a, b, and c are integers and a < b < c. S is the set of all integers

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 we have:
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 (\(\frac{11}{16}\)).
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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

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.

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 (\(\frac{11}{16}\))
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Set R has three numbers..a , b , c..so why did we take median as
(a+c)/2 shudnt it be only b??

Not so. We are told that a<b<c, but not that they are consecutive integers, it could be 3<55<79. So set R in that case would be all integers from 3 to 79 inclusive. And b not necessarily would be the median. Median would be (3+79)/2=41.

Hope it's clear.
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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 (\(\frac{11}{16}\)).
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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.
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Another quick question, so for this question we're assuming that the medium is equal to mean. I thought the only way for that to happen is if there is no skewness in the set, but it doesn't say that anywhere. Is there any sort of general rule to tell if medium = mean?

Thanks so much Bunuel

Red part is not correct: we cannot 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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For any evenly spaced set (aka AP) the arithmetic mean (average) is equal to the median (consecutive integers are evenly spaced set).
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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?

The median (mean) of the integers from a to c, inclusive is always (a+c)/2 (if you have some additional info you can obtain this value in another way but this way is ALWAYS true).

Consider two sets: {1, 2, 3} and {3, 4, 5, 6, 7, 8, 9} --> combined set {1, 2, 3, 4, 5, 6, 7 8, 9}

As you've written the median (mean) of combined set should be (2+6)/2=4, which is wrong as median of combined set is 5.

Hope it's clear.
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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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Bunnel, if i take set S as 3,6,8 and set Q as 8,14,16 whats wrong with it? satisfy questions requirement and are not in AP.
We cant apply consecutive integers formula then.

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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Hi.. So for this question only way to solve is by knowing the fact that median will lie between the 2 and eliminating the options... right?

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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