You Can Tell Whether a Fraction Will Ever Stop Dividing

Yes — look at the denominator's prime factors. If they are only 2s and 5s, the decimal stops; if anything else is there, it repeats for ever.

So $\frac{3}{8}$ terminates and $\frac{1}{3}$ does not, and you know that before dividing. This page covers everything in the ICSE Class 7 Mathematics chapter's second part: terminating and recurring decimals, converting recurring decimals to fractions, rounding and significant figures, and word problems.

A terminating decimal ends after a finite number of digits. A non-terminating recurring decimal goes on for ever with a group of digits repeating.

The test uses the denominator in lowest terms. If its only prime factors are 2 and 5, the decimal terminates; otherwise it recurs.

- $\frac{3}{8}$: denominator $8=2^3$, so it terminates — $0.375$.
- $\frac{7}{20}$: denominator $20=2^2\times 5$, so it terminates — $0.35$.
- $\frac{1}{3}$: denominator 3, so it recurs — $0.333\dots$
- $\frac{5}{6}$: denominator $6=2\times 3$, so it recurs — $0.8333\dots$

Bar notation marks the repeating group with a bar over it:

$$\frac{1}{3}=0.\overline{3},\frac{4}{11}=0.\overline{36},\frac{1}{7}=0.\overline{142857}$$

Recurring decimals come in two kinds. A pure recurring decimal repeats from immediately after the point, as $0.\overline{3}$. A mixed one has some non-repeating digits first:

$$\frac{5}{6}=0.8\overline{3}$$

where only the 3 repeats.

The step people skip is reducing first. $\frac{6}{30}$ looks as though it should recur because of the 3, but it reduces to $\frac{1}{5}=0.2$, which terminates. The test only works on the lowest-terms denominator.

There are two short rules, one for each kind.

Pure recurring decimal. Write the repeating group as the numerator, and as many 9s as there are repeating digits as the denominator:

$$0.\overline{7}=\frac{7}{9}$$

$$0.\overline{36}=\frac{36}{99}=\frac{4}{11}$$

$$0.\overline{123}=\frac{123}{999}=\frac{41}{333}$$

Mixed recurring decimal. Subtract the non-repeating part from the whole group of digits, and use as many 9s as there are repeating digits followed by as many 0s as there are non-repeating digits:

$$0.1\overline{6}=\frac{16-1}{90}=\frac{15}{90}=\frac{1}{6}$$

$$0.2\overline{45}=\frac{245-2}{990}=\frac{243}{990}=\frac{27}{110}$$

Check the first one by dividing back: $1÷6=0.1666\dots$, which is $0.1\overline{6}$. Correct.

Count the digits carefully, since the denominator depends entirely on the counts. For $0.2\overline{45}$ there are two repeating digits and one non-repeating digit, so the denominator is $99$ followed by one zero — $990$. Getting that count wrong is the only real difficulty in the method.

Rounding to decimal places counts digits after the point. Look at the next digit: if it is 5 or more, round up; if less, leave it.

Take $3.4567$:

- To 2 decimal places: the next digit is 6, so round up — $3.46$.
- To 1 decimal place: the next digit is 5, so round up — $3.5$.
- To 3 decimal places: the next digit is 7, so round up — $3.457$.

Rounding to significant figures counts digits from the first non-zero digit onwards. Leading zeros are never significant, but zeros between digits are.

- $0.004567$ to 2 significant figures: the first significant digit is 4, so we keep 4 and 5, and the next digit 6 rounds up — $0.0046$.
- $34567$ to 3 significant figures: keep 3, 4 and 5; the next digit 6 rounds up — $34600$.
- $2.0378$ to 3 significant figures: keep 2, 0 and 3; the next digit 7 rounds up — $2.04$.

A shopkeeper rounding ₹121.25 to the nearest rupee gives ₹121, since the next digit is 2.

The distinction to hold on to is where you start counting. Decimal places count from the point; significant figures count from the first non-zero digit. So $0.004567$ has 2 significant figures as $0.0046$ but would be $0.00$ to two decimal places — the same number, two completely different answers.

Identify the operation, keep the units consistent, and round only at the end.

Cost. If $2.5$ kg of dal costs ₹48.50 per kg:

$$2.5\times 48.50=₹121.25$$

Rate from total. If $3.5$ m of cloth costs ₹294, then 1 m costs

$$294÷3.5=2940÷35=₹84$$

Average. Three bags weigh $2.5$ kg, $3.75$ kg and $4.25$ kg. The total is $10.5$ kg, so the average is

$$\frac{10.5}{3}=3.5\text{ kg}$$

Capacity. A tank holds $45.5$ L and is emptied into bottles of $0.75$ L each:

$$45.5÷0.75=4550÷75=60.67$$

so 60 full bottles can be filled, with some left over.

That last answer shows why the context matters as much as the arithmetic. You cannot fill $0.67$ of a bottle and call it filled, so the answer to how many full bottles is 60 — rounding down, regardless of the usual rounding rule.