Is there any simpler trick to solve this question?

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.

While the answers above are more mathematically sophisticated, the poster wanted a simple method, so in that spirit ...

Start by finding a common denominator, as mentioned above:

.99999999(1.0003) - .99999991(1.0001)
-------------------------------------
________(1.0003)(1.0001)


Then notice that in the numerator, you have a number ending with 7 minus a number ending in 1, so whatever the numerator is, it ends in a 6.

In the denominator, you have a number ending in 3.

6/3 = 2, so the answer ends in 2, and only D works.

Not a foolproof method, by any means, but hey!

[email protected] wrote:0.99999999/1.0001- 0.99999991/1.0003

approximating - 0.99999999 and 0.99999991 to 0.99

I get

0.99/(1.0001) - (0.99/1.0003)

= 0.99 [1.0003 - 1.0001]/ (1.0001*1.0003)
= 0.99*2*10^-4/(approx 1).
Ans = 2*10^-4

Is this approach the right way to go ?