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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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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. _________________

Re: average of four distinct positive integers [#permalink]

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04 Sep 2009, 22:42

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?
_________________

--------------------------------------------------------------- Check-out the following: --------------------------------------------------------------- 1. Math Divisibility Test

Re: average of four distinct positive integers [#permalink]

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04 Sep 2009, 23:12

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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--------------------------------------------------------------- Check-out the following: --------------------------------------------------------------- 1. Math Divisibility Test

Re: If the average of four distinct positive integers is 60, how [#permalink]

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30 Mar 2013, 02:59

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.

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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