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Intern  Joined: 11 Sep 2006
Posts: 39
(1001^2 - 999^2)/(101^2 - 99^2) =  [#permalink]

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3 00:00

Difficulty:   5% (low)

Question Stats: 84% (01:29) correct 16% (01:54) wrong based on 116 sessions

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$$\frac{1001^2-999^2}{101^2-99^2}=$$

A. 10
B. 20
C. 40
D. 80
E. 100

Originally posted by Googlore on 20 Sep 2006, 14:35.
Last edited by Bunuel on 15 Oct 2019, 08:47, edited 3 times in total.
Renamed the topic, edited the question and added the OA.
SVP  Joined: 01 May 2006
Posts: 1506
Re: (1001^2 - 999^2)/(101^2 - 99^2) =  [#permalink]

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1
1
We should apply : a^2-b^2 = (a+b)*(a-b)

[(1001^2) - (999^2)]/[(101^2)-(99^2)]
= (1001+999)*(1001-999) / [(101+99)*(101-99)]
= (2000)*2 / [200*2]
= 10
Director  Joined: 23 Jun 2005
Posts: 648
GMAT 1: 740 Q48 V42 Re: (1001^2 - 999^2)/(101^2 - 99^2) =  [#permalink]

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I arrived at the same answer as Fig in a different way.
Write it as
(a+b)^2 = a^2 +2ab_b^2 -----> 1
(a-b)^2 = a^2-2ab+b^2. -----> 2
When you subtract 1 from 2, you get 4ab. Applying this
(1000+1)^2 -(1000-1)^2/(100+1)^2 - (100-1)^2
4000/400 = 10
Math Expert V
Joined: 02 Sep 2009
Posts: 58446

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$$\frac{1001^2-999^2}{101^2-99^2}=$$

A. 10
B. 20
C. 40
D. 80
E. 100

$$\frac{1001^2-999^2}{101^2-99^2}=\frac{(1001-999)(1001+999)}{(101-99)(101+99)}=\frac{2*2000}{2*200}=10$$

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Senior Manager  Joined: 08 Nov 2010
Posts: 273
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Re: (1001^2 - 999^2)/(101^2 - 99^2) =  [#permalink]

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damn. So simple. Thanks m8.
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Re: (1001^2 - 999^2)/(101^2 - 99^2) =  [#permalink]

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Googlore wrote:
$$\frac{1001^2-999^2}{101^2-99^2}=$$

A. 10
B. 20
C. 40
D. 80
E. 100

Both the numerator and the denominator are a difference of squares. Using the given information, we have:

(1001 - 999)(999 + 1001)/[(101 - 99)(101 + 99)]

(2)(2000)/[(2)(200)] = 10

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If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Re: (1001^2 - 999^2)/(101^2 - 99^2) =   [#permalink] 16 Oct 2019, 19:35
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