Author 
Message 
TAGS:

Hide Tags

Moderator
Joined: 10 May 2010
Posts: 826

If the average of four distinct positive integers is 60 [#permalink]
Show Tags
08 Sep 2012, 10:32
3
This post was BOOKMARKED
Question Stats:
26% (01:56) correct
74% (01:25) wrong based on 125 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 sum of two largest integers is 190. (2) The median of the four integers is 50. CORRECT QUESTION WITH A SOLUTION: m8q28138554.html#p1120013
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
The question is not can you rise up to iconic! The real question is will you ?
Last edited by walker on 16 Jul 2013, 20:18, edited 3 times in total.
Renamed the topic and edited the question.



Moderator
Joined: 10 May 2010
Posts: 826

Re: M8  Q28 [#permalink]
Show Tags
08 Sep 2012, 10:35
From Statement 1: a + b + c + d = 240 c + d = 190 a+ b = 50 We can have a situation where c < 50 say c = 40 and d = 150. Here the no of integers less than 50 = 3. Or we can have c > 50 c = 90 Here the no of integers less than 50 = 2. Not Sufficient IMO the answer should be B
_________________
The question is not can you rise up to iconic! The real question is will you ?
Last edited by AbhiJ on 08 Sep 2012, 23:09, edited 1 time in total.



Intern
Joined: 30 Dec 2010
Posts: 10

Re: M8  Q28 [#permalink]
Show Tags
08 Sep 2012, 10:54
If the average of four distinct positive integers is 60, how many integers of these four are less than 50?
(1) The sum of two largest integers is 190.
(2) The median of the four integers is 50.
The answer should be C. Explanation: From the main statement, it is given that a+b+c+d = 240, now if you look at the statement (1) , it says that c+d = 190 ( assuming that c and d are two largest integers out of 4 positive integers. which also means that a+b = 240190 = 50 from statment (2), it says that median of the four integers is 50. which means that (b+c)/2 = 50 ( assuming that a<b<c<d or assume that b and c are the middle two integers. so, it means b+c = 100 Now if you combine both statements, you will find that a+b = 50 and b+c = 100 now, if a+b = 50 so the largest value of b = 49 and if so then then c > 50 because b+c = 100. and d of course would be greater than 50 from the statement c+d = 190. So, we can deduce that both statements together are sufficient to answer ( ans : c & d) this question. So, the answer will be C.



Senior Manager
Joined: 11 May 2011
Posts: 361
Location: US

Re: M8  Q28 [#permalink]
Show Tags
08 Sep 2012, 11:54
My answer is C as well.
_________________
 What you do TODAY is important because you're exchanging a day of your life for it! 



Manager
Joined: 02 Jun 2011
Posts: 107

Re: M8  Q28 [#permalink]
Show Tags
08 Sep 2012, 21:39
1
This post received KUDOS
1
This post was BOOKMARKED
AbhiJ wrote: If the average of four distinct positive integers is 60, how many integers of these four are less than 50?
(1) The sum of two largest integers is 190. Means that two of the numbers are greator than 60. avg be 95. so we need to have two nos. whose avg should be 25. so rest two no. should always be less than 50. therefore it give answer to above question. (2) The median of the four integers is 50. Median is 50 means middle two no. avg is 50. so one of them should be greator than 50 other less than 50. so from rest two no. one no. should be less than 50 and other more than 50. therefore this also gives the answer. For me 'D' is the right choice.



Current Student
Joined: 25 Jun 2012
Posts: 141
Location: United States
GMAT 1: 700 Q47 V40 GMAT 2: 740 Q48 V44
GPA: 3.48

Re: M8  Q28 [#permalink]
Show Tags
09 Sep 2012, 00:03
AbhiJ wrote: From Statement 1: a + b + c + d = 240 c + d = 190 a+ b = 50
We can have a situation where c < 50 say c = 40 and d = 150.
Here the no of integers less than 50 = 3.
Or we can have c > 50 c = 90
Here the no of integers less than 50 = 2.
Not Sufficient
IMO the answer should be B I got the same thing



