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If m is the average (arithmetic mean) of the first 10

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If m is the average (arithmetic mean) of the first 10 [#permalink]

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If m is the average (arithmetic mean) of the first 10 positive multiples of 5 and if M is the median of the first 10 positive multiples of 5, what is the value of M – m ?

(A) –5
(B) 0
(C) 5
(D) 25
(E) 27.5
[Reveal] Spoiler: OA
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New post 27 Feb 2012, 13:43
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BANON wrote:
If m is the average (arithmetic mean) of the first 10 positive multiples of 5 and if M is the median of the first 10 positive multiples of 5, what is the value of M – m ?

(A) –5
(B) 0
(C) 5
(D) 25
(E) 27.5


10 sec approach:

The first 10 positive multiples of 5 is an evenly spaced set. One of the most important properties of evenly spaced set (aka arithmetic progression) is: in any evenly spaced set the arithmetic mean (average) is equal to the median.

Hence M=m --> M-m=0.

Answer: B.

For more on this issue check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.
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Re: If m is the average [#permalink]

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New post 27 Feb 2012, 14:08
1st 10 multiples of 5: 5,10,15,20,25,30,35,40,45,50

mean = 275/10 = 27.5 = m


median = (25+30)/2 = 22.5 = M

M-m= -5

where i am doing wrong???
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New post 27 Feb 2012, 14:15
right now i am banging my head to the wall. :oops:
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Re: If m is the average (arithmetic mean) of the first 10 [#permalink]

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As it is a Evenly spaced set, the MEAN and MEDIAN will be same for the set.

So, the MEAN or MEDIAN for such sets will be -> Avg. of First term and last term.
i.e, (5+50)/2 in this case.
Blindly we can say answer is zero, because the Q asks the difference between MEAN and MEDIAN for Evenly spaced set(Basic Thumb rule)... :-D
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Re: If m is the average (arithmetic mean) of the first 10 [#permalink]

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New post 24 Jun 2013, 20:30
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BANON wrote:
If m is the average (arithmetic mean) of the first 10 positive multiples of 5 and if M is the median of the first 10 positive multiples of 5, what is the value of M – m ?

(A) –5
(B) 0
(C) 5
(D) 25
(E) 27.5



Responding to a pm:

m = mean = (5 + 10 + 15 + ....+ 50)/10 = 27.5

M = median
Median of 10 numbers will be the average of the middle two numbers i.e. 5th and the 6th numbers. 5th number = 25, 6th number = 30. Median = (25+30)/2 = 27.5

m - M = 0

This solution is the simplest I could think of which uses nothing but the definition of mean and median.
Notice that actually, you will do far less to arrive at the answer.

Mean of an arithmetic progression is the middle value in case there are odd number of terms and average of middle 2 values if there are even number of terms.
Median of an arithmetic progression is the middle term in case there are odd number of terms and average of middle 2 values if there are even number of terms.
So basically, they are both same in case of an arithmetic progression.

I would suggest you to check out the following posts. They discuss these concepts in detail:

http://www.veritasprep.com/blog/2012/04 ... etic-mean/
http://www.veritasprep.com/blog/2012/05 ... eviations/
http://www.veritasprep.com/blog/2012/05 ... on-median/
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Re: If m is the average (arithmetic mean) of the first 10 [#permalink]

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New post 25 Jun 2013, 12:52
Just out of curiosity, Why are we not taking 0 as the first multiple of 5. After all, 0 should be the first multiple of any number :)
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Re: If m is the average (arithmetic mean) of the first 10 [#permalink]

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pavan2185 wrote:
Just out of curiosity, Why are we not taking 0 as the first multiple of 5. After all, 0 should be the first multiple of any number :)


Yes, 0 is a multiple of every integer, except 0 itself. But the question talks about positive integers and 0 is neither positive nor negative.

Hope it's clear.
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Re: If m is the average (arithmetic mean) of the first 10 [#permalink]

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New post 11 Sep 2014, 00:45
BANON wrote:
If m is the average (arithmetic mean) of the first 10 positive multiples of 5 and if M is the median of the first 10 positive multiples of 5, what is the value of M – m ?

(A) –5
(B) 0
(C) 5
(D) 25
(E) 27.5



for the evenly spaced set

, Mean = median
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Re: If m is the average (arithmetic mean) of the first 10 [#permalink]

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Re: If m is the average (arithmetic mean) of the first 10 [#permalink]

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New post 06 Aug 2016, 03:01
Using the rule => For any AP series the mean = median
So the difference = 0
Additionally
Mean = Median = A(V) = First term+last term /2 = 55/2 = 27.5
Hence=> Mean = median = 27.5


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Re: If m is the average (arithmetic mean) of the first 10   [#permalink] 06 Aug 2016, 03:01
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