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quantumliner
(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
Hi quantumliner
Correct choice of numbers would be 4,6,8,10 (0 is not positive - a,b,c,d are all positive numbers). Although it does not change the methodology. Just wanted to highlight. Thanks for the solution
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quantumliner
(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

The question says:

Quote:
Which of the following CANNOT be the median of the four positive integers a, b, c, and d ?

While evaluating option D you have taken 0 as one of the numbers. But zero is neither positive nor negative integer. Can you explain?
Let me know if I am making some mistake here :)
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daftypatty
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

I understand why A, B, and C are not the correct answers. But, could someone explain D & E?

Note that a, b, c and d are positive integers.

The median will be the average of the second and third terms. If the second and third terms are equal, median will be equal to each.

But the median cannot be greater than 3 terms. a + b + d will be greater than all 3 because they are all positive integers. So (E) is definitely not possible.
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VeritasPrepKarishma
daftypatty
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

I understand why A, B, and C are not the correct answers. But, could someone explain D & E?

Note that a, b, c and d are positive integers.

The median will be the average of the second and third terms. If the second and third terms are equal, median will be equal to each.

But the median cannot be greater than 3 terms. a + b + d will be greater than all 3 because they are all positive integers. So (E) is definitely not possible.

Hi VeritasPrepKarishma :)

Is there a way we can eliminate option D without picking numbers?
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daftypatty
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

Where is this question from? It has two correct answers. a+b+d cannot be the median, as Karishma explains above, but (a+b+c)/2 cannot be the median either, if our numbers are all positive.

I won't go through a complete proof, since it's not the kind of thing you need to do on the GMAT, but if you wanted to prove it you'd consider two cases:

- first assume that two out of the three letters a, b and c are the two "middle numbers". Set the median equal to (a+b+c)/2. You'll find one letter must equal zero, which we know is impossible, so answer D cannot be the median in this case;

- only one possibility remains: assume that the fourth letter, d, is one of the two "middle numbers", and assume some other letter, say c, is the other "middle number" (it won't matter which one - you could repeat the proof identically using any letter). Again set the median equal to (a + b + c)/2. You'll find that d = a + b. But that means, if a and b are positive, that d is larger than both a and b, and if that's true, d and c cannot be the two "middle numbers" in the set. So we can't make answer choice D the median in this case either.

That second case gets a bit technical to explain fully, so I've just outlined it in enough detail to hopefully help anyone attempting it on their own. I feel like there's probably an easier way to prove this that I might not be seeing so early in the morning, but the question is flawed unless there's a typo among the answer choices.

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