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

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

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New post 28 Mar 2018, 03:10
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C
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Question Stats:

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

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New post 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

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

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New post 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...
(Start with the repeating portion.)

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

Answer C

(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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New post 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
Answer C

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

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

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

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New post 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

Answer: C
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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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