Free math worksheets

- Name the repeating decimal with a variable. - Shift the decimal point by one place so the repeating tails align. - Subtract the aligned equations.

1. Let \(x=0.\overline{7}\). 2. Then \(10x=7.\overline{7}\). 3. Subtract: \(10x-x=7\), so \(9x=7\). 4. Therefore, \(x=\frac{7}{9}\).

\(0.\overline{7}=\frac{7}{9}\)

- Count the digits in the repeating block. - Use a power of ten that shifts one complete block. - Check whether the resulting numerator and denominator share a factor.

1. Let \(x=0.\overline{34}\). 2. Since the period has \(2\) digits, \(100x=34.\overline{34}\). 3. Subtracting gives \(100x-x=34\), so \(99x=34\). 4. Thus, \(x=\frac{34}{99}\), which is already in lowest terms.

\(0.\overline{34}=\frac{34}{99}\)

- Treat every position in the repeating block as significant, including the zero. - Shift by the full length of the period. - Align the repeating tails before subtracting.

1. Let \(x=0.\overline{052}\). 2. The period has \(3\) digits, so \(1000x=52.\overline{052}\). 3. Subtracting gives \(999x=52\). 4. Therefore, \(x=\frac{52}{999}\), which is in lowest terms.

\(0.\overline{052}=\frac{52}{999}\)

- The repeating tail must line up after shifting. - Subtract the original decimal from the shifted one. - Reduce the resulting improper fraction.

1. Let \(x=2.\overline{6}\). 2. Then \(10x=26.\overline{6}\). 3. Subtract: \(10x-x=24\), so \(9x=24\). 4. Therefore, \(x=\frac{24}{9}=\frac{8}{3}\).

\(2.\overline{6}=\frac{8}{3}\)

- Temporarily work with the positive magnitude. - Shift by the full repeating block. - Apply the sign after reducing the fraction.

1. First convert the positive decimal. Let \(y=0.\overline{18}\). 2. Then \(100y=18.\overline{18}\), so \(99y=18\). 3. Thus, \(y=\frac{18}{99}=\frac{2}{11}\). 4. Restore the negative sign: \(-0.\overline{18}=-\frac{2}{11}\).

\(-0.\overline{18}=-\frac{2}{11}\)

- Count all digits in the repeated block. - Choose the shift that returns the same block to the same positions. - Verify that subtraction would cancel every repeating digit.

1. The period has \(3\) digits, so the correct equation is C: \(1000x=305.\overline{305}\). 2. Subtracting the original equation gives \(999x=305\). 3. Therefore, \(x=\frac{305}{999}\), which is in lowest terms.

Equation C; \(0.\overline{305}=\frac{305}{999}\)

- Convert the decimal before comparing numerators. - Reduce the fraction to a denominator of \(11\). - Match the reduced form to the given expression.

1. Let \(x=0.\overline{54}\). 2. Then \(100x=54.\overline{54}\), so \(99x=54\). 3. Thus, \(x=\frac{54}{99}=\frac{6}{11}\). 4. Therefore, \(n=6\).

\(n=6\)

- Apply the same variable method used for any one-digit period. - Subtract the two aligned equations. - Interpret the solved value as an exact equality.

1. Let \(x=0.\overline{9}\). 2. Then \(10x=9.\overline{9}\). 3. Subtracting gives \(9x=9\). 4. Thus, \(x=1\), so \(0.\overline{9}=1\) exactly.

\(0.\overline{9}=1\)

- Shift past the nonrepeating zero. - Use the exact value of a repeating string of nines. - Express the result in both fraction and decimal form.

1. Let \(x=0.0\overline{9}\). 2. Then \(10x=0.\overline{9}=1\). 3. Therefore, \(x=\frac{1}{10}\). 4. Hence, \(0.0\overline{9}=0.1\) exactly.

\(0.0\overline{9}=\frac{1}{10}=0.1\)

- Convert the repeating representation before comparing. - Reduce the fraction obtained from the subtraction method. - Different representations can name the same number.

1. Let \(x=0.\overline{81}\). 2. Then \(100x=81.\overline{81}\), so \(99x=81\). 3. Thus, \(x=\frac{81}{99}=\frac{9}{11}\). 4. The two expressions represent the same rational number.

The claim is false. \(0.\overline{81}=\frac{9}{11}\).

- Convert both repeating decimals separately. - Notice whether the resulting denominators match. - Perform the exact subtraction after conversion.

