Free Math Worksheets

- Find the least common multiple of \(10\) and \(25\). - Determine the scale factor from each original denominator to the common denominator. - Multiply each numerator by the same factor used for its denominator.

1. The least common multiple of \(10\) and \(25\) is \(50\), so the least common denominator is \(50\). 2. Rewrite \(\frac{3}{10}\): \(\frac{3\times5}{10\times5}=\frac{15}{50}\). 3. Rewrite \(\frac{4}{25}\): \(\frac{4\times2}{25\times2}=\frac{8}{50}\).

The least common denominator is \(50\). The fractions are \(\frac{15}{50}\) and \(\frac{8}{50}\).

- Start with a unit fraction that is less than \(\frac{3}{4}\). - Subtract that unit fraction from \(\frac{3}{4}\). - Check whether the remainder is another unit fraction.

1. Choose a unit fraction less than \(\frac{3}{4}\), such as \(\frac{1}{2}\). 2. Find the difference: \(\frac{3}{4}-\frac{1}{2}=\frac{3}{4}-\frac{2}{4}=\frac{1}{4}\). 3. The remainder is a different unit fraction, so \(\frac{3}{4}=\frac{1}{2}+\frac{1}{4}\).

\(\frac{3}{4}=\frac{1}{2}+\frac{1}{4}\)

- A unit fraction has a numerator of \(1\). - Find a common denominator before adding fractions with unlike denominators. - Consider unit fractions whose denominators are factors of \(6\).

1. Look for two unit fractions that can be rewritten with denominator \(6\). 2. Since \(\frac{1}{2}=\frac{3}{6}\) and \(\frac{1}{3}=\frac{2}{6}\), \(\frac{1}{2}+\frac{1}{3}=\frac{3}{6}+\frac{2}{6}=\frac{5}{6}\).

\(\frac{5}{6}=\frac{1}{2}+\frac{1}{3}\)

- Find a common multiple of \(12\) and \(18\). - Rewrite both fractions with the same denominator. - Check whether the final numerator and denominator have a common factor.

1. The least common denominator of \(12\) and \(18\) is \(36\). 2. Rewrite the fractions: \(\frac{5}{12}=\frac{15}{36}\) and \(\frac{7}{18}=\frac{14}{36}\). 3. Add: \(\frac{15}{36}+\frac{14}{36}=\frac{29}{36}\). The fraction is already in simplest form.

\(\frac{29}{36}\)

- Rewrite the fractions in each sum with common denominators. - Calculate each sum before comparing the pair. - Simplify or rewrite the results so they are easy to compare.

1. For a), \(\frac{3}{10}+\frac{2}{5}=\frac{3}{10}+\frac{4}{10}=\frac{7}{10}\). The other sum is \(\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\). Since \(\frac{7}{10}<\frac{3}{4}\), the second sum is greater. 2. For b), \(\frac{5}{6}+\frac{1}{12}=\frac{10}{12}+\frac{1}{12}=\frac{11}{12}\), and \(\frac{2}{3}+\frac{1}{4}=\frac{8}{12}+\frac{3}{12}=\frac{11}{12}\). The sums are equal.

a) \(\frac{1}{4}+\frac{1}{2}\) is greater. b) The two sums are equal; each is \(\frac{11}{12}\).

- A common denominator must be a multiple of both original denominators. - Test each proposed number for divisibility by \(6\) and \(10\). - Use the same scale factor for each numerator and denominator.

1. A common denominator must be divisible by both \(6\) and \(10\). 2. The numbers \(30\) and \(60\) are divisible by both. The numbers \(20\) and \(45\) are not. 3. The least common denominator is \(30\). 4. Rewrite the fractions: \(\frac{1}{6}=\frac{5}{30}\) and \(\frac{3}{10}=\frac{9}{30}\).

\(30\) and \(60\) are suitable. Using denominator \(30\), the fractions are \(\frac{5}{30}\) and \(\frac{9}{30}\).

- Find the least number divisible by all three denominators. - Determine the scale factor for each denominator. - Multiply each numerator by the same factor used for its denominator.

1. The least common multiple of \(10,15,\) and \(6\) is \(30\). 2. \(\frac{3}{10}=\frac{3\times3}{10\times3}=\frac{9}{30}\). 3. \(\frac{4}{15}=\frac{4\times2}{15\times2}=\frac{8}{30}\). 4. \(\frac{1}{6}=\frac{1\times5}{6\times5}=\frac{5}{30}\).

