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The average (arithmetic mean) of the even integers from 0 to

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The average (arithmetic mean) of the even integers from 0 to  [#permalink]

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New post 27 Oct 2009, 20:44
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The average (arithmetic mean) of the even integers from 0 to 100 inclusive is how much greater than the average (arithmetic mean) of the even integers from 0 to 50 inclusive?

A) 24
B) 25
C) 26
D) 48
E) 50
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Re: Average of even integers  [#permalink]

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New post 27 Oct 2009, 21:02
11
10
Navigator wrote:
The average (arithmetic mean) of the even integers from 0 to 100 inclusive is how much greater than the average (arithmetic mean) of the even integers from 0 to 50 inclusive?

A) 24
B) 25
C) 26
D) 48
E) 50

The answer is:
Is there a quick approach to calculate (without writing down every single even integer)?


TIPS:

A. In any evenly spaced set (aka symmetric distribution, aka arithmetic progression): Mean=Median=(First term + Last term)/2
B. The sum of the elements in a set equals the mean multiplied by the number of items in the set.

So, according to A mean of a set of even numbers from 0 to 100 =(0+100)/2=50 and mean of a set of even numbers from 0 to 50 =(0+50)/2=25

Difference=50-25=25

Answer: B.

SOME MORE TIPS:

C. The product of n consecutive integers is always divisible by n! Given 6*7*8*9, we have n = 4 consecutive integers. The product of 6*7*8*9, therefore, is divisible by 4! = 4*3*2*1 = 24.
D. Sum of n consecutive integers and divisibility. There are two cases, depending upon whether n is odd or even:
If n is odd, the sum of the integers is always divisible by n. Given 6+7+8, we have n = 3 consecutive integers. The sum of 6+7+8, therefore, is divisible by 3.
If n is even, the sum of the integers is never divisible by n. Given 6+7+8+9, we have n = 4 consecutive integers. The sum of 6+7+8+9, therefore, is not divisible by 4.

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Re: Average of even integers  [#permalink]

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New post 27 Oct 2009, 22:10
AM of even integers from 2 to 100 = 51

AM of even integers from 2 to 50 = 26

So the difference = 51-26 = 25

I will go with option B
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Re: Average of even integers  [#permalink]

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New post 27 Oct 2009, 22:25
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deepakraam wrote:
AM of even integers from 2 to 100 = 51

AM of even integers from 2 to 50 = 26

So the difference = 51-26 = 25

I will go with option B


I think that even though the answer is right, the solution is not. We are told: "even integers from 0 to 100 and 0 to 50 inclusive" Which means that mean would be in first case (0+100)/2=50 and in second (0+50)/2=25. It didn't affect the answer but it's still incorrect.
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Re: Average of even integers  [#permalink]

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New post 28 Oct 2009, 06:26
Thanks for the detailed approach and additional tips. Very helpful!
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Re: Average of even integers  [#permalink]

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New post 28 Oct 2009, 11:30
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Bunuel wrote:
I think that even though the answer is right, the solution is not. We are told: "even integers from 0 to 100 and 0 to 50 inclusive" Which means that mean would be in first case (0+100)/2=50 and in second (0+50)/2=25. It didn't affect the answer but it's still incorrect.


Right, 0 is considered an even integer. It doesn't contribute to the sum but does bring down the average. He got the right answer because he ignored 0 in both the 0-50 and the 0-100 averages. Since they were subtracted the two errors canceled out :-D Lucky You!
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Re: Average of even integers  [#permalink]

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New post 28 Oct 2009, 16:47
Bunuel, thanks for prompt explanation and useful tips. It greatly helps to study math section :)
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Re: Average of even integers  [#permalink]

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New post 29 Oct 2009, 18:42
The average (arithmetic mean) of the even integers from 0 to 100 inclusive is how much greater than the average (arithmetic mean) of the even integers from 0 to 50 inclusive?

A) 24
B) 25
C) 26
D) 48
E) 50

My answer

Sum of first N even no is = N(N+1)

The average (arithmetic mean) of the even integers from 0 to 100 inclusive = (100*101)/100 = 101

The average (arithmetic mean) of the even integers from 0 to 50 inclusive = (50*51)/50 = 51

Difference = 101 - 51 = 50

So my answer is E
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Re: Average of even integers  [#permalink]

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New post 30 Oct 2009, 17:33
IMO B,
it can not be E
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Re: Average of even integers  [#permalink]

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New post 30 Oct 2009, 18:32
hgp2k wrote:
swatirpr wrote:
The average (arithmetic mean) of the even integers from 0 to 100 inclusive is how much greater than the average (arithmetic mean) of the even integers from 0 to 50 inclusive?

A) 24
B) 25
C) 26
D) 48
E) 50

My answer

Sum of first N even no is = N(N+1)

The average (arithmetic mean) of the even integers from 0 to 100 inclusive = (100*101)/100 = 101

This is where you are going wrong. The number of even integers from 0 to 100, inclusive is 51 and not 100.
The average (arithmetic mean) of the even integers from 0 to 50 inclusive = (50*51)/50 = 51
Same as above. Number of even integers is 26 and not 50
Difference = 101 - 51 = 50

So my answer is E


Oops!!

Thanks so much.
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Re: Average of even integers  [#permalink]

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New post 17 May 2011, 22:47
first + last terms / 2 = average

25
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Re: The average (arithmetic mean) of the even integers from 0 to  [#permalink]

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New post 19 Apr 2014, 23:32
For series

0+2+4+6+..................+100

According to arithmetic progression:

First term + (n-1) *common difference = Last term

0+(n-1)*2=100 => n=51

Sum = n/2 (First term + last term) = 25*51

Mean = 25*51/52= 25

Similarly for series

0+2+4......................+50

mean = 50

hence difference is 25
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Re: The average (arithmetic mean) of the even integers from 0 to  [#permalink]

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New post 12 Dec 2016, 10:08
Nice one .
Here is what i did in this one -->
Both series are in AP
Hence mean = median = average of the first and the last terms

Average of first set => 0+100/2 =>50
Average of the second set => 0+50/2 => 25

Hence difference = 25
Hence B

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Re: The average (arithmetic mean) of the even integers from 0 to  [#permalink]

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Re: The average (arithmetic mean) of the even integers from 0 to &nbs [#permalink] 26 Jan 2018, 18:56
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