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a, b, and c are integers and a < b < c. S is the set of all [#permalink]

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06 Feb 2012, 03:37

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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, 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?

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 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 \(R=\frac{a+c}{2}=\frac{c*\frac{3}{8}+c}{2}=c*\frac{11}{16}\)

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?

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

For any evenly spaced set (aka AP) the arithmetic mean (average) is equal to the median (consecutive integers are evenly spaced set). _________________

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.

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

For any evenly spaced set (aka AP) the arithmetic mean (average) is equal to the median (consecutive integers are evenly spaced set).

Bunuel,

how do we know that they are evenly spaced. The a<b< c can be 1<2<3 or random 4<78<125 (not evenly spaced). Am i missing something?

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

For any evenly spaced set (aka AP) the arithmetic mean (average) is equal to the median (consecutive integers are evenly spaced set).

Bunuel,

how do we know that they are evenly spaced. The a<b< c can be 1<2<3 or random 4<78<125 (not evenly spaced). Am i missing something?

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.

Re: a, b, and c are integers and a < b < c. S is the set of all [#permalink]

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10 Jul 2013, 02:58

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.

Re: a, b, and c are integers and a < b < c. S is the set of all [#permalink]

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10 Jul 2013, 03:09

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abhinawster wrote:

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?

Re: a, b, and c are integers and a < b < c. S is the set of all [#permalink]

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08 Oct 2014, 23:02

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Re: a, b, and c are integers and a < b < c. S is the set of all [#permalink]

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12 Nov 2015, 07:47

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a, b, and c are integers and a < b < c. S is the set of all [#permalink]

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06 Dec 2015, 06:52

nonameee 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?

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.

Bunuels solution looks easier.. but may be my approach will help some of you

1.\(\frac{(b+c)}{2} = \frac{7}{8}c\) after some calculations we get\(\frac{c}{b}=\frac{4}{3}\) and using this ratio let's say c=8, b=6 2. Let's make the same for a&b: \(\frac{(a+b)}{2} = \frac{3}{4}b\) after some calculations we get b=2a, and from (1) we know that b=6 than a=3 Our numbers in set R look like this 3,4,5,6,7,8. The Median of the set is equal to \(\frac{(8+3)}{2}\) = \(\frac{11}{2}\) What fraction of c is the median of set R --> \(\frac{11}{2}=x*8\) (which is C) x\(=11/16\) Answer C _________________

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Re: a, b, and c are integers and a < b < c. S is the set of all [#permalink]

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10 Feb 2016, 01:30

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smathur1291 wrote:

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

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