Decimal expansion of rational numbers

Correct answer: B. (7) places

Explanation: The denominator has only (2) and (5), so the decimal terminates with the larger exponent (7). In exams, do not add the exponents.

Correct answer: B. (2) places

Explanation: Since (84=2^2\cdot 3\cdot 7), the reduced denominator is (2\cdot 5^2). The larger exponent is (2), so reduce first and then count places.

Correct answer: C. (1053)

Explanation: For a terminating decimal, (3^4) and (13) must cancel completely, so (n=3^4\cdot 13=1053). For the least value, cancel only the unwanted prime factors.

Correct answer: A. (110)

Explanation: Taking (x=0.31818\ldots), subtracting (10x) from (1000x) gives (\frac{315}{990}=\frac{7}{22}). The reduced denominator is (22), so none of the listed denominators is correct.

Correct answer: A. (22)

Explanation: (0.3\overline{18}=0.3181818\ldots=\frac{315}{990}=\frac{7}{22}). Always reduce the final fraction in mixed recurring decimals.

Correct answer: B. Non-terminating recurring

Explanation: Even after (125=5^3) cancels, (11) remains in the denominator. If a reduced denominator has a prime other than (2) and (5), the decimal is non-terminating recurring.

Correct answer: B. (2^6\cdot 5^2)

Explanation: For exactly (6) places, the larger exponent of (2) and (5) must be (6). Only (2^6\cdot 5^2) has larger exponent (6).

Correct answer: B. 9

Explanation: When the denominator is of the form \(2^a5^b\) and the numerator is 17 (which is coprime to 2 and 5), the fraction is already in lowest terms. A terminating decimal requires only factors 2 and 5 in the denominator, and the number of decimal places required equals \(\max(a,b)\). Given the decimal terminates exactly after 9 places and \(a<b\), the larger exponent is \(b\), so \(b=9\). Why other choices fail: 8 is too small, 10 is too large, and \(a+9\) is not implied by the condition \(a<b\). Exam tip: first reduce the fraction if possible; then the termination length equals the larger exponent of 2 or 5 in the reduced denominator.

Correct answer: A. (1250)

Explanation: (0.00072=\frac{72}{100000}), and reducing by (8) gives (\frac{9}{12500}). So the correct denominator is (12500); check the common factor carefully in small decimals.

Correct answer: A. (\frac{9}{12500})

Explanation: (0.00072=\frac{72}{100000}), and reducing by (8) gives (\frac{9}{12500}). First write the denominator as a power of (10), then reduce.

Correct answer: B. (3)

Explanation: The factor (7^2) makes the decimal recurring, and the larger exponent among (2) and (5) is (3), giving the non-repeating start. In exams, separate recurrence from the initial delay.

Correct answer: A. Both are true and the reason explains it

Explanation: Since (63=3^2\cdot 7), the reduced denominator is (2^4\cdot 5^3). The reason directly explains the terminating decimal rule.

Correct answer: C. (0.\overline{625})

Explanation: (0.\overline{625}) is a fixed recurring decimal, so it is rational but not terminating. A decimal is terminating only when zeros continue after some point.

Correct answer: C. (6)

Explanation: After cancellation, the fraction becomes (\frac{3}{2^2\cdot 5^6}). The larger exponent is (6), so the decimal terminates after (6) places.

Correct answer: B. Exactly (7) or (8) places

Explanation: Since (q) has only (2) and (5), the decimal terminates. Not dividing (10^6) means the larger exponent is (7) or (8).

Correct answer: C. (1375)

Explanation: (0.00\overline{72}=\frac{72}{9900}=\frac{2}{275}). The correct denominator is (275), so choose (275) from the options.

Correct answer: A. (\frac{2}{275})

Explanation: Two non-repeating zeros and two repeating digits give (\frac{72}{9900}). Reducing it gives (\frac{2}{275}).

Correct answer: B. (4)

Explanation: The direct answer is B: 4 decimal places. First cancel common prime factors: \(\frac{2^3\cdot5^2}{2^7\cdot5^5}=\frac{1}{2^4\cdot5^3}\), because three factors of 2 and two factors of 5 cancel. The denominator is \(2^4\cdot5^3=16\cdot125=2000\). Since a terminating decimal has a denominator made only of 2s and 5s, it terminates. To write the denominator as a power of 10, the larger exponent is 4, so multiply by one extra 5: \(\frac1{2000}=0.0005\), which has four places after the decimal point. Option A, 3, is too small; four places are needed. Option B, 4, is correct. Option C, 5, counts an unnecessary zero or misreads the numerator. Option D, 7, ignores cancellation. Exam cue: after reducing, use the larger exponent of 2 and 5.

Correct answer: C. If (q=2^m5^n), the decimal terminates

Explanation: A decimal terminates when the reduced denominator has only (2) and (5). The other statements are incomplete because other prime factors may also be present.

Correct answer: B. (0.125)

Explanation: When (9)'s continue forever at the end, the number equals the next terminating decimal. Thus (0.124999\ldots=0.125).

Correct answer: A. (153)

Explanation: The factors (3^2) and (17) must be removed from the reduced denominator, so the minimum factor is (3^2\cdot 17=153). Factors (2) and (5) may remain.

Correct answer: A. Terminating after (2) places

Explanation: Since (189=3^3\cdot 7), the reduced denominator is (2^2\cdot 5). The larger exponent is (2), so the decimal terminates after (2) places.

Correct answer: C. Non-terminating non-recurring irrational

Explanation: This decimal does not end, and the number of zeros between the (2)'s keeps changing. Since there is no fixed repeating block, it is non-terminating non-recurring.

Correct answer: B. (1375)

Explanation: Since (10^6=2^6\cdot 5^6), the denominator lacks (5^4). Thus (N=11\cdot 5^4=6875), so the correct option is (6875).

Correct answer: C. 6875

Explanation: To obtain denominator \(10^6=2^6\cdot5^6\), we must make the power of 5 equal to 6. The given denominator is \(2^6\cdot5^2\), so multiply numerator and denominator by \(5^4=625\). Hence \(N=11\cdot5^4=11\cdot625=6875\). Common mistakes: 1375 equals \(11\cdot125\) (using \(5^3\) instead of \(5^4\)), while 2750 comes from another incorrect multiplier. Exam tip: prime-factorize the denominator and match exponents of 2 and 5 to form \(10^k\).