Decimal expansion of rational numbers

Step 1: A reduced denominator must contain only (2) and (5). Step 2: (80=2^4\cdot 5), so it is possible. (48), (84), and (98) contain primes like (3) or (7). Step 3: If lowest form is given, check the prime factors of the denominator directly.

Step 1: (\frac{99}{9900}) reduces to (\frac{1}{100}) because (9900\div 99=100). Step 2: The reduced denominator is (100=2^2\cdot 5^2), so the decimal terminates. Step 3: With large numbers, check reduction carefully by division.

Step 1: (\frac{27}{990}=\frac{3}{110}). Step 2: (110=2\cdot 5\cdot 11), so (11) remains in the denominator. Hence the decimal is non-terminating recurring. Step 3: If a reduced denominator has a prime other than (2) or (5), it will not terminate.

Step 1: The reduced denominator has only powers of (2) and (5). Step 2: The number of decimal places equals the larger exponent. Since (r>s), the larger exponent is (r). Step 3: Remember (\max(r,s)) for decimal places.

Step 1: (0.375=\frac{375}{1000}). Step 2: Dividing numerator and denominator by (125) gives (\frac{3}{8}). So the denominator is (8). Step 3: Always reduce after converting a decimal to a fraction.

Step 1: (0.\overline{25}) has a repeating block, so it is rational. Step 2: It is not terminating because the decimal does not end. (\sqrt{2}) and (\pi) are irrational. Step 3: Rational numbers can be terminating or non-terminating recurring.

Step 1: (56=2^3\cdot 7). Step 2: For a terminating decimal, (7) must not remain in the reduced denominator. Therefore (m) must contain (7). Step 3: Cancel all denominator primes other than (2) and (5).

Step 1: (3125=5^5). Step 2: The denominator has power (0) of (2) and power (5) of (5). So the decimal terminates after (5) places. Step 3: Remembering (3125=5^5) helps in quick factorisation.

Step 1: In (x=0.21818\ldots), the non-repeating part is (2) and the repeating part is (18). Step 2: First use (10x=2.1818\ldots), then (1000x=218.1818\ldots) so the recurring parts align. Step 3: Choose powers of (10) based on the lengths of the non-repeating and repeating parts.

Step 1: The denominator is (2^6\cdot 5^3), and the fraction is in lowest form because (37) does not cancel. Step 2: The larger exponent is (6), so the decimal terminates after (6) places. Step 3: Do not add the exponents for decimal places.

Step 1: Rational numbers have either terminating or non-terminating recurring decimals. Step 2: (\sqrt{5}) is irrational, so its decimal is non-terminating non-recurring. Step 3: To identify non-terminating non-recurring decimals, look for irrational numbers.

Step 1: (3^0=1), so there is actually no factor (3) in the denominator. Step 2: The denominator is (2^2\cdot 5^3), containing only (2) and (5). Hence the decimal terminates. Step 3: Do not get confused by a zero exponent.

Step 1: The repeating block (142857) has (6) digits. Step 2: For a purely recurring decimal, before reducing, the denominator has the same number of (9)'s. So it is (999999). Step 3: The number of repeating digits tells the number of (9)'s.

Step 1: (16=2^4), so (2^4) cancels from the denominator. Step 2: The reduced denominator is (2^3\cdot 5^4). The larger exponent is (4), so the decimal terminates after (4) places. Step 3: Include powers hidden in the numerator during cancellation.

Step 1: The block (36) repeats in the decimal. Step 2: A recurring decimal is always rational, but it is not terminating. So it is rational and non-terminating recurring. Step 3: When a repeating block appears, identify the number as rational.

Step 1: The denominator has (7), and the numerator (1) cannot cancel it. Step 2: The reduced denominator has (7) besides (2) and (5), so the decimal is non-terminating recurring. Step 3: Having (2) and (5) in the denominator does not guarantee termination.

Step 1: We need (10^5=2^5\cdot 5^5). Step 2: The denominator (2^3\cdot 5^5) lacks (2^2). So multiply numerator and denominator by (2^2=4). Hence (N=4p). Step 3: To make (10^k), multiply by the missing prime power.

Step 1: (35=5\cdot 7). Step 2: The factor (5) and one (7) cancel, but one (7) remains. The reduced denominator is (2^2\cdot 7). So the decimal is non-terminating recurring. Step 3: After partial cancellation, check what factor remains.

Step 1: (0.0048=\frac{48}{10000}). Step 2: The greatest common factor of (48) and (10000) is (16), so (\frac{48}{10000}=\frac{3}{625}). The denominator is (625). Step 3: Even for small decimals, reduce to lowest form.

Step 1: A non-terminating decimal of a rational number is recurring. Step 2: This happens when the reduced denominator has at least one prime factor other than (2) and (5). So option (C) is correct. Step 3: (2) or (5) may also be present, but some other prime must remain.

Step 1: To make the denominator (10^k=2^k5^k), both exponents must be made equal. Step 2: The required (k) equals the larger exponent. So the number of decimal places is (\max(a,b)). Step 3: This rule is frequently tested in terminating decimal questions.

Step 1: (44=2^2\cdot 11). Step 2: After cancellation, the denominator becomes (2\cdot 5=10). So the decimal terminates after (1) place. Since that exact statement is not listed, the given options contain an issue. Step 3: Complete your calculation before trusting the options.

Step 1: (44=2^2\cdot 11). Step 2: Cancelling (2^2\cdot 11) from (2^3\cdot 5\cdot 11) leaves (2\cdot 5=10). So the decimal terminates after (1) place. Step 3: Count decimal places only after complete cancellation.

Step 1: The digit (6) repeats, so the decimal is recurring. Step 2: Every recurring decimal is rational, but this one does not terminate. Hence it is rational and non-terminating recurring. Step 3: A bar over digits shows the repeating part.

Step 1: (81=3^4), so (3^4) cancels completely from the denominator. Step 2: The reduced denominator is (2^4\cdot 5^2). The larger exponent is (4), so the decimal terminates after (4) places. Step 3: First check cancellation of prime factors other than (2) and (5).