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I FOUND THIS QUESTION IN ONE OF THE PAPERS....CAN U PLS HELP

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I FOUND THIS QUESTION IN ONE OF THE PAPERS....CAN U PLS HELP

by nolf » Tue Jan 06, 2009 1:38 am
Hi

So here is the question...
a,b,c are integers and given 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.... 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) 38
B) 1/2
C) 11/16
d) 5/7
E) 3/4
thanks
Harsha
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by ronniecoleman » Tue Jan 06, 2009 3:32 am
11/16


:wink:
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by bluementor » Tue Jan 06, 2009 3:46 am
The point to note here is that we are dealing with consecutive integers. In a set of consecutive integers:

median = (first integer + last integer)/2

So from the given information:

Median(set S) = (b + a)/2 = 3b/4
a = b/2 (eq. 1)

Median(set Q) = (c + b)/2 = 7c/8
b = 3c/4 (eq. 2)

Using equations 1 and 2, we can solve for:

Median of set R = (a + c)/2
=(b/2 + c)/2
=(3c/8 + c)/2
=11c/16 . Choose C.

-BM-
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by piyush_nitt » Tue Jan 06, 2009 5:01 am
bluementor wrote:The point to note here is that we are dealing with consecutive integers. In a set of consecutive integers:

median = (first integer + last integer)/2

So from the given information:

Median(set S) = (b + a)/2 = 3b/4
a = b/2 (eq. 1)

Median(set Q) = (c + b)/2 = 7c/8
b = 3c/4 (eq. 2)

Using equations 1 and 2, we can solve for:

Median of set R = (a + c)/2
=(b/2 + c)/2
=(3c/8 + c)/2
=11c/16 . Choose C.

-BM-
Integers are "Consecutive intergers" not mentioned in the Q.

It could be anything Isn't it?

Please explain?
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by bluementor » Tue Jan 06, 2009 8:29 am
piyush_nitt wrote:
bluementor wrote:The point to note here is that we are dealing with consecutive integers. In a set of consecutive integers:

median = (first integer + last integer)/2

So from the given information:

Median(set S) = (b + a)/2 = 3b/4
a = b/2 (eq. 1)

Median(set Q) = (c + b)/2 = 7c/8
b = 3c/4 (eq. 2)

Using equations 1 and 2, we can solve for:

Median of set R = (a + c)/2
=(b/2 + c)/2
=(3c/8 + c)/2
=11c/16 . Choose C.

-BM-
Integers are "Consecutive intergers" not mentioned in the Q.

It could be anything Isn't it?

Please explain?
Actually, you are right. I have made an assumption that this set will not contain the same integers. What we are sure is that the question states "all integers from a to b inclusive", but I have assumed that these integers within this set are all unique to each other i.e. no repetitions.

For example, for set S, if a=4 and b=7, then set S is {4, 5, 6, 7} and I'm assuming it will not be {4, 4, 5, 6, 6, 6, 7}, etc.

If my assumption is right, then we are essentially dealing with consecutive integers. Otherwise, I'm not sure how to solve this problem.

-BM-
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