Recurring Decimals to Fractions

Multiply by a power of ten so that two copies of the decimal have matching recurring tails.

1. Let x=0.444\ldots. One digit recurs, so multiply by 10: 10x=4.444\ldots.
2. Both numbers have the same tail of 4s after the decimal point, so subtracting removes it: 10x-x=4.444\ldots-0.444\ldots.
3. This gives 9x=4, so x=\frac{4}{9}, which cannot be simplified.
4. \frac{4}{10} is wrong: it is 0.4, a decimal that stops.

Multiply by a power of ten so that two copies of the decimal have matching recurring tails.

1. Let x=0.272727\ldots. A block of two digits recurs, so multiply by 100: 100x=27.272727\ldots.
2. Subtract to remove the recurring tail: 100x-x=27.2727\ldots-0.2727\ldots, so 99x=27.
3. So x=\frac{27}{99}. Divide the numerator and denominator by 9 to get \frac{3}{11}.

Multiply by a power of ten so that two copies of the decimal have matching recurring tails.

1. Let x=0.1666\ldots. Only the 6 recurs; the 1 does not.
2. Multiply by 10 to move the 1 past the decimal point: 10x=1.666\ldots. Multiply by 100 to get one more 6 in front: 100x=16.666\ldots.
3. These two have the same tail, so subtract: 100x-10x=16.666\ldots-1.666\ldots, giving 90x=15.
4. So x=\frac{15}{90}. Divide by 15: x=\frac{1}{6}.

Multiply by a power of ten so that two copies of the decimal have matching recurring tails.

1. Let x=0.0888\ldots. The 0 does not recur, but the 8 does.
2. Write two multiples with the same tail: 10x=0.888\ldots and 100x=8.888\ldots.
3. Subtract: 100x-10x=8.888\ldots-0.888\ldots, giving 90x=8.
4. So x=\frac{8}{90}. Divide by 2: x=\frac{4}{45}.

Multiply by a power of ten so that two copies of the decimal have matching recurring tails.

1. Let x=1.222\ldots. One digit recurs, so multiply by 10: 10x=12.222\ldots.
2. Subtract to remove the tail: 10x-x=12.222\ldots-1.222\ldots, giving 9x=11.
3. So x=\frac{11}{9}. Check: it is greater than 1, as 1.222\ldots is.

Choose two multiples whose digits after the decimal point match.

1. The block 38 has two digits, so multiply by 100: 100x=38.3838\ldots, which has the same tail as x=0.3838\ldots.
2. Subtracting: 100x-x=38.3838\ldots-0.3838\ldots=38.
3. 10x-x does not work: 10x=3.8383\ldots has a different tail from x, so subtracting leaves 3.4545\ldots, still recurring.
4. So the correct equation is 100x-x=38, that is 99x=38.

Compare 1000x with 10x to cancel the repeating two-digit block.

1. Let x=0.2131313\ldots. The 2 does not recur; the block 13 does.
2. Multiply by 10 to move the 2 past the decimal point: 10x=2.131313\ldots. Then by 100 more to move one whole block: 1000x=213.131313\ldots.
3. Subtract: 1000x-10x=213.1313\ldots-2.1313\ldots, giving 990x=211.
4. So x=\frac{211}{990}. 211 is prime, so this is in its simplest form.

Move both non-recurring digits before the decimal point, then move one further place.

1. Let x=0.045555\ldots. The digits 0 and 4 do not recur; the 5 does.
2. Move both non-recurring digits past the decimal point: 100x=4.5555\ldots. Move one more place: 1000x=45.5555\ldots.
3. Subtract: 1000x-100x=45.555\ldots-4.555\ldots, giving 900x=41.
4. So x=\frac{41}{900}. 41 is prime and does not divide 900, so this is in its simplest form.

The integer part must be included when multiplying the decimal by powers of ten.

1. Let x=2.3181818\ldots. The 3 does not recur; the block 18 does.
2. Then 10x=23.181818\ldots and 1000x=2318.181818\ldots have the same tail.
3. Subtract: 1000x-10x=2318.18\ldots-23.18\ldots, giving 990x=2295.
4. So x=\frac{2295}{990}. Dividing the numerator and denominator by 45 gives \frac{51}{22}.
5. \frac{23}{10}+\frac{18}{99} is wrong: it is 2.3+0.1818\ldots=2.4818\ldots, because the 18 starts in the hundredths, not the tenths.

The zero is part of the repeating three-digit block, so move the decimal point three places.

