Decimal representation

For \(\frac{9}{32}\), the denominator is \(32=2^5\). In lowest form, a rational number has a terminating decimal only when its denominator has prime factors \(2\) and/or \(5\) only. Hence, \(\frac{9}{32}=0.28125\) terminates. The denominators of \(\frac{5}{18}\) and \(\frac{11}{27}\) contain the factor \(3\), and \(\frac{7}{21}=\frac{1}{3}\) also has a factor \(3\) in its reduced denominator, so they are non-terminating recurring decimals. Exam tip: Reduce the fraction first, then factorise its denominator.

In 0.6666\ldots, the digit 6 repeats forever in a fixed pattern, so it is a recurring decimal. In 0.121221222\ldots and 0.1010010001\ldots, the lengths of the digit groups keep changing, while in 0.123456789\ldots the digits keep increasing; none has a fixed repeating block. Exam tip: in a recurring decimal, a digit or a fixed group of digits repeats indefinitely.

\(\frac{1}{3}=0.333\ldots\), where the digit 3 repeats forever. Hence, its decimal expansion is non-terminating recurring. In contrast, \(\frac{1}{8}=0.125\), \(\frac{5}{2}=2.5\), and \(\frac{7}{20}=0.35\) are terminating decimals. Exam tip: In lowest form, if a denominator has a prime factor other than 2 or 5, the decimal expansion is non-terminating recurring.

For \(\frac{13}{40}\), the denominator is \(40=2^3\times5\). A fraction in lowest terms has a terminating decimal only when its denominator has no prime factors other than 2 and 5. Hence, \(\frac{13}{40}=0.325\) is terminating. In contrast, \(\frac{7}{15}\) has 3 in its denominator, and \(\frac{5}{21}\) has 3 and 7, so their decimals are non-terminating recurring. Exam tip: First reduce the fraction to lowest terms, then factorise the denominator.

Since \(125=5^3\), the denominator of \(\frac{11}{125}\) in lowest form has only the prime factor \(5\). Therefore, its decimal expansion terminates. In fact, \(\frac{11}{125}=\frac{88}{1000}=0.088\). A non-terminating recurring decimal occurs when the denominator has a prime factor other than \(2\) or \(5\). Exam tip: first reduce the fraction, then check the prime factors of its denominator.

The correct answer is \(\frac{5}{12}\). A fraction in lowest terms has a terminating decimal only when the prime factors of its denominator are 2 and/or 5. Here, \(12=2^2\times3\), and the factor 3 is present, so its decimal expansion is non-terminating recurring. In contrast, 20, 25, and 125 have only 2 and/or 5 as prime factors. Exam tip: First reduce the fraction to lowest terms, then check the denominator's prime factors.

For \(\frac{7}{12}\), the denominator has prime factorisation \(12=2^2\times3\). In lowest form, a fraction has a non-terminating recurring decimal when its denominator contains a prime factor other than 2 or 5. Hence, \(\frac{7}{12}=0.58\overline{3}\). In contrast, 16, 40, and 125 have only 2 and/or 5 as prime factors, so their decimal expansions terminate. Exam tip: first reduce the fraction, then check the prime factors of its denominator.

\(\frac{11}{50}=0.22\), so its decimal expansion terminates. A rational number \(\frac{p}{q}\) has a terminating decimal expansion only when, in lowest terms, the prime factors of \(q\) are only 2 and/or 5. Here, \(50=2\times5^2\). In contrast, \(24\), \(18\), and \(27\) contain 3 as a prime factor, so their decimal expansions are non-terminating recurring. Exam tip: First reduce the fraction, then check the prime factors of its denominator.

For \(\frac{13}{15}\), the prime factorisation of the denominator is \(15=3\times5\). A rational number has a terminating decimal only when, in lowest terms, its denominator has no prime factors other than \(2\) and \(5\). Since the denominator here contains \(3\), its decimal expansion is non-terminating recurring. The denominators in the other options contain only factors of \(2\) and/or \(5\). Exam tip: First reduce the fraction to lowest terms, then check the prime factors of its denominator.

(2500\times4=10000), so ( \frac{109}{2500}=\frac{436}{10000}=0.0436 ). Making the denominator a power of (10) is a fast method.

(0.00078125=\frac{78125}{100000000}=\frac{1}{1280}). Counting decimal places and simplifying the fraction is important.

The denominator has (17), a factor other than (2) and (5). So the rational number will have a non-terminating recurring decimal.

After 45, the block 07 repeats as 07, 07, 07, ... . Therefore, the bar must be placed only over the recurring block 07: \(0.45\overline{07}\). In option B, only 7 is treated as repeating, but 0 also occurs before every repeating 7. Exam tip: group the decimal digits and identify the shortest block that repeats.

The direct answer is option A, 3.205181818… . A bar over digits means exactly those digits repeat forever. The non-repeating part is 205, and the repeating block is 18. Therefore write 3.205 followed by 18 repeatedly: 3.20518181818… . Option A is correct. Option B, 3.20518518…, does not repeat the block 18 continuously. Option C, 3.20518, stops after one occurrence and is not the stated recurring decimal. Option D, 3.20520518…, repeats or inserts the wrong digits before the recurring block. Do not treat the bar as covering 205; it covers only 18.

Direct answer: option A, 0.690476190476... . Divide 29 by 42. Since 29<42, write 0 and a decimal point, then use 290: 42 goes 6 times, remainder 38. Bring down 0 to get 380: 42 goes 9 times, remainder 2. Bring down 0 to get 20: digit 0, remainder 20. Bring down 0 to get 200: digit 4, remainder 32. Bring down 0 to get 320: digit 7, remainder 26. Bring down 0 to get 260: digit 6, remainder 8. Bring down 0 to get 80: digit 1, remainder 38. The remainder 38 has appeared before, so the block 904761... repeats after the initial 6, giving 0.690476190476... . Option A matches this division. Option B, 0.692929..., has different digits. Option C is too large; option D is far too small and does not represent 29/42. A repeated remainder is the reason the decimal is recurring.

( \frac{13}{17}=0.764705\ldots ), which is less than (0.7648). View all numbers in decimal form for comparison.

The tenths and hundredths digits are equal, so the thousandths digit must satisfy (a>8). So the smallest value is (9).

The number of zeros between (8)'s keeps changing, so there is no fixed recurring part. Such a decimal is non-terminating non-recurring.

(512) is a power of (2), so the decimal terminates and ( \frac{41}{512}=0.080078125 ). The power-of-(10) method also works for large denominators.

(0.00140625=\frac{140625}{100000000}=\frac{9}{6400}). Always write the final answer in simplest form.

(84=2^2\times3\times7) has (3) and (7). Such a rational number in simplest form gives a non-terminating recurring decimal.

The first (3) is in tenths and the last (3) is in millionths. Zero-value places need not be written separately.

Multiplying both numbers by (10000) gives (5390\div7=770). Removing decimals makes division easier.

Before subtracting, align the decimal points and write 9.002 as 9.00200. Then \(9.00200-5.18725=3.81475\), so the correct answer is \(3.81475\). An option such as \(3.82475\) results from an error in aligning decimal places or borrowing. Exam tip: Add zeros at the end when needed before subtracting decimals.

Writing equal decimal places gives (0.14062500+0.00781250+0.00015625=0.14859375). Align decimal points while adding.