Problem Solving

Is there an easy way to solve the below question:

0.99999999/1.0001- 0.99999991/1.0003

1) 10^-8
2) 3(10^-8)
3) 3(10^-4)
4) 2(10^-4)
5) 10^-4

Thank you,
Prerna

0.99999999/1.0001 - .99999991/1.0003 =

10^-8

3(10^-8)

3(10^-4)

2(10^-4)

10^-4

(x+y)(x-y) = x² - y².
In the identity above, x+y and x-y are called CONJUGATES.

It is possible to rephrase decimals as follows:
1.01 = 1 + .01.
.99 = 1 - .01.

Notice that (1 + .01) and (1 - .01) are CONJUGATES:
= (1 + .01)(1 - .01)
= 1² - (.01)²
= 1 - .0001
= .9999.
Notice also that the product of these conjugates (.9999) is ALMOST IDENTICAL to one of the numerators in the problem above (.99999999).

The two DENOMINATORS in the problem above can be rephrased as follows:
1.0001 = 1 + .0001
1.0003 = 1 + .0003.

In order for these two denominators to CANCEL OUT, the two NUMERATORS are almost certainly composed of the following sets of CONJUGATES:
(1 + .0001)(1 - .0001)
(1 + .0003)(1 - .0003).

Thus:
0.99999999/1.0001 - .99999991/1.0003

= [(1 + .0001)(1 - .0001) / (1 + .0001)] - [(1 + .0003)(1 - .0003) / (1 + .0003)]

= (1 - .0001) - (1 - .0003)

= .0002

= 2 * 10^(-4).

The correct answer is D.