1. Let x=0.084084\ldots. The block 084 has three digits (the zero is part of it), so multiply by 1000: 1000x=84.084084\ldots.
2. Subtract: 1000x-x=84.084\ldots-0.084\ldots, giving 999x=84.
3. So x=\frac{84}{999}. Divide the numerator and denominator by 3 to get \frac{28}{333}.

Convert the decimal to a fraction by subtracting multiples with matching recurring tails.

1. Let x=0.372727\ldots. Then 1000x=372.727\ldots and 10x=3.727\ldots.
2. Subtract to get 990x=369, so x=\frac{369}{990}.
3. Divide both parts by 9: x=\frac{41}{110}. Therefore p=41.

Convert the recurring decimal first, then use a common denominator.

1. If x=0.363636\ldots, subtraction of x from 100x gives 99x=36. Thus x=\frac{4}{11}.
2. The common denominator for 11 and 4 is 44.
3. The sum is \frac{16}{44}+\frac{11}{44}=\frac{27}{44}.

Write enough decimal digits to find the first position at which the values differ.

1. The bars give a=0.616161\ldots and b=0.611111\ldots.
2. The tenths and hundredths agree, but the thousandths are 6 for a and 1 for b.
3. Therefore a>b.

Convert the recurring decimal to a fraction before subtracting it.

1. First write 0.1\overline{8} as a fraction. Let y=0.1888\ldots, so 10y=1.888\ldots and 100y=18.888\ldots.
2. Subtract: 100y-10y=18.888\ldots-1.888\ldots, giving 90y=17, so y=\frac{17}{90}.
3. Now solve x+\frac{17}{90}=\frac{1}{2} by subtracting \frac{17}{90} from both sides: x=\frac{45}{90}-\frac{17}{90}=\frac{28}{90}.
4. Divide by 2: x=\frac{14}{45}.

Convert each decimal separately, then subtract using the given denominator.

1. Convert each decimal. Let x=0.5454\ldots, so 100x=54.5454\ldots. Subtract: 100x-x=54, so 99x=54 and x=\frac{54}{99}=\frac{6}{11}.
2. Let y=0.2777\ldots, so 10y=2.777\ldots and 100y=27.777\ldots. Subtract: 100y-10y=25, so 90y=25 and y=\frac{25}{90}=\frac{5}{18}.
3. 198=11\times18, so write both over 198: \frac{6}{11}=\frac{108}{198} and \frac{5}{18}=\frac{55}{198}.
4. The difference is \frac{108-55}{198}=\frac{53}{198}, so k=53.

Find the exact fraction represented by the known decimal, then multiply it by the given quotient.

1. The known decimal is \frac{24}{99}=\frac{8}{33}.
2. The equation gives d=\frac{5}{2}\times\frac{8}{33}=\frac{20}{33}.
3. Multiply top and bottom by 3 to get d=\frac{60}{99}, which is 0.\overline{60}.

After converting the decimals, divide by a fraction by multiplying by its reciprocal.

1. Let u=1.0272727\ldots. Then 1000u-10u=1027.2727\ldots-10.2727\ldots=1017.
2. Therefore u=\frac{1017}{990}=\frac{113}{110}. Also 0.\overline{18}=\frac{18}{99}=\frac{2}{11}.
3. The quotient is \frac{113}{110}\times\frac{11}{2}=\frac{113}{20}.

Convert the multiplier to a fraction, then use the inverse operation.

1. If x=0.424242\ldots, then 100x-x=42, so x=\frac{42}{99}=\frac{14}{33}.
2. If the original number is N, then \frac{14}{33}N=56.
3. Divide by 14 and multiply by 33: N=\frac{56\times33}{14}=132.

Convert both decimals to fractions with the same denominator before simplifying the ratio.

1. For x=0.1232323\ldots, subtracting 10x from 1000x gives 990x=122, so x=\frac{122}{990}.
2. For y=0.2131313\ldots, the same method gives 990y=211, so y=\frac{211}{990}.
3. Both ratio terms can be multiplied by 990, giving 122:211.
4. These integers have no common factor greater than 1, so the ratio is already in its simplest form.

Convert both decimals to fractions and simplify the denominator before dividing.

1. For x=0.2454545\ldots, 1000x-10x=245.4545\ldots-2.4545\ldots=243. Hence x=\frac{243}{990}=\frac{27}{110}.
2. The other recurring decimal is \frac{18}{99}=\frac{2}{11}, so the denominator is 1-\frac{2}{11}=\frac{9}{11}.
3. Divide by multiplying by the reciprocal: \frac{27}{110}\times\frac{11}{9}=\frac{3}{10}.