Decimal Computation with [ Number Properties ] GMAT, GRE

$\frac{0.99999999}{1.0001}-\frac{0.99999991}{1.0003}=$




(A) 10^(-8)
(B) 3*10^(-8)
(C) 3*10^(-4)
(D) 2*10^(-4)
(E) 10^(-4)




Both  0.99999999/1.0001 and 0.99999991/1.0003 look intimidating, but once you recognize the pattern a question like this can be handled relatively easily. 

Let's take the analogy of difference of squares. where x^2 - y^2 = 0 becomes (x + y)(x - y) = 0 . 

The question above is a situation in which we can use the idea of the difference of two squares to our advantage. 

First, let’s first use this concept with a few easier whole numbers. For instance, let’s say we were asked:

999,999/1,001 – 9,991/103 = ?

We could rewrite this as:

(1,000,000 – 1)/1,001 – (10,000 – 9)/103

(1000 + 1)(1000 – 1)/1,001 – (100 – 3)(100 + 3)/103

(1,001)(999)/1,001 – (97)(103)/103

999 – 97 = 902

It might seem a roundabout way of doing things but on test day, when you're in a timed environment, you'll be completing the calculus quickly and accurately with difference of squares. 

Let's use the same approach for the decimalized version.

0.99999999/1.0001 – 0.99999991/1.0003

[(1 – 0.00000001)/1.0001] – [(1 – 0.00000009)/1.0003]

[(1 – 0.0001)(1 + 0.0001)/1.0001] – [(1 – 0.0003)(1 + 0.0003)]

Try not to make any mistakes with the number of decimal places in our values. Since 0.00000001
has 8 decimal places, the decimals in the factors of the numerator of the first set of brackets must each have 4 decimal places (their square root of the factors of 10). Similarly, since 0.00000009 has 8 decimal places, the decimals in the factors of the numerator of the second set of brackets must each have 4 decimal places. 

Going further:

[(1 – 0.0001)(1 + 0.0001)/1.0001] – [(1 – 0.0003)(1 + 0.0003)]

[(0.9999)(1.0001)/1.0001] – [(0.9997)(1.0003)/1.0003]

0.9999 – 0.9997

0.0002

Converting this to scientific notation:

2 x 10^-4

D