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# If X = 0.(81), where the brackets around the 2-digit sequence 81 indic

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Math Expert
Joined: 02 Sep 2009
Posts: 56371
If X = 0.(81), where the brackets around the 2-digit sequence 81 indic  [#permalink]

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28 Mar 2018, 03:10
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Difficulty:

45% (medium)

Question Stats:

56% (01:13) correct 44% (01:44) wrong based on 100 sessions

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If X = 0.(81), where the brackets around the 2-digit sequence 81 indicates the sequence repeats indefinitely, what is the value of 99*X?

(A) 80.(19)
(B) 80.19
(C) 81
(D) 81.81
(E) 81.(81)

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Posts: 1073
Re: If X = 0.(81), where the brackets around the 2-digit sequence 81 indic  [#permalink]

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28 Mar 2018, 03:29
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Bunuel wrote:
If X = 0.(81), where the brackets around the 2-digit sequence 81 indicates the sequence repeats indefinitely, what is the value of 99*X?

(A) 80.(19)
(B) 80.19
(C) 81
(D) 81.81
(E) 81.(81)

We'll break our complex calculation (x99) into small, easy bits (x100 - x1)
This is a Precise approach.

100*X would be 81.818181818181 with infinite 81 after the decimal point.
That is, 100*X = 81.(81)
So 99*X 100*X - X = 81.(81) - 0.(81) = 81

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If X = 0.(81), where the brackets around the 2-digit sequence 81 indic  [#permalink]

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31 Mar 2018, 00:33
Bunuel wrote:
If X = 0.(81), where the brackets around the 2-digit sequence 81 indicates the sequence repeats indefinitely, what is the value of 99*X?

(A) 80.(19)
(B) 80.19
(C) 81
(D) 81.81
(E) 81.(81)

One way to approach these questions is to set up a simple subtraction in which the repeating part is positioned to be subtracted from itself.

A repeating decimal is some fraction, x
Let x = .818181...

Essentially, multiply the repeating decimal by some power of 10 so that you can subtract the repeating portion away.

This decimal repeats every two numbers.
Use 100*x to move the decimal two places to the right.
Thus 100x = 81.81818181....

Subtract x from 100x, and
Subtract 0.8181... from 81.8181...

100x = 81.818181...
___-x =_-0.818181...
__99x = 81

(The fraction, x, would be $$\frac{81}{99}=\frac{9}{11}$$)

For related topic and slightly different method see Bunuel how to convert a recurring decimal to fraction.

The method I used (multiply by a power of 10) is explained HERE.
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Re: If X = 0.(81), where the brackets around the 2-digit sequence 81 indic  [#permalink]

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31 Mar 2018, 01:31
Bunuel wrote:
If X = 0.(81), where the brackets around the 2-digit sequence 81 indicates the sequence repeats indefinitely, what is the value of 99*X?

(A) 80.(19)
(B) 80.19
(C) 81
(D) 81.81
(E) 81.(81)
X=.(81); 100X=81.(81); 100X=81+.(81);100X=81+X;X=81/99

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Joined: 15 Apr 2017
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Location: United States
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Re: If X = 0.(81), where the brackets around the 2-digit sequence 81 indic  [#permalink]

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04 Apr 2018, 17:43
This is the weirdest question ever considering that the question asks about .81 repeating and the answer is in whole format. I mean the real answer is 80.999 repeating.

I think this question is incorrectly worded
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Joined: 12 Sep 2017
Posts: 298
Re: If X = 0.(81), where the brackets around the 2-digit sequence 81 indic  [#permalink]

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19 Feb 2019, 09:15
X = 0.(81)

99(0.81818181..).... 81/99 = 9/11.

99(9/11)

9 * 9 = 81

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Joined: 14 Oct 2015
Posts: 6991
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Re: If X = 0.(81), where the brackets around the 2-digit sequence 81 indic  [#permalink]

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21 Feb 2019, 18:06
Bunuel wrote:
If X = 0.(81), where the brackets around the 2-digit sequence 81 indicates the sequence repeats indefinitely, what is the value of 99*X?

(A) 80.(19)
(B) 80.19
(C) 81
(D) 81.81
(E) 81.(81)

We see that 100X = 81.(81). Now, subtract X = 0.(81) from 100X = 81.(81), obtaining:

99x = 81

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Re: If X = 0.(81), where the brackets around the 2-digit sequence 81 indic   [#permalink] 21 Feb 2019, 18:06
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