Decimal representation

( \frac{59}{256}=0.23046875 ), so (0.23046875-0.10546875=0.125). Convert the fraction into decimal and subtract.

In \(0.0747474\ldots\), the first digit after the decimal point is \(0\), and then the block \(74\) repeats: \(0,74,74,74,\ldots\). Hence, the bar is placed only over \(74\), giving \(0.0\overline{74}\). In \(0.\overline{074}\), the entire block \(074\) would repeat, producing a different decimal. Exam tip: identify the shortest repeating block before placing the bar.

In a recurring decimal, a fixed block of \(p\) digits repeats. Here, the gaps of zeros between 1s are 1, 2, 3, … and keep increasing, so no fixed period exists; hence the number is irrational. Exam tip: using only 0 and 1 does not make a number rational.

To make (2^{11}) into (10^{11}), we multiply by (5^{11}). So there will be at most (11) decimal places.

The denominator is \\(2^6\times5^9\\). A terminating decimal is formed by making the denominator a power of 10, so the powers of 2 and 5 must be made equal. There are six factors of 2 and nine factors of 5; therefore, multiply by \\(2^3\\) to obtain \\(2^9\times5^9=10^9\\). This requires at most nine decimal places.

The same conclusion follows from the standard rule: the maximum number of places is the larger exponent in the simplified denominator. Here, \\(\max(6,9)=9\\), so option B is correct. The value 15 would incorrectly add the exponents. Although a special numerator may cancel factors and produce fewer places, the maximum possible number remains nine.

The number of zeros between (6)'s increases, so there is no fixed repetition. Such a decimal is non-terminating non-recurring.

To convert a decimal into a percentage, multiply it by 100. Here, \(0.0064 \times 100 = 0.64\), so the percentage is \(0.64\%\). Option C results from shifting the decimal point one place too far. Exam tip: when multiplying a decimal by 100, move the decimal point two places to the right.

The direct answer is option C, 4882.8125. Multiplication by a power of 10 shifts the decimal point to the right. The number 10,000,000 has seven zeros, so the decimal point moves seven places right in 0.00048828125. Count carefully: after seven places the number becomes 4882.8125. Another safe method is to write 0.00048828125 × 10,000,000 and cancel the decimal shift: 0.00048828125 × 10 = 0.0048828125, ×100 = 0.048828125, ×1000 = 0.48828125, and continuing to ×10,000,000 gives 4882.8125. Option A, 48.828125, represents a shift that is too short by two places. Option B, 488.28125, is also too small because the decimal has not moved far enough. Option C, 4882.8125, has the correct seven-place shift. Option D, 48828.125, is too large because the decimal has been moved one place too far. The useful exam cue is: multiplying by 10, 100, or 1000 moves the decimal right by 1, 2, or 3 places respectively; count the zeros.

( \frac{47}{128}=0.3671875 ), so (0.3675) is greater than (0.3671875) and less than (0.368). Convert boundary values into decimals.

The direct answer is option C, 1/32000. The decimal has eight digits after the decimal point, so 0.00003125 = 3125/100000000. Now simplify by dividing numerator and denominator by 3125: 3125/3125 = 1, and 100000000/3125 = 32000. Hence the fraction in lowest terms is 1/32000. Option A, 1/3200, is ten times too large. Option B, 1/3125, is also larger and does not equal the decimal. Option C is correct because it is the exact simplified fraction. Option D, 3125/1000000, equals 1/320, not the given small decimal; it also uses the wrong denominator for the number of decimal places. Memory cue: eight decimal places mean denominator 10^8 before cancellation.

The denominator has (29), a factor other than (2) and (5). So the rational number will have a non-terminating recurring decimal.

The first three digits after the decimal point, 000, do not repeat. After them, the block 91 repeats: 0.000 91 91 91... Hence, the bar must be placed only over 91, giving \(0.000\overline{91}\). In \(0.00\overline{091}\), the block 091 is treated as repeating, which does not match the given decimal. Exam tip: identify the smallest repeating block before placing the bar.

For \(\frac{21}{160}\), the denominator is \(160=2^5\times5\). A rational number has a terminating decimal only when its simplified denominator contains only factors \(2\) and/or \(5\). The denominators \(66,90,84\) also contain \(3\) or \(7\). Exam tip: simplify first, then factorise the denominator.

