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

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(1001^2-999^2)/(101^2-99^2)= [#permalink]  20 Sep 2006, 13:35
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Difficulty:

5% (low)

Question Stats:

86% (01:50) correct 14% (01:00) wrong based on 56 sessions
$$\frac{1001^2-999^2}{101^2-99^2}=$$

A. 10
B. 20
C. 40
D. 80
E. 100
[Reveal] Spoiler: OA

Last edited by Bunuel on 08 Jan 2014, 04:00, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
SVP
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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
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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
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(1001^2-999^2)/(101^2-99^2)= [#permalink]  09 May 2010, 17:55
$$\frac{1001^2-999^2}{101^2-99^2}=$$

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

Q10.jpg [ 8.23 KiB | Viewed 1682 times ]

Math Expert
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Re: what is the strategy to be used? [#permalink]  09 May 2010, 18:05
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Expert's post
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Hi,

I'm sure there is a strategy to use in order to solve the question...I dont think that the GMAT expects me to do the calculation!!

You should apply the following formula: $$x^2-y^2=(x-y)(x+y)$$

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

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Re: what is the strategy to be used? [#permalink]  12 Dec 2010, 00:09
damn. So simple. Thanks m8.
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Re: Is there a faster way to calculate this that I'm missing... [#permalink]  07 Jan 2014, 22:49
Googlore wrote:
Is there a faster way to calculate this that I'm missing...

[(1001^2) - (999^2)]/[(101^2)-(99^2)]

Remember this formula a² - b² = (a+b) (a-b)

Applying this formula,

= (1001+999) x (1001-999)/ (101+99) x (101-99)

= 2000 x 2/ 200 x 2

= 10
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Kindly consider for kudos if my post was helpful!

Re: Is there a faster way to calculate this that I'm missing...   [#permalink] 07 Jan 2014, 22:49
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