Decimal expansions — Class 10 CBSE Mathematics MCQs with Solutions

Solution: $\frac{1}{6}$ has denominator $6 = 2 \times 3$. Since 3 is a prime factor (not 2 or 5), the decimal will not terminate. Computing: $1 \div 6 = 0.1\overline{6}$ (the digit 6 repeats). In contrast, $\frac{1}{8} = \frac{1}{2^3}$ terminates as $0.125$, and $\frac{1}{5}$ terminates as $0.2$.

Solution: The condition for a terminating decimal is that the denominator (in lowest terms) has only 2 and 5 as prime factors. Since $q = 2^3 \times 5^2$ satisfies this, the decimal will terminate. The number of decimal places will be $\max(3, 2) = 3$ (we need to make the powers of 2 and 5 equal, then convert to a power of 10). For example, $\frac{1}{200} = \frac{1}{2^3 \times 5^2} = \frac{5}{2^3 \times 5^3} = \frac{5}{1000} = 0.005$.

Solution: In long division of $p \div q$, each step produces a remainder less than $q$. There are exactly $q$ possible remainders: {0, 1, 2, ..., $q-1$}. Since we perform more division steps than there are possible remainders, by the pigeonhole principle, some remainder must repeat. Once a remainder repeats, the sequence of quotient digits repeats from that point onward. This guarantees periodicity for any rational number (or termination if remainder becomes 0).

Solution: By definition, an irrational number cannot be expressed as $\frac{p}{q}$. We proved earlier that every rational number has a terminating or repeating decimal. Therefore, irrational numbers must have decimals that are neither terminating nor repeating. For example, $\pi \approx 3.14159265...$ and $\sqrt{2} \approx 1.41421356...$ show no repeating pattern and never terminate.