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More on Mean & Median

by theCodeToGMAT » Fri Sep 20, 2013 2:31 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)1/2 (C)11/16 (D)5/7 (E)3/4

OA [spoiler][C][/spoiler]
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by Uva@90 » Fri Sep 20, 2013 3:13 am
theCodeToGMAT 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 [spoiler][C][/spoiler]
Hi Rahul,
Are you sure OA is C because I am getting OA B.

Here is how I did,
S Includes set of integers from a to b inclusive and median is (3/4)(b)

Let us take the value of b as '4'(for simplicity)
Median of Set S will be now '3'
so a can be any number less than 3, let us assume 2

S = {2,3,4}

Similarly Set q includes b,c, inclusive and median (7/8)(c)
Let us take the value of c ad '8'
median of set Q will be now '7

Q = {4,7,8}

Set includes all the integers from a to c inclusive

So,
R = {2,3,4,7,8}
hence median of Set R is 4

to be find: what fraction of C is median of set R
4=X*(8) => X=1/2

hence OA is B

Experts Correct me if I am wrong.

Thanks in advance.
Regards
Uva.
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by theCodeToGMAT » Fri Sep 20, 2013 3:24 am
Yes, indeed the answer is [spoiler][C][/spoiler]

My Steps to Solution:

S --> A TO B
Q --> B TO C

Since A-->B---->C are in AP
So, Median = Mean [Mean = Median doesn't mean AP]

For A --> B
(a+b)/2 = 3b/4
2a = b

For B ---> C
(b+c)/2 = 7c/8
4b=3c
b = 3c/4
a = 3c/8


Now, for A -----> C

(a + c)/2
= (3c/8 + c)/2
= (11c/16)

So, [spoiler][C][/spoiler]
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by ganeshrkamath » Fri Sep 20, 2013 3:29 am
theCodeToGMAT 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 [spoiler][C][/spoiler]
Since S is a set of consecutive integers, median = mean.
Mean = (b+a)/2 = (3/4)b
(b+a) = (3/2)b
a = b/2

Similarly, solving for set Q
(c+b)/2 = (7/8)c
(c+b) = (7/4)c
b = (3/4)c

Now,
median of set R = mean of set R = (a+c)/2
= (b/2 + c)/2
= (3c/8 + c)/2
= (11c/8)/2
= (11/16)c

Choose C

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by ganeshrkamath » Fri Sep 20, 2013 3:32 am
Uva@90 wrote:Similarly Set q includes b,c, inclusive and median (7/8)(c)
Let us take the value of c ad '8'
median of set Q will be now '7

Q = {4,7,8}
I think you've gone wrong here. The question says all integers from b to c, inclusive.
In your case, Q should've included the numbers 5 and 6.

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by GMATGuruNY » Fri Sep 20, 2013 4:48 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 bnpetteway » Fri Sep 20, 2013 4:50 pm
I did the same setup as ganesh and I got to that same answer.
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