Average and median

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heshamelaziry: Legendary Member; Posts: 869; Joined: Wed Aug 26, 2009 3:49 pmFollowed by: 3 members; Location: California; Thanked: 13 times

by heshamelaziry » Tue Dec 01, 2009 5:59 pm

Matt and four of his friends sold 60 tickets to a bowl game. If Matt sold 20 tickets and ecah of his friends sold at least one ticket, did Matt sell more tickets than the sum of three of his friends?

A- The average (arithmetic mean) of number of tickets sold is 12.

B- the median of the number of tickets sold is 30.

I do not have OA.

Thanks everyone.

Last edited by heshamelaziry on Tue Dec 01, 2009 10:46 pm, edited 1 time in total.

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Source: Beat The GMAT — Data Sufficiency | Tue Dec 01, 2009 5:59 pm

papgust: Community Manager; Posts: 1537; Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members; Thanked: 653 times

by papgust » Tue Dec 01, 2009 10:38 pm

IMO its B

We know that Matt sold 20 tickets. So the rest 40 tickets must have been sold by his other friends. We also know that each of his friends sold at least 1 ticket.

Did Matt sell more tickets than three of his friends? (In this question, i find it a little ambiguous whether it means that "Did Matt sell more tickets than each of his 3 friends?" or "Did Matt sell more tickets than sum of his 3 friends?")

I assume that the second one is what is meant.

(1) Average is 12. Which means that sum of the tickets sold = 12 * 5 = 60. we know that Matt sold 20, so 60-20 = 40. 40 must have been sold by 4 friends.

However, we can have any kind of combination of tickets sold among Matt's four friends. Some are,
36, 1, 1, 1 (YES, Matt sold more tickets than 3 of his friends)
10, 10, 10, 10 (NO, Matt did not sell more tickets than 3 of his friends)

Insufficient.

(2) Median = 30. which means 3rd position out of 5. We know that Matt sold 20 tickets. One of his 4 friends sold 30 tickets (from stmt II)
Other 3 friends have clearly sold only remaining 10 tickets, which means that Matt sold more tickets than 3 of his friends

Sufficient.

Hence B

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heshamelaziry: Legendary Member; Posts: 869; Joined: Wed Aug 26, 2009 3:49 pmFollowed by: 3 members; Location: California; Thanked: 13 times

by heshamelaziry » Tue Dec 01, 2009 10:44 pm

papgust wrote:IMO its B

We know that Matt sold 20 tickets. So the rest 40 tickets must have been sold by his other friends. We also know that each of his friends sold at least 1 ticket.

Did Matt sell more tickets than three of his friends? (In this question, i find it a little ambiguous whether it means that "Did Matt sell more tickets than each of his 3 friends?" or "Did Matt sell more tickets than sum of his 3 friends?")

I assume that the second one is what is meant.

(1) Average is 12. Which means that sum of the tickets sold = 12 * 5 = 60. we know that Matt sold 20, so 60-20 = 40. 40 must have been sold by 4 friends.

However, we can have any kind of combination of tickets sold among Matt's four friends. Some are,
36, 1, 1, 1 (YES, Matt sold more tickets than 3 of his friends)
10, 10, 10, 10 (NO, Matt did not sell more tickets than 3 of his friends)

Insufficient.

(2) Median = 30. which means 3rd position out of 5. We know that Matt sold 20 tickets. One of his 4 friends sold 30 tickets (from stmt II)
Other 3 friends have clearly sold only remaining 10 tickets, which means that Matt sold more tickets than 3 of his friends

Sufficient.

Hence B

This is how I see this: x1, 20, 30, x2, x3 now what values could be for x2 and x3 considering that they have to be more than 30 and that the total is 60 tickets ? I think I answered B or C but i think the answer is E

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papgust: Community Manager; Posts: 1537; Joined: Mon Aug 10, 2009 6:10 pmFollowed by: 252 members; Thanked: 653 times

by papgust » Tue Dec 01, 2009 11:00 pm

Interesting point! When you try to order it based on the median, then the order looks quite confusing. As there are only 60 tickets, we already have 20+30=50 and 30 as the median. so, x2 and x3 do not take meaningful values.

However, i looked at only from logical point of view. If there 50 tickets already sold, then the remaining 10 must have been sold by other 3 friends which is clearly less than Matt's.

I'm not too sure of what to answer now. Good point, hesham. Experts/others, pls provide your reasoning!

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Wed Dec 02, 2009 11:46 pm

heshamelaziry wrote:Matt and four of his friends sold 60 tickets to a bowl game. If Matt sold 20 tickets and ecah of his friends sold at least one ticket, did Matt sell more tickets than the sum of three of his friends?

A- The average (arithmetic mean) of number of tickets sold is 12.

B- the median of the number of tickets sold is 30.

I do not have OA.

Thanks everyone.

Either the question or statement (2) is wrong.

We know that collectively the 5 of them sold 60 tickets. We also know (without any further info) that the average number of tickets sold is simply:

sum of terms / # of terms = 60/5 = 12

Statement (1) is completely useless, since it repeats information that we already knew. Since (1) brings absolutely nothing to the table, the answer must be either B or E (choice C is only correct if each statement contributes something useful).

(2) tells us that the median # of tickets sold is 30. However, that's impossible.

The medium of a set with an odd number of terms is simply the middle term in the set, so our set must be:

{w, x, 30, y, z}

in which w and x are less than or equal to 30 and y and z are each greater than or equal to 30. Since the minimum values for w and x are 1, even if we let y=z=30, we have 92 tickets sold. However, according to the stem, only 60 tickets were sold.

Accordingly, this question makes no sense and there is no correct answer.

If statement (2) told us that the median number of tickets sold is 3, the answer would be B and the question would be fine.

What's the source?

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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