1. Convert \(0.\overline{7}=\frac{7}{9}\). 2. Convert \(0.\overline{2}=\frac{2}{9}\). 3. Subtract: \(\frac{7}{9}-\frac{2}{9}=\frac{5}{9}\).

\(\frac{5}{9}\)

- Form the unreduced block-over-nines fraction. - Look for a common factor shared with \(99\). - Explain the denominator change as part of reducing an equivalent fraction.

1. A two-digit period gives \(0.\overline{48}=\frac{48}{99}\). 2. The numerator and denominator share a factor of \(3\). 3. Dividing both by \(3\) gives \(\frac{48}{99}=\frac{16}{33}\). 4. No further common factor remains.

\(0.\overline{48}=\frac{16}{33}\). The denominator becomes \(33\) because both \(48\) and \(99\) are divided by \(3\).

- What distinguishes a repeating decimal from a nonrepeating decimal? - To convert a repeating block of two digits, multiply by \(100\) and subtract the original number. - A pattern governed by a rule is not necessarily periodic.

1. The decimal for \(x\) repeats: \(x = 1.\overline{23}\). Therefore, \(x\) is rational. 2. The decimal for \(y\) is nonterminating and nonrepeating because the number of consecutive \(2\)s keeps increasing. Therefore, \(y\) is irrational. 3. Let \(x = 1.\overline{23}\). Then \(100x = 123.\overline{23}\). Subtracting gives \(99x = 122\), so \(x = \frac{122}{99}\). The numerator and denominator have no common factor, so the fraction is in lowest terms.

1. \(x\) is rational, and \(y\) is irrational. 2. The decimal for \(x\) repeats, while the decimal for \(y\) is nonterminating and nonrepeating. 3. \(x = \frac{122}{99}\)

- Separate the nonrepeating part from the repeating part. - Create two shifted equations whose repeating tails match. - Reduce after solving for the variable.

1. Let \(x=0.1\overline{6}\). 2. Shift past the nonrepeating digit: \(10x=1.\overline{6}\). 3. Shift one more place: \(100x=16.\overline{6}\). 4. Subtract: \(100x-10x=15\), so \(90x=15\). 5. Thus, \(x=\frac{15}{90}=\frac{1}{6}\).

\(0.1\overline{6}=\frac{1}{6}\)

- Count the nonrepeating digits before the bar. - Build two equations that differ by one complete period. - Simplify the final fraction carefully.

1. Let \(x=0.42\overline{3}\). 2. Shift past the two nonrepeating digits: \(100x=42.\overline{3}\). 3. Shift one more digit: \(1000x=423.\overline{3}\). 4. Subtract: \(1000x-100x=381\), so \(900x=381\). 5. Therefore, \(x=\frac{381}{900}=\frac{127}{300}\).

\(0.42\overline{3}=\frac{127}{300}\)

- Account separately for the prefix and the two-digit period. - Align the repeating blocks before subtracting. - Expect an improper fraction because the decimal is greater than \(1\).

1. Let \(x=1.2\overline{45}\). 2. Shift past the one nonrepeating digit: \(10x=12.\overline{45}\). 3. Shift two additional places: \(1000x=1245.\overline{45}\). 4. Subtract: \(1000x-10x=1233\), so \(990x=1233\). 5. Thus, \(x=\frac{1233}{990}=\frac{137}{110}\).

\(1.2\overline{45}=\frac{137}{110}\)

- Compare the length of the repeating block with the number of places shifted. - The infinite tails must match exactly before subtraction. - Reduce the fraction after correcting the setup.

1. Multiplying by \(10\) shifts only one digit, so the repeating tails do not align. 2. The period has \(2\) digits, so use \(100x=27.\overline{27}\). 3. Subtracting \(x=0.\overline{27}\) gives \(99x=27\). 4. Therefore, \(x=\frac{27}{99}=\frac{3}{11}\).

Riley shifted by too few places. The correct conversion is \(0.\overline{27}=\frac{3}{11}\).

- Write both shifted decimals explicitly. - Check that the repeating parts line up. - Subtract the finite parts after the tails cancel.

1. \(10x=4.\overline{2}\) and \(100x=42.\overline{2}\). 2. Subtracting gives \(100x-10x=42.\overline{2}-4.\overline{2}=38\). 3. Thus, \(90x=38\). 4. Therefore, \(x=\frac{38}{90}=\frac{19}{45}\).

The box contains \(38\), and \(0.4\overline{2}=\frac{19}{45}\).

- Relate the number of repeating digits to a power of \(10\). - Notice what coefficient remains after subtracting the original variable. - Check for a common factor only after forming the fraction.

