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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
A) 24
B) 25
C) 26
D) 48
E) 50
27 Oct 2009, 22:02
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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
The answer is: Is there a quick approach to calculate (without writing down every single even integer)?
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.
General Discussion
27 Oct 2009, 23:10
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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
AM of even integers from 2 to 50 = 26
So the difference = 51-26 = 25
I will go with option B
Lucky You! 
