Author 
Message 
TAGS:

Hide Tags

Senior DS Moderator
Joined: 27 Oct 2017
Posts: 906
Location: India
Concentration: International Business, General Management
GPA: 3.64
WE: Business Development (Energy and Utilities)

Tom started a company with 5 of his friends from the college. They mad
[#permalink]
Show Tags
Updated on: 07 Oct 2018, 06:34
Question Stats:
58% (01:23) correct 42% (01:13) wrong based on 47 sessions
HideShow timer Statistics
Tom started a company with 5 of his friends from the college. They made a software and sold it. Tom sold 15, and his friends each sold at least 1. Did Tom sell more than at least 3 of his friends? 1) The median number of software sold by 5 friends is 14. 2) The average number of software sold by 5 friends is 9. Weekly Quant Quiz #3 Question No 10
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Win Tests/ Prep Courses Weekly Quant Quiz Contest Weekly Quant Quiz Questions Direct Download SC: Confusable words All you need for Quant, GMAT PS Question Directory,GMAT DS Question Directory Error log/Key Concepts Combination Concept: Division into groups Question of the Day (QOTD) Free GMAT CATS
Originally posted by gmatbusters on 06 Oct 2018, 10:42.
Last edited by gmatbusters on 07 Oct 2018, 06:34, edited 2 times in total.
Renamed the topic and edited the question.



Senior DS Moderator
Joined: 27 Oct 2017
Posts: 906
Location: India
Concentration: International Business, General Management
GPA: 3.64
WE: Business Development (Energy and Utilities)

Re: Tom started a company with 5 of his friends from the college. They mad
[#permalink]
Show Tags
06 Oct 2018, 10:43
Official Explanation Statement:The median number of sells is 14, so you get (14 sells in maximum), (14 sells in maximum), (14 sells), ( ), ( ). Thus, Tom always sold 15 more than 3 of his friends, hence yes it is sufficient. Statement 2:Did Toll sell more than at least 3 of his friends? for answer to be No, it should be (1 sell), (1 sell), (15 sells), (15 sells), (15sells). However, in this way, the total sum becomes 1+1+15+15+15, so the total average of sells cannot be 9, hence it is impossible. Thus, it is always at least (1 sell), (1 sell), (13 sells), (15 sells), (15 sells), and the total average of software sold is 9. Therefore, Tom sold 15 sold more than at least 3 of his friends, hence yes, it is sufficient. The answer is D. Answer: D
_________________
Win Tests/ Prep Courses Weekly Quant Quiz Contest Weekly Quant Quiz Questions Direct Download SC: Confusable words All you need for Quant, GMAT PS Question Directory,GMAT DS Question Directory Error log/Key Concepts Combination Concept: Division into groups Question of the Day (QOTD) Free GMAT CATS



Intern
Joined: 06 Feb 2018
Posts: 16

Re: Tom started a company with 5 of his friends from the college. They mad
[#permalink]
Show Tags
06 Oct 2018, 10:53
St 1 suggests median sold is 14
i.e 3 sold at max 14 which is sufficient to say Tom sold more than at least three
St 2 says av = 9
This means all 5 sold total of 45
Assume two friends sold at 1 and 1, this leaves 43 left to be sold.
3x = 43 i.e average is less than 15 i.e at least 1 has to sell less than 15
Sufficient
D



Intern
Joined: 09 Oct 2017
Posts: 14
Location: India

Re: Tom started a company with 5 of his friends from the college. They mad
[#permalink]
Show Tags
06 Oct 2018, 10:53
D
Attachments
image.jpg [ 1.64 MiB  Viewed 356 times ]



Intern
Joined: 06 Feb 2018
Posts: 16

Re: Tom started a company with 5 of his friends from the college. They mad
[#permalink]
Show Tags
06 Oct 2018, 10:54
St 1 suggests median sold is 14
i.e 3 sold at max 14 which is sufficient to say Tom sold more than at least three
St 2 says av = 9
This means all 5 sold total of 45
Assume two friends sold at 1 and 1, this leaves 43 left to be sold.
3x = 43 i.e average is less than 15 i.e at least 1 has to sell less than 15
Sufficient
D



RC Moderator
Joined: 24 Aug 2016
Posts: 401
Location: Canada
Concentration: Entrepreneurship, Operations

Re: Tom started a company with 5 of his friends from the college. They mad
[#permalink]
Show Tags
06 Oct 2018, 10:54
2) suff .... as 452= 43/3 <15 1) Suff say the ascending number is A B C D E & X median 14 ==> C+D= 28 if Tom is at D, he sold more than ABC , And he can be at E & X ... in all cases he sold more than ABC He can not be at C as then d will be 13 ( not possible)....... Thus Ans D
_________________
Please let me know if I am going in wrong direction. Thanks in appreciation.



