Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 02 Oct 2009
Posts: 64

a, b, and c are integers and a < b < c. S is the set of all integers
[#permalink]
Show Tags
18 Sep 2009, 00:15
Question Stats:
51% (03:04) correct 49% (03:02) wrong based on 270 sessions
HideShow timer Statistics
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}
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
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]
Show Tags
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 (\(\frac{11}{16}\)).
_________________




Manager
Joined: 11 Sep 2009
Posts: 112

Re: a, b, and c are integers and a < b < c. S is the set of all integers
[#permalink]
Show Tags
Updated on: 10 Feb 2019, 00:36
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
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.




Math Expert
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]
Show Tags
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 (\(\frac{11}{16}\))
_________________



Manager
Joined: 02 Oct 2009
Posts: 64

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



Math Expert
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]
Show Tags
27 Oct 2009, 22:59
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.
_________________



Intern
Joined: 22 Nov 2010
Posts: 1

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



Math Expert
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]
Show Tags
15 Dec 2010, 03:19
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 (\(\frac{11}{16}\)).
_________________



Director
Joined: 23 Apr 2010
Posts: 501

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



Manager
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]
Show Tags
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



Math Expert
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]
Show Tags
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
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.
_________________



Math Expert
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]
Show Tags
06 Feb 2012, 03:39
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).
_________________



Director
Joined: 23 Apr 2010
Posts: 501

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



Math Expert
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]
Show Tags
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.
_________________



Math Expert
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]
Show Tags
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.
_________________



Intern
Joined: 13 Apr 2013
Posts: 13
Location: India
Concentration: Operations, Strategy
GPA: 3.5
WE: Operations (Transportation)

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



Math Expert
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]
Show Tags
10 Jul 2013, 02:09
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.
_________________



Intern
Joined: 09 Feb 2016
Posts: 1

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



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10477
Location: Pune, India

Re: a, b, and c are integers and a < b < c. S is the set of all integers
[#permalink]
Show Tags
10 Feb 2016, 00:30
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 Answer (C)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10477
Location: Pune, India

Re: a, b, and c are integers and a < b < c. S is the set of all integers
[#permalink]
Show Tags
07 Mar 2016, 22:00
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) Answer (C)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >




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



Go to page
1 2
Next
[ 30 posts ]

