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 #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#