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Median

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Median [#permalink] New post 23 Oct 2011, 00:57
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) ½ (C) 11/16 (D) 5/7 (E) 3/4
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Re: Median [#permalink] New post 26 Oct 2011, 08:40
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You can solve this problem algebraically, or you can plug in numbers.

For the algebraic approach, you need to recognize that sets S and Q consist of consecutive integers. We know that the median of a set of consecutive integers is equal to the mean, which can be found by averaging the first and last integers in the set.

Since S runs from a to b, its mean/median = (a+b)/2

Since Q runs from b to c, its mean/median=(b+c)/2

So now we can use the provided information about the medians to create two equations:

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

From here we just need to do some manipulations to get to our answer:

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

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

Since R runs from a to c, let’s combine these results to find a in terms of c:

a=(3c/4)/2
a=3c/8

Now we can plug these values in to find the median of R. The mean/median of R=(a+c)/2=(3c/8 + c)/2=(11c/8)/2=11c/16

The answer is C.

However, it might be easier to plug in numbers here. We just have to think it over a little bit. If these sets consist of integers, and the medians are fractions of the end numbers, there are only so many possibilities.

Let’s look at set S first. If the median is (3/4)b, we know that b must be a multiple of 4. What are our possibilities?

b=4, median=3, set S runs from 2-4 (there is one term after the median, so there must be one term before)
b=8, median=6, set S runs from 4-8 (there are two terms after the median, so there must be two terms before)

If we want additional possibilities, we can just keep expanding this pattern:
a/last term/b = 2/3/4 or 4/6/8 or 6/9/12 or 9/12/16, etc. Each valid set is just a multiple of the first one.

We can do the same thing for set Q. Since the median is (7/8)c, we know that c must be a multiple of 8. So our possibilities for b/median/c are as follows:

6/7/8
12/14/16
18/21/24, etc.

Since both sets contain b, we need values for sets S and Q that have the same value for b. The first overlapping value is 12. So we can choose 6-12 for set S and 12-16 for set Q. Now we have our set R.

Since set R runs from a to c, or 6 to 16, its median is 11. c=16, so median/c=11/16.

_________________


Dmitry Farber | Manhattan GMAT Instructor | New York


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Re: Median   [#permalink] 26 Oct 2011, 08:40
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