\(\frac{9}{30},\frac{8}{30},\frac{5}{30}\)

- Break the description into two separate calculations. - Rename mixed numbers when needed before subtracting. - Use a common denominator for the final addition.

1. Find the difference: \(7\frac{1}{8}-2\frac{3}{4}=7\frac{1}{8}-2\frac{6}{8}=4\frac{3}{8}\). 2. Find the sum: \(3\frac{5}{6}+1\frac{1}{2}=3\frac{5}{6}+1\frac{3}{6}=5\frac{1}{3}\). 3. Add the results: \(4\frac{3}{8}+5\frac{1}{3}=9\frac{17}{24}\).

\(9\frac{17}{24}\)

- Compare the approximate sizes of the fractions in Tim's work. - What must happen to a number when you add a positive amount to it? - Recall the rule for adding fractions with different denominators.

1. Sarah's size check is valid. Using denominator \(21\), \(\frac{2}{3}=\frac{14}{21}\) and \(\frac{3}{7}=\frac{9}{21}\), so \(\frac{3}{7}<\frac{2}{3}\). 2. Adding the positive fraction \(\frac{1}{4}\) to \(\frac{2}{3}\) must produce a value greater than \(\frac{2}{3}\). Therefore Tim's result cannot be correct. 3. Tim incorrectly added the numerators and denominators separately. Fractions with different denominators must first be rewritten with a common denominator before their numerators can be added.

Sarah is correct because adding a positive fraction to \(\frac{2}{3}\) must make the value larger, while \(\frac{3}{7}<\frac{2}{3}\). Tim's error was adding the numerators and denominators separately instead of using equivalent fractions with a common denominator.

- Think about what the denominator tells you about the size of each part. - Are thirds and fifths the same size? - Find a common denominator, then add the equivalent fractions.

1. The denominator tells the size of the equal parts of a whole. Thirds and fifths are different-sized parts, so their numerators do not count the same unit. 2. Use denominator \(15\): \(\frac{2}{3}=\frac{10}{15}\) and \(\frac{1}{5}=\frac{3}{15}\). 3. Now the parts are the same size, so add the numerators: \(\frac{10}{15}+\frac{3}{15}=\frac{13}{15}\).

A common denominator is needed because the fractions must describe equal-sized parts before those parts can be counted together. The sum is \(\frac{13}{15}\).

- Subtract a unit fraction such as \(\frac{1}{3}\) or \(\frac{1}{4}\) from \(\frac{2}{5}\). - For part 2), try splitting the remainder into two unit fractions. - Make sure no denominator is repeated within a representation.

1. For two unit fractions, subtract \(\frac{1}{3}\): \(\frac{2}{5}-\frac{1}{3}=\frac{6}{15}-\frac{5}{15}=\frac{1}{15}\). Therefore, \(\frac{2}{5}=\frac{1}{3}+\frac{1}{15}\). 2. For three unit fractions, begin with \(\frac{1}{4}\): \(\frac{2}{5}-\frac{1}{4}=\frac{8}{20}-\frac{5}{20}=\frac{3}{20}\). Since \(\frac{3}{20}=\frac{1}{10}+\frac{1}{20}\), one representation is \(\frac{2}{5}=\frac{1}{4}+\frac{1}{10}+\frac{1}{20}\).

1) \(\frac{2}{5}=\frac{1}{3}+\frac{1}{15}\) 2) \(\frac{2}{5}=\frac{1}{4}+\frac{1}{10}+\frac{1}{20}\)

- For part a), use the numerator to determine the number of twelfths. - For part b), look for unit fractions that can be rewritten with denominator \(12\). - Add using a common denominator to check your answer.

1. For a), \(\frac{5}{12}=\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}\). 2. For b), one choice is \(\frac{1}{4}+\frac{1}{6}\). Using a common denominator, \(\frac{1}{4}+\frac{1}{6}=\frac{3}{12}+\frac{2}{12}=\frac{5}{12}\). 3. Another valid choice is \(\frac{1}{3}+\frac{1}{12}=\frac{4}{12}+\frac{1}{12}=\frac{5}{12}\).

a) \(\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}\) b) For example, \(\frac{1}{4}+\frac{1}{6}\) or \(\frac{1}{3}+\frac{1}{12}\).

