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If m is the average (arithmetic mean) of the first 10 [#permalink]
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27 Feb 2012, 14:40
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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
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Re: If m is the average [#permalink]
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27 Feb 2012, 14:43
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Re: If m is the average [#permalink]
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27 Feb 2012, 15:08
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
Mm= 5
where i am doing wrong???



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Re: If m is the average [#permalink]
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27 Feb 2012, 15:13
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
Mm= 5
where i am doing wrong??? Arithmetic: (25+30)/2 = 55/2 = 27.5.
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Re: If m is the average [#permalink]
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27 Feb 2012, 15:15
right now i am banging my head to the wall.



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Re: If m is the average (arithmetic mean) of the first 10 [#permalink]
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20 Jun 2013, 23:00
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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)...



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Re: If m is the average (arithmetic mean) of the first 10 [#permalink]
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24 Jun 2013, 21:30
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 ... eticmean/http://www.veritasprep.com/blog/2012/05 ... eviations/http://www.veritasprep.com/blog/2012/05 ... onmedian/
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Re: If m is the average (arithmetic mean) of the first 10 [#permalink]
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25 Jun 2013, 13:52
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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25 Jun 2013, 14:04
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Re: If m is the average (arithmetic mean) of the first 10 [#permalink]
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11 Sep 2014, 01:45
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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24 Sep 2015, 20:58
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Re: If m is the average (arithmetic mean) of the first 10 [#permalink]
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15 Dec 2016, 05:18
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 Mm= Zero Hence B
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Re: If m is the average (arithmetic mean) of the first 10 [#permalink]
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05 Jan 2017, 15:15
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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09 Jan 2017, 10:42
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. Thus the answer is 0. Answer: B
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Re: If m is the average (arithmetic mean) of the first 10 [#permalink]
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01 Apr 2017, 09:17
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+(n1)d) (where a=first term and d=difference) M=(Sum of 5th and 6th term)/2 M=(25+30)/2= 27.5 Therefore Mm= 27.527.5=0



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Re: If m is the average (arithmetic mean) of the first 10 [#permalink]
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01 Apr 2017, 09:19
JeffTargetTestPrep wrote: 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. Thus the answer is 0. Answer: B That is one of the smart ways to solve such questions. Thanks.



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Re: If m is the average (arithmetic mean) of the first 10 [#permalink]
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11 Jun 2017, 06:58
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! Sent from my Moto G (4) using GMAT Club Forum mobile app




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