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by anksbhandari » Thu Jul 17, 2008 9:15 am
Please Explain,

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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what number represents smiley face? please provide. thanks.

by hakyology » Thu Jul 17, 2008 10:13 am
can't do it without real smiley face number, i think.
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Re: Inequalities:

by parallel_chase » Thu Jul 17, 2008 10:32 am
anksbhandari wrote:Please Explain,

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
From next time kindly disable the smilies before posting the question.

Is the answer 11/16?
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by VP_Tatiana » Thu Jul 17, 2008 10:41 am
Hi Parallel,

I was unable to solve your problem because one of the numbers was mistyped and displayed a smiley face instead! Could you please retype your question? Thanks!

Tatiana
Tatiana Becker | GMAT Instructor | Veritas Prep
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by parallel_chase » Thu Jul 17, 2008 10:45 am
VP_Tatiana wrote:Hi Parallel,

I was unable to solve your problem because one of the numbers was mistyped and displayed a smiley face instead! Could you please retype your question? Thanks!

Tatiana
(7/8)c

The code for the smily is /8)

I guess the answer is 11/16, let me know what you think.
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I agree.

by hakyology » Thu Jul 17, 2008 11:14 am
I agree. I got 11/16.
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by ildude02 » Thu Jul 17, 2008 2:05 pm
With 7/8, I got 1/2 as the answer. Can you please tell how you got 11/16? I solved it using numbers. So it wasn't the fastest way.
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Repeat questions

by anksbhandari » Sat Jul 19, 2008 2:58 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
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The figure behind : (7/8)C

by anksbhandari » Sat Jul 19, 2008 3:00 am
The fraction behind smiling face is (7/8) C
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OA

by anksbhandari » Sat Jul 19, 2008 3:05 am
( The fraction behind smiliy is (7/8) C)

OA is option C 6/11
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Re: OA

by parallel_chase » Sat Jul 19, 2008 3:09 am
anksbhandari wrote:( The fraction behind smiliy is (7/8) C)

OA is option C 6/11
Option C is 11/16.

To disable the smily, while posting just check the option " diable the smily"
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by ildude02 » Sat Jul 19, 2008 8:52 am
Can someone solve the problem? I wanted to see if my apprach is different and if there is an easier way to solve then by choosing numbers.
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by parallel_chase » Sat Jul 19, 2008 11:52 am
ildude02 wrote:Can someone solve the problem? I wanted to see if my apprach is different and if there is an easier way to solve then by choosing numbers.
S [a-b] = 3/4 b

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

Q [b-c] = 7/8 c

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

b=3/4 c

Therefore,

a=3/8 c
b= 3/4 c

R [a-c] = ?

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

Hence C is the answer.

let me know if you have any questions.
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by ildude02 » Sat Jul 19, 2008 12:28 pm
Thanks for your explanation. I assume you based it on the theory that Average of consecutive integers is equal to Median ? Not sure how I missed it, damn!
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