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

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Intern
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(1001^2 - 999^2)/(101^2 - 99^2) =  [#permalink]

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New post Updated on: 15 Oct 2019, 08:47
5
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

82% (01:31) correct 18% (01:47) wrong based on 147 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.
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Re: (1001^2 - 999^2)/(101^2 - 99^2) =  [#permalink]

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New post 20 Sep 2006, 14:38
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
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Re: (1001^2 - 999^2)/(101^2 - 99^2) =  [#permalink]

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New post 20 Sep 2006, 21:15
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
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New post 09 May 2010, 19:05
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New post 12 Dec 2010, 01:09
1
damn. So simple. Thanks m8.
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Re: (1001^2 - 999^2)/(101^2 - 99^2) =  [#permalink]

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New post 16 Oct 2019, 19:35
1
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

Answer: A
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Re: (1001^2 - 999^2)/(101^2 - 99^2) =   [#permalink] 16 Oct 2019, 19:35
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