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# M20-14

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Math Expert
Joined: 02 Sep 2009
Posts: 51121

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16 Sep 2014, 00:08
1
8
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Difficulty:

85% (hard)

Question Stats:

45% (01:15) correct 55% (01:32) wrong based on 198 sessions

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If the average of four positive integers is 30, how many integers of these four are greater than 20 ?

(1) One of the integers is 98

(2) The median of the four integers is less than 12

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:08
2
5
Official Solution:

The average of four positive integers is 30, means that the sum of these four positive integers is $$4*30=120$$.

(1) One of the integers is 98. If one of the integers is 98 then the sum of the other three is $$120-98=22$$. Now, since all integers are positive then no integer among these three could be greater than 20 (the greatest value of one of them can only be 20 if the other two are 1 and 1). So, there is only one integer greater than 20, the fourth one, which is 98. Sufficient.

(2) The median of the four integers is less than 12. If the set is $$\{1, 1, 1, 117 \}$$ then there is only one integer greater than 20 but if the set is $$\{1, 1, 21, 97 \}$$ then there are two integers greater than 20. Not sufficient.

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Joined: 29 Nov 2014
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26 Apr 2016, 05:32
I think this is a high-quality question and I don't agree with the explanation. Is 0 not a positive integer?
If yes then ,
one of the possibilities for statement 1 can be
0 0 22 98
In this case statement 1 will also not be sufficient.
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Joined: 02 Sep 2009
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26 Apr 2016, 05:35
aravind8024 wrote:
I think this is a high-quality question and I don't agree with the explanation. Is 0 not a positive integer?
If yes then ,
one of the possibilities for statement 1 can be
0 0 22 98
In this case statement 1 will also not be sufficient.

0 is NOT a positive integer, it's neither positive nor negative.
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Joined: 13 Jan 2016
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13 Aug 2016, 23:49
I don't think the explanation is quite right.
Shouldn't the right answer be E and not A?

The sum of other three numbers divided by 4 [ (x1 + x2 + x3)/4 ] has to be equal to 5.5
Or, sum of the other three numbers has to be equal to 22, which can be true for (1,1,20) as well as (1,2,19).
So we cant say how many numbers are greater than 20.

Edit: I was wrong. The question says 'greater than 20'. I included 20 as a possible solution.
A is correct.
Manager
Joined: 12 Aug 2015
Posts: 118
Location: India
Concentration: General Management, Strategy
GMAT 1: 690 Q50 V32
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11 Nov 2016, 07:44
Hi Bunuel,

Can you pls explain solution for 2. I cannot understand how do you come up with two sets:{1, 1, 1, 117 \} and {1, 1, 21, 97 \}

Bunuel wrote:
Official Solution:

The average of four positive integers is 30, means that the sum of these four positive integers is $$4*30=120$$.

(1) One of the integers is 98. If one of the integers is 98 then the sum of the other three is $$120-98=22$$. Now, since all integers are positive then no integer among these three could be greater than 20 (the greatest value of one of them can only be 20 if the other two are 1 and 1). So, there is only one integer greater than 20, the fourth one, which is 98. Sufficient.

(2) The median of the four integers is less than 12. If the set is $$\{1, 1, 1, 117 \}$$ then there is only one integer greater than 20 but if the set is $$\{1, 1, 21, 97 \}$$ then there are two integers greater than 20. Not sufficient.

Manager
Joined: 12 Aug 2015
Posts: 118
Location: India
Concentration: General Management, Strategy
GMAT 1: 690 Q50 V32
GPA: 3.38

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11 Nov 2016, 07:44
Hi Bunuel,

Can you pls explain solution for 2. I cannot understand how do you come up with two sets:{1, 1, 1, 117 \} and {1, 1, 21, 97 \}

Bunuel wrote:
Official Solution:

The average of four positive integers is 30, means that the sum of these four positive integers is $$4*30=120$$.

(1) One of the integers is 98. If one of the integers is 98 then the sum of the other three is $$120-98=22$$. Now, since all integers are positive then no integer among these three could be greater than 20 (the greatest value of one of them can only be 20 if the other two are 1 and 1). So, there is only one integer greater than 20, the fourth one, which is 98. Sufficient.

(2) The median of the four integers is less than 12. If the set is $$\{1, 1, 1, 117 \}$$ then there is only one integer greater than 20 but if the set is $$\{1, 1, 21, 97 \}$$ then there are two integers greater than 20. Not sufficient.

Intern
Status: Don't watch the clock,Do what it does, Keep Going.
Joined: 10 Jan 2017
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02 Oct 2017, 07:09
1. If one integer is 98 then it is obvious that the rest three are smaller than 20.
2. Median will be sum of 2nd and 3rd number divided by 2 ,ofcourse after they are arranged in ascending order. {1,2,3,114},{ 2,5,8,105} etc and so on There are alot of posiibilities here .Insufficient.
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Joined: 16 Dec 2016
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18 Jan 2018, 21:38
There are many ways to get a2+a3+a4=22 ,and it may or may not require one of the number to be 20, so 20 can be used 0 time or 1 time...so statement 1 is not sufficient...
Math Expert
Joined: 02 Sep 2009
Posts: 51121

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18 Jan 2018, 22:25
sushanttinku wrote:
There are many ways to get a2+a3+a4=22 ,and it may or may not require one of the number to be 20, so 20 can be used 0 time or 1 time...so statement 1 is not sufficient...

The questions asks: how many integers of these four are greater than 20? We are given that the integers are positive and know from (1) that x + y + z = 22. None of the x, y, and z can be more than 20 here, because the least sum in this case would be 21 + 1 + 1 = 23. So, only one integer out of four is more than 20, the one which is given to be 98.

Hope it's clear.
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Re: M20-14 &nbs [#permalink] 18 Jan 2018, 22:25
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# M20-14

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