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Qns on Median

by skyline77 » Mon Sep 01, 2008 6:02 am
Does anyone has a clue to solve this?

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
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by mals24 » Mon Sep 01, 2008 6:31 am
consider the numbers 1, 3, 7.

the median of a set of numbers from 1 to 3 inclusive = {1, 2, 3} = 2
another way to find it is take the average of the end points ie (1+3)/2 = 2

The median of a set of numbers from 3 to 7 inclusive = {3,4,5,6,7} = (3+7)/2=5

now coming back to the question.

it says a<b<c = intergers

S = all integers from a to b incl
median = (3/4)b
Q = all integers from b to c incl
median = (7/8 )c
R = all integers from a to c incl

median of S = avg of its end points = (a+b)/2= (3/4)b

solving it you will get a = b/2 or b=2a----- equ 1

median of Q = avg of its end points = (b+c)/2 = (7/8 )c

Solving it you will get b = (3/4)c or 4b = 3c------equ 2

multiply equ 1 with 4 you'll get 4b = 8a-------equ 3

equating equ 2 and 3 we get 3c=8a
hence a = (3/8 )c

median of R = avg of its end points = (a+c)/2 = [(3/8 )c+c]/2 = (11/16)c

We need to find what fraction of c is the median of R = [(11/16)c]/c = 11/16

Hence the answer IMO should be C

Tell me if you have any doubts.
Nice ques btw took a long time to solve it. Basically since the question asks us about c we should try and get all the other variables in terms of c.
Last edited by mals24 on Mon Sep 01, 2008 6:33 am, edited 2 times in total.
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Re: Qns on Median

by sudhir3127 » Mon Sep 01, 2008 6:31 am
skyline77 wrote:Does anyone has a clue to solve this?

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
IMO C..

We need to know that

"In a set of consecutive integers the median of the set in equal to the mean of the set."

Median of set S = Mean

Median of S = (a+b)/2

(3/4)b = (a+b)/2

therefore a= b/2....................................................1

median of set Q= (b+c)/2

7c/8= (b+c)/2

therefore c=4b/3 ..................................................2

thus

Median of set R=(a+c)/2

Adding 1 and 2 we get

(b/2+4b/3)/2

=11b/12

which is

11/12*b

from 2 we know b= 3c/4

thus its 11/12*3c/4
hence

11c/16. C

Hope that helps..do let me know if u still have doubts..
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Re: Qns on Median

by skyline77 » Mon Sep 01, 2008 5:20 pm
sudhir3127 wrote:
skyline77 wrote:Does anyone has a clue to solve this?

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
IMO C..

We need to know that

"In a set of consecutive integers the median of the set in equal to the mean of the set."

Median of set S = Mean

Median of S = (a+b)/2

(3/4)b = (a+b)/2

therefore a= b/2....................................................1

median of set Q= (b+c)/2

7c/8= (b+c)/2

therefore c=4b/3 ..................................................2

thus

Median of set R=(a+c)/2

Adding 1 and 2 we get

(b/2+4b/3)/2

=11b/12

which is

11/12*b

from 2 we know b= 3c/4

thus its 11/12*3c/4
hence

11c/16. C

Hope that helps..do let me know if u still have doubts..
Thank you so much. I was struck halfway thru. I will try again and if any problem i will check with you again.
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by karthikpandian19 » Thu Jun 07, 2012 12:08 am
Mean = Median for cons. intg

3b/4 = (a +b)/2
3b=2a+2b
b=2a

7c/8=(b + c)/2
7c=4b+4c
c=4b/3 = 8a/3

Median = (a + c)/2 ; Substitute value of "a" from prev eqn: 11c/16
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by GMATGuruNY » Thu Jun 07, 2012 2:36 am
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) ½ (C) 11/16 (D) 5/7 (E) ¾
Let c=16.
Median of Q = (7/8)*c = (7/8)*16 = 14.
Since 14 is halfway between b and 16, b=12.
Median of S = (3/4)*b = (3/4)*12 = 9.
Since 9 is halfway between a and 12, a=6.

The median of set R -- which is composed of all of the integers from a=6 to c=16, inclusive -- is the average of 6 and 16:
(6+16)/2 = 11.

(Median of R)/c = 11/16.

