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by sanju09 » Thu Dec 24, 2015 2:42 am
oquiella wrote: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

Answer: D


Please help simplify quickest way possible.
A 0 is missing in the first denominator, anyway:

All the numerators and denominators here are close to 1, but numerators are far more closer to 1 than do the denominators. Let's round numerators like

0.99999999/1.0001 - 0.99999991/1.0003

= 1/1.0001 - 1/1.0003

= {(1.0003 - 1.0001)}/{(1.0001)(1.0003)}

= (0.0002)/{(1.0001)(1.0003)}

= definitely 2 times some negative power of 10

= [spoiler]only (D)[/spoiler]
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by Brent@GMATPrepNow » Thu Dec 24, 2015 6:36 am
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
One approach is to combine the fractions and then use some approximation.

First combine the fractions by finding a common denominator.
(9999.9999)/(10001) - (9999.9991)/(10003)
= (9999.9999)(10003)/(10001)(10003) - (9999.9991)(10001) /(10003)(10001)
= [(10003)(9999.9999) - (10001)(9999.9991)] / (10001)(10003)
= [(10003)(10^4) - (10001)(10^4)] / (10^4)(10^4) ... (approximately)
= [(10003) - (10001)] / (10^4) ... (divided top and bottom by 10^4)
= 2/(10^4)
= 2*10^(-4)
= D

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by Brent@GMATPrepNow » Thu Dec 24, 2015 6:37 am
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
Another approach is to recognize that both 9999.9999 and 9999.9991 can be rewritten as differences of squares.

First, 0.99999999 = 1 - 0.00000001
= (1 - 0.0001)(1 + 0.0001)

Similarly, 9999.9991 = 1 - 0.00000009
= (1 - 0.0003)(1 + 0.0003)

Original question: 0.99999999/1.0001 - 0.99999991/1.0003
= (1 - 0.0001)(1 + 0.0001)/(1.0001) - (1 - 0.0003)(1 + 0.0003)/(1.0003)
= (1 - 0.0001)(1.0001)/(1.0001) - (1 - 0.0003)(1.0003)/(1.0003)
= (1 - 0.0001) - (1 - 0.0003)
= 1 - 0.0001 - 1 + 0.0003
= 0.0002
= 2 x 10^(-4) = D

Cheers,
Brent
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