Math Expert
Joined: 02 Sep 2009
Posts: 39640

Re: M8  Q28 [#permalink]
Show Tags
09 Sep 2012, 03:42
AbhiJ wrote: If the average of four distinct positive integers is 60, how many integers of these four are less than 50?
(1) The sum of two largest integers is 190.
(2) The median of the four integers is 50. THIS IS THE OLD VERSION OF THE QUESTION. REVISED VERSION OF THE QUESTION IS BELOW: If the average of four distinct positive integers is 60, how many integers of these four are less than 50?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 \(0<a<b<c<d\). 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 \(b<50\), 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 \(median=\frac{b+c}{2}=50\). Since given that \(b<c\) then \(b<50<c\), so both \(a\) and \(b\) are less than 50. Sufficient. Answer: D.
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 27 Feb 2007
Posts: 68

Re: If the average of four distinct positive integers is 60 [#permalink]
Show Tags
09 Sep 2012, 11:23
Hi Bunuel,
Isn’t the first data in Statement (1) of the problem (median 51) redundant? Because, even if we just know that the sum of the two largest integers is 190, we can answer the question – that there are 2 integers less than 50.
Regards rakesh.id



Current Student
Joined: 25 Jun 2012
Posts: 141
Location: United States
GMAT 1: 700 Q47 V40 GMAT 2: 740 Q48 V44
GPA: 3.48

Re: M8  Q28 [#permalink]
Show Tags
09 Sep 2012, 14:12
Bunuel wrote: AbhiJ wrote: If the average of four distinct positive integers is 60, how many integers of these four are less than 50?
(1) The sum of two largest integers is 190.
(2) The median of the four integers is 50. THIS IS THE OLD VERSION OF THE QUESTION. REVISED VERSION OF THE QUESTION IS BELOW: If the average of four distinct positive integers is 60, how many integers of these four are less than 50?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 \(0<a<b<c<d\). 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 \(b<50\), 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 \(median=\frac{b+c}{2}=50\). Since given that \(b<c\) then \(b<50<c\), so both \(a\) and \(b\) are less than 50. Sufficient. Answer: D. Thanks Bunuel. That makes a lot more sense



Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)

Re: M8  Q28 [#permalink]
Show Tags
09 Sep 2012, 15:13
Bunuel wrote: AbhiJ wrote: If the average of four distinct positive integers is 60, how many integers of these four are less than 50?
(1) The sum of two largest integers is 190.
(2) The median of the four integers is 50. THIS IS THE OLD VERSION OF THE QUESTION. REVISED VERSION OF THE QUESTION IS BELOW: If the average of four distinct positive integers is 60, how many integers of these four are less than 50?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 \(0<a<b<c<d\). 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 \(b<50\), 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 \(median=\frac{b+c}{2}=50\). Since given that \(b<c\) then \(b<50<c\), so both \(a\) and \(b\) are less than 50. Sufficient. Answer: D. (1) The median of the three largest integers is 51  where did this come from? I don't see it in the question...
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7440
Location: Pune, India

Re: If the average of four distinct positive integers is 60 [#permalink]
Show Tags
09 Sep 2012, 22:24
I think there are two versions of this question: Version 1: If the average of four distinct positive integers is 60, how many integers of these four are less than 50? (1) The sum of two largest integers is 190. (2) The median of the four integers is 50. Here the answer is (B) Here statement 1 is not sufficient. Let me show by taking 2 cases: Case 1: Numbers are 24, 26, 30, 160 Case 2: Numbers are 20, 30, 90, 100 Number of numbers which are less than 50 are different in these two cases. So statement 1 alone is not sufficient. Version 2: 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. Answer (D) Here statement 1 is sufficient too. Since median of the three largest integers is 51, the middle of the three largest integers must be 51. So the numbers are <51, <51, 51, >51 Since the sum of the two largest numbers is 190, the largest number must be 190  51 = 139 i.e. 79 more than the average of 60. Since 51 is 9 less than 60, we need to balance out another 70 in the two smallest numbers to get an average of 60. The numbers must be all distinct. Can one of the two smallest numbers be equal to 50? No! If it is equal to 50, it will balance out 10 of the 70 and we will need to balance out 60 from the smallest number. That will make the smallest number = 0 but all numbers must be positive. So, the two smallest numbers must be less than 50. (The concept of mean I am using here is very simple and intuitive. I have explained it in detail here: http://www.veritasprep.com/blog/2012/04 ... eticmean/)
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15942

Re: If the average of four distinct positive integers is 60 [#permalink]
Show Tags
25 Feb 2017, 16:25
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: If the average of four distinct positive integers is 60
[#permalink]
25 Feb 2017, 16:25







