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decimal number simplification

Expert replies
Source: — Problem Solving |

by lathmanu » Sun Jun 03, 2012 3:57 am
sohailmbaprep wrote:Q) (0.99999999/1.0001) - (0.99999991/1.0003) ... how to simplify this (answer shld be in terms of some power raised to 10)
(0.99999999) = 1 - 0.00000001 = 1 - 10^-8 = (1 - 10^-4)(1 + 10^4)
(1.0001) = 1 + 0.0001 = 1+10^4

(0.99999991) = 1 - 0.00000009 = 1 - (3^2*10^-8) = {1 - (3*10^-4)}{1 + (3*10^4)}
(1.0003) = 1 + 0.0003 = 1+(3*10^4)

Using the above mentioned terms
(0.99999999/1.0001) = (1 - 10^-4)

(0.99999991/1.0003) = {1 - (3*10^-4)}

Therefore
(0.99999999/1.0001) - (0.99999991/1.0003) = (1 - 10^-4) - {1 - (3*10^-4)}
1 - 10^-4 -1 + 3*10^-4 = 2*10^-4 Ans
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by GMATGuruNY » Sun Jun 03, 2012 4:32 am
0.99999999/1.0001 - 0.99999991/1.0003 =

10^-8
3(10^-8)
3(10^-4)
2(10^-4)
10^-4
(9999.9999)/10001 - (9999.9991)/10003

= [10003(9999.9999) - 10001(9999.9991)] / (10001)(10003)

≈ [10003(10^4) - 10001(10^4)] / (10^4)(10^4)

≈ (10003-10001)/(10^4)

≈ 2(10^-4).

The correct answer is D.
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by Scott@TargetTestPrep » Sat Jul 18, 2015 7:04 pm
sohailmbaprep wrote:Q) (0.99999999/1.0001) - (0.99999991/1.0003) ... how to simplify this (answer shld be in terms of some power raised to 10)
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

Solution:

When first looking at this problem, we must consider the fact that 0.99999999/1.0001 and 0.99999991/1.0003 are both pretty nasty-looking fractions. However, this is a situation in which we can use the idea of the difference of two squares to our advantage. To make this idea a little clearer, let's first illustrate the 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

Notice how cleanly the denominators canceled out in this case. Even though the given problem has decimal numbers, we can follow the same approach.

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)]

When converting this using the difference of squares, we must be very careful 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. 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. Let's continue to simplify.

[(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 to match the answer choices, we have:

2 x 10^-4

Answer: D

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by Jim@StratusPrep » Sun Jul 19, 2015 3:43 am
Hope this drawing helps a bit...
Attachments
OG 13 PS 199.pdf
(4.44 MiB) Downloaded 209 times
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by nikhilgmat31 » Wed Jul 29, 2015 5:24 am
(0.99999999) = 1 - 0.00000001 = 1 - 10^-8 = (1 - 10^-4)(1 + 10^4)

(1.0001) = 1 + 0.0001 = 1+10^4

(0.99999991) = 1 - 0.00000009 = 1 - 9*10^-8 = {1 - (3*10^-4)}{1 + (3*10^4)}
(1.0003) = 1 + 0.0003 = 1+(3*10^4)

Using the above mentioned terms
(0.99999999/1.0001) = (1 - 10^-4)

(0.99999991/1.0003) = {1 - (3*10^-4)}

Therefore
(0.99999999/1.0001) - (0.99999991/1.0003) = (1 - 10^-4) - {1 - (3*10^-4)}

1 - 10^-4 -1 + 3*10^-4 = 2*10^-4

Answer is 2 * 10 ^ -4
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