GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 12 Dec 2018, 01:48

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• ### The winning strategy for 700+ on the GMAT

December 13, 2018

December 13, 2018

08:00 AM PST

09:00 AM PST

What people who reach the high 700's do differently? We're going to share insights, tips and strategies from data we collected on over 50,000 students who used examPAL.
• ### GMATbuster's Weekly GMAT Quant Quiz, Tomorrow, Saturday at 9 AM PST

December 14, 2018

December 14, 2018

09:00 AM PST

10:00 AM PST

10 Questions will be posted on the forum and we will post a reply in this Topic with a link to each question. There are prizes for the winners.

# M08-28

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 51115

### Show Tags

15 Sep 2014, 23:38
3
17
00:00

Difficulty:

95% (hard)

Question Stats:

39% (01:22) correct 61% (01:39) wrong based on 136 sessions

### HideShow timer Statistics

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.

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 51115

### Show Tags

15 Sep 2014, 23:38
1
2
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.

_________________
Intern
Joined: 16 Jan 2016
Posts: 1

### Show Tags

07 Feb 2016, 09:17
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.
Math Expert
Joined: 02 Sep 2009
Posts: 51115

### Show Tags

07 Feb 2016, 09:19
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...
_________________
Intern
Joined: 13 Apr 2015
Posts: 32

### Show Tags

11 Feb 2016, 08:02
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
Intern
Joined: 28 Dec 2015
Posts: 39

### Show Tags

04 Jun 2016, 23:01
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??
Math Expert
Joined: 02 Sep 2009
Posts: 51115

### Show Tags

05 Jun 2016, 03:06
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
_________________
Manager
Joined: 21 Sep 2015
Posts: 77
Location: India
GMAT 1: 730 Q48 V42
GMAT 2: 750 Q50 V41

### Show Tags

02 Jul 2016, 22:44
I think this is a high-quality question and I agree with explanation.
_________________

Appreciate any KUDOS given !

Senior Manager
Joined: 31 Mar 2016
Posts: 385
Location: India
Concentration: Operations, Finance
GMAT 1: 670 Q48 V34
GPA: 3.8
WE: Operations (Commercial Banking)

### Show Tags

06 Jul 2016, 04:06
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 30 Aug 2015
Posts: 1

### Show Tags

02 Oct 2016, 19:46
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?
Math Expert
Joined: 02 Sep 2009
Posts: 51115

### Show Tags

03 Oct 2016, 06:42
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.
_________________
Intern
Joined: 02 Oct 2017
Posts: 23

### Show Tags

11 Feb 2018, 19:42
Bunuel

What if the numbers are

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

Then clearly (2) would be insufficient.

Thanks,

ucb2k7
Math Expert
Joined: 02 Sep 2009
Posts: 51115

### Show Tags

11 Feb 2018, 20:58
1
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.
_________________
Intern
Joined: 08 Jan 2018
Posts: 4

### Show Tags

21 Apr 2018, 21:38
Why can't any number be zero?
Math Expert
Joined: 02 Sep 2009
Posts: 51115

### Show Tags

22 Apr 2018, 01:21
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.
_________________
Manager
Joined: 31 Jul 2017
Posts: 191
Location: India
GMAT 1: 500 Q47 V15
GPA: 3.4
WE: Information Technology (Computer Software)

### Show Tags

06 Sep 2018, 21:50
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
_________________

If it helps you please press Kudos!

Thank You
Sudhanshu

Re: M08-28 &nbs [#permalink] 06 Sep 2018, 21:50
Display posts from previous: Sort by

# M08-28

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.