Problem on mean..

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Problem on mean..

by MI3 » Sat Jul 16, 2011 3:19 am
Q. The mean of (54,820)^2 and (54,822)^2 =
(A) (54,821)^2 (B) (54,821.5)^2 (C) (54,820.5)^2 (D) (54,821)^2 + 1 (E) (54,821)^2 - 1

IMO - C, am I correct?
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by Frankenstein » Sat Jul 16, 2011 3:27 am
Hi,
(a+b)^2 + (a-b)^2 = 2(a^2+b^2)
Use the same here
(54,820)^2 + (54,822)^2 = (54,821-1)^2 + (54,821+1)^2 =2[(54,821)^2 + 1^2]
So, average is 1/2[(54,820)^2 + (54,822)^2] = (54,821)^2 + 1

Hence, D
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by ikaplan » Sat Jul 16, 2011 4:29 am
Thanks for the formula Frankenstein- I just updated my flash cards ;)
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