The correct answer is C.
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by karthikpandian19 » Thu Jun 07, 2012 6:37 am
plugging the numbers really helps.......most of your explanation is dealing with plugging nos...thanks mitch
GMATGuruNY 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) ½ (C) 11/16 (D) 5/7 (E) ¾
Let c=16.
Median of Q = (7/8)*c = (7/8)*16 = 14.
Since 14 is halfway between b and 16, b=12.
Median of S = (3/4)*b = (3/4)*12 = 9.
Since 9 is halfway between a and 12, a=6.

The median of set R -- which is composed of all of the integers from a=6 to c=16, inclusive -- is the average of 6 and 16:
(6+16)/2 = 11.

(Median of R)/c = 11/16.

The correct answer is C.
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by 1947 » Thu Jun 07, 2012 10:52 pm
Mitch...you have a number plugging way for all questions....now the question is how can we inculcate such approach...
I always approach the questions by regular algebra way....and then you have a quicker way to solve this.
GMATGuruNY 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) ½ (C) 11/16 (D) 5/7 (E) ¾
Let c=16.
Median of Q = (7/8)*c = (7/8)*16 = 14.
Since 14 is halfway between b and 16, b=12.
Median of S = (3/4)*b = (3/4)*12 = 9.
Since 9 is halfway between a and 12, a=6.

The median of set R -- which is composed of all of the integers from a=6 to c=16, inclusive -- is the average of 6 and 16:
(6+16)/2 = 11.

(Median of R)/c = 11/16.

The correct answer is C.
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by hey_thr67 » Fri Jun 08, 2012 6:51 am
I am little bit confused over mean and median.

What is the median of following numbers ?

1 2 6 7 4 9 10

median is 7 but mean is not 7.

So, you see median is not at the halfway.

So either the language of this question is not correct or I have my understanding wrong. Going by this definition, question is invalid.
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by GMATGuruNY » Fri Jun 08, 2012 7:12 am
1947 wrote:Mitch...you have a number plugging way for all questions....now the question is how can we inculcate such approach...
I always approach the questions by regular algebra way....and then you have a quicker way to solve this.
GMATGuruNY 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) ½ (C) 11/16 (D) 5/7 (E) ¾
Let c=16.
Median of Q = (7/8)*c = (7/8)*16 = 14.
Since 14 is halfway between b and 16, b=12.
Median of S = (3/4)*b = (3/4)*12 = 9.
Since 9 is halfway between a and 12, a=6.

The median of set R -- which is composed of all of the integers from a=6 to c=16, inclusive -- is the average of 6 and 16:
(6+16)/2 = 11.

(Median of R)/c = 11/16.

The correct answer is C.
Many questions that can be solved algebraically can also be solved by alternate methods -- specifically, by plugging in numbers or by plugging in the answers.

When we plug in our own values, we should choose numbers that are multiples of the divisors in the problem.
The goal is to avoid having to manipulate fractions.
Since the greatest value here is c, and the median of set Q = (7/8)c, c should be a multiple of 8.
Looking at the answer choices, we can see that the greatest denominator that is a multiple of 8 is 16, implying that 16 will be a good number to plug in for c.
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by karthikpandian19 » Fri Jun 08, 2012 7:15 am
Dude,

Median is 6

before finding median arrange them in ascending order

1 2 4 6 7 9 10


Now the median is the middle number 6
hey_thr67 wrote:I am little bit confused over mean and median.

What is the median of following numbers ?

1 2 6 7 4 9 10

median is 7 but mean is not 7.

So, you see median is not at the halfway.

So either the language of this question is not correct or I have my understanding wrong. Going by this definition, question is invalid.
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by GMATGuruNY » Fri Jun 08, 2012 7:23 am
hey_thr67 wrote:I am little bit confused over mean and median.

What is the median of following numbers ?

1 2 6 7 4 9 10

median is 7 but mean is not 7.

So, you see median is not at the halfway.

So either the language of this question is not correct or I have my understanding wrong. Going by this definition, question is invalid.
From the problem:
S is the set of ALL the integers from a to b, inclusive.
Q is the set of ALL the integers from b to c, inclusive.
R is the set of ALL the integers from a to c, inclusive.

ALL of the integers between two values = a set of CONSECUTIVE INTEGERS.
For example, all of the integers between 12 and 16, inclusive = {12,13,14,15,16}.
With consecutive integers, the median = the average = (biggest+smallest)/2 = the value HALFWAY between the biggest and the smallest.
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by hey_thr67 » Fri Jun 08, 2012 8:48 am
Ok I got the consecutive part. But what if we have even numbers 12-17 then median will be 14.5. Ok I got it since median is to be a integer. we have to have odd number of numbers in the list. right ?
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