1. Let \(x=0.\overline{124}\). 2. A \(3\)-digit shift gives \(1000x=124.\overline{124}\). 3. Subtracting gives \((1000-1)x=124\), so \(999x=124\). 4. Therefore, \(x=\frac{124}{999}\). 5. Since \(124\) and \(999\) share no common factor greater than \(1\), the fraction is already in lowest terms.

The denominator \(999\) comes from \(1000-1\). Thus, \(0.\overline{124}=\frac{124}{999}\).

- Align the repeating nines using two shifts. - Reduce the resulting fraction fully. - Check whether the reduced denominator produces a terminating decimal.

1. Let \(x=1.24\overline{9}\). 2. \(100x=124.\overline{9}\) and \(1000x=1249.\overline{9}\). 3. Subtracting gives \(900x=1125\). 4. Thus, \(x=\frac{1125}{900}=\frac{5}{4}\). 5. Therefore, the equivalent terminating decimal is \(1.25\).

\(1.24\overline{9}=\frac{5}{4}=1.25\)

- Replace the repeating decimal with an exact fraction first. - Isolate the unknown after the conversion. - Use a common denominator for the remaining subtraction.

1. Convert \(0.\overline{4}=\frac{4}{9}\). 2. Subtract to isolate \(x\): \(x=\frac{5}{6}-\frac{4}{9}\). 3. Using denominator \(18\), \(x=\frac{15}{18}-\frac{8}{18}=\frac{7}{18}\).

\(x=\frac{7}{18}\)

- Express both repeating decimals as fractions. - Find their sum before applying the meaning of average. - Keep the calculation exact.

1. Convert \(0.\overline{3}=\frac{1}{3}\) and \(0.\overline{6}=\frac{2}{3}\). 2. Their sum is \(\frac{1}{3}+\frac{2}{3}=1\). 3. Divide by \(2\): the average is \(\frac{1}{2}\).

\(\frac{1}{2}\)

- Convert the repeating side length before using the shape information. - Use the number of equal sides in the frame. - State both requested exact values with units.

1. Convert \(0.\overline{27}=\frac{27}{99}=\frac{3}{11}\). 2. An equilateral triangle has \(3\) equal sides. 3. The perimeter is \(3\cdot\frac{3}{11}=\frac{9}{11}\,\text{m}\).

Side length: \(\frac{3}{11}\,\text{m}\) Perimeter: \(\frac{9}{11}\,\text{m}\)

- Convert the repeating time to an exact fraction first. - Use the standard relationship between hours and minutes. - Check whether the converted number of minutes is reasonable for more than one hour.

1. Let \(x=1.\overline{6}\). Then \(10x-x=15\), so \(9x=15\). 2. Thus, \(x=\frac{15}{9}=\frac{5}{3}\) hours. 3. Multiply by \(60\) minutes per hour: \(\frac{5}{3}\cdot60=100\) minutes.

a) \(\frac{5}{3}\) hours b) \(100\) minutes

- Convert the repeating decimal before comparing all three forms. - Look for two representations of the same value. - Compare the remaining decimal by place value.

1. Convert \(0.\overline{45}=\frac{45}{99}=\frac{5}{11}\). 2. The first two numbers are equal. 3. \(\frac{5}{11}=0.4545\ldots<0.46\). 4. Therefore, the two equal values come before \(0.46\).

\(0.\overline{45}=\frac{5}{11}<0.46\)

- Use two shifts because the decimal has a prefix and a period. - Align the repeating tails exactly. - Compare only after obtaining the reduced fraction.

1. Let \(x=0.1\overline{8}\). 2. Then \(10x=1.\overline{8}\) and \(100x=18.\overline{8}\). 3. Subtracting gives \(90x=17\). 4. Therefore, \(x=\frac{17}{90}\), so the values are equal.

\(0.1\overline{8}=\frac{17}{90}\)

- Count the prefix digits and the repeating digits separately. - Build the smaller shift first, then move one full period farther. - Reduce the improper fraction at the end.

1. Let \(x=2.03\overline{12}\). 2. Shift past the \(2\) nonrepeating digits: \(100x=203.\overline{12}\). 3. Shift one full period farther: \(10000x=20312.\overline{12}\). 4. Subtracting gives \(9900x=20109\). 5. Thus, \(x=\frac{20109}{9900}=\frac{6703}{3300}\).

\(2.03\overline{12}=\frac{6703}{3300}\)

- Shift past both zeros before aligning a complete period. - Compare the two powers of \(10\) used in the subtraction. - Reduce only after forming the exact fraction.

