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I just had this one on a practice test and haven't seen it

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mathec
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I just had this one on a practice test and haven't seen it [#permalink] New post   30 Aug 2006, 19:42
I just had this one on a practice test and haven't seen it addressed so I thought I'd post a solution that allows one to solve it quickly. Sorry if it's a repeat. The question is:

( 1001^2 - 999^2 ) / (101^2 0 999^2 )

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

My Soln:

Expand the stem to:

[(1000 + 1)(1000 + 1) - (1000 - 1)(1000-1)] / [(100+1)(100+1) - (100-1)(100-1)]

the numerator = (1,000)(1,000) + 2,000 + 1 - (1,000)(1,000) + 2,000 - 1
which = 4,000

the denominator = (100)(100) + 200 + 1 - (100)(100) + 200 - 1
which = 400

So 4,000/400 = 10 ANS A



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reaching07
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Re: I just had this one on a practice test and haven't seen it [#permalink] New post   30 Aug 2006, 22:07
This is a great one. I was curious about the initial problem stating "(101^2 0 999^2 )" in the denominator. Just want to make sure whether it's (101^2-99^2) or not. Given the solution, I do think 999 was just a typo. I like this one.
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Re: I just had this one on a practice test and haven't seen it [#permalink] New post   30 Aug 2006, 23:18
mathec wrote:
I just had this one on a practice test and haven't seen it addressed so I thought I'd post a solution that allows one to solve it quickly. Sorry if it's a repeat. The question is:

( 1001^2 - 999^2 ) / (101^2 0 999^2 )

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

My Soln:

Expand the stem to:

[(1000 + 1)(1000 + 1) - (1000 - 1)(1000-1)] / [(100+1)(100+1) - (100-1)(100-1)]

the numerator = (1,000)(1,000) + 2,000 + 1 - (1,000)(1,000) + 2,000 - 1
which = 4,000

the denominator = (100)(100) + 200 + 1 - (100)(100) + 200 - 1
which = 400

So 4,000/400 = 10 ANS A

It was discussed many times
( 1001^2 - 999^2 ) / (101^2 - 999^2 )
[(1000 + 1)(1000 + 1) - (1000 - 1)(1000-1)]---the correct formula is
(1001-999)*(1001+999)=2*(2000)IMHO it is much more easier
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Re: I just had this one on a practice test and haven't seen it [#permalink] New post   31 Aug 2006, 06:31
Thanks reaching07, although I don't have access to the test anymore it must have been a typo. Yurik79- Can you please explain the "correct formula." Thanks.
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mathec
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Re: I just had this one on a practice test and haven't seen it [#permalink] New post   31 Aug 2006, 06:32
Also, I did a search using the Google search feature and came up with nothing for this topic. Can you tell me where it's located?
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Yurik79
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Re: I just had this one on a practice test and haven't seen it [#permalink] New post   31 Aug 2006, 06:50
mathec wrote:
Thanks reaching07, although I don't have access to the test anymore it must have been a typo. Yurik79- Can you please explain the "correct formula." Thanks.

(a^2-b^2)=(a-b)(a+b)
hope this helps
Good explanation here
https://www.gmatclub.com/phpbb/viewtopic ... 01+2+999+2
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mathec
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Re: I just had this one on a practice test and haven't seen it [#permalink] New post   31 Aug 2006, 07:26
Ahh, of course. Me making things more complicated again.
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Re: I just had this one on a practice test and haven't seen it [#permalink] New post   31 Aug 2006, 16:58
a^2-b^2=(a-b)(a+b)

1001^2-999^2= (1001-999)(1001+999)=2(2000)

101^2-99^2=(101-99)(101+99)=2(200)

=10



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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Re: I just had this one on a practice test and haven't seen it [#permalink]
31 Aug 2006, 16:58
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