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# Statistics

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Manager
Status: Berkeley Haas 2013
Joined: 23 Jul 2009
Posts: 192
Followers: 1

Kudos [?]: 32 [3] , given: 16

Statistics [#permalink]  19 Aug 2009, 11:04
3
KUDOS
00:00

Difficulty:

(N/A)

Question Stats:

25% (05:44) correct 75% (00:00) wrong based on 5 sessions
a,b, and c are integers and a<b<c. S is the set of all integers from a to b, inclusive. Q is the set of all integers from b to c, inclusive. The median of set S is (3/4)*b. The median of set Q is (7/8)*c. If R is the set of all integers from a to c, inclusive, what fraction of c is the median of set R?
(A) 3/8
(B) 1/2
(C) 11/16
(D) 5/7
(E) 3/4

Explanations welcome. Found it worth sharing
Current Student
Joined: 12 Jun 2009
Posts: 1847
Location: United States (NC)
Concentration: Strategy, Finance
Schools: UNC (Kenan-Flagler) - Class of 2013
GMAT 1: 720 Q49 V39
WE: Programming (Computer Software)
Followers: 22

Kudos [?]: 211 [0], given: 52

Re: Statistics [#permalink]  19 Aug 2009, 11:25
is it C?

I kinda drawn it out like:

3< 4.5(M) < 6 < 7(M) <8
I assumed C was 8 and since M(median) is 7, b has to be 6 and then figured out a being 3. Now add them all and found to be 33/6 = 5.5 average and then found which answer multiplies by C(8) gives me 5.5 Obviously not an efficient way of doing it but at least it got some kinda of result...

Kudos for the question!
_________________

Manager
Joined: 11 Sep 2009
Posts: 129
Followers: 5

Kudos [?]: 236 [4] , given: 6

Re: Statistics [#permalink]  18 Sep 2009, 00:15
4
KUDOS

The key to this problem is remembering that the median for a consecutive set of numbers is equivalent to its mean. For example, the mean and median of a set consisting of x, x+1, x+2, ... ,y will always be (x+y)/2.

For set S, consisting of numbers (a, a+1,...,b), the median is given to be 3/4*b:

(a+b)/2 = (3/4)*b
a = b/2

For set Q, consisting of numbers (b, b+1,...,c), the median is given to be 7/8*c:

(b+c)/2 = (7/8)*c
b = (3/4)*c

For set R, consisting of numbers (a, a+1,...c), the median needs to be found:
a = b/2 = (3/4*c)/2 = (3/8)*c

Median = (a + c)/2 = (3/8*c + c)/2 = (11/8)*c/2 = (11/16)*c
Manager
Joined: 10 Jul 2009
Posts: 129
Location: Ukraine, Kyiv
Followers: 2

Kudos [?]: 66 [0], given: 60

Re: Statistics [#permalink]  18 Sep 2009, 05:08
very good question, goldgoldandgold!
Clear and easy explanation, AKProdigy87!
Thank you too, guys. Kudos.
_________________

Never, never, never give up

Intern
Joined: 11 Oct 2010
Posts: 27
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: Statistics [#permalink]  05 Nov 2010, 00:31
tough question to solve in real-time... looks easy when u check soln
Intern
Joined: 22 Nov 2010
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: Statistics [#permalink]  15 Dec 2010, 02:57
I have a doubt here...the question does not say anywhere that the numbers are consecutive...then is it feasible to assume the mean=median property? or is plugging numbers a better option?
Math Expert
Joined: 02 Sep 2009
Posts: 28767
Followers: 4588

Kudos [?]: 47320 [0], given: 7121

Re: Statistics [#permalink]  15 Dec 2010, 03:19
Expert's post
rishabh26m wrote:
I have a doubt here...the question does not say anywhere that the numbers are consecutive...then is it feasible to assume the mean=median property? or is plugging numbers a better option?

Given that S is the set of all integers from a to b, inclusive, Q is the set of all integers from b to c, inclusive and R is the set of all integers from a to c, inclusive, so sets S, Q and R have to be consecutive integers sets. For any set of consecutive integers (generally for any evenly spaced set) median (also the mean) equals to the average of the first and the last terms.

So given:
Median of $$S=\frac{a+b}{2}=b*\frac{3}{4}$$ --> $$b=2a$$;

Median of $$Q=\frac{b+c}{2}=c*\frac{7}{8}$$ --> $$b=c*\frac{3}{4}$$ --> $$2a=c*\frac{3}{4}$$ --> $$a=c*\frac{3}{8}$$;

Median of $$R=\frac{a+c}{2}=\frac{c*\frac{3}{8}+c}{2}=c*\frac{11}{16}$$

Answer: C ($$\frac{11}{16}$$).
_________________
Re: Statistics   [#permalink] 15 Dec 2010, 03:19
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