Aimathic
Login | English | Deutsch

Grade 5 math worksheets

Build your own math worksheets from 35,000+ problems for grades 3 to 12, from fractions to AP Calculus. Every problem comes with step-by-step solutions.

Add fractions with unlike denominators

Learning goals

  • Finds a common denominator for unlike fractions.
  • Rewrites fractions as equivalent fractions with the common denominator.
  • Adds numerators while preserving the denominator.
  • Simplifies the result when appropriate.
  • Checks the sum using benchmarks or estimation.

Click problems to add them to your worksheet.

Add fractions with unlike denominators 1
Compare the shaded parts in the two same-size circles. a) Explain why \(\frac{1}{2}\) can be renamed as \(\frac{2}{4}\) using the circles. b) Complete \(\frac{1}{2}+\frac{1}{4}=\frac{\square}{4}\).
Figure for problem 558950

Hints

- Compare the shaded amount in the two same-size circles. - Ask how many fourths cover the same amount as one half. - Then add the extra fourth.

Solution

1. Circle a) shades \(1\) of \(2\) equal parts, while circle b) shades \(2\) of \(4\) equal parts. The shaded amounts are the same size, so \(\frac{1}{2}=\frac{2}{4}\). 2. Add one more fourth: \(\frac{2}{4}+\frac{1}{4}=\frac{3}{4}\).

Answer

a) The same amount is shaded in both circles, so \(\frac{1}{2}=\frac{2}{4}\). b) \(\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\).
Add fractions with unlike denominators 2
Which rewrite is correct for adding \(\frac{1}{3}+\frac{1}{6}\) using sixths? A) \(\frac{1}{3}=\frac{1}{6}\) B) \(\frac{1}{3}=\frac{2}{6}\) C) \(\frac{1}{3}=\frac{3}{6}\) Choose the correct rewrite, then find the sum.

Hints

- Check whether each rewrite keeps the value of \(\frac{1}{3}\) unchanged. - Picture one whole split into \(6\) equal parts and ask how many of those parts make one third. - After both fractions use sixths, add the number of sixths and simplify.

Solution

1. Choice B is correct because \(\frac{1}{3}=\frac{2}{6}\). 2. Add the sixths: \(\frac{2}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}\).

Answer

B) \(\frac{1}{3}=\frac{2}{6}\), so \(\frac{1}{3}+\frac{1}{6}=\frac{1}{2}\).
Add fractions with unlike denominators 3
What addition equation do the two segments show? At what fraction does the second segment end?
Figure for problem 558952

Hints

- Use the tick spacing to think in fourths. - How many fourths are equal to one half? - Count forward from the end of the first segment.

Solution

1. On a line divided into fourths, \(\frac{1}{2}=\frac{2}{4}\). 2. Starting at \(\frac{1}{4}\) and moving another \(\frac{2}{4}\) ends at \(\frac{3}{4}\). 3. Therefore \(\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\).

Answer

The second segment ends at \(\frac{3}{4}\), so \(\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\).
Add fractions with unlike denominators 4
For \(\frac{2}{3}+\frac{1}{5}\), explain why the fractions must be rewritten with a common denominator before their numerators can be added. Then calculate the sum.

Hints

- 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.

Solution

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: \(\frac{10}{15}+\frac{3}{15}=\frac{13}{15}\).

Answer

A common denominator is needed because the fractions must describe equal-sized parts before those parts can be counted together. The sum is \(\frac{13}{15}\).
Add fractions with unlike denominators 5
A unit fraction has numerator \(1\), such as \(\frac{1}{2}\), \(\frac{1}{3}\), or \(\frac{1}{10}\). Ancient Egyptians represented fractions as sums of different unit fractions. Write \(\frac{3}{4}\) as a sum of two different unit fractions. Briefly explain your method.

Hints

- Start with a unit fraction less than the given fraction. - Subtract it to find what remains. - Check whether the remainder is also a unit fraction and whether the denominators differ.

