Decimal expansion of rational numbers

Correct answer: B. Non-terminating recurring with (4) initial non-repeating digits

Explanation: Since (17) remains, the decimal is non-terminating recurring. The larger exponent in (2^4\cdot 5^4) gives (4) initial non-repeating digits.

Correct answer: A. Only (2) and (5) can occur

Explanation: For a terminating decimal, the reduced denominator (q) can contain only (2) and (5). In (q^4), powers increase but no new prime factor appears.

Correct answer: B. Non-terminating recurring

Explanation: For a rational number, the decimal expansion terminates exactly when the denominator in lowest terms contains only the primes 2 and 5. Factors 2 and 5 can be combined to make a power of 10, so division ends. A remaining prime such as 17 prevents this and produces a repeating pattern of digits instead.

After cancellation, \(2^5\) removes part of \(2^9\), and one factor 17 removes part of \(17^2\). The denominator becomes \(2^4\cdot 5^2\cdot 17\). Because 17 remains, the denominator is not of the required form. Thus the decimal expansion is non-terminating recurring, so option B follows. Option A would be possible only if every factor other than 2 and 5 had cancelled.

Correct answer: B. (64)

Explanation: (0.015625=\frac{15625}{1000000}=\frac{1}{64}). Convert a terminating decimal to a fraction and reduce the denominator.

Correct answer: A. (\frac{1}{6400})

Explanation: (0.00015625=\frac{15625}{100000000}), and reducing by (15625) gives (\frac{1}{6400}). Do not forget to cancel common factors in large denominators.

Correct answer: C. (\frac{1}{448})

Explanation: (448=2^6\cdot 7), so (6) non-repeating digits appear before the recurring part. For comparison, check the larger power of (2) and (5).

Correct answer: B. 1375

Explanation: Core idea: \(10^8=2^8\cdot 5^8\). The given denominator has \(2^8\) but only \(5^5\), so it is short by \(5^3\). Multiply numerator and denominator by \(5^3=125\) to obtain denominator \(10^8\). Thus \(N=11\times125=1375\).
Why other options are wrong: 275 equals \(11\times25\) (wrong if you multiply by \(5^2\) instead of \(5^3\)). 2750 is simply twice the correct N (a mistake from an extra factor 2). 6875 equals \(11\times625\) (would result from multiplying by \(5^4\)).
Exam tip: Compare prime-power factors of the denominator with \(10^n\); multiply numerator by the missing power of 2 or 5 to convert to denominator \(10^n\).

Correct answer: D. (16)

Explanation: For exactly (3) places, the larger exponent must be (3). Since (16=2^4), it terminates after (4) places.

Correct answer: B. (0.01)

Explanation: The answer is B, 0.01. Write the number as 0.00999… . The repeating 9s after the thousandths place complete the next hundredth: 0.00999… = 0.01000… = 0.01. Equivalently, 0.00999… = 0.009 + 0.00099… and the total reaches one hundredth. A, 0.009, is only the part before the infinite tail, so it is smaller. B is correct because 0.01 is the exact terminating representation. C, 9/999, equals 1/111, approximately 0.009009…, so it is different. D, 99/1000, equals 0.099, which is much larger and has the wrong place value. Memory cue: leading zeros do not change the rule; 0.00999… carries to 0.01.

Correct answer: B. (273)

Explanation: The factors (3), (7), and (13) must be removed from the reduced denominator, so the minimum factor is (3\cdot 7\cdot 13=273). Factors (2) and (5) may remain.

Correct answer: C. (5) places

Explanation: Since (750=2\cdot 3\cdot 5^3), the reduced denominator is (2^5\cdot 5^2). The larger exponent is (5), so the decimal terminates after (5) places.

Correct answer: C. (1100)

Explanation: (0.124545\ldots=\frac{1245-12}{9900}=\frac{1233}{9900}=\frac{137}{1100}). Always reduce the final fraction in mixed recurring decimals.

Correct answer: A. (3773)

Explanation: The factors (7^3) and (11) must be removed from the reduced denominator, so (n=7^3\cdot 11=3773). For the least value, do not cancel (2) and (5).

Correct answer: B. (4)

Explanation: Since (19^2) remains, the decimal is non-terminating recurring, and the larger exponent among (2) and (5) is (4). In such questions, separate recurrence from the initial delay.

Correct answer: B. (3125)

Explanation: For exactly (6) places, the larger exponent of (2) and (5) must be (6). Since (3125=5^5), it gives only (5) decimal places.

Correct answer: B. 368

Explanation: Core idea: write denominators with the same prime factors. Since \(10^9=2^9\cdot5^9\) and the given denominator is \(2^5\cdot5^9\), multiply denominator by \(2^4\) to get \(2^9\). Multiply the numerator by the same factor: \(N=23\times2^4=23\times16=368\). The closest distractor 184 corresponds to multiplying by \(2^3=8\) (one factor of 2 short), hence incorrect; 736 and 1472 result from using larger powers of 2. Exam tip: compare exponents of 2 and 5 in the denominator and multiply numerator/denominator to equalize them to powers of 10.

Correct answer: A. (37)

Explanation: (0.\overline{216}=\frac{216}{999}=\frac{8}{37}). For a purely recurring decimal, first use a denominator of (9)'s and then reduce fully.