- Find the least common denominator of \(2\) and \(6\). - Think about how the numerator of \(\frac{2}{3}\) tells the number of copies of \(\frac{1}{3}\).

1. The least common denominator of \(2\) and \(6\) is \(6\). Rewrite \(\frac{1}{2}\) as \(\frac{3}{6}\): \(\frac{3}{6}+\frac{1}{6}=\frac{4}{6}=\frac{2}{3}\). 2. With equal unit fractions, the denominator stays the same and the numerator tells how many copies to add. No search for compatible denominators is needed.

1) \(\frac{1}{2}+\frac{1}{6}=\frac{3}{6}+\frac{1}{6}=\frac{4}{6}=\frac{2}{3}\) 2) Equal unit fractions can be written directly from the numerator and denominator, so no search for different denominators is required.

- Find a common denominator for the fractions in each sum. - Add the numerators after rewriting the fractions. - Simplify the result and compare it with \(\frac{2}{9}\).

1. For a), use denominator \(45\): \(\frac{1}{5}+\frac{1}{45}=\frac{9}{45}+\frac{1}{45}=\frac{10}{45}=\frac{2}{9}\). 2. For b), use denominator \(90\): \(\frac{1}{9}+\frac{1}{10}+\frac{1}{90}=\frac{10}{90}+\frac{9}{90}+\frac{1}{90}=\frac{20}{90}=\frac{2}{9}\).

a) \(\frac{9}{45}+\frac{1}{45}=\frac{10}{45}=\frac{2}{9}\) b) \(\frac{10}{90}+\frac{9}{90}+\frac{1}{90}=\frac{20}{90}=\frac{2}{9}\)

- Use the numerator to write \(\frac{2}{3}\) as repeated thirds. - Replace one third with the equivalent sum given in the problem. - Check that the three denominators in part 2) are different.

1. Two equal unit fractions give \(\frac{2}{3}=\frac{1}{3}+\frac{1}{3}\). 2. Replace one \(\frac{1}{3}\) with \(\frac{1}{4}+\frac{1}{12}\): \(\frac{2}{3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{12}\).

1) \(\frac{2}{3}=\frac{1}{3}+\frac{1}{3}\) 2) \(\frac{2}{3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{12}\)

- Rewrite unit fractions with denominator \(12\). - For part b), try decomposing one unit fraction from part a) into two smaller unit fractions. - Check each sum using a common denominator.

1. For a), \(\frac{1}{3}+\frac{1}{4}=\frac{4}{12}+\frac{3}{12}=\frac{7}{12}\). Another valid answer is \(\frac{1}{2}+\frac{1}{12}\). 2. For b), \(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}=\frac{4}{12}+\frac{2}{12}+\frac{1}{12}=\frac{7}{12}\).

a) For example, \(\frac{1}{3}+\frac{1}{4}\) or \(\frac{1}{2}+\frac{1}{12}\). b) For example, \(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}\).

- Find the least common multiple of the denominators in each set. - Determine the scale factor for each original denominator. - Multiply each numerator and denominator by the same factor.

1. For a), the least common denominator is \(24\): \(\frac{5}{6}=\frac{20}{24}\) and \(\frac{7}{8}=\frac{21}{24}\). 2. For b), the least common denominator is \(36\): \(\frac{2}{9}=\frac{8}{36}\) and \(\frac{5}{12}=\frac{15}{36}\). 3. For c), the least common denominator is \(12\): \(\frac{3}{4}=\frac{9}{12}\), \(\frac{1}{6}=\frac{2}{12}\), and \(\frac{2}{3}=\frac{8}{12}\).

a) \(\frac{20}{24}\) and \(\frac{21}{24}\) b) \(\frac{8}{36}\) and \(\frac{15}{36}\) c) \(\frac{9}{12},\frac{2}{12},\frac{8}{12}\)

- Find a common denominator for each sum. - Add the rewritten numerators. - Simplify each result before comparing it with \(\frac{2}{3}\).

1. For A), \(\frac{1}{2}+\frac{1}{6}=\frac{3}{6}+\frac{1}{6}=\frac{4}{6}=\frac{2}{3}\). 2. For B), \(\frac{1}{2}+\frac{1}{4}+\frac{1}{12}=\frac{6}{12}+\frac{3}{12}+\frac{1}{12}=\frac{10}{12}=\frac{5}{6}\), so it is not equal to \(\frac{2}{3}\). 3. For C), \(\frac{1}{2}+\frac{1}{10}+\frac{1}{15}=\frac{15}{30}+\frac{3}{30}+\frac{2}{30}=\frac{20}{30}=\frac{2}{3}\).

