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MGMAT Cat II

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MGMAT Cat II

by mdavis » Mon Feb 09, 2009 9:24 am
I am having trouble understanding the notation here. Here is the question:

What is the median value of the set R, if for every term in the set, Rn = Rn–1 + 3?

(1) The first term of set R is 15.

(2) The mean of set R is 36.

OA: B

I do not quite understand how to interpret Rn = Rn–1 + 3? I thought that if for example, taking (1) here, if R is 15, would that not be R15=(15-1)+3=17?

The MGMAT Cat explanation says that it is a spread of 3. So I am not looking at this equation correctly. Could someone point out what I am doing wrong? Thanks so much!
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by BuckeyeT » Mon Feb 09, 2009 9:42 am
I think this belong in the data sufficiency section. But, I can offer the explanation as I see it.

The statement, for set R, is "what is the median value if Rn=Rn-1+3"

The MGMAT Cat is correct is mentioning a spread of 3. Let's suppose we pick any number Rn in the set. Rn=Rn-1+3 can be rewritten as Rn-1=Rn-3. So, we know that Rn-3 is also in the set. As an example, let's say 15 is a randomly chosen value in the set. As we just mentioned 15-3=12 is also a value in the set (Rn-1=12).

Because there is a 3 point spread in values, we know that the mean will be either a fraction or an integer. It will be a fraction when there are an even number of values in the set (example, set{2,5} = (2+5)/2 = 7/2). It will be an integer when there are an odd number of values in the set (example, set{2,5,8} = (2+5+8)/3 = 15/5 or 5). By this second example, we also realize that if the number of values in the set is odd, the mean is equal to the median.

Unfortunately, the (1) information only tells us that 15 is the first term in the set. Is it the only set member? How many more are there? Insufficient.

(2) Tells us the mean of the set is 36. From our work above, we know that an integer mean tells us that the set has an odd number of set members. And moreso, the mean is equal to the median in these situations. So, (2) is sufficient.

The DS answer would be B.
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by mdavis » Mon Feb 09, 2009 10:21 am
Whoops, thought I was in DS. I am not understanding the equation you reworked. You said that it could be, Rn-1=Rn-3. So if it was 15, would that not be 14=12? That seems like a 2 point spread. What exactly am I doing here? It seems like I am plugging in numbers incorrectly or something.
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by BuckeyeT » Mon Feb 09, 2009 10:48 am
Sorry about the cluttered writing in my original. And, it's hard to illustrate sub writing like R(sub n).

R(sub n) represents a certain value in set R. R(sub n -1) represents another value in set R. Just like R(sub n+1) would represent another value in set R.

Since the original statement says R(sub n) = R(sub n-1) + 3, let's think of these as two different values within set R.

R(sub n) = x
R(sub n-1) = y

So to rewrite the original statement, x = y + 3. This can be rewritten as y = x - 3.

So if R(sub n-1) = 15, then y = 15 (from our substitution above).

Then, x = y + 3 can be seen as x = 15 + 3 or x = 18.

Your only issue is accidently mixing your sub values with actual values.
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by mdavis » Mon Feb 09, 2009 11:28 am
Ahh, that makes so much sense now. I need to remember these things! Thanks so much for your help. I really appreciate it.
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by sureshbala » Mon Feb 09, 2009 12:16 pm
The question says that each term is 3 more than the previous term i.e the given series is an arithmetic series with common difference.

Satement 1: Since we do not know the number of terms this is not sufficient.

Statement 2: In an Arithmetic series since mean = median, statemnt 2 is sufficient.
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