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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If m is the average (arithmetic mean) of the first 10

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Intern  Joined: 20 Feb 2012
Posts: 28
If m is the average (arithmetic mean) of the first 10  [#permalink]

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Question Stats: 86% (01:23) correct 14% (01:54) wrong based on 1056 sessions

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 62619
Re: If m is the average  [#permalink]

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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.

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

Hope it helps.
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Intern  Joined: 20 Feb 2012
Posts: 28
Re: If m is the average  [#permalink]

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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???
Math Expert V
Joined: 02 Sep 2009
Posts: 62619
Re: If m is the average  [#permalink]

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BANON wrote:
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???

Arithmetic: (25+30)/2 = 55/2 = 27.5.
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Intern  Joined: 20 Feb 2012
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Re: If m is the average  [#permalink]

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right now i am banging my head to the wall. Intern  Joined: 03 Dec 2012
Posts: 3
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)... Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10254
Location: Pune, India
Re: If m is the average (arithmetic mean) of the first 10  [#permalink]

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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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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 Math Expert V
Joined: 02 Sep 2009
Posts: 62619
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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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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Joined: 12 Aug 2015
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GRE 1: Q169 V154 Re: If m is the average (arithmetic mean) of the first 10  [#permalink]

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The first 10 multiples of 5 would be an AP series with D=5

Rule => For any Ap series => Mean = Median = Average of the first and the last term.
Hence we don't need to calculate the mean or the median as they would be equal=> m=M

Hence M-m= Zero
Hence B

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

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The first 10 positive multiples of 5 will form an evenly spaced set (0 is not included as it's not positive), and with any evenly spaced set, the median equals the mean, so their difference will be zero.
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Re: If m is the average (arithmetic mean) of the first 10  [#permalink]

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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

To answer this question we can use the following rule:

When we have an evenly spaced set of numbers, the mean and the median are equal. Recall that in an evenly spaced set of numbers there is a common difference between consecutive terms in the set. For example, consecutive integers, consecutive odd integers, consecutive even integers, and consecutive multiples of any given number are all examples of evenly spaced sets.

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

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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

m=(5+10+15+20+25+30+.......+50)/10= 275/10=27.5 (Formula to find sum: nth term=a+(n-1)d) (where a=first term and d=difference)
M=(Sum of 5th and 6th term)/2
M=(25+30)/2= 27.5

Therefore M-m= 27.5-27.5=0
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Re: If m is the average (arithmetic mean) of the first 10  [#permalink]

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Bunuel wrote:
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.

Thank you!

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

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VeritasPrepKarishma Bunuel

Quote:
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.

I hope arithmetic mean and average are the same.

For no is arithmetic progression, does not average = (first term + last term) / 2

I got to same answer using this approach.
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Re: If m is the average (arithmetic mean) of the first 10  [#permalink]

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VeritasPrepKarishma Bunuel

Quote:
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.

I hope arithmetic mean and average are the same.

For no is arithmetic progression, does not average = (first term + last term) / 2

I got to same answer using this approach.

Yes, arithmetic mean and average are the same.

Average = Sum/Number of elements

For an arithmetic prog,
Average = (first term + last term) / 2
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Re: If m is the average (arithmetic mean) of the first 10  [#permalink]

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BANON wrote:
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???

Not directly relateed by when a questsion asks for the first __ multiples, wouldn't that exclude the number itself so in this case.... it would start at 10? 10,15,20.....
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Re: If m is the average (arithmetic mean) of the first 10  [#permalink]

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arcticTO
Quote:
1st 10 multiples of 5: 5,10,15,20,25,30,35,40,45,50

Quote:
Not directly relateed by when a questsion asks for the first __ multiples, wouldn't that exclude the number itself so in this case.... it would start at 10? 10,15,20....

I am not an expert, but here are my two cents.
The SHORT ans to your query is NO, we can not exclude the no itself since the smallest (first) multiple of that no will
be when the no is multiplied by 1.

Say first multiple of 10 is always 10* 1 = 10
Next ones will !0 *2 = 20 ...

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

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VeritasPrepKarishma wrote:
VeritasPrepKarishma Bunuel

Quote:
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.

I hope arithmetic mean and average are the same.

For no is arithmetic progression, does not average = (first term + last term) / 2

I got to same answer using this approach.

Yes, arithmetic mean and average are the same.

Average = Sum/Number of elements

For an arithmetic prog,
Average = (first term + last term) / 2

Hi VeritasPrepKarishma hope my questions finds you well what is difference between this formula For an arithmetic prog, Average = (first term + last term)/2 and this one $$x_n = a + d(n−1)$$ ?

happy weekend  Re: If m is the average (arithmetic mean) of the first 10   [#permalink] 24 Feb 2018, 04:03

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