Decimal expansion of rational numbers

Step 1: In lowest form, the denominator contains only powers of (2) and (5). Step 2: So the decimal expansion is terminating. The number of decimal places can be up to the larger exponent, (3). Step 3: In exams, always reduce the fraction first.

Step 1: (1250=2\cdot 5^4), and the fraction is already in lowest form. Step 2: The powers of (2) and (5) are (1) and (4), so the decimal has (4) places. Step 3: Use the larger exponent to find the terminating decimal length quickly.

Step 1: (\frac{63}{140}) is in lowest form because (63) and (140) have no common factor. Step 2: (140=2^2\cdot 5\cdot 7), so the denominator has (7). 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 denominator has only powers of (2) and (5), so the decimal terminates. Step 2: The number of decimal places equals the larger of (a) and (b). For exactly (6) places, (\max(a,b)=6). Step 3: In such questions, use the larger exponent, not the sum.

Step 1: For a terminating decimal, the reduced denominator must contain only (2) and (5). Step 2: In option (D), the denominator contains (3), and after reducing (\frac{25}{2^2\cdot 3\cdot 5}), the factor (3) remains. So it will not terminate. Step 3: Always consider possible cancellation before deciding.

Step 1: (\frac{18}{225}=\frac{2}{25}). Step 2: Since (25=5^2), the decimal terminates and has (2) decimal places. Step 3: Do not decide from the original denominator before reducing the fraction.

Step 1: (180=2^2\cdot 3^2\cdot 5). Step 2: For a terminating decimal, (3^2) must cancel completely from the denominator. So (n) must contain (3^2). Step 3: Focus on removing denominator primes other than (2) and (5).

Step 1: The denominator is (2^2\cdot 5^3). Step 2: Since (10^k=2^k\cdot 5^k), the powers must become equal. The larger power is (3), so (k=3). Step 3: To form (10^k), make the powers of (2) and (5) equal.

Step 1: (\frac{21}{42}=\frac{1}{2}). Step 2: The reduced denominator is (2), so the decimal terminates. In the other options, factors like (3) or (7) do not cancel completely. Step 3: Such questions test whether you reduce the fraction first.

Step 1: (0.00048=\frac{48}{100000}). Step 2: Reducing gives (\frac{48}{100000}=\frac{3}{625}), and (625=5^4). Step 3: The number of decimal digits does not always give the final denominator; reduce first.

Step 1: A denominator with (3) along with (2) or (5) gives a non-terminating recurring decimal. Step 2: (\frac{1}{6}=\frac{1}{2\cdot 3}), so the recurring part starts earliest. The others have (2^2), (5), or (2\cdot 5), causing a longer non-repeating start. Step 3: In mixed denominators, powers of (2) and (5) show how much the recurring part is delayed.

Step 1: (2^4\cdot 5^4=10^4). Step 2: A reduced denominator of (10^4) gives a decimal terminating after (4) places. The numerator condition assures no hidden further reduction. Step 3: If the reduced denominator is (10^k), think of (k) decimal places.

Step 1: The denominator is (2^m5^n), and the power of (5) is larger. Step 2: To make (10^n=2^n5^n), the power of (2) must be increased to (n). So multiply by (2^{n-m}). Step 3: First identify which prime power is short.

Step 1: The decimal type is decided by the denominator in lowest form. Step 2: If the reduced denominator has only factors (2) and (5), the decimal terminates. The other statements are incomplete because factors like (3) or (7) may also be present. Step 3: Be careful with words like always.

Step 1: The numerator (77=7\cdot 11), so the factor (7) in the denominator cancels. Step 2: The reduced denominator becomes (2^3\cdot 5^2), containing only (2) and (5). Hence the decimal terminates. Step 3: In tricky questions, extra denominator factors may cancel with the numerator.

Step 1: (\frac{14}{350}=\frac{1}{25}), so it actually terminates. Step 2: The other listed fractions also reduce to denominators containing only (2) and (5). Therefore none of them is non-terminating recurring. Step 3: If a requested option does not appear, recheck every simplification carefully.

Step 1: Reduce each option. (\frac{45}{90}=\frac{1}{2}), (\frac{36}{96}=\frac{3}{8}), and (\frac{28}{175}=\frac{4}{25}), so they terminate. Step 2: (\frac{26}{195}=\frac{2}{15}), and the denominator still has (3), so it is non-terminating recurring. Step 3: Reducing every option is the safest method.

Step 1: Since (\frac{p}{q}) is already in lowest form, check (q) directly. Step 2: (72=2^3\cdot 3^2), which contains (3). So the decimal is non-terminating recurring. Step 3: If lowest form is stated, do not overthink the numerator.

Step 1: The reduced denominator has (11) along with (2) and (5). Step 2: Such a rational number has a non-terminating recurring decimal. Non-terminating non-recurring decimals are linked with irrational numbers. Step 3: A rational non-terminating decimal is always recurring.

Step 1: (0.\overline{27}=\frac{27}{99}). Step 2: (\frac{27}{99}=\frac{3}{11}), so the reduced denominator is (11). Step 3: First form a denominator with (9)'s according to the repeating block, then reduce.