Decimal expansion of rational numbers

The direct answer is B, non-terminating recurring. Factor the numerator: \\(55=5\\cdot11\\). The denominator is \\(2^2\\cdot5^3\\cdot11^2\\). Cancel one 5 and one 11. The reduced denominator becomes \\(2^2\\cdot5^2\\cdot11\\). A reduced denominator permits a terminating decimal only if all its prime factors are 2 or 5. Since 11 remains, termination is impossible. This is still a rational number, so its decimal is non-terminating but repeating. Option A, terminating, ignores the remaining 11. Option B is correct. Option C, non-terminating non-recurring, is associated with irrational numbers, whereas this is a quotient of integers. Option D, terminating after two places, is unsupported and also contradicted by 11. Exam cue: cancel common factors first; one leftover prime other than 2 or 5 is enough to decide.

For exactly (8) places, the larger exponent of (2) and (5) must be (8). Only (2^8\cdot 5^3) satisfies this.

A rational number in lowest terms has a terminating decimal iff its denominator's prime factors are only 2 and/or 5. Here 31 is coprime to 2 and 5, so the denominator remains \(2^a5^b\). The number of decimal places for termination equals \(\max(a,b)\). Given that this equals 10 and that \(b>a\), the larger exponent must be \(b\), so \(b=10\). The distractor \(a+10\) is incorrect because the termination length is not a sum involving \(a\) but the maximum of the exponents. Exam tip: always reduce the fraction first and use "termination length = max(exponents of 2 and 5 in denominator" ).

(0.00084=\frac{84}{100000}), and reducing by (4) gives (\frac{21}{25000}). Even for small decimals, check the greatest common factor carefully.

The factor (17) makes the decimal recurring, and the larger exponent among (2) and (5) is (6), giving the initial non-repeating part. Understand recurrence and delay separately.

Since (169=13^2), the reduced denominator is (2^3\cdot 5^4). Therefore the reason correctly explains the terminating decimal rule.

The direct answer is C, 7 decimal places. Cancel the common factor \\(3^4\\) from numerator and denominator. For powers of 5, \\(5^2/5^5=1/5^3\\). Therefore the reduced fraction has denominator \\(2^7\\cdot5^3\\). It contains only 2 and 5, so the decimal terminates. To express the denominator as a power of 10, the larger exponent, 7, determines the number of places; the three factors of 5 are paired with three of the seven factors of 2, and the remaining four factors of 2 are supplied with 5s. Option A, 3, considers only the power of 5. Option B, 5, is neither the required maximum nor the result of the cancellation. Option C, 7, is correct. Option D, non-terminating, is wrong because no prime factor other than 2 and 5 remains. Reduce first, then compare exponents.

Since (q) has only (2) and (5), the decimal terminates. Not dividing (10^8) means the larger exponent is (9) or (10).

Two non-repeating zeros and two repeating digits give (\frac{54}{9900}). Reducing it gives (\frac{3}{550}).

The direct answer is B, 5 decimal places. Cancel powers with the same base: \\(2^5/2^{10}=1/2^5\\) and \\(5^2/5^6=1/5^4\\). The reduced denominator is \\(2^5\\cdot5^4\\). Since it contains only 2 and 5, the decimal terminates. The larger exponent is 5, so five decimal places are sufficient and determine the terminating length in this question. Option A, 4, uses only the exponent of 5 and misses one factor of 2. Option B, 5, is correct. Option C, 6, incorrectly adds or otherwise misreads the exponents. Option D, 10, uses the original denominator exponent of 2 without cancellation. Do not count all original powers; first subtract exponents when the same bases occur in numerator and denominator. Memory cue: reduce, check only 2 and 5, then take the larger exponent.

The answer is B, 0.47. Infinitely repeating 9s make the decimal equal to the next terminating decimal: 0.46999… = 0.47000… = 0.47. To reason carefully, 0.46 is followed by 0.00999…, and that repeating part equals 0.01; therefore 0.46 + 0.01 = 0.47. A, 0.46, ignores the positive repeating tail and is too small. B is correct because it is the exact equivalent value. C, 469/999, produces a different repeating pattern, 0.469469…, not 0.46999…. D, 4699/10000, equals 0.4699 and stops before reaching 0.47. Exam cue: a final infinite string of 9s causes carrying, not ordinary rounding based on a later unknown digit.

