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a, b, and c are integers and a < b < c. S is the set of all
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Updated on: 20 Aug 2012, 00:50
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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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Originally posted by KocharRohit on 27 Oct 2009, 22:18.
Last edited by Bunuel on 20 Aug 2012, 00:50, edited 1 time in total.
Edited the question and added the OA.




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Re: Median of a combined interval
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06 Feb 2012, 02:49
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 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 \(R=\frac{a+c}{2}=\frac{c*\frac{3}{8}+c}{2}=c*\frac{11}{16}\) Answer: C (\(\frac{11}{16}\)).
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a, b, and c are integers and a < b < c. S is the set of all
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06 Feb 2012, 02:37
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.




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Re: Statistics..
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27 Oct 2009, 22:39
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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Re: Statistics..
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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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27 Oct 2009, 22:59



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Re: Statistics..
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27 Oct 2009, 23:01
Thanks Buddy..much clear now..



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a, b, and c are integers and a < b < c. S is the set of all
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12 Jan 2010, 04:46
for nos between b and c ,7c/8 is the median..means c has to be a multiple of 4 or 8... but for 3b/4 to be a median, c has to be a multiple of 8 ..... lets take it as 8.. median=7*8/7=7.... set is 6,7,8...this makes b=6...median=6*3/4= 9/2.... so the set is 3,4,5,6.. set Q is 3,4,5,6,7,8.. the median is 11/2,which is 11/16 of 8 or c.... we can try the same with other multiple of 8 also..
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Re: Median advanced and tricky
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16 Jan 2010, 11:54
i fails to understand how can the median between two number a and b can be a+b/2.It depends on number of integer between the range so median between 1 6 and 7 cannot be 4 it will be 6 .so please calcify the approach in details



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Re: Median advanced and tricky
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16 Jan 2010, 12:43
arghya05 wrote: i fails to understand how can the median between two number a and b can be a+b/2.It depends on number of integer between the range so median between 1 6 and 7 cannot be 4 it will be 6 .so please calcify the approach in details Well, from some point you are right: median divides the sample in two parts: Equal number of observations in each. It doesn't necessarily mean (a+b)/2 But you should also consider that, in set S, there are " all integers from a to b, inclusive", and in set Q there are " all integers from b to c, inclusive" So these sets are sets of consecutive integers. In this case Mean=Median=(a+b)/2



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Re: Median of a combined interval
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06 Feb 2012, 02:52
Bunuel, I know the solution that you've given (I've read it in some of your previous posts).
But could you please explain where is the mistake in my solution?
Thank you.



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Re: Median of a combined interval
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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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Re: Median of a combined interval
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06 Feb 2012, 03:23
Can someone please explain the mistake in my original solution in the first post? Thanks a lot.



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Re: Median of a combined interval
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06 Feb 2012, 03:34



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Re: Median of a combined interval
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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: Median of a combined interval
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06 Feb 2012, 04:20
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: Median of a combined interval
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06 Feb 2012, 04:22
Yes, thanks a lot. I got it.



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Re: Median of a combined interval
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03 Dec 2012, 10:14
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?



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Re: Median of a combined interval
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04 Dec 2012, 03:15
aditi2013 wrote: 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: Median of a combined interval &nbs
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