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Arithmetic - Operations on Rational Numbers

Expert replies
Source: — Problem Solving |

by Brent@GMATPrepNow » Sat Apr 09, 2016 6:55 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 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
Brent Hanneson - Creator of GMATPrepNow.com
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by Brent@GMATPrepNow » Sat Apr 09, 2016 6:56 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 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

Cheers,
Brent
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by [email protected] » Sat Apr 09, 2016 7:40 am
Thanks Brent, the first approach is easier for me though a bit tricky. Being able to recognise 1 - 10^-8 as difference of two squares is a bit high level but I now understand it.

Thanks again.
AYO
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by GMATGuruNY » Sun Apr 10, 2016 3:24 am
0.99999999/1.0001 - .99999991/1.0003 =

10^-8

3(10^-8)

3(10^-4)

2(10^-4)

10^-4
Approach 1:
In each answer choice, exactly ONE DIGIT is a positive integer.
Try to determine the value of this digit.

Let x = .99999999/1.0001
Then:
1.0001x = .99999999
10001x = 9999.9999.
The digits in red imply that the rightmost digit of x must be 9, since 1*9 = 9.

Let y = .99999991/1.0003
Then:
1.0003y = .99999991
10003y = 9999.9991.
The digits in red imply that the rightmost digit of y must be 7, since 3*7 = 21.

Thus:
0.99999999/1.0001 - .99999991/1.0003
= x-y
= (value with a rightmost digit of 9) - (value with a rightmost digit of 7)
= value with a rightmost digit of 2.
The correct answer choice must include a digit of 2.

The correct answer is D.

Approach 2:
(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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by Matt@VeritasPrep » Mon Apr 11, 2016 1:18 pm
Let's say that the number want is x.

.99999999/1.0001 - .99999991/1.0003 = x

(.99999999*1.0003 - .99999991*1.0001)/(1.0001*1.0003) = x

.99999999*1.0003 - .99999991*1.0001 = 1.0001*1.0003*x

(long decimal ending in 7) - (long decimal ending in 1) = (long decimal ending in 3) * x

The left hand side will end in a 6, so the right hand side must end in a 6. For that to happen, x must end in a 2, so we're set with D.
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