The factors (3^4) and (19) must be removed from the reduced denominator, so the minimum factor is (81\cdot 19=1539). Factors (2) and (5) may remain.

The direct answer is option C: non-terminating recurring. To decide the decimal type, first reduce the fraction by cancelling common factors. Since 245 = 5 × 7², the numerator cancels one factor 5 and two factors 7 from the denominator 2² × 5² × 7³. The reduced denominator is therefore 2² × 5 × 7. A rational number has a terminating decimal only when, after reduction, its denominator has no prime factors except 2 and 5. Here the factor 7 remains, so the decimal cannot terminate. Because the number is rational, its decimal digits must eventually repeat; hence it is non-terminating recurring. Option A is wrong because a factor 7 remains, so it does not end after two places. Option B is wrong for the same reason; it does not end after three places. Option C is correct because the decimal continues and repeats. Option D is wrong because every rational number has either a terminating or a recurring decimal, never a non-recurring one. Memory cue: reduced denominator containing only 2s and 5s means terminating; any other prime factor means recurring.

This decimal does not end, and the number of zeros keeps changing. Since there is no fixed repeating block, it is non-terminating non-recurring.

Reason: \(10^8=2^8\cdot5^8\). The given denominator has \(5^8\) but only \(2^4\). To convert the denominator to \(2^8\cdot5^8\) multiply numerator and denominator by \(2^4=16\). Thus \(N=37\times16=592\). About distractors: C (1184) corresponds to multiplying by 32 (\(2^5\)) mistakenly; A (296) is half of the correct value (\(37\times8\)) and often comes from dividing instead of multiplying; D (23125) would result from incorrectly multiplying by \(5^4=625\) (\(37\times625\)). Exam tip: factorise \(10^n\) as \(2^n\cdot5^n\) and balance the powers of 2 and 5 to find the factor for the numerator quickly.

The reduced denominator is (10^7), so the decimal terminates exactly after (7) places. If the denominator is reduced, do not assume further cancellation.

Since (243=3^5), the reduced denominator is (2^5\cdot 5^4). The larger exponent is (5), so the decimal terminates after (5) places.

At most (9) decimal places means the fraction can be written with denominator (10^9). Therefore the reduced denominator must divide (10^9).

(0.\overline{063}=\frac{63}{999}), and reducing by (9) gives (\frac{7}{111}). An initial zero inside the repeating block is also counted as a digit.

Both (7^0) and (19^0) equal (1), so the effective denominator is (2^5\cdot 5^2). The larger exponent is (5), so the decimal terminates after (5) places.

The factor (41) makes the decimal recurring, and the larger exponent of (2) and (5) is (7), giving the non-repeating start. In mixed denominators, the larger exponent gives the delay.

(0.046875=\frac{46875}{1000000}), and reducing gives (\frac{3}{64}). Convert the decimal to a fraction and reduce fully.

The direct answer is option B: non-terminating recurring. First factor the numerator: 320 = 2⁶ × 5. The denominator is 2⁷ × 5³ × 11. Cancelling 2⁶ and one factor 5 leaves 2 × 5² × 11 in the denominator. For a rational fraction in lowest terms, a terminating decimal is possible only if the denominator contains no primes other than 2 and 5. The factor 11 remains, so the decimal does not terminate. Since the fraction is rational, its continuing decimal digits must repeat in a cycle, so it is non-terminating recurring. Option A is wrong because the denominator is not made only of 2s and 5s. Option B is correct because 11 remains after reduction. Option C is wrong because a rational number cannot have a non-terminating non-recurring decimal. Option D is wrong because the presence of 11 prevents termination after one place, and in fact prevents termination at any finite place. Memory cue: always cancel first, then inspect the remaining denominator; 2 and 5 only means terminating.

Since (62=2\cdot 31), the factor (31) cancels and the reduced denominator is (2^3\cdot 5^3). If an extra prime appears, check cancellation first.

(0.\overline{36}=\frac{36}{99}) and (0.\overline{63}=\frac{63}{99}), so their sum is (1). The sum of two recurring decimals can be terminating.