Solution

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

Answer

\(\frac{3}{4}=\frac{1}{2}+\frac{1}{4}\)
Add fractions with unlike denominators 6
A unit fraction has numerator \(1\), such as \(\frac{1}{2}\), \(\frac{1}{3}\), or \(\frac{1}{10}\). Write \(\frac{5}{6}\) as the sum of two different unit fractions.

Hints

- Each unit fraction must have numerator \(1\). - Look for different denominators that can be changed into sixths. - Check the sum by using a common denominator.

Solution

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

Answer

\(\frac{5}{6}=\frac{1}{2}+\frac{1}{3}\)
Add fractions with unlike denominators 7
Calculate the sum and write the answer in simplest form: \(\frac{5}{12}+\frac{7}{18}\)

Hints

- 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.

Solution

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.

Answer

\(\frac{29}{36}\)
Add fractions with unlike denominators 8
Which sum is greater in each part? Compare the results. a) \(\frac{3}{10}+\frac{2}{5}\) or \(\frac{1}{4}+\frac{1}{2}\) b) \(\frac{5}{6}+\frac{1}{12}\) or \(\frac{2}{3}+\frac{1}{4}\)

Hints

- 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.

Solution

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.

Answer

a) \(\frac{1}{4}+\frac{1}{2}\) is greater. b) The two sums are equal; each is \(\frac{11}{12}\).
Add fractions with unlike denominators 9
Panels a) and b) show the two addends. Panels c) and d) are candidate twelfths models for panel a), and panels e) and f) are candidate twelfths models for panel b). Exactly one candidate in each pair has the same shaded length as its original bar. a) Read the two addends. b) Choose the matching twelfths model from c) or d) for panel a), and from e) or f) for panel b). For each choice, state how many twelfths have the same length as the original shaded part. c) Use the selected common-unit models to find the sum.
Figure for problem 554434

Hints

- Compare the shaded length in each original bar with the two candidate bars assigned to it. - The correct candidate must cover the same amount of the whole even though it uses smaller equal parts. - Add only after both shaded amounts have been expressed with the same-sized parts.

Solution

1. Panel a) shows \(\frac{1}{3}\), and panel b) shows \(\frac{1}{4}\). 2. In the twelfths candidates, panel d) shades the same length as \(\frac{1}{3}\), so \(\frac{1}{3}=\frac{4}{12}\). Panel e) shades the same length as \(\frac{1}{4}\), so \(\frac{1}{4}=\frac{3}{12}\). 3. Using same-sized parts, \(\frac{4}{12}+\frac{3}{12}=\frac{7}{12}\).

Answer

a) \(\frac{1}{3}\) and \(\frac{1}{4}\) b) d) matches panel a): \(\frac{4}{12}\). e) matches panel b): \(\frac{3}{12}\). c) \(\frac{7}{12}\)
Add fractions with unlike denominators 10
Panels a) and b) show the two addends. Panels c) through f) all use the same \(5\times6\) grid. Panels c) and d) are candidates for panel a); panels e) and f) are candidates for panel b). a) Read the two original fractions. b) Select the matching common-grid model from each pair. For each selected grid, state whether its shaded region is made of full rows or full columns and explain how that matches the original denominator. c) State the two equivalent fractions in thirtieths and find the sum in simplest form.
Figure for problem 554435

Hints

- In a \(5\times6\) grid, full rows naturally show fifths and full columns naturally show sixths. - Inspect the orientation and number of complete shaded rows or columns in each candidate. - Once both fractions are expressed as thirtieths, add same-sized cells.

Solution

1. Panel a) shows \(\frac{2}{5}\), and panel b) shows \(\frac{1}{6}\). 2. Panel d) shades \(2\) full rows out of \(5\), so it preserves the fifths structure and represents \(\frac{2}{5}=\frac{12}{30}\). Panel e) shades \(1\) full column out of \(6\), so it preserves the sixths structure and represents \(\frac{1}{6}=\frac{5}{30}\). 3. Therefore \(\frac{12}{30}+\frac{5}{30}=\frac{17}{30}\).

