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Expert replies
Source: — Problem Solving |

by Anju@Gurome » Mon Apr 22, 2013 3:59 am
cpay3245 wrote:0.99999999/1.0001 - .99999991/1.0003 =
As some of the options are pretty close (C, D, and E) I would not go for estimation as any wrong decision at any stage of the estimation can lead me to a wrong answer.

So, I'll post the algebraic approach.
Let us assume, x = 0.0001 = 10^(-4)
x² = 0.00000001 = 10^(-8)

We can write,
  • 0.99999999 = 1 - 0.00000001 = 1 - 10^(-8) = 1 - x²
    1.0001 = 1 + 0.0001 = 1 + x
    0.99999991 = 1 - 0.00000009 = 1 - (0.0003)² = 1 - (3x)²
    1.0003 = 1 + 0.0003 = 1 + 3x
So, (0.99999999 / 1.0001) - (0.99999991 / 1.0003)
= [(1 - x²)/(1 + x)] - [(1 - (3x)²)/(1 + 3x)]
= [(1 - x)(1 + x)/(1 + x)] - [(1 - 3x)(1 + 3x))/(1 + 3x)]
= [(1 - x)] - [(1 - 3x)]
= 1 - x -1 + 3x
= 3x - x
= 2x
= 2*(0.0001)
= 0.0002
= 2 * 10^(-4)

The correct answer is D.
Anju Agarwal
Quant Expert, Gurome

Backup Methods : General guide on plugging, estimation etc.
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by Brent@GMATPrepNow » Mon Apr 22, 2013 6:39 am
cpay3245 wrote:(9999.9999)/(10001) - (9999.9991)/(10003) =

A. 10^-8

B. 3(10^-8)

C. 3(10^-4)

D. 2(10^-4)

E. 10^-4
Another 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)
= [(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

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by GMATGuruNY » Mon Apr 22, 2013 9:20 am
cpay3245 wrote: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.
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