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03 Aug 2017, 15:13
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Difficulty:
15%
(low)
Question Stats:
83% (00:56) correct
17%
(01:22)
wrong
based on 83
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Which of the following CANNOT be the median of the four positive integers a, b, c, and d ?
(A) c
(B) d
(C) (a+d)/2
(D) (a+b+c)/2
(E) a + b + d
(A) c
(B) d
(C) (a+d)/2
(D) (a+b+c)/2
(E) a + b + d
I understand why A, B, and C are not the correct answers. But, could someone explain D & E?
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03 Aug 2017, 16:22
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(A) c
If all 4 numbers are same. Ex. 1,1,1,1. Then Median = C
(B) d
If all 4 numbers are same. Ex. 1,1,1,1. Then Median = D
(C) (a+d)/2
If the 4 number are : 2, 4, 6, 8
Then median = (a+d)/2 = (b+c)/2
(D) (a+b+c)/2
If the 4 numbers are : 0, 4, 6, 8
Then median = (b+c)/2 = (a+b+c)/2
(E) a + b + d
If the four numbers are a,b,c,d and d > c > b > a
So median = (b+c)/2
Let's assume the median = a+b+d
then, a+b+d = (b+c)/2
==> 2a+2b+2d = b+c
==> 2a+b+2d - c = 0
As we know d > c, therefore 2d > c, hence 2d-c>0. So 2a+b+2d-c will result in a positive number.
Given per question, a,b,c,d are all positive numbers. Hence if the a+b+d is the median, then 2a+b+2d - c comes out to be 0, which is not possible. Hence Option E cannot be the median.
Answer is E
If all 4 numbers are same. Ex. 1,1,1,1. Then Median = C
(B) d
If all 4 numbers are same. Ex. 1,1,1,1. Then Median = D
(C) (a+d)/2
If the 4 number are : 2, 4, 6, 8
Then median = (a+d)/2 = (b+c)/2
(D) (a+b+c)/2
If the 4 numbers are : 0, 4, 6, 8
Then median = (b+c)/2 = (a+b+c)/2
(E) a + b + d
If the four numbers are a,b,c,d and d > c > b > a
So median = (b+c)/2
Let's assume the median = a+b+d
then, a+b+d = (b+c)/2
==> 2a+2b+2d = b+c
==> 2a+b+2d - c = 0
As we know d > c, therefore 2d > c, hence 2d-c>0. So 2a+b+2d-c will result in a positive number.
Given per question, a,b,c,d are all positive numbers. Hence if the a+b+d is the median, then 2a+b+2d - c comes out to be 0, which is not possible. Hence Option E cannot be the median.
Answer is E
03 Aug 2017, 23:46
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