Answer

a) \(\frac{2}{5}\) and \(\frac{1}{6}\) b) d) matches panel a) because it shades \(2\) of \(5\) full rows. e) matches panel b) because it shades \(1\) of \(6\) full columns. c) \(\frac{2}{5}=\frac{12}{30}\), \(\frac{1}{6}=\frac{5}{30}\), and the sum is \(\frac{17}{30}\).
Add fractions with unlike denominators 11
Find \(\frac{3}{8}+\frac{1}{4}\). Write the fractions with a common denominator, simplify if needed, and use a benchmark to check that your answer is reasonable.

Hints

- One denominator is a multiple of the other, so one fraction can be renamed quickly. - Add only after both fractions use the same-sized parts. - Compare your sum with \(\frac{1}{2}\) to check its size.

Solution

1. Rewrite \(\frac{1}{4}\) as \(\frac{2}{8}\). 2. Add: \(\frac{3}{8}+\frac{2}{8}=\frac{5}{8}\). 3. The sum should be greater than \(\frac{1}{2}\) because \(\frac{3}{8}+\frac{1}{8}=\frac{1}{2}\), and \(\frac{1}{4}>\frac{1}{8}\). The result \(\frac{5}{8}\) passes that check.

Answer

\(\frac{3}{8}+\frac{1}{4}=\frac{5}{8}\). Since \(\frac{3}{8}+\frac{1}{8}=\frac{1}{2}\) and \(\frac{1}{4}>\frac{1}{8}\), the sum should be greater than \(\frac{1}{2}\). The answer \(\frac{5}{8}\) passes that check.
Add fractions with unlike denominators 12
Add the difference of \(7\frac{1}{8}\) and \(2\frac{3}{4}\) to the sum of \(3\frac{5}{6}\) and \(1\frac{1}{2}\).

Hints

- Break the description into two separate calculations first. - In the subtraction, compare the fractional parts after giving them a common denominator. Will you need to regroup one whole? - Use a common denominator when you add the two results.

Solution

1. Find the difference. Rewrite \(2\frac{3}{4}\) as \(2\frac{6}{8}\). Since \(\frac{1}{8}<\frac{6}{8}\), regroup \(7\frac{1}{8}\) as \(6\frac{9}{8}\). Then \(6\frac{9}{8}-2\frac{6}{8}=4\frac{3}{8}\). 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: \(4\frac{3}{8}+5\frac{1}{3}=9\frac{17}{24}\).

Answer

\(9\frac{17}{24}\)
Add fractions with unlike denominators 13
Tim and Sarah are solving \(\frac{2}{3}+\frac{1}{4}\). Tim writes \(\frac{2+1}{3+4}=\frac{3}{7}\). Sarah says, “That cannot be right. \(\frac{3}{7}\) is smaller than \(\frac{2}{3}\), but we are adding a positive amount.” Explain why Sarah's reasoning makes sense and identify Tim's mistake.

Hints

- 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.

Solution

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.

Answer

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.
Add fractions with unlike denominators 14
Jakob writes \(2 \frac{3}{4}+1 \frac{1}{2}=3 \frac{4}{6}=3 \frac{2}{3}\). Explain why the answer must be wrong without immediately redoing the whole calculation. Identify the error and give the correct mixed-number answer.

Hints

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

Solution

1. Estimate first: \(2 \frac{3}{4}\) is close to \(3\), and adding \(1 \frac{1}{2}\) must give a result greater than \(4\). Jakob's answer \(3 \frac{2}{3}\) is less than \(4\), so it cannot be correct. 2. Jakob 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 \(3+1 \frac{1}{4}=4 \frac{1}{4}\).

Answer

The result must be greater than \(4\), so \(3 \frac{2}{3}\) cannot be correct. Jakob added numerators and denominators separately. The correct answer is \(4 \frac{1}{4}\).
Add fractions with unlike denominators 15
A fraction can be written as a sum of unit fractions. 1) Use a common denominator to verify that \(\frac{1}{2}+\frac{1}{6}=\frac{2}{3}\). 2) Explain why writing \(\frac{2}{3}\) as \(\frac{1}{3}+\frac{1}{3}\) is easier than searching for two different unit fractions.

