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Average of 4 distinct positive integers is 60. How many of

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Average of 4 distinct positive integers is 60. How many of [#permalink] New post 05 Sep 2007, 16:04
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D
E

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Average of 4 distinct positive integers is 60. How many of the integers are less than 50?

1. One of the integers is 200
2. Median of the 4 integers is 50


The OA is D but I disagree
I know we can solve it with stem 1
However, I disagree with B being able to tell us anything
Am I not correct that when there is an even number of terms, the median is the average of the middle 2 terms?
Here is the reason I disagree with stem two being an answer

We could have numbers that look like this and still have a median of 50 and an average of 60
50 50 50 90

That would leave 0 numbers below 50

We could also have 30 48 52 110

In that instance the median is 50 but we have two numbers below 50

So with stem 2..we can't tell how many numbers are below 50 ..

Am I way out in left field?
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 [#permalink] New post 05 Sep 2007, 16:33
I got A.

St1:
Total = 240
If one of the integers is 200, then 40 needs to be shared among the remaining 3. But it doesn't matter how we split it out because all three are going to be less than 50. Sufficient.

St2:
Our set could be {10,40,60,130} ---> median = 50 and # of integers less than 50 = 2
Our set could be {50,50,50,90} ----> median = 50 and # of integers less than 50 = 0

Insufficient.

Ans A
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 [#permalink] New post 05 Sep 2007, 16:36
ywilfred wrote:
I got A.

St1:
Total = 240
If one of the integers is 200, then 40 needs to be shared among the remaining 3. But it doesn't matter how we split it out because all three are going to be less than 50. Sufficient.

St2:
Our set could be {10,40,60,130} ---> median = 50 and # of integers less than 50 = 2
Our set could be {50,50,50,90} ----> median = 50 and # of integers less than 50 = 0

Insufficient.

Ans A


Yep I agree B is insufficient..I thought I was going crazy when the OA said D...I guess the author made a mistake...
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 [#permalink] New post 05 Sep 2007, 16:50
jamesrwrightiii wrote:
ywilfred wrote:
I got A.

St1:
Total = 240
If one of the integers is 200, then 40 needs to be shared among the remaining 3. But it doesn't matter how we split it out because all three are going to be less than 50. Sufficient.

St2:
Our set could be {10,40,60,130} ---> median = 50 and # of integers less than 50 = 2
Our set could be {50,50,50,90} ----> median = 50 and # of integers less than 50 = 0

Insufficient.

Ans A


Yep I agree B is insufficient..I thought I was going crazy when the OA said D...I guess the author made a mistake...


Oh hang on... the question says the integers must be distinct, so {50,50,50 90} is not valid. So the author is right. OA must be D. Because there are only 4 elements in the set. So the median is always going ot be the average of the second and third number. Since (50+50)/2 is not going to be valid, then it must be variations such as (10,90), (30,70).... etc. In any case, # of integers less than 50 must always be 2.
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 [#permalink] New post 05 Sep 2007, 17:12
That is what I missed..the word disticint...:(
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Re: DS Median [#permalink] New post 05 Sep 2007, 20:31
jamesrwrightiii wrote:
Average of 4 distinct positive integers is 60. How many of the integers are less than 50?

1. One of the integers is 200
2. Median of the 4 integers is 50


The OA is D but I disagree
I know we can solve it with stem 1
However, I disagree with B being able to tell us anything
Am I not correct that when there is an even number of terms, the median is the average of the middle 2 terms?
Here is the reason I disagree with stem two being an answer

We could have numbers that look like this and still have a median of 50 and an average of 60
50 50 50 90

That would leave 0 numbers below 50

We could also have 30 48 52 110

In that instance the median is 50 but we have two numbers below 50

So with stem 2..we can't tell how many numbers are below 50 ..

Am I way out in left field?


S1: obviously suff.

S2: S/#=A

S/4=60 S=240. lets say the four integers are w,x,y,z. The median is 50. So x+y/2=50 So x+y=100. Thus w+z has to be 140.

Since X,Y must be the middle numbers here. W,Z must be the outer numbers.

Ok so we know that W<X<Y<Z. So we can have 48+49+51+92.

Lets try other numbers: 2+3+97+138.

From this it can be seen that w<50 or the numbers wont work. x must also be less than 50 b/c the numbers are distinct.

Work it out u can't have w or x greater than 50 or this doesnt work.

So B is Sufficient.

Answer is D
Re: DS Median   [#permalink] 05 Sep 2007, 20:31
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