Decimal expansion of rational numbers

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.

Step 1: Let (x=0.4777\ldots). Step 2: (10x=4.777\ldots) and (100x=47.777\ldots), so (90x=43) and (x=\frac{43}{90}). Step 3: Separate the non-repeating and repeating parts before multiplying.

Step 1: The number of decimal places is the larger of (a) and (b). Step 2: Since (a<b), the larger exponent is (b). For exactly (4) places, (b=4). Step 3: When a comparison is given, identify the larger exponent first.

Step 1: Reduce the options. Step 2: (\frac{121}{363}=\frac{1}{3}), whose denominator is (3), so the decimal is non-terminating recurring. The other options reduce to denominators with only (2) and (5). Step 3: Check the lowest form of every option first.

Step 1: (10^8=2^8\cdot 5^8). Step 2: The denominator (2^r5^s) must divide (10^8), so (r\leq 8) and (s\leq 8). Step 3: When converting to denominator (10^k), remember the divisor condition.

Step 1: (128=2^7), so (\frac{7}{128}) has (7) decimal places. Step 2: (625=5^4), (40=2^3\cdot 5), and (160=2^5\cdot 5), giving (4), (3), and (5) places. Step 3: For comparison, factorise the denominators quickly.

Step 1: (0.000125=\frac{125}{1000000}). Step 2: Dividing both by (125) gives (\frac{1}{8000}). So the denominator is (8000). Step 3: Even when a decimal has many zeros, reduce the fraction fully.

Step 1: (18=2\cdot 3^2). Step 2: After cancellation, the denominator becomes (2\cdot 5^4). The larger exponent is (4), so the decimal terminates after (4) places. Step 3: Carefully cancel prime powers present in the numerator.

Step 1: For exactly (5) decimal places, the larger exponent of (2) and (5) in the reduced denominator must be (5). Step 2: (3125=5^5), so it gives (5) places. (250) and (40) give fewer places, while (1600=2^6\cdot 5^2) gives (6) places. Step 3: Compare the prime exponents of the denominator.

Step 1: (\frac{17}{200}) is in lowest form. Step 2: (200=2^3\cdot 5^2), so the denominator has only (2) and (5). Hence the decimal terminates, and the reason explains the assertion. Step 3: In assertion-reason questions, check whether the reason truly explains the assertion.

Step 1: (0.125=\frac{125}{1000}). Step 2: Reducing gives (\frac{1}{8}), and (8=2^3). So the prime factorisation of (q) is (2^3). Step 3: Convert the decimal to a fraction and then reduce the denominator.

Step 1: A rational number has either a terminating decimal or a non-terminating recurring decimal. Step 2: (2.01001000100001\ldots) has no fixed repeating block. So it cannot be rational. Step 3: Decide by checking repetition, not merely by seeing a long decimal.

Step 1: (7^0=1), so there is no actual factor (7) in the denominator. Step 2: The denominator is (2^5\cdot 5^3), so the decimal terminates with (5) places. Step 3: Do not get confused by a zero exponent.

Step 1: (55=5\cdot 11). Step 2: After cancellation, the denominator becomes (2\cdot 5\cdot 11). Since (11) remains, the decimal is non-terminating recurring. Step 3: After partial cancellation, always check the remaining factors.

Step 1: (0.\overline{6}=\frac{2}{3}) and (0.\overline{3}=\frac{1}{3}). Step 2: Their sum is (1), whose decimal (1.0) is terminating. Step 3: The sum of recurring decimals can sometimes be terminating.

Step 1: (126=2\cdot 3^2\cdot 7) and (1575=3^2\cdot 5^2\cdot 7). Step 2: After cancellation, we get (\frac{2}{25}). Since (25=5^2), the decimal terminates after (2) places. Step 3: Prime factorisation helps with larger numbers.