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# Trick to solve this?

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17 Nov 2009, 06:39
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Question Stats:

100% (02:06) correct 0% (00:00) wrong based on 3 sessions

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hi

could anyone suggest if there is any trick to get this ans without actual calculation

$$\frac{1001^2 - 999^2}{101^2 - 99^2}$$
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Re: Trick to solve this? [#permalink]

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17 Nov 2009, 06:48
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fireplayer wrote:
hi

could anyone suggest if there is any trick to get this ans without actual calculation

$$\frac{1001^2 - 999^2}{101^2 - 99^2}$$

$$x^2-y^2=(x-y)*(x+y)$$

Applying this to the problem we get:

$$\frac{1001^2 - 999^2}{101^2 - 99^2}=\frac{(1001-999)*(1001+999)}{(101-99)*(101+99)}=\frac{2*2000}{2*200}=10$$
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Re: Trick to solve this? [#permalink]

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17 Nov 2009, 09:10
Alternate -

Numerator can be written as (1000+1)^2 -(1000-1)^2
with this the 1000^2 and 1^2 cancel out so we are left with 4*1000

Similarly for denominator we get 4*100

dividing we get 10.
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Questions with exponents from sample exam [#permalink]

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22 Nov 2009, 14:23
Hi all,

I encountered this difficult question on a practice exam and was hoping someone could explain an efficient way to solve it.

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

In other words, 1001 squared minus 999 squared, all over 101 squared minus 99 squared.

I can't find an easy way to cancel these exponents.

Thanks.
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Re: Questions with exponents from sample exam [#permalink]

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22 Nov 2009, 14:43
socal_gmat wrote:
(1001)^2-(999)^2 / (101)^2 - (99)^2

dude, it's very very simple. Apply a^2 - b^2 = (a+b)(a-b) formula for both numerator and denominator. The answer should be 10.
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22 Nov 2009, 15:21
Hi all,

What is a fast way to work through this problem solving question:

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

Thanks.
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23 Nov 2009, 08:06
socal_gmat wrote:
Hi all,

What is a fast way to work through this problem solving question:

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

Thanks.

The numerators and denominators are of the form a^2-b^2 = (a+b)(a-b)

So

(1001+999)(1001-999)/(101+99)(101-99)
2000*2/200*2 = 10

Re: Complicated exponent question   [#permalink] 23 Nov 2009, 08:06
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