Hints

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

Solution

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.

Answer

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.
Add fractions with unlike denominators 16
The fraction \(\frac{2}{9}\) can be written as a sum of different unit fractions. a) Verify that \(\frac{2}{9}=\frac{1}{5}+\frac{1}{45}\). b) Verify that \(\frac{2}{9}=\frac{1}{9}+\frac{1}{10}+\frac{1}{90}\).

Hints

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

Solution

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 \(90\): \(\frac{1}{9}+\frac{1}{10}+\frac{1}{90}=\frac{10}{90}+\frac{9}{90}+\frac{1}{90}=\frac{20}{90}=\frac{2}{9}\).

Answer

a) \(\frac{9}{45}+\frac{1}{45}=\frac{10}{45}=\frac{2}{9}\) b) \(\frac{10}{90}+\frac{9}{90}+\frac{1}{90}=\frac{20}{90}=\frac{2}{9}\)
Add fractions with unlike denominators 17
Consider the fraction \(\frac{7}{12}\). a) Find two different unit fractions whose sum is \(\frac{7}{12}\). b) Write \(\frac{7}{12}\) as a sum of three different unit fractions.

Hints

- Rewrite candidate unit fractions as twelfths. - For part b), try splitting a unit fraction from part a) into two different ones. - Check that your sum is exactly seven twelfths.

Solution

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

Answer

a) For example, \(\frac{1}{3}+\frac{1}{4}\) or \(\frac{1}{2}+\frac{1}{12}\). b) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}\)
Add fractions with unlike denominators 18
Which of the following sums of different unit fractions are equal to \(\frac{2}{3}\)? Verify each sum. A) \(\frac{1}{2}+\frac{1}{6}\) B) \(\frac{1}{2}+\frac{1}{4}+\frac{1}{12}\) C) \(\frac{1}{2}+\frac{1}{10}+\frac{1}{15}\)

Hints

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

Solution

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), \(\frac{1}{2}+\frac{1}{10}+\frac{1}{15}=\frac{15}{30}+\frac{3}{30}+\frac{2}{30}=\frac{20}{30}=\frac{2}{3}\).

Answer

A) and C) are equal to \(\frac{2}{3}\).
Add fractions with unlike denominators 19
Calculate each pair of sums using common denominators. Then compare the exact values. For each part, state the exact difference between the greater and smaller sums. a) \(A=\frac{4}{11}+\frac{2}{7}\) and \(B=\frac{4}{11}+\frac{3}{7}\) b) \(C=\frac{1}{2}+\frac{1}{5}\) and \(D=\frac{1}{2}+\frac{1}{6}\) c) \(E=\frac{2}{3}+\frac{1}{3}\) and \(F=\frac{4}{9}+\frac{5}{9}\)

Hints

- For each unlike-denominator sum, rename the fractions using a common denominator before adding. - Compare the exact sums only after you have calculated them. - To find how much greater one sum is, subtract the smaller exact sum from the larger one.

Solution

1. For a), \(A=\frac{28}{77}+\frac{22}{77}=\frac{50}{77}\) and \(B=\frac{28}{77}+\frac{33}{77}=\frac{61}{77}\). Thus \(B>A\), and the difference is \(\frac{11}{77}=\frac{1}{7}\). 2. For b), \(C=\frac{5}{10}+\frac{2}{10}=\frac{7}{10}\). Also, \(D=\frac{3}{6}+\frac{1}{6}=\frac{4}{6}=\frac{2}{3}\). Using denominator \(30\), \(\frac{7}{10}-\frac{2}{3}=\frac{21}{30}-\frac{20}{30}=\frac{1}{30}\), so \(C>D\). 3. For c), \(E=\frac{3}{3}=1\) and \(F=\frac{9}{9}=1\), so the values are equal and their difference is \(0\).