1. Let \(x=0.00\overline{72}\). 2. Then \(100x=0.\overline{72}\) and \(10000x=72.\overline{72}\). 3. Subtracting gives \(9900x=72\). 4. Thus, \(x=\frac{72}{9900}=\frac{2}{275}\). 5. The two prefix places contribute two factors of \(10\) to the unreduced denominator.

\(0.00\overline{72}=\frac{2}{275}\)

- Write the shifted decimals and compare the positions of the repeating digits. - The smaller multiplier must move past the entire nonrepeating prefix. - Subtract only equations with identical infinite tails.

1. The decimal has two nonrepeating digits, so \(x\) and \(1000x\) do not have aligned repeating tails. 2. Use \(100x=58.\overline{3}\) and \(1000x=583.\overline{3}\). 3. Subtracting gives \(900x=525\). 4. Therefore, \(x=\frac{525}{900}=\frac{7}{12}\).

The repeating tails align for \(1000x-100x\), not \(1000x-x\). Thus, \(0.58\overline{3}=\frac{7}{12}\).

- Convert the repeating probability before using the number of plays. - Reduce the fraction to make the multiplication simpler. - Interpret the product in the context of wins.

1. Convert \(0.\overline{36}=\frac{36}{99}=\frac{4}{11}\). 2. Multiply by \(55\): \(55\cdot\frac{4}{11}=5\cdot4=20\). 3. The exact fraction is \(\frac{4}{11}\), corresponding to \(20\) wins out of \(55\) plays.

Probability: \(\frac{4}{11}\) Expected wins out of \(55\): \(20\)

- Rewrite the repeating scale factor exactly. - Preserve the units attached to the factor. - Look for cancellation when applying the factor to the actual length.

1. Convert \(0.\overline{24}=\frac{24}{99}=\frac{8}{33}\). 2. Multiply by \(33\,\text{ft}\): \(33\cdot\frac{8}{33}=8\). 3. The model length is \(8\,\text{in}\).

a) \(\frac{8}{33}\,\text{in}\) per foot b) \(8\,\text{in}\)

- Convert the repeating decimal to lowest terms. - Focus on what must happen to the denominator for the product to be a whole number. - Use the smallest multiplier that provides all needed factors.

1. Convert \(0.\overline{15}=\frac{15}{99}=\frac{5}{33}\). 2. The product is \(n\cdot\frac{5}{33}\). 3. Since \(5\) and \(33\) are relatively prime, the smallest positive \(n\) that cancels the denominator is \(33\). 4. The product is \(33\cdot\frac{5}{33}=5\).

\(n=33\), and the product is \(5\).

- A three-digit repeating block first gives a denominator of \(999\). - Test each option by reducing its block-over-nines fraction. - Stop once the reduced fraction matches exactly.

1. Convert choice B: \(0.\overline{378}=\frac{378}{999}\). 2. Divide numerator and denominator by \(27\): \(\frac{378}{999}=\frac{14}{37}\). 3. Therefore, choice B is the required decimal.

B. \(0.\overline{378}\)

- Notice the effect of placing one zero before the repeating block. - Convert each decimal with its correct place value. - Compare the fractions using a quotient.

1. \(A=\frac{27}{99}=\frac{3}{11}\). 2. The extra nonrepeating zero shifts the value one place right, so \(B=\frac{27}{990}=\frac{3}{110}\). 3. \(\frac{A}{B}=\frac{3/11}{3/110}=10\). 4. Therefore, \(A\) is \(10\) times as large as \(B\).

\(A=\frac{3}{11}\), \(B=\frac{3}{110}\), and \(A\) is \(10\) times as large as \(B\).

- Keep the leading zero in the first repeating block. - Convert both two-digit periods using the same initial denominator. - Look at how the numerators combine.

1. \(0.\overline{09}=\frac{9}{99}=\frac{1}{11}\). 2. \(0.\overline{90}=\frac{90}{99}=\frac{10}{11}\). 3. Their sum is \(\frac{1}{11}+\frac{10}{11}=1\). 4. The repeating blocks add to \(99\), matching the unreduced denominator \(99\).

\(0.\overline{09}+0.\overline{90}=1\)

- Treat the zero as part of the two-digit period. - Subtract the whole original decimal after shifting one complete block. - Check whether the final numerator shares a factor with \(99\).

1. Let \(x=4.\overline{05}\). 2. Then \(100x=405.\overline{05}\). 3. Subtracting the original equation gives \(99x=401\). 4. Therefore, \(x=\frac{401}{99}\), which is in lowest terms.

