The average 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
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gmat740
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Let x = 54820
so we have to find the average of x^2 and (x+2)^2
thus, [x^2 and (x+2)^2]/2
Now when you expand (x+2)^2, you get,
= 2[x^2 +2x +2]/2 = x^2 +2x +2 = (x+1)^2 +1
But x =54,820
so x+1 = 54821
thus answer is (54821)^2 + 1
IMO D
so we have to find the average of x^2 and (x+2)^2
thus, [x^2 and (x+2)^2]/2
Now when you expand (x+2)^2, you get,
= 2[x^2 +2x +2]/2 = x^2 +2x +2 = (x+1)^2 +1
But x =54,820
so x+1 = 54821
thus answer is (54821)^2 + 1
IMO D
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dumb.doofus
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Nice approach gmat740..
I thought it in this way.. almost similar to yours..
a^2 + b^2 = (a - b)^2 + 2ab ------------- (1)
Now, if x = 54821,
then a = x - 1, b = x + 1
(x+1)(x-1) = x^2 -1 --------------- (2)
so (a^2 + b^2)/2 = [4 + 2(x^2-1)]/2
= 1 + x^2
= 1 + 54821^2 i.e. D
I thought it in this way.. almost similar to yours..
a^2 + b^2 = (a - b)^2 + 2ab ------------- (1)
Now, if x = 54821,
then a = x - 1, b = x + 1
(x+1)(x-1) = x^2 -1 --------------- (2)
so (a^2 + b^2)/2 = [4 + 2(x^2-1)]/2
= 1 + x^2
= 1 + 54821^2 i.e. D
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