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mgmat 2

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by resilient » Wed May 21, 2008 4:10 pm
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.

qa is b
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by luvaduva » Wed May 21, 2008 4:38 pm
I originally did the problem quickly just assuming that all the numbers in R were evenly spaced integers(potentially a bad assumption looking back hehe) which means that the mean = median. And if we know the first term and the last term we can calculate the mean quickly by using:

(first + last)/2

(1) NS we are given the first term, but we don't know anything about the last.

(2) We are given the mean so from what I said above we can calculate the median. Sufficient.

Looking back at the problem, I realized that the above also applies to not only consectuively placed integers, but also consecutively spaced real too(I just tested with basic fractions).

Good to know.
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by netigen » Wed May 21, 2008 11:12 pm
https://www.beatthegmat.com/set-rules-pl ... 11062.html

Ans is B as median = mean in this case
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mgmat 2

by resilient » Sat May 24, 2008 9:53 pm
in statement two we are given the average but how do I connect this to the median value. still not seeing the whole picture although I learned a nice trick for mean.
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Re: mgmat 2

by 4meonly » Sat Feb 07, 2009 2:39 am
resilient wrote: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.

qa is b
Need exhaustive explanation. I feel myself stupid today :shock:
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by 4meonly » Sat Feb 07, 2009 2:42 am
Median=Mean in case of a set with equal distribution (1,2,3,4) or (2,5,8,11,14)
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by sureshbala » Sat Feb 07, 2009 4:55 am
The terms in the given set are in Arithmetic Progression.

Statement 1: Since we do not know the number of terms in the set we can calculate the median.

Statement 2: Given that mean is 36. For every Arithmetic series, Mean = Median. So statement 2 is alone is sufficient.
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by dixitsandeep » Tue Oct 23, 2012 11:21 am
Nobody here seems to have got right answer.

B is answer when you assume that the set cannot contain negative numbers.

To exclude that assumption you need to know first term of the set.

And that's why answer is C.
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by himallik » Fri Apr 26, 2013 1:54 pm
Answer is still B.

Take an evenly spaced set with 3 as common difference. { -6, -3, 0, 3, 6, 9, 12, 15 }.


Median = (3+6)/2 = 4.5
Mean = (-6-3-0+3+6+9+12+15)/8 = 4.5


So it is the property of evenly spaced set. Mean = Median. It doesn't matter if it has only positive or both positive and negative numbers in it.

Thanks,
Mallik
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by srcc25anu » Fri Apr 26, 2013 3:42 pm
Set is basically an equally spaced set with interval of 3 between 2 consecutive terms.

St1: First Term = 15
we do not know the number of elements in the set hence we cannot determine the median.
If set R has 2 terms {15, 18} median = 16.5
but if Set R has 3 terms {15, 18, 20} then median = 18
Not sufficient

St2: Mean of R = 36
Since this is an equally spaced set, mean = median
hence median = 36
SUFFICIENT

ANS: B
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by Matt@VeritasPrep » Sun Apr 28, 2013 7:38 pm
Just to distill all the technical explanations given here into one takeaway phrase:

"In an evenly spaced set, the mean is equal to the median!"

Since the terms in this set are {x, x+3, x+3+3, x+3+3+3, etc.} it's evenly spaced, so (B) does it.
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by masoom j negi » Fri Dec 21, 2018 9:06 pm
The function defines that the terms in Set R are in A.P. with common difference 3.
Statement 1.We can't determine the median of an A.P. with just its first term and Common difference. Last term or the total no of terms is also required. Hence, Insufficient.
Statement 2. In A.P. the mean of the series is equal to the median of the series. Hence, Sufficient.
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