Arithmatic Operations on rational numbers

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by Anurag@Gurome » Thu Jul 19, 2012 2:52 am
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by iqbalrafi » Thu Jul 19, 2012 6:00 am
Thanks Anurag!!! But its very similar to the solution given in the OG. Isn't there any other way???

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by brainturner » Thu Jul 19, 2012 8:18 am
a^2-b^2=(a+b)(a-b)

(0.99999999/1.0001) - (0.99999991/1.0003)

=(1-10^-8/1+10^-4) - (1-9*10^-8/1+3*10^-4)

=((1+10^-4)*(1-10^-4))/1+10^-4) - ((1+3*10^-4)*(1-3*10^-4))/1+3*10^-4)

= (1-10^-4)-(1-3*10^-4)

= [spoiler]2*10^-4 Answer D[/spoiler]

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by GMATGuruNY » Thu Jul 19, 2012 3:02 pm
iqbalrafi wrote:Hi everyone! This is a PS question(PS 199) from OG 13th. The solution given there seems very complex to me. Can anyone solve this in an easier way??

(0.99999999/1.0001) - (0.99999991/1.0003) =

A. 10^-8
B. 3(10^-8)
C. 3(10^-4)
D. 2(10^-4)
E. 10^-4
(9999.9999)/(10001) - (9999.99991)(10003)

= (10^-4)(99999999/10001) - (10^-4)(99999991/10003)

= (10^-4) (99999999/10001 - 99999991/10003)

The correct answer must be C, D or E, implying three possibilities:
99999999/10001 - 99999991/10003 = 3.
99999999/10001 - 99999991/10003 = 2.
99999999/10001 - 99999991/10003 = 1.

Since their difference is an integer, the quotients here are almost certainly integer values themselves.
Focus on the UNITS DIGITS:
The units digit of 99999999/10001 must be 9, since 10001 must be multiplied by a units digit of 9 to yield 99999999.
The units digit of 99999991/10003 must be 7, since 10003 must be multiplied by a units digit of 7 to yield 99999991.
Since 9-7=2:
99999999/10001 - 99999991/10003 = 2.

The correct answer is D.
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