Decimal expansion of rational numbers

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).

Step 1: The reduced denominator contains only powers of (2) and (5). Step 2: The number of decimal places equals the larger exponent. Here the larger exponent is (6). Step 3: For terminating decimals, do not add the exponents.

Step 1: (75=3\cdot 5^2). Step 2: Cancelling (3\cdot 5^2) from the denominator leaves (2^3). So the decimal terminates after (3) places. Step 3: Always complete cancellation before counting decimal places.

Step 1: The denominator has only powers of (2) and (5). Step 2: The number of decimal places is the larger of (4) and (n). For exactly (7) places, (n=7). Step 3: When the word exactly appears, match the larger exponent carefully.

Step 1: (\frac{39}{2600}=\frac{3}{200}). Step 2: (200=2^3\cdot 5^2), so the larger exponent is (3). The decimal terminates after (3) places. Step 3: Do not conclude from the denominator before reducing.

Step 1: (0.00\overline{27}=0.00272727\ldots). Step 2: Converting gives (\frac{27}{9900}=\frac{3}{1100}). Hence (q=1100). Step 3: Include the zeros before the repeating block carefully in the denominator.

Step 1: (91=7\cdot 13), so the factor (13) in the denominator cancels. Step 2: The reduced denominator is (2^2\cdot 5), containing only (2) and (5). Hence the decimal terminates. Step 3: An extra prime factor may cancel with the numerator.

Step 1: (735=3\cdot 5\cdot 7^2). Step 2: For a terminating decimal, (3) and (7^2) must not remain in the reduced denominator. So (m) must contain (3\cdot 7^2=147). Step 3: The factor (5) may remain, but (3) and (7) must cancel.

Step 1: (9=3^2), so the reduced denominator contains the prime factor (3). Step 2: If a reduced denominator has a prime other than (2) and (5), the decimal is non-terminating recurring. Step 3: Break composite factors into primes first.

Step 1: This decimal does not terminate. Step 2: The number of zeros between the (1)'s keeps changing, so there is no fixed repeating block. Hence it is non-terminating non-recurring. Step 3: Do not call a decimal rational unless a repeating block is present.

Step 1: (0.\overline{09}=\frac{09}{99}=\frac{9}{99}). Step 2: (\frac{9}{99}=\frac{1}{11}), so the reduced denominator is (11). Step 3: If a zero is part of the repeating block, count it as a digit.