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Combination of sets

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Combination of sets

by Anindya Madhudor » Mon Dec 03, 2012 11:50 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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by eaakbari » Mon Dec 03, 2012 12:39 pm
IMO C.

Since all sets contain consecutive integers, we can conclude that its an EVENLY spaced set where median = mean.

For S , median = a+b/2 = b*3/4
which results simplifies
b = 2a

For Q , median = c+b/2 = b*7/8
which results simplifies
4b = 3c

Combining the above two, we can arrive
8a = 3c

Now Set R
Mean = (a+c)/2
Substituting above equation and simplifying

11/16 *c

Hence C

P.S.
Anindya, you pose quite challenging questions every now and then. Whats the source?
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by Anindya Madhudor » Mon Dec 03, 2012 1:06 pm
Thanks for your response.

The problems I have posted in the past couple of days are from a problem set i found in a file sharing site. It claims to contain 700-800 level questions. But the document contains no answers. I can share that with you if you want. Please send me your e-mail address.

Let's get back to the problem in hand.

I found the answer to be 3/4. Again, I don't have the OA.

Here is what i did.

I picked some numbers to fit the problem.

Let a=4, b=8, c=16. Set S has all numbers from 4 to 8 appearing once. Set Q has all numbers appearing from 8 to 16 appearing once, except 15, which appears 5 times. Then median of set S is 6 and median of set Q is 14.

When you combine Q and S, the median is 12, which is 3/4 of c.

Let me know if you see anything wrong in my thinking.
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by GMATGuruNY » Mon Dec 03, 2012 1:12 pm
Anindya Madhudor 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
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.

Since R is the set of all the integers from a to c, inclusive, R = {6,7,8,9,10,11,12,13,14,15,16}.
Median of R = 11.

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

The correct answer is C.
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by eaakbari » Mon Dec 03, 2012 1:20 pm
Your picked numbers are NOT fitting into the below criteria
The median of set S is (3/4)b. The median of set Q is (7/8)c.
Besides I think picking number's is a long and convoluted solution to this problem.
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by Anindya Madhudor » Mon Dec 03, 2012 1:24 pm
I agree that picking numbers is not an efficient way of solving this problem. Thanks for your response.
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by subhakam » Tue Jan 08, 2013 9:06 pm
Question for eaakbari - what made you choose Median = Mean (a+c)/2 instead of just picking b? When we have Set R = {a, b, c} we have odd number of terms; isn't the median = middle value for consecutive integers when the # of terms are odd?

If we do pick b, then the answer is 3/4 but if we do what you suggested - Median = Mean (first+last term)/2 we get 11/16.

Please help - could not understand what made you switch and not fall for the trap answer (3/4)

Many thanks!!
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by eaakbari » Wed Jan 09, 2013 10:14 am
subhakam wrote:Question for eaakbari - what made you choose Median = Mean (a+c)/2 instead of just picking b? When we have Set R = {a, b, c} we have odd number of terms; isn't the median = middle value for consecutive integers when the # of terms are odd?

If we do pick b, then the answer is 3/4 but if we do what you suggested - Median = Mean (first+last term)/2 we get 11/16.

Please help - could not understand what made you switch and not fall for the trap answer (3/4)

Many thanks!!
Subhakam,

Evenly spaced sets: As the name implies, each element in the set is a equal distance from the previous element, i.e. an Arithmetic progression
Consecutive integers, multiples of any number 'n' are examples of evenly spaced sets.

Remember the following about all evenly spaced sets
5.15 Evenly Spaced Sets:

i. Mean and Median are equal
ii. Mean and Median of the set = Average (First + Last terms)
iii. Sum of the elements = Mean of the set * Number of elements in the set

Applying (ii) Median = Mean (a+c)/2.

Now as for your question as to why picking b is wrong and why a difference of answers: You have misread the question.
a,b,c are NOT mentioned as consecutive integers.
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by bnpetteway » Wed Jan 09, 2013 3:59 pm
@ GuruNY, why did you choose 16 for C?

Edit: That's okay I see why. Could it work if I were to choose 8 instead or are we runningg the risk of b being smaller than a?
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by subhakam » Wed Jan 09, 2013 4:48 pm
Thank you very much Eaakbari! While I did not indeed see them listed as consecutive integers, I was assuming it:)
Now the question does not state evenly spaced integers explicitly either. Did you reach that conclusion before hand or only after you calculated and reached the result b= 2a (that implies the set is evenly spaced with a multiple of 2)

All help greatly appreciated!!
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by GMATGuruNY » Fri Jan 11, 2013 8:53 am
bnpetteway wrote:@ GuruNY, why did you choose 16 for C?

Edit: That's okay I see why. Could it work if I were to choose 8 instead or are we runningg the risk of b being smaller than a?
Let c=8.
Median of Q = (7/8)c = (7/8)8 = 7.

Since 7 is halfway between b and 8, b=6.
Median of S = (3/4)6 = 4.5.

Since 4.5 is halfway between a and 6, a=3.

Since R is the set of all the integers from a to c, inclusive, R = {3,4,5,6,7,8}.
Median of R = (5+6)/2 = 5.5.

Thus:
(Median of R)/c = 5.5/8 = 11/16.

The correct answer is C.
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