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The average (arithmetic mean) of the even integers from 0 to
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27 Oct 2009, 21:44
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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
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27 Oct 2009, 22:02
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=5025=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
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27 Oct 2009, 23:10
AM of even integers from 2 to 100 = 51
AM of even integers from 2 to 50 = 26
So the difference = 5126 = 25
I will go with option B



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Re: Average of even integers
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27 Oct 2009, 23:25
deepakraam wrote: AM of even integers from 2 to 100 = 51
AM of even integers from 2 to 50 = 26
So the difference = 5126 = 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
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28 Oct 2009, 07:26
Thanks for the detailed approach and additional tips. Very helpful!



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Re: Average of even integers
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28 Oct 2009, 12:30
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 050 and the 0100 averages. Since they were subtracted the two errors canceled out Lucky You!
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Re: Average of even integers
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28 Oct 2009, 17:47
Bunuel, thanks for prompt explanation and useful tips. It greatly helps to study math section
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Re: Average of even integers
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29 Oct 2009, 19:42
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
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30 Oct 2009, 18:33
IMO B, it can not be E
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Re: Average of even integers
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30 Oct 2009, 19:32
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
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17 May 2011, 23:47
first + last terms / 2 = average
25



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Re: The average (arithmetic mean) of the even integers from 0 to
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20 Apr 2014, 00:32
For series
0+2+4+6+..................+100
According to arithmetic progression:
First term + (n1) *common difference = Last term
0+(n1)*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
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12 Dec 2016, 11:08
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
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11 Oct 2019, 00:03
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Re: The average (arithmetic mean) of the even integers from 0 to
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