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gabriel: Legendary Member; Posts: 986; Joined: Wed Dec 20, 2006 11:07 amFollowed by: 1 member; Location: India; Thanked: 51 times

by gabriel » Wed May 28, 2008 11:33 am

A very nice DS question, just solved it and it was fun

. Will post the solution and the answer later. Please do not give one word answers, post the solution you used to get to your answer.

Q.) Six numbers are randomly selected and placed within a set. If the set has a range of 16, a median of 6, a mean of 7 and a mode of 7, what is the greatest of the six numbers?

(1) The sum of the two smallest numbers is one-fifth of the sum of the two greatest numbers

(2) The middle two numbers are 5 and 7.

Cheers.

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Source: Beat The GMAT — Data Sufficiency | Wed May 28, 2008 11:33 am

netigen: Legendary Member; Posts: 631; Joined: Mon Feb 18, 2008 11:57 pmFollowed by: 3 members; Thanked: 29 times

by netigen » Wed May 28, 2008 1:46 pm

From the question we know that the numbers will be of the form

A B 5 7 7 F

Mode = 7 so 7 has to occur more than once
Median = 6 and since the number of numbers is even the median will not be in the set so it will be average of 3rd and 4th number

(19+a+b+f) = 7 x 6 = 42
a+b+e = 42-19 = 23

from option A, a+b = (e+7)/5
hence, A is sufficient to answer the question.

B is insufficient and redundant information.

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gabriel: Legendary Member; Posts: 986; Joined: Wed Dec 20, 2006 11:07 amFollowed by: 1 member; Location: India; Thanked: 51 times

by gabriel » Fri May 30, 2008 5:05 am

netigen wrote:From the question we know that the numbers will be of the form

A B 5 7 7 F

Mode = 7 so 7 has to occur more than once
Median = 6 and since the number of numbers is even the median will not be in the set so it will be average of 3rd and 4th number

(19+a+b+f) = 7 x 6 = 42
a+b+e = 42-19 = 23

from option A, a+b = (e+7)/5
hence, A is sufficient to answer the question.

B is insufficient and redundant information.

The answer is A indeed. This question is good because it tests rarely seen concepts like mode and median. Nice job.

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aatech: Master | Next Rank: 500 Posts; Posts: 192; Joined: Tue Apr 29, 2008 7:41 am; Thanked: 14 times

by aatech » Fri May 30, 2008 6:05 am

netigen wrote:From the question we know that the numbers will be of the form

A B 5 7 7 F

Mode = 7 so 7 has to occur more than once
Median = 6 and since the number of numbers is even the median will not be in the set so it will be average of 3rd and 4th number

(19+a+b+f) = 7 x 6 = 42
a+b+e = 42-19 = 23

from option A, a+b = (e+7)/5
hence, A is sufficient to answer the question.

B is insufficient and redundant information.

How did you select the set to be A B 5 7 7 F ... Stmt 2 says mid 2 nos are 5 and 7... the stem does not say so... ???

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netigen: Legendary Member; Posts: 631; Joined: Mon Feb 18, 2008 11:57 pmFollowed by: 3 members; Thanked: 29 times

by netigen » Fri May 30, 2008 9:12 am

Mode = 7 so 7 has to occur more than once
Median = 6 and since the number of numbers is even the median will not be in the set so it will be average of 3rd and 4th number

The bold part above explains why 5 and 7 are the 3rd and 4th numbers in the set.

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aatech: Master | Next Rank: 500 Posts; Posts: 192; Joined: Tue Apr 29, 2008 7:41 am; Thanked: 14 times

by aatech » Fri May 30, 2008 9:22 am

Got it.. thanks

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jasonc: Master | Next Rank: 500 Posts; Posts: 127; Joined: Sat Apr 19, 2008 1:02 am; Thanked: 12 times

by jasonc » Fri May 30, 2008 3:03 pm

netigen wrote:Mode = 7 so 7 has to occur more than once
Median = 6 and since the number of numbers is even the median will not be in the set so it will be average of 3rd and 4th number

The bold part above explains why 5 and 7 are the 3rd and 4th numbers in the set.

Theres actually a minor error more to your median statement.

"Since the number of elements is even, the median will not be in the set" is slightly misleading. The statement should be "since the number of elements is even, the median will be the average of the 3rd & 4th number." Meaning the 3rd & 4th number could be 5 & 7, or 6 & 6. However, since we know that the mode is 7, and thus the # of occurances of 7 > # of occurances of 6, if the 3rd & 4th numbers were 6 & 6, then we would need 3 7's. Since we only have 6 elements in the set, we can't possibly have 3 7's, therefore the 3rd & 4th elements have to be 5 & 7 respectively.

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