Decimal expansions — Class 10 CBSE Mathematics MCQs with Solutions

Solution: Here $3125 = 5^5$, which is of the form $2^m \times 5^n$ (with $m = 0$, $n = 5$). Hence $\frac{13}{3125}$ has a terminating decimal expansion; in fact $\frac{13}{3125} = 0.00416$.

Solution: $8 = 2^3$, $125 = 5^3$ and $25 = 5^2$ are all of the form $2^m \times 5^n$, so those three expansions terminate. But $12 = 2^2 \times 3$ contains the prime $3$, and $\frac{7}{12}$ is already in simplest form, so its decimal expansion is non-terminating recurring.

Solution: Write the denominator as a power of $10$: $\frac{17}{8} = \frac{17}{2^3} = \frac{17 \times 5^3}{10^3} = \frac{2125}{1000} = 2.125$. So the expansion terminates after exactly $3$ decimal places.

Solution: The number of decimal places equals the larger of the exponents of $2$ and $5$ in the denominator, i.e. $\max(3, 2) = 3$. Indeed $\frac{23}{200} = \frac{23 \times 5}{1000} = \frac{115}{1000} = 0.115$.

Solution: $\frac{1}{3} \times \frac{3}{10} = \frac{1}{10} = 0.1$, which terminates after exactly one decimal place. Multiplying by $\frac{1}{10}$ gives $\frac{1}{30}$, which is still non-terminating, while $\frac{3}{100}$ gives $0.01$, terminating after two places, not one.

Solution: Multiply numerator and denominator by $2^4$ to equalise the powers: $\frac{77 \times 2^4}{2^7 \times 5^7} = \frac{1232}{10^7} = 0.0001232$. Since $77 = 7 \times 11$ shares no factor with $10$, the expansion terminates after $\max(3, 7) = 7$ decimal places.

Solution: A non-terminating recurring decimal always represents a rational number, so it can be written as $\frac{p}{q}$. A fraction in simplest form has a terminating expansion exactly when $q = 2^m \times 5^n$. Since this expansion does not terminate, $q$ must contain at least one prime factor other than $2$ and $5$.