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

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Manager
Joined: 26 Dec 2018
Posts: 142
Location: India
If the digits 53 in the decimal 0.00053 repeat indefinitely, what is t  [#permalink]

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31 Dec 2018, 04:07
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45% (medium)

Question Stats:

62% (01:40) correct 38% (01:42) wrong based on 118 sessions

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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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Joined: 09 Mar 2018
Posts: 1004
Location: India
Re: If the digits 53 in the decimal 0.00053 repeat indefinitely, what is t  [#permalink]

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

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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:

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

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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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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e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2888
If the digits 53 in the decimal 0.00053 repeat indefinitely, what is t  [#permalink]

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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

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Joined: 02 Aug 2009
Posts: 7737
Re: If the digits 53 in the decimal 0.00053 repeat indefinitely, what is t  [#permalink]

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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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Posts: 262
If the digits 53 in the decimal 0.00053 repeat indefinitely, what is t  [#permalink]

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

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