Decimal representation

We need to divide 313 by 128000. Since 128000 = 128 × 1000 and 313 ÷ 128 = 2.4453125, division by the additional factor 1000 gives 0.0024453125. The same result follows by long division or by writing 313/128000 and shifting the decimal appropriately. Thus option B is correct. Option A is ten times too large, which corresponds to placing the decimal one position too far to the right. Option C is ten times too small, with one extra zero after the decimal. Option D does not represent the given fraction at all. The denominator has only factors 2 and 5, so a terminating decimal is expected.

(0.00003125=\frac{3125}{100000000}=\frac{1}{32000}). Count decimal places and simplify the fraction.

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.

A fraction in lowest terms has a terminating decimal exactly when the denominator contains no prime factors other than 2 and 5. To express q = 2⁹ × 5¹² as a power of 10 multiplied by a remaining factor, pair nine 2s with nine 5s to form 10⁹. Five 5s remain, so the denominator can be converted to 10¹² by supplying three additional factors of 2 in the numerator’s decimal scaling. In general, the required number of places is the larger exponent, max(9, 12) = 12. Therefore option D is correct. Cancellation in p may reduce the actual number, but 12 is the maximum possible.

The governing rule is that a rational number has a decimal expansion that terminates or eventually repeats a fixed block of digits. In option C, the block 45 repeats forever, so 0.\overline{45} = 45/99 = 5/11, which is a ratio of integers and therefore rational. The other displayed decimals are described as non-terminating and non-repeating. Such decimal expansions cannot be written as p/q and are irrational. The important test is not merely whether the decimal continues indefinitely: some endless decimals are rational when they repeat, while non-repeating endless decimals are irrational. Therefore option C is the only valid answer.

A non-terminating, non-recurring decimal continues without end and never settles into a fixed repeating cycle. Option B, 0.4040040004…, has an increasing and changing number of zeros between successive 4s, so no fixed block repeats. It is therefore non-terminating and non-recurring, representing an irrational number. Option A repeats the block 27 and is rational. Option C terminates and equals 1/8. Option D repeats 8 and equals 8/9, so it is also rational. The correct test is to inspect both features—whether the expansion ends and whether its digits eventually repeat. Only option B satisfies both required conditions.

The governing rule is that a rational number p/q, written in lowest terms, has a terminating decimal expansion exactly when the prime factors of q are only 2 and/or 5. Option D is 7/16, and 16 = 2⁴, so its decimal expansion terminates: 7/16 = 0.4375. Options A, B, and C each simplify formally to 1, provided their denominators are non-zero, so they also have terminating values; however, as written, they contain irrational or transcendental expressions rather than a direct rational fraction. Therefore this question is ambiguous if all expressions are treated algebraically. Among the listed direct fraction forms, D is the intended and clearly justified answer, because its denominator visibly satisfies the terminating-decimal criterion.

The governing concept is the decimal representation of rational and irrational numbers. A terminating decimal or a non-terminating decimal with a repeating block is rational, because it can be expressed as a fraction of integers. Therefore, 2.47000... terminates, while 2.474747... and 2.471471471... repeat definite blocks and are also rational. In option D, the digits continue indefinitely without one fixed block repeating regularly. Hence it represents an irrational number. The ellipsis alone does not make a number irrational; the essential test is whether the decimal terminates or eventually repeats.

\(\frac{13}{40}\) 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.