Manager
Joined: 16 Sep 2011
Posts: 83

Re: Tom started a company with 5 of his friends from the college. They mad
[#permalink]
Show Tags
06 Oct 2018, 10:56
Tom started a company with 5 of his friends from the college. They made a software and sold it. Tom sold 15, and his friends each sold at least 1. Did Tom sell more than at least 3 of his friends?
1) The median number of software sold by 5 friends is 14. 2) The average number of software sold by 5 friends is 9.
Option A: median sold is 14
so x y 14 a b ...which means a can be 15 or b can be 15 so either way Tom has sold at least more than 3 of his friends
hence it Is sufficient
Option b: average of software = 9 so total = 45 Tom sold =15 so rest = 30.. so split can be 5,7,8, 10 or 1,2,3, 23 so it is not sufficient
A is the answer



Manager
Joined: 02 Apr 2018
Posts: 50

Re: Tom started a company with 5 of his friends from the college. They mad
[#permalink]
Show Tags
06 Oct 2018, 10:57
B:
s1 doesn't give you enough information, just that 2 people have sold 14 and 15. we dont know if the remaining three friends are greater or less than that
s2 if the average is 9 and this statment weren't true then at least 3 people would've sold 15 and the last 2 could've sold 1. this would give you an average greater than 9 thus is sufficient because no tom didn't sell more than at least 3 friends.



PS Forum Moderator
Joined: 16 Sep 2016
Posts: 314

Re: Tom started a company with 5 of his friends from the college. They mad
[#permalink]
Show Tags
06 Oct 2018, 10:58
According to me option (D) should be correct. Please find my solution attached. Regards, G
Attachments
Q10.jpeg [ 114.13 KiB  Viewed 342 times ]



Intern
Joined: 16 Aug 2018
Posts: 2

Re: Tom started a company with 5 of his friends from the college. They mad
[#permalink]
Show Tags
06 Oct 2018, 11:01
A.
Since the median is 14, then the 1st and 2nd entries are at least 1 each. These three add up to 16, which is greater than what Tom sold. Definite answer of No. Sufficient. As for Statement 2, the average of 9 software sold yields 45 software sold by all friends. The list for this could be {1,1,1,15,27} which gives us a Yes, or it could be {6, 6, 6, 12, 15} which gives us a No. Therefore, Statement 2 is IS.



Senior Manager
Joined: 08 Jun 2013
Posts: 452
Location: India
GPA: 3.82
WE: Engineering (Other)

Re: Tom started a company with 5 of his friends from the college. They mad
[#permalink]
Show Tags
06 Oct 2018, 11:16
Tom started a company with 5 of his friends from the college. They made a software and sold it. Tom sold 15, and his friends each sold at least 1. Did Tom sell more than at least 3 of his friends? 1) The median number of software sold by 5 friends is 14. 2) The average number of software sold by 5 friends is 9. No of software's Tom sold = T and No of software's his 5 friends sold => a, b, c, d, e (say) then T = 15 and \(a, b, c, d, e >= 1\) St 1 : The median number of software sold by 5 friends is 14. let's say\(a<=b<=c<=d<=e\) then c =14 so \(a<=b<=c= 14\) Sufficient St 2 : The average number of software sold by 5 friends is 9. a + b + c + d + e = 9*5 = 45 Now if 3 friend's of Tom sold equal to or more number of software's then Tom > i.e. say \(c, d, e >= 15\) c + d + e = 45 minimum so a = b = 0 > Not possible as his friends each sold at least 1 software. So at least 3 of Tom's friends have sold less number of software's than Tom. Sufficient Ans : D
_________________
It seems Kudos button not working correctly with all my posts...
Please check if it is working with this post......
is it?....
Anyways...Thanks for trying



BSchool Forum Moderator
Joined: 23 May 2018
Posts: 253
Location: Pakistan
GPA: 3.4

Re: Tom started a company with 5 of his friends from the college. They mad
[#permalink]
Show Tags
06 Oct 2018, 12:03
The answer is D because; 1) Median number sold is 14. This means that there are two numbers less than or equals to 14. 14 being less than 15, Tom has sold more softwares than at least of his 3 friends. 2) Average of the 5 is 9. Therefore 45 softwares sold where everyone sold at least one. Assuming that 2 sold 15 softwares each, the remaining 15 will add up to the total of the remaining 3. Again making Tom having sold more softwares than at least 3 of his friends.
_________________
If you can dream it, you can do it.
Practice makes you perfect.
Kudos are appreciated.



Intern
Joined: 05 Oct 2018
Posts: 3

Re: Tom started a company with 5 of his friends from the college. They mad
[#permalink]
Show Tags
06 Oct 2018, 14:08
1) The median is 14 in a group of 5, so 14 and two other values below 14 were sold. Therefore, Tom sold more (15) than at least 3 of his friends. Sufficient 2) The average is 9 in a group of 5. It is easy to see a scenario where Tom sold more than at least 3 of his friends (all sold 9 for example). So, the question is whether there is a scenario where he sold more than at most 2 of his friends. Can we reach an average of 9 if 3 of his friends sold the same number as him? (3*15+2*1) / 5 > 9. So, no it is impossible for him to sell more than only 2 of his friends. This answer is also sufficient.
D



Intern
Joined: 14 Mar 2018
Posts: 4

Re: Tom started a company with 5 of his friends from the college. They mad
[#permalink]
Show Tags
07 Oct 2018, 06:03
is answer d
1. 15, 15 0r 14, any three numbers equal or less than 14 2. 15 , cant be 15 as left 4 can be total 30 .




Re: Tom started a company with 5 of his friends from the college. They mad &nbs
[#permalink]
07 Oct 2018, 06:03