Answer

a) \(A=\frac{50}{77}\), \(B=\frac{61}{77}\); \(B\) is greater by \(\frac{1}{7}\). b) \(C=\frac{7}{10}\), \(D=\frac{2}{3}\); \(C\) is greater by \(\frac{1}{30}\). c) \(E=F=1\); the difference is \(0\).
Add fractions with unlike denominators 20
Which expression is closer to \(1\)? First calculate each expression, then find its distance from \(1\). Expression A: \(\frac{2}{5}+\frac{1}{2}\) Expression B: \(\frac{3}{4}+\frac{1}{8}\)

Hints

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

Solution

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 \(1\).

Answer

Expression A is closer to \(1\). Its distance is \(\frac{1}{10}\), compared with \(\frac{1}{8}\) for Expression B.
Add fractions with unlike denominators 21
The bar shows one whole divided into three colored unit-fraction amounts. Make three different four-term unit-fraction decompositions of \(1\) by choosing one colored term and splitting it into two equal unit fractions. For each decomposition, state which original unit fraction you split.
Figure for problem 540717

Hints

- Read the colored segments as fractions of the twelve equal sections. - Split just one unit fraction at a time, keeping the other parts unchanged. - Check that every new expression has four unit fractions and totals one whole.

Solution

1. The blue part is \(\frac{6}{12}=\frac{1}{2}\). Split it into \(\frac{1}{4}+\frac{1}{4}\) to get \(1=\frac{1}{4}+\frac{1}{4}+\frac{1}{3}+\frac{1}{6}\). 2. The orange part is \(\frac{4}{12}=\frac{1}{3}\). Split it into \(\frac{1}{6}+\frac{1}{6}\) to get \(1=\frac{1}{2}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}\). 3. The green part is \(\frac{2}{12}=\frac{1}{6}\). Split it into \(\frac{1}{12}+\frac{1}{12}\) to get \(1=\frac{1}{2}+\frac{1}{3}+\frac{1}{12}+\frac{1}{12}\).

Answer

Split \(\frac{1}{2}\): \(\frac{1}{4}+\frac{1}{4}+\frac{1}{3}+\frac{1}{6}\). Split \(\frac{1}{3}\): \(\frac{1}{2}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}\). Split \(\frac{1}{6}\): \(\frac{1}{2}+\frac{1}{3}+\frac{1}{12}+\frac{1}{12}\).
Add fractions with unlike denominators 22
Start with one whole. Split it into \(\frac{1}{2}+\frac{1}{2}\). Next, replace one half with \(\frac{1}{3}+\frac{1}{6}\). Finally, replace the sixth with \(\frac{1}{12}+\frac{1}{12}\). a) List the fractions left after all the replacements. b) How many fractions are in your final list? c) Verify their total.

Hints

- Replace a term only when the instructions say that it splits. - Keep any fraction that was not replaced. - For the final total, write every fraction in twelfths.

Solution

1. One of the original halves stays as \(\frac{1}{2}\). 2. The other half becomes \(\frac{1}{3}+\frac{1}{6}\). That sixth becomes two twelfths. 3. The final fractions are \(\frac{1}{2},\frac{1}{3},\frac{1}{12},\frac{1}{12}\), making four terms. 4. Their total is \(\frac{6}{12}+\frac{4}{12}+\frac{1}{12}+\frac{1}{12}=1\).

Answer

a) \(\frac{1}{2},\frac{1}{3},\frac{1}{12},\frac{1}{12}\) b) \(4\) c) \(\frac{6}{12}+\frac{4}{12}+\frac{1}{12}+\frac{1}{12}=1\)
Add fractions with unlike denominators 23
Add \(\frac{1}{4}+\frac{2}{9}+\frac{5}{6}\). Use the least common denominator, and write the result as a mixed number in simplest form.

Hints

- Find one denominator that all three denominators divide into evenly. - Rename each fraction without changing its value. - After adding, decide whether the result should be written as a mixed number.

