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Statistics Problem 1 - Median

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Statistics Problem 1 - Median

by surajgarg » Wed Jul 21, 2010 8:34 pm
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 t0 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 later. I would like to know how to approach this problem.
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by 4GMAT_Mumbai » Wed Jul 21, 2010 8:48 pm
Hi,

Interesting Problem ... Thanks !!!

Median of series from b to c is (7/8)th of c

Think about it ... In that series, c is the last term. 7c/8 is the middle term. What would be the first term?

It would be 6c/8 ... This is because, 7c/8 is the middle term of a series running from 6c/8 to 8c/8 (aka c).

So, b = 6c / 8.

Come to the first series from a to b. Median is 3b / 4.

We know that b = 6c / 8.

Substituting b in the median; we get median of first series = (3/4) * (6c / 8) = 9c / 16.

So, collating the things;

Median of first series = 9c / 16

b = 6c / 8

c = c (trivial).

All of these are integers. So, let us assume that c = 16 (so that 9c / 16 is an integer).

Median of first series = 9c / 16 = 9

b = 6c / 8 = 12

c = c (trivial) = 16.

Hence, the series is a=6,7,8,(First Median)9,10,11,b=12,13,(Second Median)14,15,c=16.

Hence, overall median = 11.

Ans. c

Is it correct ... Hoping to learn better methods.

Thanks.
Naveenan Ramachandran
4GMAT, Dadar(W) & Ghatkopar(W), Mumbai
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by outreach » Wed Jul 21, 2010 10:46 pm
https://www.beatthegmat.com/hard-median- ... 29024.html
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by sumanr84 » Thu Jul 22, 2010 12:35 am
Best answer from the link given by outreach,

Rule in action : Mean and Median for consecutive set of integer are same.

The Answer is 11/16.

The median of Set S which has all the values from a to b (inclusive) is (a+b)/2.

So (a+b)/2 = 3b/4 --> b = 2a

Similarly

The median of Set Q which has all the values from b to c (inclusive) is (b+c)/2.

So (b+c)/2 = 7/8c ---> c = 4b/3

Substitute b = 2a --> c = 8a/3

The median of a to c is ie a to 8a/3 id (8a/3 + a )/2

This gives the value = 11a/6.

we want in terms of c
substitite a= 3c/8

Gives the answer 11c/16.
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by surajgarg » Thu Jul 22, 2010 12:41 am
sumanr84 wrote:Best answer from the link given by outreach,

Rule in action : Mean and Median for consecutive set of integer are same.

The Answer is 11/16.

The median of Set S which has all the values from a to b (inclusive) is (a+b)/2.

So (a+b)/2 = 3b/4 --> b = 2a

Similarly

The median of Set Q which has all the values from b to c (inclusive) is (b+c)/2.

So (b+c)/2 = 7/8c ---> c = 4b/3

Substitute b = 2a --> c = 8a/3

The median of a to c is ie a to 8a/3 id (8a/3 + a )/2

This gives the value = 11a/6.

we want in terms of c
substitite a= 3c/8

Gives the answer 11c/16.
Hi Suman,

The Mean = Median rule does work for consecutive integers. But the question does not state that the integers are consecutive.

However, in this particular problem the rule worked, even though it was an assumption that the the integers are consecutive.
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by sumanr84 » Thu Jul 22, 2010 12:54 am
surajgarg wrote:
Hi Suman,

The Mean = Median rule does work for consecutive integers. But the question does not state that the integers are consecutive.

However, in this particular problem the rule worked, even though it was an assumption that the the integers are consecutive.
Question does mention that S is the set of all integers from a to b, inclusive. Q is the set of all integers from b to c.

all integers from a to b - It has to be consecutive right ?

Even if its is not consecutive,in order to find MEDIAN, first you need to arrange all integers in increasing order before you go and find the middle.
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by surajgarg » Thu Jul 22, 2010 12:58 am
sumanr84 wrote:
surajgarg wrote:
Hi Suman,

The Mean = Median rule does work for consecutive integers. But the question does not state that the integers are consecutive.

However, in this particular problem the rule worked, even though it was an assumption that the the integers are consecutive.
Question does mention that S is the set of all integers from a to b, inclusive. Q is the set of all integers from b to c.

all integers from a to b - It has to be consecutive right ?

Damn my eyes or my brain or whatever. All integers does imply consecutive integers. Thanks for the pointer :)

One very important take away from many such problems is to be very very careful of words in the question, especially if it is a lengthy one. (Importance of verbal in the quant section :D)

It can make or break one's approach to the problem.
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by sumanr84 » Thu Jul 22, 2010 1:38 am
surajgarg wrote:
One very important take away from many such problems is to be very very careful of words in the question, especially if it is a lengthy one. (Importance of verbal in the quant section :D)

It can make or break one's approach to the problem.
I completely agree with suraj. I have done numerous mistakes due to eye-slippage. GMAT tricks here..;-)
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