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

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I just had this one on a practice test and haven't seen it [#permalink]

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30 Aug 2006, 18:42
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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30 Aug 2006, 21:07
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: 1001^2 - 999^2 .... [#permalink]

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30 Aug 2006, 22:18
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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31 Aug 2006, 05:31
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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31 Aug 2006, 05:32
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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31 Aug 2006, 05:50
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
http://www.gmatclub.com/phpbb/viewtopic ... 01+2+999+2
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31 Aug 2006, 06:26
Ahh, of course. Me making things more complicated again.
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31 Aug 2006, 15:58
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
31 Aug 2006, 15:58
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