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If the average of four distinct positive integers is 60, how

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If the average of four distinct positive integers is 60, how [#permalink] New post 04 Sep 2009, 17:50
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If the average of four distinct positive integers is 60, how many integers of these four are smaller than 50?

(1) One of the integers is 200.
(2) The median of the four integers is 50.
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Re: average of four distinct positive integers [#permalink] New post 04 Sep 2009, 18:33
Let I1,I2,I3 & I4 are the distinct positive integers in ascending order.

=> (I1 + I2 + I3 + I4) /4 = 60

=> I1 + I2 + I3 + I4 = 240

1.) I4 = 200

=> I1 + I2 + I3 = 40 , hence, 3 integers are less than 50 .. sufficient..

2.) median = 50
=> (I2 + I3)/2 = 50
=> I2 + I3 = 100

and, I1 + I4 = 100

All 4 can be 50 , and tow can be grater than 50 and 2 can be less than 50..nothing is certain..hence, insufficient..

Ans. A
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Re: average of four distinct positive integers [#permalink] New post 04 Sep 2009, 19:55
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If the average of four distinct positive integers is 60, how many integers of these four are smaller than 50?

Note: we have four distinct positive integers and x1+x2+x3+x4=240.

(1) One of the integers is 200 --> x1+x2+x3+200=240 --> x1+x2+x3=40, hence three integers are less than 50. Sufficient.

(2) The median of the four integers is 50 --> Median of this set would be the average of middle numbers: x2+x3=100 --> x2<50 (as integers are distinct). x1<x2, hence we have two integers less than 50. Sufficient.

Answer: D.

BUT there is a problem with this question: from (1) we got that there are 3 integers less than 50 and from (2) we got that there are 2 integers less than 50. In DS statements never contradict so either of the statement should be changed.
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Re: average of four distinct positive integers [#permalink] New post 04 Sep 2009, 21:42
We have two equations:
I2 + I3 = 100 and I1 + I4 = 140
[strike]50 + 50 = 100 and 40 + 100 = 140 => 1 number is less than 50
50 + 50 = 100 and 50 + 90 = 140 => 0 numbers less than 50
40 + 60 = 100 and 40 + 100 = 140 => 2 numbers are less than 50[/strike]

Eliminated because we need distinct 4 numbers
40 + 60 = 100 and 50 + 90 = 140 => 1 number is less than 50
40 + 60 = 100 and 30 + 110 = 140 => 2 numbers are less than 50

Where am I thinking wrong?
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Re: average of four distinct positive integers [#permalink] New post 04 Sep 2009, 22:01
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sharkk wrote:
We have two equations:
I2 + I3 = 100 and I1 + I4 = 140
[strike]50 + 50 = 100 and 40 + 100 = 140 => 1 number is less than 50
50 + 50 = 100 and 50 + 90 = 140 => 0 numbers less than 50
40 + 60 = 100 and 40 + 100 = 140 => 2 numbers are less than 50[/strike]

Eliminated because we need distinct 4 numbers
40 + 60 = 100 and 50 + 90 = 140 => 1 number is less than 50
40 + 60 = 100 and 30 + 110 = 140 => 2 numbers are less than 50

Where am I thinking wrong?


The sum of the second and third biggest numbers (as the median is average of the second and third largest numbers average) must be 100, average=50 second large<50<third large. First one<second<50. In your post you have sum of smallest and third biggest sum 100 and smallest and biggest sum 140 which is wrong.
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Re: average of four distinct positive integers [#permalink] New post 04 Sep 2009, 22:12
Excellent point. Now I got it. Many thanks for your help.
Now, I see there will always 2 numbers less than 50. Wow!!!
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Re: If the average of four distinct positive integers is 60, how [#permalink] New post 05 Mar 2012, 01:39
I guess this problem is a poorly constructed one.

Option:1 is giving number of values less than 50 as: 3
But, Option:2 is giving number of values less than 50 as: 2

As far as I know, on the GMAT, both option choices should lead to same output.

Am I missing something?

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Re: If the average of four distinct positive integers is 60, how [#permalink] New post 05 Mar 2012, 01:48
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nravi549 wrote:
I guess this problem is a poorly constructed one.

Option:1 is giving number of values less than 50 as: 3
But, Option:2 is giving number of values less than 50 as: 2

As far as I know, on the GMAT, both option choices should lead to same output.

Am I missing something?

Cheers!


You are right. Check this: if-the-average-of-four-distinct-positive-integers-is-60-how-83523.html#p626142
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Re: If the average of four distinct positive integers is 60, how [#permalink] New post 30 Mar 2013, 01:59
i believe that (A) should be the answer. the problem with secoand statement is that 50, 50, 50, 90 can be a case where median is sum of the middle two terms that is 50. so, none of the terms are less than 50 here.


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Re: If the average of four distinct positive integers is 60, how [#permalink] New post 30 Mar 2013, 02:37
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cyberjadugar wrote:
i believe that (A) should be the answer. the problem with secoand statement is that 50, 50, 50, 90 can be a case where median is sum of the middle two terms that is 50. so, none of the terms are less than 50 here.


regards,


We are given that: "the average of four distinct positive integers is..."
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Re: If the average of four distinct positive integers is 60, how   [#permalink] 30 Mar 2013, 02:37
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