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1 0.000001 =

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when you will do subtraction , you will get = 0.999999 or 999999 * $$10^{-6}$$
now you need to find factors of 999999 = 999 * 1001
(for example you need factors of 99 = 9 * 11 , 9999 = 99 * 101 and so on if 9 is repeated even times...)

so in answer choices option C , (1.001) * (0.999) is answer
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Bunuel wrote:
1 − 0.000001 =

A. (1.01)(0.99)
B. (1.11)(0.99)
C. (1.001)(0.999)
D. (1.111)(0.999)
E. (1.0101)(0.0909)

NEW question from GMAT® Quantitative Review 2019

(PS00918)

1^2-(0.001)^2= (1+0.001) (1-0.001)= (1.001)(0.999)

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Bunuel wrote:
1 − 0.000001 =

A. (1.01)(0.99)
B. (1.11)(0.99)
C. (1.001)(0.999)
D. (1.111)(0.999)
E. (1.0101)(0.0909)

NEW question from GMAT® Quantitative Review 2019

(PS00918)

We have, 1−0.000001=0.999999

A. (1.01)(0.99)=(1+0.01)(0.99)=0.99+0.0099=0.9999
B. (1.11)(0.99)=(1+0.1+0.01)(0.99)=0.99+0.099+0.0099=1.0989
C. (1.001)(0.999)=(1+0.001)(0.999)=0.999+0.000999=0.999999
Ans. (C)
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Bunuel wrote:
1 − 0.000001 =

A. (1.01)(0.99)
B. (1.11)(0.99)
C. (1.001)(0.999)
D. (1.111)(0.999)
E. (1.0101)(0.0909)

NEW question from GMAT® Quantitative Review 2019

(PS00918)

$$1 = 1^2$$

And $$0.000001 = \frac{1}{1,000,000} = \frac{1}{1,000^2} = (\frac{1}{1,000})^2 = (0.001)^2$$

So, 1 and 0.000001 are SQUARES
This means 1 - 0.000001 is a difference of squares

We can factor a difference of squares as follows: x² - y² = (x + y)(x - y)

Likewise....
1 - 0.000001 = 1² - (0.001)²
= (1 + 0.001)(1 - 0.001)
= (1.001)(0.999)

Cheers,
Brent

Originally posted by BrentGMATPrepNow on 05 Aug 2018, 08:15.
Last edited by BrentGMATPrepNow on 10 Dec 2020, 12:49, edited 1 time in total.
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Bunuel wrote:
1 − 0.000001 =

A. (1.01)(0.99)
B. (1.11)(0.99)
C. (1.001)(0.999)
D. (1.111)(0.999)
E. (1.0101)(0.0909)

Let’s first re-express 0.000001 in exponential form:

1 - 10^-6

Factoring the resulting expression as a difference of squares, we have:

(1 - 10^-3)(1 + 10^-3)

(1 - 0.001)(1 + 0.001) = 0.999 x 1.001

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chetan2u wrote:
Bunuel wrote:
1 − 0.000001 =

A. (1.01)(0.99)
B. (1.11)(0.99)
C. (1.001)(0.999)
D. (1.111)(0.999)
E. (1.0101)(0.0909)

NEW question from GMAT® Quantitative Review 2019

(PS00918)

Two ways..

1) algebraic..
$$1-0.000001=1^2-(0.001)^2=(1+0.001)(1-0.001)=1.001*0.999$$

2) choices
1-0.000001=0.999999
So the answer will have 6 decimal point.
A and B have 2+2=4 decimal points so eliminate
E has 4+4=8, so out
Left C and D
D is 1.111*0.999 the multiple cannot have all 9s as 11*99=1098 and also it will surely be >1.... eliminate

Left is C
1.001*0.999 will be <1 and will give all 9s

Hi

Per the purple

-- how do you know the multiple does not have all 9's [ i am assuming you did not actually calculate out on paper that the resulting number will NOT have all 9's]

Per the red colored font
-- how could you be sure the resulting number MUST BE > 1 (without calculating or pulling out a calculator) --- i thought it could be below 1 or just above 1

can you tell me, how is your thinking -- regarding how is this definitely > 1 ?

Thank you !
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Bunuel wrote:
1 − 0.000001 =

A. (1.01)(0.99)
B. (1.11)(0.99)
C. (1.001)(0.999)
D. (1.111)(0.999)
E. (1.0101)(0.0909)

NEW question from GMAT® Quantitative Review 2019

(PS00918)

Well, So start with, we know the difference between 1 and 0.000001 will have 6 digits AFTER the decimal point. So, use the same concept and apply to the options.

A. (1.01)(0.99) - here, multiplying a number having 2 digits after the decimal point to the same will give us answer with 4 digits after the decimal point.

Similarly, applying the same to all options, we see that the answer to Option C will have 6 digits after the decimal point.

Hope that helps!
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chetan2u wrote:
Bunuel wrote:
1 − 0.000001 =

A. (1.01)(0.99)
B. (1.11)(0.99)
C. (1.001)(0.999)
D. (1.111)(0.999)
E. (1.0101)(0.0909)

NEW question from GMAT® Quantitative Review 2019

(PS00918)

Two ways..

1) algebraic..
$$1-0.000001=1^2-(0.001)^2=(1+0.001)(1-0.001)=1.001*0.999$$

2) choices
1-0.000001=0.999999
So the answer will have 6 decimal point.
A and B have 2+2=4 decimal points so eliminate
E has 4+4=8, so out
Left C and D
D is 1.111*0.999 the multiple cannot have all 9s as 11*99=1098 and also it will surely be >1.... eliminate

Left is C
1.001*0.999 will be <1 and will give all 9s

I also used exactly same thought process but It’s not Practical though. If answer choices were closer enough? Pls help
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GMATPrepNow wrote:
Bunuel wrote:
1 − 0.000001 =

A. (1.01)(0.99)
B. (1.11)(0.99)
C. (1.001)(0.999)
D. (1.111)(0.999)
E. (1.0101)(0.0909)

NEW question from GMAT® Quantitative Review 2019

(PS00918)

1 = 1²
0.000001 = 1/1,000,000 = 1/1,000² = (1/1,000)² = = (0.001)²

So, 1 and 0.000001 are SQUARES
This means 1 - 0.000001 is a difference of squares

We can factor a difference of squares as follows: x² - y² = (x + y)(x - y)

Likewise....
1 - 0.000001 = 1² - (0.001)²
= (1 + 0.001)(1 - 0.001)
= (1.001)(0.999)

Cheers,
Brent

OMG! Thanks! Understood I didn’t see that hidden [a^2 - b^2] formula
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Hi ScottTargetTestPrep

By looking at the question, we instantly know that 1-0.000001 = 0.999999. But how did you determine that we have to go the a^2-b^2 route to get to the answer as 0.999999 is not presented in the option?
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Vegita wrote:
Hi ScottTargetTestPrep

By looking at the question, we instantly know that 1-0.000001 = 0.999999. But how did you determine that we have to go the a^2-b^2 route to get to the answer as 0.999999 is not presented in the option?

We should always take a quick look at answer choices before we attempt to solve a question. We know from the answer choices that we should express our answer as a product of two numbers, and we should notice that some of the answer choices are in the form of (1 + x)(1 - x). The first thing that comes to mind at this point is to use the difference of two squares. If you were able to determine that the answer is 0.999999 but using the difference of two squares did not come to your mind, you could always multiply the two numbers in the answer choices. It will take a little longer, but since the factors have a lot of 1's and 0's, it won't be that bad.
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