Solution

1. The least common denominator of \(4\), \(9\), and \(6\) is \(36\). 2. Rewrite: \(\frac{1}{4}=\frac{9}{36}\), \(\frac{2}{9}=\frac{8}{36}\), and \(\frac{5}{6}=\frac{30}{36}\). 3. Add: \(\frac{9+8+30}{36}=\frac{47}{36}=1\frac{11}{36}\).

Answer

\(1\frac{11}{36}\)
Add fractions with unlike denominators 24
Compare the two sums \(A=\frac{2}{3}+\frac{7}{10}\) and \(B=\frac{3}{4}+\frac{3}{5}\). Find both sums and determine which is greater and by how much.

Hints

- Find each sum using a convenient common denominator for its pair of fractions. - Then compare the two resulting values using a common denominator.

Solution

1. \(A=\frac{20}{30}+\frac{21}{30}=\frac{41}{30}\). 2. \(B=\frac{15}{20}+\frac{12}{20}=\frac{27}{20}\). 3. Rewrite with denominator \(60\): \(\frac{82}{60}-\frac{81}{60}=\frac{1}{60}\). 4. Therefore \(A\) is greater by \(\frac{1}{60}\).

Answer

\(A\) is greater than \(B\) by \(\frac{1}{60}\).
Add fractions with unlike denominators 25
How close is \(2\frac{1}{3}+1\frac{5}{8}\) to \(4\)? Find the exact sum and then find the fraction needed to reach \(4\).

Hints

- Use equivalent fractions for the two fractional parts. - Notice whether the fractional sum creates another whole. - Compare the final sum with \(4\) to find the remaining gap.

Solution

1. Use denominator \(24\): \(\frac{1}{3}=\frac{8}{24}\) and \(\frac{5}{8}=\frac{15}{24}\). 2. Add: \(2\frac{8}{24}+1\frac{15}{24}=3\frac{23}{24}\). 3. The amount needed to reach \(4\) is \(4-3\frac{23}{24}=\frac{1}{24}\).

Answer

The sum is \(3\frac{23}{24}\), which is \(\frac{1}{24}\) below \(4\).
Add fractions with unlike denominators 26
Find the fraction \(x\) that makes \(\frac{3}{10}+x=\frac{4}{5}\) true. Write \(x\) in simplest form and check it by addition.

Hints

- Think of the unknown as the amount needed to reach the fraction on the right. - Rewrite the fractions so the pieces have the same size before comparing or combining them. - Use the original addition equation to check your result.

Solution

1. Rewrite \(\frac{4}{5}=\frac{8}{10}\). 2. The missing addend is \(\frac{8}{10}-\frac{3}{10}=\frac{5}{10}=\frac{1}{2}\). 3. Check: \(\frac{3}{10}+\frac{1}{2}=\frac{3}{10}+\frac{5}{10}=\frac{8}{10}=\frac{4}{5}\).

Answer

\(x=\frac{1}{2}\).
Add fractions with unlike denominators 27
Find the whole number \(n\) that makes \(\frac{3}{8}+\frac{n}{12}=\frac{7}{8}\) true. Explain how equivalent fractions help you find the missing numerator.

Hints

- First determine how much must be added to \(\frac{3}{8}\) to reach \(\frac{7}{8}\). - Then rewrite that missing fraction with denominator \(12\). - Check the completed addition equation.

Solution

1. The missing addend is \(\frac{7}{8}-\frac{3}{8}=\frac{4}{8}=\frac{1}{2}\). 2. Rewrite \(\frac{1}{2}\) with denominator \(12\): \(\frac{1}{2}=\frac{6}{12}\). 3. Therefore \(n=6\). 4. Check: \(\frac{3}{8}+\frac{6}{12}=\frac{3}{8}+\frac{1}{2}=\frac{7}{8}\).

Answer

\(n=6\).
Add fractions with unlike denominators 28
Add \(\frac{5}{12}+\frac{7}{18}\). Then compare the sum with \(\frac{3}{4}\) and state by how much the sum is greater or less.

