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# M08-28

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:38
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95% (hard)

Question Stats:

40% (02:03) correct 60% (02:25) wrong based on 97 sessions

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If the average of four distinct positive integers is 60, how many integers of these four are less than 50?

(1) The median of the three largest integers is 51 and the sum of two largest integers is 190.

(2) The median of the four integers is 50.

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16 Sep 2014, 00:38
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Official Solution:

It's almost always better to express the average in terms of the sum: the average of four distinct positive integers is 60, means that the sum of four distinct positive integers is $$4*60=240$$.

Say four integers are $$a$$, $$b$$, $$c$$ and $$d$$ so that $$0 \lt a \lt b \lt c \lt d$$. So, we have that $$a+b+c+d=240$$.

(1) The median of the three largest integers is 51 and the sum of two largest integers is 190. The mdian of $$\{b,c,d\}$$ is 51 means that $$c=51$$. Now, if $$b=50$$, then only $$a$$, will be less than 50, but if $$b \lt 50$$, then both $$a$$ and $$b$$, will be less than 50. But we are also given that $$c+d=190$$. Substitute this value in the above equation: $$a+b+190=240$$, which boils down to $$a+b=50$$. Now, since given that all integers are positive then both $$a$$ and $$b$$ must be less than 50. Sufficient.

(2) The median of the four integers is 50. The median of a set with even number of terms is the average of two middle terms, so $$\text{median}=\frac{b+c}{2}=50$$. Since given that $$b \lt c$$ then $$b \lt 50 \lt c$$, so both $$a$$ and $$b$$ are less than 50. Sufficient.

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07 Feb 2016, 10:17
1
In statement # 2:

What about the case when both 'b' and 'c' are 50?

The problem does not state b > c anywhere.

So we can have 2 valid set of elements:

Set 1: 1, 48, 52, 139 -> 2 elements are less than 50
Set 2: 1, 50, 50, 139 -> 1 element is less than 50.

Insufficient -> Answer must be A.
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07 Feb 2016, 10:19
2
Krabathor wrote:
In statement # 2:

What about the case when both 'b' and 'c' are 50?

The problem does not state b > c anywhere.

So we can have 2 valid set of elements:

Set 1: 1, 48, 52, 139 -> 2 elements are less than 50
Set 2: 1, 50, 50, 139 -> 1 element is less than 50.

Insufficient -> Answer must be A.

Please read the question carefully: If the average of four distinct positive integers is 60...
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11 Feb 2016, 09:02
a1+a2+a3+a4=240
Statement 1:
a3=51
a3+a4=194
a4=143
so a1+a2=46 so both ae less than 50

Statement 2
a2+a3/2=50
a2+a3=100
a1+a4=140
since a2 and a3 are taken for median so a1 is less than a2 and a3.Hence less than 50
Sufficient

So D
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05 Jun 2016, 00:01
Bunuel wrote:
If the average of four distinct positive integers is 60, how many integers of these four are less than 50?

(1) The median of the three largest integers is 51 and the sum of two largest integers is 190.

(2) The median of the four integers is 50.

Dear bunuel,

Can you post more of such questions??
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05 Jun 2016, 04:06
Ashishsteag wrote:
Bunuel wrote:
If the average of four distinct positive integers is 60, how many integers of these four are less than 50?

(1) The median of the three largest integers is 51 and the sum of two largest integers is 190.

(2) The median of the four integers is 50.

Dear bunuel,

Can you post more of such questions??

Search in Statistics and Sets problems: search.php?search_id=tag&tag_id=34
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02 Jul 2016, 23:44
I think this is a high-quality question and I agree with explanation.
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06 Jul 2016, 05:06
I think this is a high-quality question and I agree with explanation.
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02 Oct 2016, 20:46
I don't know if I'm reading the same question that the explanation...

If the average of four distinct positive integers is 60, how many integers of these four are less than 50?
(1) The median of the three largest integers is 51 and the sum of two largest integers is 190.
(2) The median of the four integers is 50.

(1) Sufficient --> 'a','b' and 'c'> 50 so 3 integers are less than 50.
(2) Insufficient-> 'a','b' are less than 50, 'c' could be more o less than 50. So answer could be 2 or 3

Can anyone explain me what I'm missing?
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03 Oct 2016, 07:42
GiancarloGV wrote:
I don't know if I'm reading the same question that the explanation...

If the average of four distinct positive integers is 60, how many integers of these four are less than 50?
(1) The median of the three largest integers is 51 and the sum of two largest integers is 190.
(2) The median of the four integers is 50.

(1) Sufficient --> 'a','b' and 'c'> 50 so 3 integers are less than 50.
(2) Insufficient-> 'a','b' are less than 50, 'c' could be more o less than 50. So answer could be 2 or 3

Can anyone explain me what I'm missing?

(2) The median of the four integers is 50. The median of a set with even number of terms is the average of two middle terms, so $$\text{median}=\frac{b+c}{2}=50$$. Since given that $$b \lt c$$ then $$b \lt 50 \lt c$$, so both $$a$$ and $$b$$ are less than 50. Sufficient.
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11 Feb 2018, 20:42
Bunuel

What if the numbers are

x1,x2,x3,x4 where x2=50 and x3=51?

Then clearly (2) would be insufficient.

Thanks,

ucb2k7
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11 Feb 2018, 21:58
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ucb2k7 wrote:
Bunuel

What if the numbers are

x1,x2,x3,x4 where x2=50 and x3=51?

Then clearly (2) would be insufficient.

Thanks,

ucb2k7

(2) says that the median is 50. If x2=50 and x3=51, the median is 50.5 NOT 50.
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21 Apr 2018, 22:38
Why can't any number be zero?
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22 Apr 2018, 02:21
1
RCF wrote:
Why can't any number be zero?

If the average of four distinct positive integers is 60, how many integers of these four are less than 50?

0 is neither positive nor negative.
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06 Sep 2018, 22:50
Summ of all numbers = 60 * 4 = 240

1) x , 51,L ==> L =190 - 51 = 139

51 & 139 add up to 190

Remaining 2 numbers must add up to = 240 - 190 = 50

Possible combinations

50 ------ 0 & 50 ... Not possible because it is given that numbers are positive & distinct integers

50 ----- 1 & 49

50 ----- 2 & 48.

2) 2 middle term be x & y

Median = (x + y)/2 = 50.

x + y = 100

x & y are distinct ==> either of one is greater than 50 & other is less than 50.

==> Overall 2 terms less than 50 & 2 term greater than 50.

Sufficient.

Ans- D
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03 Jul 2019, 00:49
I think this is a high-quality question and I agree with explanation.
Re M08-28   [#permalink] 03 Jul 2019, 00:49
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# M08-28

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