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# (1.00001)(0.99999) - (1.00002)(0.99998) =

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Joined: 10 Sep 2012
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24 Nov 2012, 11:39
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Difficulty:

75% (hard)

Question Stats:

56% (02:08) correct 44% (01:58) wrong based on 504 sessions

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$$(1.00001)(0.99999) - (1.00002)(0.99998) =$$

A. 0

B. $$10^{-10}$$

C. $$3(10^{-10})$$

D. $$10^{-5}$$

E. $$3(10^{-5})$$

The solution suggests to turn all the expressions into scientific notation and then simplify but I was wondering if anyone here had any mind boggling tricks to do this in a more elegant manner?
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Re: (1.00001)(0.99999) - (1.00002)(0.99998) =  [#permalink]

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24 Nov 2012, 12:33
17
5
anon1 wrote:
(1.00001)(0.99999) - (1.00002)(0.99998) =

A. 0
B. 10^-10
C. 3(10^-10)
D. 10^-5
E. 3(10^-5)

The solution suggests to turn all the expressions into scientific notation and then simplify but I was wondering if anyone here had any mind boggling tricks to do this in a more elegant manner?

Apply $$a^2-b^2=(a+b)(a-b)$$.

$$1.00001*0.99999-1.00002*0.99998=(1+0.00001)(1-0.00001)-(1+0.00002)(1-0.00002)=$$
$$=1^2-0.00001^2-1^2+0.00002^2=0.00002^2-0.00001^2$$.

Next, $$0.00002^2-0.00001^2=(0.00002+0.00001)(0.00002-0.00001)=0.00003*0.00001=3*10^{-5}*10^{-5}=3*10^{-10}$$.

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Re: (1.00001)(0.99999) - (1.00002)(0.99998) =  [#permalink]

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24 Nov 2012, 12:41
14
1
anon1 wrote:
(1.00001)(0.99999) - (1.00002)(0.99998) =

A. 0
B. 10^-10
C. 3(10^-10)
D. 10^-5
E. 3(10^-5)

The solution suggests to turn all the expressions into scientific notation and then simplify but I was wondering if anyone here had any mind boggling tricks to do this in a more elegant manner?

$$= (1 + 0.00001)(1 - 0.00001) - (1 + 0.00002)(1 - 0.00002)$$

$$= (1 + 10^{-5})(1 - 10^{-5}) - (1 + 2*10^{-5})(1 - 2*10^{-5})$$

$$= 1 - 10^{-10} - 1 + 4*10^{-10}$$

$$=3*10^{-10}$$

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Re: (1.00001)(0.99999) - (1.00002)(0.99998) =  [#permalink]

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24 Nov 2012, 12:28
1
(1.00001)(0.99999) - (1.00002)(0.99998) = (100001/100000)*(99999/100000) - (100002/100000)*(99998/100000) =
Right here from the get go we see that the denominator of the expression above is 100000*100000 = 10^(-10) - Denominator, and we live it right there and focus on the numerator.
100001*99999 = (100000+1)*99999=100000*99999+99999
100002*99998 = (100000+2)*99998 = 100000*99998+2*99998
= 100000*99999+99999-100000*99998-2*99998 = 100000+99999-2*99998 = 100000+99999-2*(99999-1)=100000+99999-2*99999+2=100000-99999+2=3 - Numerator
Or, 3*(10^(-10) = 3/10^(10)
The numbers might look intimidating and cumbersome, but a little manipulation goes a long way. Or, once you got the denominator you can cross out some answer choices and guess in case you running out of time.
Please, feel free to correct me, if I went awry
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Re: (1.00001)(0.99999) - (1.00002)(0.99998) =  [#permalink]

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Updated on: 27 Aug 2014, 03:12
6
1
I wrote it in a simple way:

$$(1 + \frac{1}{10^5}) (1 - \frac{1}{10^5}) - (1 + \frac{2}{10^5}) (1 - \frac{2}{10^5})$$

$$= 1 - \frac{1}{10^{10}} - (1 - \frac{4}{10^{10}})$$

$$= \frac{4}{10^{10}} - \frac{1}{10^{10}}$$

$$= \frac{3}{10^{10}}$$

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Originally posted by PareshGmat on 30 May 2013, 02:51.
Last edited by PareshGmat on 27 Aug 2014, 03:12, edited 3 times in total.
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Re: (1.00001)(0.99999) - (1.00002)(0.99998) =  [#permalink]

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30 May 2013, 02:59
somalwar wrote:
I wrote it in a simple way:
(1 + 1/10^5) (1 - 1/10^5) - (1 + 2/10^5) (1 + 2/10^5) =
= 1 - 1/10^10 - (1 - 4/10^10) =
= 4/10^10 - 1/10^10
= 3/10^10

That's correct. One typo though: should be - instead of +.
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Re: (1.00001)(0.99999) - (1.00002)(0.99998) =  [#permalink]

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17 Nov 2016, 15:31
In short, this is a very messy way of dealing with factoring.

