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In the first step of a scientific experiment, 400 milliliters of a certain solution were divided among four empty containers. What was the median volume of solution among the containers?
(1) The range of the volumes in the four containers was twice the difference between the greatest volume and the average volume.
(2) The container with the greatest volume of solution had 50 milliliters more than the median volume.
In the first step of a scientific experiment, 400 milliliters of a certain solution were divided among four empty containers. What was the median volume of solution among the containers?
(1) The range of the volumes in the four containers was twice the difference between the greatest volume and the average volume.
(2) The container with the greatest volume of solution had 50 milliliters more than the median volume.
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26 Dec 2024, 10:00
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Bunuel
Manhattan Prep Official Explanation:
This Data Sufficiency problem is testing Statistics. Carefully jot down the givens:
400 mL total
4 empty containers: __ __ __ __
The question asks for the value of the median, which will be the average of the middle two volumes once they are listed in increasing order. Rephrase to ask for the sum of the middle two volumes.
Statement (1): SUFFICIENT. Since the total is 400 mL, the average volume is 100 mL. Therefore
range = 2(max – 100)
max – min = 2 × max – 200
200 = max + min
Since the maximum and the minimum sum to 200, the middle two values must also sum to 200 and the median is 100.
Statement (2): INSUFFICIENT. Test cases to prove that this is not sufficient.
Case #1: 50, 100, 100, 150. This satisfies the statement, as 150 is 50 greater than the median of 100.
Case #2: 20, 110, 110, 160. This also satisfies the statement, as 160 is 50 greater than the median of 110.
Since these two valid cases result in different medians, this statement is not sufficient.
The correct answer is (A): Statement (1) alone is sufficient, but statement (2) is not sufficient.
General Discussion
26 Dec 2024, 09:36
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Let the volumes be a,b,c and d such that a<b<c<d
Median volume. = (b+c)/2
We already know a+b+c+d = 400
(1) The range of the volumes in the four containers was twice the difference between the greatest volume and the average volume.
Range = d-a
Greatest volume = d
Average volume = (a+b+c+d)/4
d-a = 2*(d-(a+b+c+d)/4)
d-a=(4d-(a+b+c+d))/2
2(d-a) = 3d -a -(b+c) => b+c = d+a
We know a+b+c+d = 400 => 2(b+c) = 400 => b+c = 200 => (b+c)/2 = 100
Hence Statement (1) is sufficient.
(2) The container with the greatest volume of solution had 50 milliliters more than the median volume.
d = 50 + (b+c)/2
a+b+c+d = 400
If we substitute d in the above equation, we still don't a so we can't determine b+c. => Insufficient
Statement (1) is sufficient, but statement (2) is not sufficient => (A) is the answer.
Median volume. = (b+c)/2
We already know a+b+c+d = 400
(1) The range of the volumes in the four containers was twice the difference between the greatest volume and the average volume.
Range = d-a
Greatest volume = d
Average volume = (a+b+c+d)/4
d-a = 2*(d-(a+b+c+d)/4)
d-a=(4d-(a+b+c+d))/2
2(d-a) = 3d -a -(b+c) => b+c = d+a
We know a+b+c+d = 400 => 2(b+c) = 400 => b+c = 200 => (b+c)/2 = 100
Hence Statement (1) is sufficient.
(2) The container with the greatest volume of solution had 50 milliliters more than the median volume.
d = 50 + (b+c)/2
a+b+c+d = 400
If we substitute d in the above equation, we still don't a so we can't determine b+c. => Insufficient
Statement (1) is sufficient, but statement (2) is not sufficient => (A) is the answer.
