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

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Manager
Joined: 02 Oct 2009
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a, b, and c are integers and a < b < c. S is the set of all integers  [#permalink]

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18 Sep 2009, 00:15
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51% (03:04) correct 49% (03:02) wrong based on 270 sessions

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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) 3/8
(B) 1/2
(C) 11/16
(D) 5/7
(E) 3/4

\sqrt{I solved it lke this...
for set S (a+b)/ 2 = (3/4 ) B
for set Q (b+c)/2 = (7c)/8
after solving we got
3c = 4b = 8a
median for set R is b so
4b = 3c
b = (3c)/4..

got E ..plasde tell me am i wrong somewhere?
Rohit}
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Joined: 02 Sep 2009
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Re: a, b, and c are integers and a < b < c. S is the set of all integers  [#permalink]

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06 Feb 2012, 02:49
24
17
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?

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

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

Answer: C ($$\frac{11}{16}$$).
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Re: a, b, and c are integers and a < b < c. S is the set of all integers  [#permalink]

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Updated on: 10 Feb 2019, 00:36
5
29

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

Originally posted by AKProdigy87 on 27 Oct 2009, 22:18.
Last edited by Bunuel on 10 Feb 2019, 00:36, edited 3 times in total.
Edited the question and added the OA.
##### General Discussion
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Joined: 02 Sep 2009
Posts: 64242
Re: a, b, and c are integers and a < b < c. S is the set of all integers  [#permalink]

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27 Oct 2009, 22:39
5
5
KocharRohit 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) 1/2
(C) 11/16
(D) 5/7
(E) 3/4

I solved it lke this...
for set S (a+b)/ 2 = (3/4 ) B
for set Q (b+c)/2 = (7c)/8
after solving we got
3c = 4b = 8a
median for set R is b so
4b = 3c
b = (3c)/4..

got E ..plasde tell me am i wrong somewhere?
Rohit

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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Posts: 64
Re: a, b, and c are integers and a < b < c. S is the set of all integers  [#permalink]

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27 Oct 2009, 22:46
Set R has three numbers..a , b , c..so why did we take median as
(a+c)/2 shudnt it be only b??
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Posts: 64242
Re: a, b, and c are integers and a < b < c. S is the set of all integers  [#permalink]

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27 Oct 2009, 22:59
1
KocharRohit wrote:
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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Re: a, b, and c are integers and a < b < c. S is the set of all integers  [#permalink]

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15 Dec 2010, 02:57
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?
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Posts: 64242
Re: a, b, and c are integers and a < b < c. S is the set of all integers  [#permalink]

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15 Dec 2010, 03:19
3
4
rishabh26m wrote:
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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Re: a, b, and c are integers and a < b < c. S is the set of all integers  [#permalink]

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06 Feb 2012, 02:37
5
15
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

Thank you.
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Joined: 31 Jan 2012
Posts: 67
Re: a, b, and c are integers and a < b < c. S is the set of all integers  [#permalink]

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06 Feb 2012, 03:01
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
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Posts: 64242
Re: a, b, and c are integers and a < b < c. S is the set of all integers  [#permalink]

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06 Feb 2012, 03:34
nonameee wrote:
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

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

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06 Feb 2012, 03:39
1
kys123 wrote:
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).
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Re: a, b, and c are integers and a < b < c. S is the set of all integers  [#permalink]

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06 Feb 2012, 04:11
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?
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Re: a, b, and c are integers and a < b < c. S is the set of all integers  [#permalink]

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06 Feb 2012, 04:20
2
nonameee wrote:
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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Re: a, b, and c are integers and a < b < c. S is the set of all integers  [#permalink]

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04 Dec 2012, 03:15
2
Bunuel wrote:
kys123 wrote:
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.

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

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10 Jul 2013, 01:58
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.
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Re: a, b, and c are integers and a < b < c. S is the set of all integers  [#permalink]

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10 Jul 2013, 02:09
1
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?

The same applies to set Q.

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

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09 Feb 2016, 22:39
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?
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Re: a, b, and c are integers and a < b < c. S is the set of all integers  [#permalink]

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10 Feb 2016, 00:30
2
2
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

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

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

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07 Mar 2016, 22:00
1
1
goldgoldandgold 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) 1/2
(C) 11/16
(D) 5/7
(E) 3/4

Since the question and options give all data in terms of fractions, we can assume values. The point is for which variable we should assume a value?
a < b < c
S = { a, ..., b } Median = (3/4)b
Q = {b, ..., c } Median = (7/8)c

Assume a value of c (which will lead to a value of b) such that c will get divided by 8 and 4 successively so that both medians are integers.
Say c = 32

Median of Q = (7/8)*32 = 28
b = 28 - 4 = 24

Median of S = (3/4)*24 = 18
a = 18 - 6 = 12

R = { 12, 13, 14, ... 32}
Median of R = 22 which as a fraction of c is 22/32 = (11/16)

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Re: a, b, and c are integers and a < b < c. S is the set of all integers   [#permalink] 07 Mar 2016, 22:00

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