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hard median question

by abhijeetsinghai » Thu Jan 22, 2009 6:53 am
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.11/16
c. 1/2
d.5/7


[spoiler]OA: b[/spoiler]


I don't knw how to approach this question. Please throw some light. Thanx !!
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Re: hard median question

by Brent@GMATPrepNow » Thu Jan 22, 2009 8:30 am
abhijeetsinghai wrote: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.11/16
c. 1/2
d.5/7
[spoiler]OA: b[/spoiler]

I don't knw how to approach this question. Please throw some light. Thanx !!
Here's one approach:
Set S: We know that the largest number is b and the median is 3b/4. Since the median is in the middle of the numbers of set S, we know that the smallest number in set S is 2b/4
You can see that 3b/4 (the median) lies directly in the middle 2b/4 and b (i.e., 2b/4, 3b/4, 4b/4)
So, set S ranges from b/2 to b

Set Q: Here the largest number is c and the median is 7c/8. Using similar logic from above, we can conclude that set Q ranges from 6c/8 to 8c/8, with 7c/8 squarely in the middle.

Here's the important step. 6c/8 is the smallest number in set Q. This value is equivalant to b (the largest value in set S)

So, now we have a relationship between c and b. We know that 6c/8 = b
From here, we can find two integer values for b and c that satisfy this equation amd determine how sets S and Q might look.

One solution is b=6 and c=8
For set S, we know that the largest value is 6, so keep adding numbers until we meet the criterion that the median is 3/4 b. We do the same with set S, beginning with c=8.
From here we can conclude that set S={3,4,5,6} and set Q={6,7,8}
The median of {3,4,5,6,7,8} is 5.5, which means the median is 11/16 of c

We could have chosen other numbers for b and c but the fraction remains 11/16
For example, b=12 and c=16 gives us S={6,7,8,9,10,11,12} and Q={12,13,14,15,16}
The median of {6, 7, . . . 15, 16} is 11, making it 11/16 of c
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by gaggleofgirls » Thu Jan 22, 2009 8:39 am
Here's how I worked it.

First off, for the set of all integers inclusive, this means the set of consecutive integers. With a set of consecutive integers, the median = the mean (the middle number = the average)

For The median of set S to be an integer and = 3/4 B, then B must be a multiple of 4.

For the median/mean of set Q to be an integer and = 7/8 C, then C must be a multiple of 8 AND B (the smallest number in set Q) must be the same standard deviation from the mean of Q that C is. That SD is the multiple of 8 (SD=1 when C=8, SD=2, when C=16, etc).

This sets up the relationship C-B = 2x
But this is only for set Q. x does not apply to set S. C has to be the x multiple of 8 but B can be any multiple of 4 and then we will work backwards from there to find the composition of set S.

For x = 1, C = 8. Since B <C and a multiple of 4, then B can on = 4 but ir doesn't satisfy the C-B = 2X equation.

For X = 2, C = 16 and B can be 4, 8 or 12.
16-12 = 2(2) does satisfy the equation, so B = 12

So, now we know set Q B=12, median = 7/8*16 = 14 and C = 16, so set Q (of all integers between B and C inclusive) is 12, 13, 14, 15, 16

Now we can determine set S.
B still = 12
median = 3/4 *12 = 9
Since the median is also the mean for this set (since they are consecutive integers), then a = 6
So the set S (all integers between 6 and 12 inclusive) is 6,7,8,9,10,11,12

The set R (all integers from A to C inclusive) is the set from 6 to 16, which is 6,7,8,9,10,11,12,13,14,15,16. The median of this set is 11. The fraction is 11/16.

Answer is B

Now, I need to keep working on this type ofstuff to be able to solve it in closer to 2 minutes.

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by krisraam » Thu Jan 22, 2009 8:49 am
The Answer is 11/16.

The median of Set S which has all the values from a to b (inclusive) is (a+b)/2.

So (a+b)/2 = 3b/4 --> b = 2a

Similarly

The median of Set Q which has all the values from b to c (inclusive) is (b+c)/2.

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

Substitute b = 2a --> c = 8a/3

The median of a to c is ie a to 8a/3 id (8a/3 + a )/2

This gives the value = 11a/6.

we want in terms of c
substitite a= 3c/8

gives the answer 11c/16.
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