Decimal representation

To find the multiple, divide the larger decimal by the smaller one: \(0.015625 \div 0.00015625 = 100\). Therefore, \(0.015625\) is 100 times \(0.00015625\). The ratio is also evident because the decimal point shifts two places to the right. Exam tip: For a ‘how many times’ question, divide the larger quantity by the smaller quantity.

The bar is placed only over 81. Thus, the first three digits after the decimal point, 000, do not repeat, and 81 repeats thereafter. Hence, \(0.000\overline{81}=0.000818181\ldots\). Option B has an extra zero, while option C incorrectly makes 810 the repeating block. Exam tip: only the digits under the bar repeat; digits before the bar are written once.

\(\frac{33}{154}=\frac{3}{14}\), and \(14=2\times7\). Its reduced denominator has a prime factor other than 2 or 5, so the decimal is non-terminating recurring. Exam tip: simplify the fraction before checking denominator factors.

The fraction must be simplified before its decimal type is determined. Both 99 and 363 are divisible by 33, so 99/363 = 3/11. The denominator in lowest form is consequently 11.

For a rational fraction in simplest form, a terminating decimal is possible only when the denominator has prime factors 2 and/or 5. Since 11 is a different prime factor, the decimal does not end. Dividing gives 3/11 = 0.2727..., where the digits 27 repeat forever. Therefore the expansion is non-terminating recurring, which is option C. This also shows why checking the denominator after cancellation is essential; the original denominator alone may hide the simplest structure.

In 17.374999\ldots, infinitely many 9s occur after 4. Since \(0.009999\ldots=0.010\), we get \(17.374999\ldots=17.375\). Option 17.374 is only an approximation, not an equal value. Exam tip: when infinitely many 9s follow a digit, increase that digit by 1 and remove the repeating 9s.

After the decimal point, the digits 0, 8, 0 and 5 occupy the tenths, hundredths, thousandths and ten-thousandths places, respectively. Therefore, the place value of 5 is \(\frac{5}{10000}\). Exam tip: each successive place to the right of the decimal represents one-tenth of the preceding place.

The correct answer is option C: \(0.703\overline{1}<0.703125=\frac{45}{64}\). First convert the fraction: \(\frac{45}{64}=0.703125\), because dividing 45 by 64 gives that terminating decimal. The bar in \(0.703\overline{1}\) covers only 1, so its value is \(0.7031111\ldots\). Compare the decimals from left to right: 0.703 is common, and at the next places 1 is less than 1? More precisely, after 0.703, the repeating number has 1 in the fourth decimal place, while 0.703125 has 1 there too; then compare the next digits: 1 is less than 2, so the repeating number is smaller. Option A is wrong because it places the repeating decimal as larger. Option B is wrong because it says the fraction equals the repeating decimal, but their decimal expansions differ. Option C is correct because the fraction and terminating decimal are equal and both exceed the repeating value. Option D is wrong because all three are not equal. Memory cue: a terminating decimal can be compared by writing enough repeating digits.

The correct order is \(b>a>c\), since \(0.690=0.6900\), \(0.609=0.6090\), and \(c=0.6009\). All three have 6 in the tenths place. In the hundredths place, \(b\) has 9, so it is the greatest. For \(a\) and \(c\), the hundredths digit is 0 in both, but the thousandths digit is 9 for \(a\) and 0 for \(c\); hence \(a>c\). The order \(b>c>a\) incorrectly places \(c\) above \(a\). Exam tip: add trailing zeros to compare decimals to the same number of places.

( \frac{31}{250}=0.124 ) and ( \frac{17}{640}=0.0265625 ), so the sum is (0.1505625). First convert the fractions into decimals.

( \frac{37}{128}=0.2890625 ), so (0.2890625-0.1640625=0.125). Convert the fraction into decimal and subtract.

The first digit after the decimal point, 0, occurs only once. After it, the block 96 repeats as 96, 96, 96, … . Therefore, the bar is placed only over 96: \(0.0\overline{96}\). In \(0.\overline{96}\), 0 would also repeat, while \(0.09\overline{6}\) makes only 6 recurring. Exam tip: write out a few decimal digits and identify the shortest repeating block before placing the bar.

\(\frac{21}{30}=\frac{7}{10}=0.7\), so its decimal expansion terminates. The prime-factor test is applied only after reducing the fraction; in \(\frac{7}{30}\), factor 3 remains. Exam tip: simplify first.

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

The denominator \\(2^5\times5^8\\) contains only the prime factors 2 and 5, so the fraction can have a terminating decimal after simplification. To convert the denominator into a power of 10, the five factors of 2 are paired with five factors of 5. Three factors of 5 remain, and supplying three factors of 2 gives \\(2^8\times5^8=10^8\\).

Therefore the greatest possible number of decimal places is eight, which is option B. The rule is to use the larger exponent, \\(\max(5,8)=8\\), rather than adding the exponents to get 13. Cancellation involving the numerator can make the decimal shorter, but it cannot require more than eight places under the stated simplified denominator.

The number of zeros between (8)'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: \(0.0048 \times 100 = 0.48\). Therefore, the correct answer is 0.48%. The closest distractor, 4.8%, would result from converting 0.048 into a percentage, not 0.0048. Exam tip: Move the decimal point two places to the right when converting a decimal to a percentage.

The direct answer is option C, 9765.625. Since 10000000 = 10^7, multiplying by it shifts the decimal point seven places to the right. Starting with 0.0009765625, the shift gives 9765.625, so the product is 0.0009765625 × 10000000 = 9765.625. Option A, 97.65625, shifts the decimal only five places. Option B, 976.5625, shifts it six places. Option C shifts it seven places and is correct. Option D, 97656.25, shifts it eight places and is ten times too large. A safe checking method is to estimate: a number close to 0.001 multiplied by ten million should be close to 10000, so 9765.625 is reasonable, while the smaller and larger choices are not.

( \frac{19}{32}=0.59375 ), so (0.59325) is greater than (0.593) and less than (0.59375). Convert boundary values into decimals.

In lowest terms, a denominator containing only 2 and 5 can be converted into a factor of \(10^n\), so the decimal terminates. Any other prime factor gives a recurring decimal. Exam tip: reduce the fraction first before checking its denominator.

(64000\times15625=1000000000), so ( \frac{257}{64000}=0.004015625 ). Convert the denominator into a power of (10) to get the exact decimal.

(0.000234375=\frac{234375}{1000000000}=\frac{3}{12800}). Count decimal places and write the fraction in simplest form.

The tenths digits are equal and the hundredths digit must satisfy (a<3). So (a=0,1,2), giving (3) possible values.

(5120\times1953125=10000000000), so ( \frac{191}{5120}=0.0373046875 ). Converting a large denominator into a power of (10) is an exact method.

(0.00025390625=\frac{25390625}{100000000000}=\frac{13}{51200}). Count decimal places and write the fraction in simplest form.

The denominator has (23), a factor other than (2) and (5). So the rational number will have a non-terminating recurring decimal.