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

Question

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

GMATBusters · Answer

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

hargun3045 · Answer

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

Manvi_nagpal · Answer

D

hargun3045 · Answer

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

u1983 · Answer

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

pk123 · Answer

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

lostboy · Answer

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.

Gladiator59 · Answer

According to me option (D) should be correct. Please find my solution attached.

Regards,
G

ulsbold · Answer

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.

Helium · Answer

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

MsInvBanker · Answer

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.

kylebecker9 · Answer

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

mohagar · Answer

ur style of solving is quite smart . I envy it but in positive way ...

o o

nandini14 · Answer

But the total of the other 4 friends is 30 and it can be in any ratio - could be 1,1,1,27 or 20,5,4,1 then how are we getting a definite yes or no from here

Bunuel · Answer

The total can be split in any ratio, but that does not change the logic. If the 5 friends together sold 45, then 3 of them cannot have sold 15 or more, since that would already make 45 and leave nothing for the other 2, who must sell at least 1 each. So at least 3 friends must have sold fewer than 15.

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

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