A) and C) are equal to \(\frac{2}{3}\).

- Look for an addend that is the same in both expressions. - When one addend is identical, compare only the other addends. - For fractions with numerator \(1\), how does the denominator affect the size of the fraction?

1. For a), both sums contain \(\frac{4}{11}\). Since \(\frac{3}{7}>\frac{2}{7}\), \(B>A\). 2. For b), both sums contain \(\frac{1}{2}\). Since \(\frac{1}{5}>\frac{1}{6}\), \(C>D\). 3. For c), \(E=\frac{3}{3}=1\) and \(F=\frac{9}{9}=1\), so \(E=F\).

a) \(B>A\) b) \(C>D\) c) \(E=F\)

- Calculate each sum first. - Find how much each result is below \(1\). - The smaller distance identifies the expression closer to \(1\).

1. Expression A is \(\frac{4}{10}+\frac{5}{10}=\frac{9}{10}\). Its distance from \(1\) is \(1-\frac{9}{10}=\frac{1}{10}\). 2. Expression B is \(\frac{6}{8}+\frac{1}{8}=\frac{7}{8}\). Its distance from \(1\) is \(1-\frac{7}{8}=\frac{1}{8}\). 3. Since \(\frac{1}{10}<\frac{1}{8}\), Expression A is closer to \(1\).

Expression A is closer to \(1\). Its distance is \(\frac{1}{10}\), compared with \(\frac{1}{8}\) for Expression B.

- Which fraction is above \(\frac{1}{2}\), and which is below it? - How far is each fraction from \(\frac{1}{2}\)? - Compare those two distances rather than calculating the original sum.

1. Compare \(\frac{21}{40}\) with \(\frac{1}{2}\): \(\frac{21}{40}=\frac{1}{2}+\frac{1}{40}\). 2. Compare \(\frac{19}{42}\) with \(\frac{1}{2}\): \(\frac{19}{42}=\frac{1}{2}-\frac{1}{21}\). 3. Compare the two deviations. Since \(\frac{1}{21}>\frac{1}{40}\), the amount below \(\frac{1}{2}\) is greater than the amount above \(\frac{1}{2}\). 4. Therefore the two fractions add to less than \(1\).

The sum is less than \(1\). The first fraction is \(\frac{1}{40}\) above \(\frac{1}{2}\), while the second is \(\frac{1}{21}\) below \(\frac{1}{2}\). Since \(\frac{1}{21}>\frac{1}{40}\), the deficit is larger than the excess.

- Find the difference between \(\frac{25}{49}\) and \(\frac{1}{2}\). - Find the difference between \(\frac{1}{2}\) and \(\frac{26}{53}\). - Compare the two distances to decide which side of \(1\) the sum lies on.

1. The first fraction is above \(\frac{1}{2}\): \(\frac{25}{49}-\frac{1}{2}=\frac{50}{98}-\frac{49}{98}=\frac{1}{98}\). 2. The second fraction is below \(\frac{1}{2}\): \(\frac{1}{2}-\frac{26}{53}=\frac{53}{106}-\frac{52}{106}=\frac{1}{106}\). 3. Since \(\frac{1}{98}>\frac{1}{106}\), the amount above \(\frac{1}{2}\) is greater than the amount below \(\frac{1}{2}\). 4. Therefore \(S>1\).

\(S>1\). The first fraction is \(\frac{1}{98}\) above \(\frac{1}{2}\), while the second is \(\frac{1}{106}\) below \(\frac{1}{2}\). Because \(\frac{1}{98}>\frac{1}{106}\), the sum is greater than \(1\).

- Estimate the size of the sum before calculating exactly. - Look at how the student combined the fractional parts. - Rewrite the fractions with a common denominator before adding them.

1. Estimate first: \(2 \frac{3}{4}\) is close to \(3\), and adding \(1 \frac{1}{2}\) must give a result greater than \(4\). The student's answer \(3 \frac{2}{3}\) is less than \(4\), so it cannot be correct. 2. The student incorrectly added the numerators and denominators of \(\frac{3}{4}\) and \(\frac{1}{2}\) separately. 3. Rewrite \(\frac{1}{2}\) as \(\frac{2}{4}\). Then \(\frac{3}{4}+\frac{2}{4}=\frac{5}{4}=1 \frac{1}{4}\). 4. Adding the whole-number parts gives \(3+1 \frac{1}{4}=4 \frac{1}{4}\).

