Decimal expansion of rational numbers

Correct answer: A. Terminates exactly after (5) places

Explanation: The reduced denominator is (10^5), so the decimal terminates exactly after (5) places. The numerator condition indicates no further cancellation.

Correct answer: A. (\frac{121}{2^2\cdot 5^3\cdot 11})

Explanation: In the first option, (121=11^2) cancels the denominator's (11), leaving only (2) and (5) in the denominator, so it terminates. No option is non-terminating here, so the options need rechecking.

Correct answer: C. (\frac{49}{2\cdot 5^2\cdot 7^2})

Explanation: In (\frac{49}{2\cdot 5^2\cdot 7^2}), (49=7^2) cancels completely, so it terminates. For a non-terminating recurring decimal, a factor other than (2) and (5) must remain in the reduced denominator.

Correct answer: B. Non-terminating recurring

Explanation: A rational number has a terminating decimal expansion after the fraction is reduced only when the denominator has no prime factors other than 2 and 5. If any other prime factor remains in the lowest terms, its decimal expansion is non-terminating but recurring. Therefore, cancellation must be performed before classifying the decimal; looking only at the original denominator could give a wrong conclusion.

Here, \(14=2\cdot7\), so cancellation with the numerator gives \(\frac{14}{2\cdot5^2\cdot7^2}=\frac{1}{5^2\cdot7}\). The reduced denominator still contains the prime factor 7. Hence the decimal cannot terminate and must be non-terminating recurring. Thus option B is correct. It is not non-recurring because every rational number has either a terminating or recurring decimal expansion.

Correct answer: B. (10^5)

Explanation: At most (5) decimal places means the fraction can be written with denominator (10^5). The reduced denominator must divide (10^5).

Correct answer: A. (37)

Explanation: (0.\overline{027}=\frac{27}{999}=\frac{1}{37}). An initial zero inside the repeating block is counted as a digit.

Correct answer: A. Terminating after (3) places

Explanation: Both (3^0) and (11^0) equal (1), so the effective denominator is (2^3\cdot 5^2). The larger exponent is (3), so the decimal terminates after (3) places.

Correct answer: B. (4)

Explanation: The factor (37) makes the decimal recurring, and the larger exponent of (2) and (5) is (4), giving the non-repeating start. In mixed denominators, the larger exponent gives the delay.

Correct answer: A. (\frac{3}{80})

Explanation: (0.0375=\frac{375}{10000}), and dividing by (125) gives (\frac{3}{80}). Convert the decimal to a fraction and reduce fully.

Correct answer: C. (0.58\overline{23})

Explanation: In (0.58\overline{23}), the block (23) repeats regularly, so it is rational. A fixed repeating block is a strong sign of rationality.

Correct answer: B. (\frac{38}{2^2\cdot 5^3\cdot 19})

Explanation: Since (38=2\cdot 19), the factor (19) cancels and the reduced denominator is (2\cdot 5^3). Even if an extra prime appears, check cancellation first.

Correct answer: A. Terminating

Explanation: (0.\overline{54}=\frac{54}{99}) and (0.\overline{45}=\frac{45}{99}), so their sum is (1). The sum of two recurring decimals can sometimes be terminating.

Correct answer: B. Non-terminating recurring with (2) initial non-repeating digits

Explanation: Since (9=3^2) remains, the decimal is non-terminating recurring. The larger exponent in (2^2\cdot 5^2) gives (2) initial non-repeating digits.

Correct answer: C. (q) has at least one prime other than (2) and (5)

Explanation: For a non-terminating recurring decimal, the reduced denominator has at least one prime factor other than (2) and (5). Factors (2) or (5) may also be present, but they are not enough alone.

Correct answer: B. Non-terminating recurring

Explanation: For a rational number, the decimal expansion terminates only when, in lowest terms, the denominator contains powers of 2 and 5 alone. Any remaining prime factor other than 2 or 5 makes the decimal expansion non-terminating recurring. Thus the important step is to cancel common factors completely before applying this rule.

Cancel the common factors in \(\frac{2^5\cdot7}{2^8\cdot5^2\cdot7^2}\). The result is \(\frac{1}{2^3\cdot5^2\cdot7}\), because \(2^5\) leaves \(2^3\) below and one factor 7 remains below. Since 7 is still a factor of the reduced denominator, the decimal is non-terminating recurring. Therefore option B is correct. It cannot terminate after three places, because the factor 7 prevents termination.

Correct answer: B. (16)

Explanation: (0.0625=\frac{625}{10000}=\frac{1}{16}). Convert a terminating decimal to a fraction and always reduce the denominator.

Correct answer: A. (\frac{1}{1600})

Explanation: (0.000625=\frac{625}{1000000}), and reducing by (625) gives (\frac{1}{1600}). Do not fear large denominators; cancel common factors.

Correct answer: C. (\frac{1}{112})

Explanation: (112=2^4\cdot 7), so (4) non-repeating digits appear before the recurring part. For comparison, check the larger power of (2) and (5).

Correct answer: B. (\max(r,s))

Explanation: To form (10^k=2^k5^k), both powers must reach at least the larger exponent. Therefore the minimum (k=\max(r,s)).

Correct answer: C. (\sqrt{11})

Explanation: (\sqrt{11}) is irrational, so its decimal is non-terminating non-recurring. Rational numbers are either terminating or non-terminating recurring.

Correct answer: B. 12

Explanation: Since \(10^6=2^6\cdot5^6\), compare this with the given denominator \(2^4\cdot5^6\): the factor \(2^2\) is missing. Multiply numerator and denominator by \(2^2=4\) to obtain denominator \(10^6\), so \(N=3\times4=12\). The distractor 24 would result from mistakenly multiplying by \(2^3=8\). Exam tip: to express a fraction with denominator \(10^n\), match the prime powers of 2 and 5 in the denominator to exponent \(n\).

Correct answer: B. (4) places

Explanation: Since (242=2\cdot 11^2), the reduced denominator becomes (2^2\cdot 5^4). The larger exponent is (4), so reduce first and then count decimal places.

Correct answer: C. (8) places

Explanation: The denominator has only (2) and (5), so the decimal terminates with the larger exponent (8). In exams, use the larger exponent, not the sum.

Correct answer: C. (5) places

Explanation: Since (198=2\cdot 3^2\cdot 11), the reduced denominator is (2\cdot 5^5). The larger exponent is (5), so reduce first and then count places.

Correct answer: C. (441)

Explanation: For termination, (3^2) and (7^2) must cancel completely, so (n=441). For the least value, cancel only the factors other than (2) and (5).