\(\frac{13}{40}\) is already in lowest form, and \(40=2^3\times5\). A rational number has a terminating decimal expansion when the prime factors of its denominator in lowest form are only 2 and/or 5. Therefore, \(\frac{13}{40}=0.325\). Option B is incorrect because 40 has no prime factor other than 2 and 5, so the decimal does not recur. Exam tip: first reduce the fraction to lowest terms, then factorise its denominator.

In lowest form, \(p/q\) terminates only when \(q=2^m5^n\). Since \(600=2^3\times3\times5^2\) contains 3, the decimal is non-terminating recurring, not irrational. Exam tip: first check whether the fraction is in lowest form.

The denominator is \(2^2\times5^3=500\). Converting it to a power-of-10 denominator gives \(\frac{7}{500}=\frac{14}{1000}=0.014\). Thus, there are three digits after the decimal point, so the expansion terminates after 3 decimal places. Option 2 is incorrect because \(0.014\) has three, not two, digits after the decimal point. Exam tip: if a denominator has only factors 2 and 5, the larger of their exponents gives the required number of decimal places.

\(\frac{1}{7}=0.142857142857\ldots\), in which the block 142857 repeats indefinitely. Hence, its decimal expansion is non-terminating recurring. A decimal terminates only when, in lowest form, the denominator has prime factors 2 and 5 only; here the denominator is 7. A non-terminating non-recurring decimal represents an irrational number, whereas \(\frac{1}{7}\) is rational. Exam tip: First reduce a fraction to lowest terms, then inspect the prime factors of its denominator.

\(\frac{13}{375}\) is already in lowest form, and \(375=3\times5^3\). Since the denominator also contains 3, its decimal expansion is non-terminating recurring. Exam tip: after simplification, check for only 2 and 5.

The denominator is \(2^4\times5^2\). To write it with a power of 10, multiply the numerator and denominator by \(5^2\): \(\frac{3\times5^2}{10^4}=\frac{75}{10000}=0.0075\). Hence, there are 4 digits after the decimal point. Option 2 results from considering only the power of 5, but the larger exponent, 4, is needed to form \(10^4\). Exam tip: for a denominator of the form \(2^m\times5^n\), the number of decimal places is \(\max(m,n)\).

A rational number terminates only when its denominator in lowest form has prime factors 2 and 5 only. Since \(160=2^5\times5\), \(\frac{21}{160}\) terminates. Exam tip: a remaining factor such as 3, 7, or 11 makes the decimal non-terminating.

\(\frac{7}{11}=0.\overline{63}\) is non-terminating but repeating, and it is rational because it is a ratio of integers. \(\sqrt{2}\) is non-repeating. Exam tip: every repeating decimal is rational.

In lowest form, a rational number terminates only when q=2^m×5^n. For example, 40=2^3×5. Merely being even is not sufficient; in exams, first reduce the fraction to lowest terms.

Since \(75=3\times5^2\), the denominator of \(\frac{7}{75}\) contains 3, so its decimal expansion is non-terminating recurring. In contrast, \(40=2^3\times5\). Exam tip: a reduced fraction terminates only when its denominator has factors 2 and/or 5 only.

\(\frac{5}{6}\) is in simplest form and \(6=2\times3\). A rational number has a terminating decimal expansion only if the prime factors of its denominator, in simplest form, are only \(2\) and/or \(5\). Since \(3\) is also a factor, \(\frac{5}{6}=0.8333\ldots\) is non-terminating recurring. An even denominator is not sufficient; although \(6\) is even, it contains the factor \(3\). Exam tip: first reduce the fraction to lowest terms and then factorise its denominator.

Let \(x=0.\overline{6}\). Since 6 repeats indefinitely, \(10x=6.\overline{6}\). Subtracting the original equation gives \(10x-x=6.\overline{6}-0.\overline{6}\), so \(9x=6\) and \(x=\frac{6}{9}=\frac{2}{3}\). Therefore, option B is correct. Option C, \(\frac{3}{5}\), equals the terminating decimal \(0.6\), not the recurring decimal \(0.\overline{6}\). Exam tip: a one-digit recurring decimal \(0.\overline{a}\) can be written as \(\frac{a}{9}\) and then reduced.