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26 Dec 2024, 10:58
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C
lets take the volumes to be a, b ,c ,d
a<=b<=c<=d
median = (b+c)/2
range = d - a
average = 400/4 = 100
(1)
d - a = 2(d - 100)
d + a = 200
(2)
d = median + 50
solving them both equations in (1) and (2)
d + a = 200 and d = m + 50
we get
a + m = 150
a + b + c + d = 400
150-m + b+c + m+50 = 400
median = (b+c)/2
m = 100
Median = 100
hence no single statement is suffiecient alone
need both statements, both together are sufficient
lets take the volumes to be a, b ,c ,d
a<=b<=c<=d
median = (b+c)/2
range = d - a
average = 400/4 = 100
(1)
d - a = 2(d - 100)
d + a = 200
(2)
d = median + 50
solving them both equations in (1) and (2)
d + a = 200 and d = m + 50
we get
a + m = 150
a + b + c + d = 400
150-m + b+c + m+50 = 400
median = (b+c)/2
m = 100
Median = 100
hence no single statement is suffiecient alone
need both statements, both together are sufficient
Bunuel
26 Dec 2024, 11:07
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The stem tells us that;
Then it asks us;
Let the volumes be a b c d respectively, in an increasing order.
We need to find b+c/2.
Let's look at the statements;
(1) The range of the volumes in the four containers was twice the difference between the greatest volume and the average volume.
d-a = 2{d-average}
Or, d-a = 2{d-400/4}
Or, d-a = 2{d-100}
Or, d-a = 2d-200
Or, d+a=200
Hence, b+c=200
Statement 1 is sufficient
(Eliminate BCE)
(2) The container with the greatest volume of solution had 50 milliliters more than the median volume.
d={(b+c)/2}+50
2d=b+c+100
a+b+c+d=400
Since we have no information about a,
Statement 2 is not sufficient.
(Eliminate D)
Answer is A
- In the first step of a scientific experiment, 400 milliliters of a certain solution were divided among four empty containers.
Then it asks us;
- What was the median volume of solution among the containers?
Let the volumes be a b c d respectively, in an increasing order.
We need to find b+c/2.
Let's look at the statements;
(1) The range of the volumes in the four containers was twice the difference between the greatest volume and the average volume.
d-a = 2{d-average}
Or, d-a = 2{d-400/4}
Or, d-a = 2{d-100}
Or, d-a = 2d-200
Or, d+a=200
Hence, b+c=200
Statement 1 is sufficient
(Eliminate BCE)
(2) The container with the greatest volume of solution had 50 milliliters more than the median volume.
d={(b+c)/2}+50
2d=b+c+100
a+b+c+d=400
Since we have no information about a,
Statement 2 is not sufficient.
(Eliminate D)
Answer is A
26 Dec 2024, 11:33
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let volumes be a,b,c,d
so required is b+c
1. d-a=2(d-100)
200-a=d
a+d=200
so b+c=400-200....SUFFICIENT
2. 0.5*(b+c) +50=d
50, 90,110, 150...mean=100
20,100,120,160...mean=110 ...NOT SUFFICIENT
Answer A
so required is b+c
1. d-a=2(d-100)
200-a=d
a+d=200
so b+c=400-200....SUFFICIENT
2. 0.5*(b+c) +50=d
50, 90,110, 150...mean=100
20,100,120,160...mean=110 ...NOT SUFFICIENT
Answer A