The result must be greater than \(4\), so \(3 \frac{2}{3}\) cannot be correct. The student added numerators and denominators separately. The correct answer is \(4 \frac{1}{4}\).

- For the sum, subtract a unit fraction and then decompose the remainder. - For the difference, look for a unit fraction just greater than \(\frac{3}{7}\). - Compare the number of terms and the sizes of the denominators.

1. For a), \(\frac{3}{7}-\frac{1}{3}=\frac{2}{21}\). Then \(\frac{2}{21}-\frac{1}{11}=\frac{1}{231}\). Thus, \(\frac{3}{7}=\frac{1}{3}+\frac{1}{11}+\frac{1}{231}\). 2. For b), \(\frac{1}{2}-\frac{3}{7}=\frac{7}{14}-\frac{6}{14}=\frac{1}{14}\). Thus, \(\frac{3}{7}=\frac{1}{2}-\frac{1}{14}\). 3. For c), the difference uses only two terms and has much smaller denominators.

a) One possible answer is \(\frac{3}{7}=\frac{1}{3}+\frac{1}{11}+\frac{1}{231}\). b) \(\frac{3}{7}=\frac{1}{2}-\frac{1}{14}\) c) The difference is shorter and uses smaller denominators.

- Verify each sum using a common denominator. - Compare the first unit fractions in the two methods. - Check the next larger unit fraction to determine whether it would exceed \(\frac{3}{10}\).

1. For Method A, \(\frac{1}{4}+\frac{1}{20}=\frac{5}{20}+\frac{1}{20}=\frac{6}{20}=\frac{3}{10}\). 2. For Method B, \(\frac{1}{5}+\frac{1}{10}=\frac{2}{10}+\frac{1}{10}=\frac{3}{10}\). 3. The greatest unit fraction not exceeding \(\frac{3}{10}\) is \(\frac{1}{4}\): \(\frac{1}{3}>\frac{3}{10}\), while \(\frac{1}{4}<\frac{3}{10}\). Therefore, Method A follows the greedy rule.

a) Both methods are correct. b) Method A follows the greedy rule because \(\frac{1}{4}\) is the greatest unit fraction less than \(\frac{3}{10}\).

- Test unit fractions slightly less than \(\frac{1}{4}\). - After choosing one term, subtract it from \(\frac{1}{4}\) to find the other. - The two unit fractions in each sum must be different.

1. Try unit fractions just less than \(\frac{1}{4}\). 2. Starting with \(\frac{1}{5}\), the remainder is \(\frac{1}{4}-\frac{1}{5}=\frac{1}{20}\). Thus, \(\frac{1}{4}=\frac{1}{5}+\frac{1}{20}\). 3. Starting with \(\frac{1}{6}\), the remainder is \(\frac{1}{4}-\frac{1}{6}=\frac{1}{12}\). Thus, \(\frac{1}{4}=\frac{1}{6}+\frac{1}{12}\). 4. Verify: \(\frac{1}{5}+\frac{1}{20}=\frac{4}{20}+\frac{1}{20}=\frac{1}{4}\), and \(\frac{1}{6}+\frac{1}{12}=\frac{2}{12}+\frac{1}{12}=\frac{1}{4}\).

1) \(\frac{1}{4}=\frac{1}{5}+\frac{1}{20}\) 2) \(\frac{1}{4}=\frac{1}{6}+\frac{1}{12}\)

- Start with a large unit fraction that is less than \(\frac{3}{4}\). - Subtract it to find the remainder. - Split the remainder into two different unit fractions.

1. Begin with \(\frac{1}{2}\): \(\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\). 2. Decompose the remainder: \(\frac{1}{4}=\frac{1}{5}+\frac{1}{20}\). 3. Therefore, \(\frac{3}{4}=\frac{1}{2}+\frac{1}{5}+\frac{1}{20}\). 4. Another valid answer is \(\frac{3}{4}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}\).

One possible answer is \(\frac{3}{4}=\frac{1}{2}+\frac{1}{5}+\frac{1}{20}\).