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Tom started a company with 5 of his friends from the college. They mad

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Tom started a company with 5 of his friends from the college. They mad  [#permalink]

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New post Updated on: 07 Oct 2018, 06:34
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A
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Question Stats:

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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


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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.
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Re: Tom started a company with 5 of his friends from the college. They mad  [#permalink]

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New post 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

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Re: Tom started a company with 5 of his friends from the college. They mad  [#permalink]

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New post 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
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Re: Tom started a company with 5 of his friends from the college. They mad  [#permalink]

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New post 06 Oct 2018, 10:53
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Re: Tom started a company with 5 of his friends from the college. They mad  [#permalink]

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New post 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

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Re: Tom started a company with 5 of his friends from the college. They mad  [#permalink]

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New post 06 Oct 2018, 10:54
2) suff .... as 45-2= 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
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Re: Tom started a company with 5 of his friends from the college. They mad  [#permalink]

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New post 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
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Re: Tom started a company with 5 of his friends from the college. They mad  [#permalink]

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New post 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.
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Re: Tom started a company with 5 of his friends from the college. They mad  [#permalink]

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New post 06 Oct 2018, 10:58
According to me option (D) should be correct. Please find my solution attached.

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Re: Tom started a company with 5 of his friends from the college. They mad  [#permalink]

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New post 06 Oct 2018, 11:01
1
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.
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Re: Tom started a company with 5 of his friends from the college. They mad  [#permalink]

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New post 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
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Re: Tom started a company with 5 of his friends from the college. They mad  [#permalink]

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New post 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.
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Re: Tom started a company with 5 of his friends from the college. They mad  [#permalink]

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New post 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.

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Re: Tom started a company with 5 of his friends from the college. They mad  [#permalink]

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