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

Expert replies
by imane81 » Wed Feb 03, 2010 12:18 pm
Please help addressing this one...

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

A. 10
B. 20
C. 40
D. 80
E. 100
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Source: — Problem Solving |

by swapna » Wed Feb 03, 2010 12:20 pm
Is the answer 10???

What i would do is...look at the numerator and denominator as (a^2-b^2)

In the numerator a-1001 and b-999

In the denominator a-101 and b-99

The formula for a^2-b^2 is (a+b)(a-b)

So(1001+999)(1001-999)/(101+99)(101-99)

=(2000)(2)/(200)(2)

2 gets cancelled in both numerator and denominator

2000/200=10

Hope ths helps
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by Osirus@VeritasPrep » Wed Feb 03, 2010 12:42 pm
Can someone tell me how I can figure out how to solve these problems in general? I saw something like this on my actual GMAT and I was just perplexed as to how to even go about solving it.
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by ajith » Wed Feb 03, 2010 12:45 pm
imane81 wrote:Please help addressing this one...

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

A. 10
B. 20
C. 40
D. 80
E. 100
x^2-y^2 = (x+y)*(x-y)

(1001^2-999^2)= (1001+999)(1001-999) = 2000*2
(101^2-99^2) =(101+99)(101-99) = 200*2


(1001^2-999^2)/(101^2-99^2)= 2000*2/200*2 = 10
Always borrow money from a pessimist, he doesn't expect to be paid back.
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by swapna » Wed Feb 03, 2010 12:47 pm
can u pls confirm if the answer is 10
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by thephoenix » Wed Feb 03, 2010 12:49 pm
imane81 wrote:Please help addressing this one...

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

A. 10
B. 20
C. 40
D. 80
E. 100
1001^2-999^2=(1000+1)^2-(1000-1)^2=(1000+1+1000-1)(1000+1-1000+1)=2*2000
using a^2-b^2=(a-b)(a+b)

similarly deno minator=2*200

value=2000*2/2*200=10
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by neelimareddym » Thu Feb 04, 2010 5:05 am
IMO A..

This problem is of the form 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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by granola » Tue Jun 14, 2011 11:37 am
The OA is A.
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by Brian@VeritasPrep » Tue Jun 14, 2011 3:33 pm
Great work on this thread, everyone.

Just to call attention to what swapna and neelimareddym did - I taught this to one of my students the other night, and it reiterates a critical point on GMAT quant. I don't advocate a ton of memorization but you MUST memorize and recognize the Difference of Squares rule:

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

There are problems, like this one, that you just can't solve efficiently without this rule. Recognizing and applying Difference of Squares can be crucial on quite a few GMAT problems, so if you're reading this I implore you to make a point of knowing this one and of looking for it. When you see anything in the form a^2 - b^2 there's a very high likelihood that Differe nce of Squares will come into play. And don't forget about something like 99^2 - 1. 1 is the same thing as 1^2, so remember that you can use Difference of Squares there, too.

By the time you take the test, whenever you see large numbers or squares with one subtracted for another it should be almost second-nature to check for the ability to use Difference of Squares. I've posted on here before that, to me, it's like an invisible ink decoder or one of those spy toys you get as a kid - it's an instant game-changer to take something incredibly inconvenient and put it in an easy-to-use form. Make sure you take advantage of it!
Brian Galvin
GMAT Instructor
Chief Academic Officer
Veritas Prep

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