\(4.\overline{05}=\frac{401}{99}\)

- Identify exactly which digits lie under the bar. - Compare the claimed fraction with the decimal it would actually produce. - Use separate shifts for the prefix and the period.

1. The expression \(\frac{12}{99}\) represents \(0.\overline{12}\), where both digits repeat. 2. In \(0.1\overline{2}\), only the digit \(2\) repeats. 3. Let \(x=0.1\overline{2}\). Then \(10x=1.\overline{2}\) and \(100x=12.\overline{2}\). 4. Subtracting gives \(90x=11\), so \(x=\frac{11}{90}\).

The student repeated the prefix digit incorrectly. The correct value is \(0.1\overline{2}=\frac{11}{90}\).

- Count the two prefix digits before the repeating digit. - Use two equations whose repeating tails are identical. - Compare the exact fraction with the choices.

1. Let \(x=0.07\overline{4}\). 2. Then \(100x=7.\overline{4}\) and \(1000x=74.\overline{4}\). 3. Subtracting gives \(900x=67\). 4. Thus, \(x=\frac{67}{900}\), which is choice B.

B. \(\frac{67}{900}\)

- Convert each repeating decimal independently. - Reduce before finding a common denominator. - Keep the final subtraction exact.

1. Convert \(1.\overline{3}=\frac{4}{3}\). 2. Convert \(0.\overline{81}=\frac{81}{99}=\frac{9}{11}\). 3. Subtract using denominator \(33\): \(\frac{4}{3}-\frac{9}{11}=\frac{44}{33}-\frac{27}{33}=\frac{17}{33}\).

\(\frac{17}{33}\)

- Convert a repeating two-digit block by multiplying by \(100\) and subtracting. - Compare decimals digit by digit from left to right. - Every terminating decimal is rational.

1. Since \(x = 0.121212\ldots\), \(x = \frac{12}{99} = \frac{4}{33}\). 2. The decimal for \(y\) contains blocks of \(1\)s and \(2\)s whose lengths keep increasing. Therefore, no fixed block repeats forever. The decimal is nonterminating and nonrepeating, so \(y\) is irrational. 3. Compare \(x = 0.121212\ldots\) with \(y = 0.121122\ldots\). At the fourth decimal place, \(x\) has \(2\) and \(y\) has \(1\), so \(y < x\). 4. One rational number between them is \(z = 0.1212\). Indeed, \(0.121122\ldots < 0.121200\ldots < 0.121212\ldots\).

a) \(x = \frac{4}{33}\) b) \(y\) is irrational because its increasing block lengths make its decimal expansion nonterminating and nonrepeating. c) One example is \(z = 0.1212\), since \(y < 0.1212 < x\).

- Keep the unknown digit symbolic while shifting the decimal. - Express both sides with the same denominator after conversion. - Compare the resulting numerators.

1. Let \(x=0.2\overline{d}\). 2. Then \(10x=2.\overline{d}\) and \(100x=20+d+0.\overline{d}\). 3. Subtracting gives \(90x=18+d\). 4. Since \(x=\frac{13}{45}=\frac{26}{90}\), \(18+d=26\). 5. Therefore, \(d=8\).

\(d=8\)

- Represent the two-digit block as one whole number. - Shift exactly one full period. - Keep the block symbolic until after the subtraction.

1. Let \(x=0.\overline{ab}\). 2. Then \(100x=N.\overline{ab}\). 3. Subtracting gives \(99x=N\), so \(x=\frac{N}{99}\). 4. For \(62\), \(N=62\), so \(0.\overline{62}=\frac{62}{99}\).

The rule is \(0.\overline{ab}=\frac{10a+b}{99}\). Therefore, \(0.\overline{62}=\frac{62}{99}\).

- In the first method, choose shifts that place identical repeating tails under each other. - In the second method, separate the whole, terminating, and repeating portions. - Compare the reduced results rather than their unreduced denominators.

1. a) \(1000x=3124.\overline{24}\) and \(10x=31.\overline{24}\), so \(990x=3093\). Thus \(x=\frac{3093}{990}=\frac{1031}{330}\). 2. b) Let \(y=0.0\overline{24}\). Then \(1000y-10y=24\), so \(y=\frac{24}{990}=\frac{4}{165}\). 3. \(3+\frac{1}{10}+\frac{4}{165}=\frac{990+33+8}{330}=\frac{1031}{330}\), matching part a).

a) \(\frac{1031}{330}\) b) The decomposition also gives \(\frac{1031}{330}\).