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If the digits 53 in the decimal 0.00053 repeat indefinitely, what is t

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If the digits 53 in the decimal 0.00053 repeat indefinitely, what is t  [#permalink]

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New post 31 Dec 2018, 04:07
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If the digits 53 in the decimal 0.00053 repeat indefinitely, what is the value of (10^5-10^3)(0.00053)?

A. 0
B. 0.53 repeating
C. 5.3
D. 10
E. 53

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Re: If the digits 53 in the decimal 0.00053 repeat indefinitely, what is t  [#permalink]

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New post 31 Dec 2018, 04:22
UB001 wrote:
If the digits 53 in the decimal 0.00053 repeat indefinitely, what is the value of (10^5-10^3)(0.00053)?

A. 0
B. 0.53 repeating
C. 5.3
D. 10
E. 53


(10^5-10^3)(0.00053)
when you solve the (10^5-10^3) part it will become 1000 * 99 which is multiplied with 0.000535353....
= 99*0.535353

giving an answer as 52.37 or something which will be greater than 52.

When you notice this trend the answer will be as E.
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If the digits 53 in the decimal 0.00053 repeat indefinitely, what is t  [#permalink]

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New post 31 Dec 2018, 06:46
1
UB001 wrote:
If the digits 53 in the decimal 0.00053 repeat indefinitely, what is the value of (10^5-10^3)(0.00053)?

A. 0
B. 0.53 repeating
C. 5.3
D. 10
E. 53


We know how to deal with recurring decimals. Say,

\(.000535353... = x\)
\(.535353... = 1000x\) ... (I)
\(53.535353... = 100000x\) ... (II)

(II) - (I)
\((100000 - 1000)x = 53\)
\(x = \frac{53}{(10^5 - 10^3)}\)

So we get that \(0.000535353... = \frac{53}{(10^5 - 10^3)}\)

Let's put it in our original expression:

\((10^5-10^3)(0.00053) = (10^5-10^3)*\frac{53}{(10^5 - 10^3)} = 53\)

Answer (E)

Here is a post on our Veritas Prep blog that gives details of this method:
https://www.veritasprep.com/blog/2014/0 ... fractions/
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Re: If the digits 53 in the decimal 0.00053 repeat indefinitely, what is t  [#permalink]

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New post 31 Dec 2018, 08:47
UB001 wrote:
If the digits 53 in the decimal 0.00053 repeat indefinitely, what is the value of (10^5-10^3)(0.00053)?

A. 0
B. 0.53 repeating
C. 5.3
D. 10
E. 53



(10^5-10^3)(0.00053)
can be re written as :

10^5-10^3 * 53* 10^-5

10^3(100-1)* 53/10^5

10^-2 * ( 99) * 53

~ 53 IMO E
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If the digits 53 in the decimal 0.00053 repeat indefinitely, what is t  [#permalink]

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New post 02 Jan 2019, 04:59

Solution


Given:
    • The digits 53 in the decimal 0.00053 repeat indefinitely

To find:
    • The value of \((10^5 – 10^3)(0.000535353....)\)

Approach and Working:
    • \((10^5 – 10^3)(0.000535353....) = 10^3(100 - 1)(0.000535353…) = 99 * 0.535353… = 99 * \frac{53}{99} = 53\)

Hence, the correct answer is Option E

Answer: E

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Re: If the digits 53 in the decimal 0.00053 repeat indefinitely, what is t  [#permalink]

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New post 12 Jan 2019, 01:39
UB001 wrote:
If the digits 53 in the decimal 0.00053 repeat indefinitely, what is the value of (10^5-10^3)(0.00053)?

A. 0
B. 0.53 repeating
C. 5.3
D. 10
E. 53



Of course the method has been beautifully explained by VeritasKarishma.

The type of choices too can help us in this..
\((10^5-10^3)(0.0005353..)=10^3(100-1)*0.0005353..=99*0.5353..\), should be slightly less than 100*0.5353 or 53.53
Only E fits in

Ofcourse Method is ..
\((10^5-10^3)(0.0005353..)=10^3(100-1)*0.0005353..=99*0.5353..\)
Now let \(x = 0.5353..\), so \(100x=53.5353\)..
subtract the two .. \(100x-x=53.5353...-0.5353... = > 99x=53......=> 99*(0.5353..)=53\)
E
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If the digits 53 in the decimal 0.00053 repeat indefinitely, what is t  [#permalink]

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New post 25 Feb 2019, 15:27
1
0.00053 (repeating 53 indefinitely) =

\(\frac{53}{99000}\) * \((10^5-10^3)\)

\(\frac{53}{(99*1000)}\) * \((10^3)(10^2-1)\)

\(\frac{53}{(99)(10^3)}\) * \((10^3)(99)\)

\(53\)

E
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If the digits 53 in the decimal 0.00053 repeat indefinitely, what is t   [#permalink] 25 Feb 2019, 15:27
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