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Manhattan Challenge question

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by gabriel » Sat Sep 15, 2007 12:59 am
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

Solve it people .. its a easy question ... do not be intimidated by the huge numbers ..
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by keepsmilinyaar » Sat Sep 15, 2007 2:39 am
Ans is D.

IT can be reduce to :
[(54281- 1)^2 + (54281+1)^2]/2
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by kajcha » Sat Sep 15, 2007 4:24 am
Agree D
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by samirpandeyit62 » Sat Sep 15, 2007 5:09 am
Hi Gabriel,
I did get a little scared by seeing the big nos :D , so to cope with it I replaced 54820 with x

so 54822 =x+2

so it becomes

x^2 + (x+2)^2

i.e 2x^2 + 4 + 4x

divinding this by 2 we have

x^2 + 2x + 2

Now we need to represent this in terms of answer choices wherin it should be a perfect square or square +1 / -1

Now since this expression is not a perfect square by itself so we need to convert it to a perfect square with 1 added or subtracted

we have x^2 +2x +1 +1

i.e (x+1)^2 +1

i.e 54821^2 +1

so ans should be D
Regards
Samir
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by gabriel » Sat Sep 15, 2007 5:24 am
D is my answer to .. dont know what the OA is
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by naren_nayak » Sun Sep 16, 2007 8:21 am
One could also use the process of elimination here:
(54,820)^2 should have a 0 in the unit's position & (54,822)^2 should end with a 4. The average of these two numbers will have a 2 in the unit's position. That eliminates A, B, C & E (A ends with a 1, B & C end with a .25 and E ends with a 0).
Hence the answer is D.
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by niks_01.27 » Sun Sep 16, 2007 1:44 pm
naren_nayak wrote:One could also use the process of elimination here:
(54,820)^2 should have a 0 in the unit's position & (54,822)^2 should end with a 4. The average of these two numbers will have a 2 in the unit's position. That eliminates A, B, C & E (A ends with a 1, B & C end with a .25 and E ends with a 0).
Hence the answer is D.
Nice thinking Naren!!
regards
niks...
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by 800GMAT » Mon Sep 24, 2007 10:31 am
OA given is D
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