Think (a+1)(a-1), etc.

1. We need to rewrite decimals, so they are in fractional form.

{[100,001/100,000][999,999/100,000]} - {[100,002/100,000][99998/100,000]}

2. Trick here is to recognize that we can write each of the numerator as a process applied to 100,000 (i.e. x, -, +, etc)

{[(100,000+1)(100,000-1)]/(10,000,000,000)} - {(100,000+2)(100,000-2)/(10,000,000,000)}

3. Simplify by using knowledge that (a+1)(a-1) = a^2 - 1 & (a+2)(a-2) = a^2 -4

{(10,000,000,000-1) -(10,000,000,000-4)}/(10^10) --> 3x10^(-10)
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Re: (1.00001)(0.99999) - (1.00002)(0.99998) =  [#permalink]

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10 Jan 2017, 04:35
anon1 wrote:
(1.00001)(0.99999) - (1.00002)(0.99998) =

A. 0
B. 10^-10
C. 3(10^-10)
D. 10^-5
E. 3(10^-5)

The solution suggests to turn all the expressions into scientific notation and then simplify but I was wondering if anyone here had any mind boggling tricks to do this in a more elegant manner?

1.00001*0.99999 .... ends in 9, and 1.00002*0.99998 ends in 6 , thus answer has to be in the form 3*10^-x , option c or E

but since multiplication of each term around the subtraction operation will yield 10^-x < 10^-5 thus x > 5 .... thus option c
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Re: (1.00001)(0.99999) - (1.00002)(0.99998) =  [#permalink]

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03 Nov 2017, 03:47
Does anyone know of similar question types for additional practise?
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Re: (1.00001)(0.99999) - (1.00002)(0.99998) =  [#permalink]

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03 Nov 2017, 04:25
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Re: (1.00001)(0.99999) - (1.00002)(0.99998) =  [#permalink]

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03 Nov 2017, 22:11
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Re: (1.00001)(0.99999) - (1.00002)(0.99998) =  [#permalink]

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06 Nov 2017, 17:30
anon1 wrote:
$$(1.00001)(0.99999) - (1.00002)(0.99998) =$$

A. 0

B. $$10^{-10}$$

C. $$3(10^{-10})$$

D. $$10^{-5}$$

E. $$3(10^{-5})$$

Let’s rewrite the expressions using scientific notation:

(1 + 10^-5)(1 - 10^-5) - (1 + 2 x 10^-5)(1 - 2 x 10^-5)

1^2 - (10^-5)^2 - [1^2 - 2^2 x (10^-5)^2]

1 - 10^-10 - 1 + 4 x 10^-10

3 x 10^-10

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Re: (1.00001)(0.99999) - (1.00002)(0.99998) =  [#permalink]

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28 Jan 2018, 15:00
1
Hi All,

This question actually has a really big shortcut built into it that will allow you to avoid most of the "long math":

The first part of the calculation...

(1.00001)(0.99999)

…will have 10 decimal places (5 decimal points x 5 decimal points = 10 total decimal points) and the last digit will be a 9 (1 x 9 = 9

The second part of the calculation….

(1.00002)(0.99998)

….will also have 10 decimal places (for the same reason that the first part has 10 decimal points) and the last digit will be a 6 (2 x 8 = 16)

From the answers, we know that we'll be dealing with 10 to some "negative power"; subtracting the second number from the first would give us…

._ _ _ _ _ _ _ _ _ 9
._ _ _ _ _ _ _ _ _ 6
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._ _ _ _ _ _ _ _ _ 3

So, which answer has a "3" in it and implies 10 decimal points?

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Re: (1.00001)(0.99999) - (1.00002)(0.99998) = &nbs [#permalink] 28 Jan 2018, 15:00
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