How do you find the repeating decimal 0.231 with 31 repeated as a fraction?

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Answer 1 · 2016-10-20T12:47:44

The answer is $\frac{1154}{4995}$ $1154/4995$

$0.2(31)=0.2+0.0(31)=0.2+0.031+0.00031+0.0000031+...$

Starting from the second number we have an infinite sum of a convergent geometric sequence with:

$a_1=0.031$ and $q=0.001$

So we can write the whole decimal as:

$0.2(31)=0.2+0.031/(1-0.001)=0.2+0.031/0.999=2/10+31/999=$

$=(2*999+310)/9990=(1198+310)/9990=2308/9990=1154/4995$

Answer 2 · 2016-10-20T14:35:59

$x=229/990$

let $x=0.2313131.............$

because TWO digits repeat $xx100$

we have therefore:

$100x=23.13131313131313.......$ .

& #x=0.231313131......#

Subtracting the two the repeating 31s will all cancel

$100x=23.1cancel(3131313131313.......)$ .

$x=0.2cancel(31313131......)$

we are left with

$110x-x=23.1-0.2$

$99x=22.9$

$x=(22.9)/99=229/990$