is (what could be?) the median weight of the 7 fishes.

A) 19.0
B) 19.5
C) 20.5
D) 21.5
E) 22.0


Let's first examine the 4 lightest fish.
Their average weight is 20.5 pounds
This means the TOTAL weight of the FOUR LIGHTEST fish = (4)(20.5) = 82 pounds.
So, it's possible that the 4 weights are 20.5, 20.5, 20.5 and 20.5

Notice that, AMONG THE FOUR LIGHTEST fish, 20.5 is the LIGHTEST possible weight for the heaviest fish.
If decrease one of the four weights, then we must increase another weight so that the TOTAL weight remains at 82 pounds.
For example, if we decrease one weight from 20.5 to 19.5, then we must increase another weight from 20.5 to 21.5 (to maintain the total weight of 82 pounds)
So, it's also possible that the 4 weights are: 19.5, 20.5, 20.5, 21.5

At this point, we can see that the median weight can be 21.5
We just need for the lightest fish AMONG THE FOUR HEAVIEST fish to weigh 21.5 pounds.

We're told that the four heaviest fishes have a total weight of 117 pounds
So, the 4 weights COULD be 21.5, 22.5, 30, 43

This means all seven weights are: 19.5, 20.5, 20.5, 21.5, 22.5, 30, 43

Answer: D

Cheers,
Brent

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Hello! Why don't we keep the sequence as 20,5 20,5 20,5 20,5 ? I mean why we increase one of them?
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Re: An aquarium has 7 fishes inside. The four heaviest fishes ha [#permalink]
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Samamammadova8888 wrote:
Hello! Why don't we keep the sequence as 20,5 20,5 20,5 20,5 ? I mean why we increase one of them?


This is actually a bad question (I never noticed it earlier).
It actually has more than 1 answer (see chetan2u's solution above), but it was placed in the wrong place.

I'm deleting my response.

Cheers,
Brent
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Re: An aquarium has 7 fishes inside. The four heaviest fishes ha [#permalink]
chetan2u wrote:
Zamala wrote:
On a test, I have encountered the following question. I am not entirely sure if I remember 100% of the details of the question, however, I know what was asked. Please point out if there is missing information, which is required for solving it.

Question:

An aquarium has 7 fishes inside. The four heaviest fishes have a total weight of 117 pounds. The four lightest fishes have an average weight of 20.5 pounds.
What is (what could be?) the median weight of the 7 fishes.

A) 19.0
B) 19.5
C) 20.5
D) 21.5
E) 22.0

I forgot the precise numbers of the answer choices, they were in the given range. The numbers in the questions are correct.


Hi...

Although it is explained quite nicely by Brent, let me pitch in my bit.

Median is the center value and we are given four lightest and then the four heaviest, so the common fish in the TWO groups is the median as 4th fish out of 7 is the median.
So CAN we be sure of a specific value .... NO

But we can surely think of a range..
Let the fishes in ascending order be A, B, C, D, E, F, G and we are looking for D.

1) If we take the lightest four-D, E, F, G-, the lightest out of this four, D is the median.
the largest value of D will be when all these fishes weigh the same.
So the largest value of median is \(\frac{117}{4}=28.75\)

2) If we take the heaviest four-A, B, C, D-, the heaviest out of this four, D is the median.
the smallest value of D will be when all these fishes weigh the same.
So the smallest value of median is 20.5 itself

The range is \(20.5\leq{median}\leq{28.75}\)..
Thus C, D and E , all three choices fit in..

C. 20.5 ------- 20.5, 20.5, 20.5, 20.5, 28, 30, 38.5
D. 21.5 ------- 20, 20, 20.5, 21.5, 28, 30, 37.5
E. 22 ------- 19.5, 20, 20.5, 22, 28, 30, 37

Note - If the question was asked in a multiple choice question and could have more than 1 answer, all three choices C, D and E fit in. However if we had to pick only one, the choices must be different or there may be some restrictions on the weight in initial statements.


117/4 is 29.25 *
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