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

The answer is C: 11/16.

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

Set R has three numbers..a , b , c..so why did we take median as
(a+c)/2 shudnt it be only b??

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?

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

OA:

Bunuel or someone else, where am I going wrong with this one?

Median of a combined interval will be in the middle between the median of Q and the median of S:

(\(3/4\) b + \(7/8\) c) * \(1/2\) (1)

From the formula for median of Q we get:

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

Substituting b from (2) into (1) we get:

(\(3/4\) *\(3/4\)c + \(7/8\) c) * 1/2 ==> \(23/32\) c


Please help.

Thank you.

Red part is not correct: we can not assume that as we don't know that a and c are equidistant from b. If it were so then the median would simply be b.

It should be as shown in my post: (a+c)/2.
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Bunuel, so in order to determine a median of two intervals of integers (a,b) and (b,c) (where a<b<c), you should always use the formula: (a+c)/2?

Given that "S is the set of all integers from a to b, inclusive" and "Q is the set of all integers from b to c, inclusive", which means that both S and Q are sets of consecutive integers, thus evenly spaced sets.

Hope it's clear.
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S is the set of all integers from a to b, inclusive. Say a=3 and b=8. What is set S then? S={3, 4, 5, 6, 7, 8} not {3, 6, 8}, where did 4, 5 and 7 go? Aren't they integers in the range from 3 to 8?

The same applies to set Q.

Hope it's clear.
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Another method is "assuming values".

a < b < c

Median of Q is 7/8 of c so c must be a multiple of 8. Say c = 8. Median of c = 7.
Set Q = {6, 7, 8}
Then b = 6
Median of S = (3/4) of b which is (3/4)*6 = 4.5.
Set S = {3, 4, 5, 6}

So the entire set R = {3, 4, 5, 6, 7, 8}
Median = 5.5

Median as a fraction of c: 5.5/8 = 11/16

Answer (C)
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