Decimal expansion of rational numbers

Step 1: In (5.27000\ldots), only zeros occur after a point. Step 2: So it equals (5.27) and is a terminating decimal. Step 3: Continuing zeros at the end still represent a terminating value.

Step 1: The numerator cancels (2^4\cdot 3). Step 2: The reduced denominator becomes (2^3\cdot 3\cdot 5^2). Since (3) remains, the decimal is non-terminating recurring. Step 3: A prime factor may cancel only partially.

Step 1: The purely recurring decimal (0.\overline{13}) equals (\frac{13}{99}). Step 2: The two (9)'s in the denominator match the two repeating digits. Step 3: Distinguish purely recurring decimals from mixed recurring decimals.

Step 1: (\frac{6}{375}=\frac{2}{125}). Step 2: Since (125=5^3), the decimal terminates after (3) places. Step 3: Even for small fractions, reduce to lowest form first.

Step 1: The denominator contains (2), (5), (3), and (19). Step 2: For a terminating decimal, (3) and (19) must cancel. So the minimum factor is (3\cdot 19=57). Step 3: Only (2) and (5) may remain in the denominator.

Step 1: (0.0075=\frac{75}{10000}). Step 2: Reducing gives (\frac{3}{400}), and (400=2^4\cdot 5^2). This factorisation is not present in the listed choices, so the options have an error. Step 3: Do not choose an option before writing the final denominator in prime factor form.

Step 1: (0.0075=\frac{75}{10000}). Step 2: Reducing by (25) gives (\frac{3}{400}). So the denominator is (400). Step 3: Even with many zeros in a decimal, find the greatest common factor carefully.

Step 1: View the denominator in terms of (2), (5), and other factors. Step 2: (28=2^2\cdot 7), so the power (2) of (2) gives a delay of two places before the recurring part starts. The other options give a delay of (1) or a different case. Step 3: The delay before repetition is linked to the larger power of (2) and (5).

Step 1: (625=5^4). Step 2: After cancellation, the denominator becomes (2^8\cdot 5^2). The larger exponent is (8), so the decimal terminates after (8) places. Step 3: The numerator may cancel powers of (5), but a larger power of (2) may still remain.

Step 1: (10^5=2^5\cdot 5^5). Step 2: Any divisor of it contains only powers of (2) and (5). Therefore (\frac{p}{q}) has a terminating decimal. Step 3: Being a divisor gives at most (5) places, not necessarily exactly (5).

Step 1: (0.\overline{81}=\frac{81}{99}) and (0.\overline{18}=\frac{18}{99}). Step 2: Their sum is (\frac{99}{99}=1), which is terminating. Step 3: The sum of two recurring decimals can be terminating.

Step 1: For exactly (4) decimal places, the larger power of (2) or (5) in the reduced denominator must be (4). Step 2: (625=5^4), so it gives exactly (4) places. (80=2^4\cdot 5) also gives (4) places, so the choices would need checking if only one answer is expected. Step 3: Factorise all options in such questions.

Step 1: For exactly (4) places, the larger exponent must be (4). Step 2: (16=2^4), (625=5^4), and (80=2^4\cdot 5) give exactly (4) places. (125=5^3) gives only (3) places. Step 3: For exact places, the larger exponent must match the required number.

Step 1: Let (x=2.4666\ldots). Step 2: (10x=24.666\ldots) and (100x=246.666\ldots), so (90x=222) and (x=\frac{222}{90}=\frac{37}{15}). Step 3: Align the recurring parts before subtracting.

Step 1: (98=2\cdot 7^2). Step 2: After cancellation, the denominator becomes (5\cdot 7). Since (7) remains, the decimal is non-terminating recurring. Step 3: Check whether the whole power cancels or only part of it cancels.

Step 1: (0.0008=\frac{8}{10000}). Step 2: Dividing by (8) gives (\frac{1}{1250}). Step 3: First form the denominator as a power of (10), then reduce.

Step 1: The larger power of (2) or (5) in the denominator tells the delay before the recurring part starts. Step 2: (72=2^3\cdot 3^2), so it has a delay of (3) places. The others have larger exponent (1) or (2). Step 3: Understand the initial non-repeating part in non-terminating recurring decimals.

Step 1: For a terminating decimal, the reduced denominator (q) can contain only (2) and (5). Step 2: In (q^2), the powers of the same primes increase, but no new prime factor appears. Step 3: Powers may change, but the prime types do not.

Step 1: When (9)'s continue forever at the end, the number may equal the next terminating decimal. Step 2: (0.24999\ldots=0.25). Step 3: Convert infinite repeating (9)'s into the simpler terminating form.

Step 1: For a non-terminating recurring decimal, the reduced denominator must have a prime factor other than (2) and (5). Step 2: (2^4\cdot 5\cdot 23) contains (23). Hence it gives a non-terminating recurring decimal. Step 3: Even one extra prime factor prevents termination.