1001^2)-(999^2)/(101^2)-(99^2) = 10 20 30 50 100 : Quant Question Archive [LOCKED]
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# 1001^2)-(999^2)/(101^2)-(99^2) = 10 20 30 50 100

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Senior Manager
Joined: 05 Oct 2008
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1001^2)-(999^2)/(101^2)-(99^2) = 10 20 30 50 100 [#permalink]

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24 Dec 2008, 02:57
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

(1001^2)-(999^2)/(101^2)-(99^2) =

10
20
30
50
100
Director
Joined: 01 Apr 2008
Posts: 897
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014
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Kudos [?]: 648 [1] , given: 18

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24 Dec 2008, 03:51
1
KUDOS
Simple solution: use the property a^2-b^2 = (a+b) (a-b)
=
(1001+999)(1001-999) / (101+99)(101-99)
=
(2000)(2) / (200)(2)
=
10

ans. A
Manager
Joined: 11 Apr 2008
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24 Dec 2008, 07:22
Another alternative way

(a+b)^2-(a-b)^2=4ab

so the above can be expressed as
(1000+1)^2-(1000-1)^2/(100+1)^2-(100-1)^2
=4*1000/4*100
=10
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Manager
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24 Dec 2008, 09:59
subarao wrote:
Another alternative way

(a+b)^2-(a-b)^2=4ab

so the above can be expressed as
(1000+1)^2-(1000-1)^2/(100+1)^2-(100-1)^2
=4*1000/4*100
=10

i also used the same method. answer is 10
Intern
Joined: 22 Dec 2008
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24 Dec 2008, 12:04
1
KUDOS
I have just begun looking at an MBA. I am assuming we cannot use calculators on the exam? Correct? Incorrect?
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24 Dec 2008, 21:08
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Nobody dies a virgin, life screws us all.

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28 Dec 2008, 06:14
no need of calculator...GMAT checks basics
Re: Exponents   [#permalink] 28 Dec 2008, 06:14
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