Hints

- Find a common denominator for twelfths and eighteenths. - Add the equivalent fractions before making the comparison. - Rewrite the benchmark fraction with the same denominator to find the exact gap.

Solution

1. Use denominator \(36\): \(\frac{5}{12}=\frac{15}{36}\) and \(\frac{7}{18}=\frac{14}{36}\). 2. Add: \(\frac{15}{36}+\frac{14}{36}=\frac{29}{36}\). 3. Rewrite \(\frac{3}{4}=\frac{27}{36}\). 4. The sum is greater by \(\frac{29}{36}-\frac{27}{36}=\frac{2}{36}=\frac{1}{18}\).

Answer

The sum is \(\frac{29}{36}\), which is \(\frac{1}{18}\) greater than \(\frac{3}{4}\).
Add fractions with unlike denominators 29
Is \(1\frac{3}{8}+2\frac{5}{6}\) closer to \(4\) or to \(5\)? Find the exact sum and use it to justify your answer.

Hints

- Find a common denominator for the fractional parts. - Notice whether the fractional sum creates another whole. - Compare the final mixed number with both nearby whole numbers.

Solution

1. Use denominator \(24\): \(\frac{3}{8}=\frac{9}{24}\) and \(\frac{5}{6}=\frac{20}{24}\). 2. Add: \(1\frac{9}{24}+2\frac{20}{24}=3\frac{29}{24}=4\frac{5}{24}\). 3. The sum is \(\frac{5}{24}\) above \(4\) and \(\frac{19}{24}\) below \(5\), so it is closer to \(4\).

Answer

The sum is \(4\frac{5}{24}\), which is closer to \(4\).
Add fractions with unlike denominators 30
Which sum is closer to \(1\)? Find both sums and compare their distances from \(1\). \(A=\frac{3}{8}+\frac{5}{12}\) \(B=\frac{2}{5}+\frac{4}{7}\)

Hints

- Find each sum using equivalent fractions with a common denominator. - A value’s distance from \(1\) is the amount needed to reach one whole. - Compare the two distances rather than only comparing the sums.

Solution

1. \(A=\frac{9}{24}+\frac{10}{24}=\frac{19}{24}\), so its distance from \(1\) is \(\frac{5}{24}\). 2. \(B=\frac{14}{35}+\frac{20}{35}=\frac{34}{35}\), so its distance from \(1\) is \(\frac{1}{35}\). 3. Since \(\frac{1}{35}<\frac{5}{24}\), \(B\) is closer to \(1\).

Answer

\(B\) is closer to \(1\). \(A=\frac{19}{24}\) and \(B=\frac{34}{35}\).
Add fractions with unlike denominators 31
Add \(\frac{7}{15}+\frac{5}{18}\). Then compare the sum with \(\frac{2}{3}\) and state the exact difference.

Hints

- Find a common denominator for fifteenths and eighteenths. - Add the equivalent fractions before comparing with the benchmark. - Rewrite \(\frac{2}{3}\) using the same denominator to find the exact gap.

Solution

1. Use denominator \(90\): \(\frac{7}{15}=\frac{42}{90}\) and \(\frac{5}{18}=\frac{25}{90}\). 2. Add: \(\frac{42}{90}+\frac{25}{90}=\frac{67}{90}\). 3. Rewrite \(\frac{2}{3}=\frac{60}{90}\). The sum is greater by \(\frac{67}{90}-\frac{60}{90}=\frac{7}{90}\).

Answer

The sum is \(\frac{67}{90}\), which is \(\frac{7}{90}\) greater than \(\frac{2}{3}\).
Add fractions with unlike denominators 32
Panels a) and b) show two fractions. Kai suggests renaming both using twenty-fourths, while Ren suggests using forty-eighths. a) Read each original fraction from the bars. b) Which common unit is more efficient, and why can both choices work? c) Find the sum in simplest form.
Figure for problem 554436

Hints

- Count shaded parts and total equal parts in each bar before doing any arithmetic. - A common unit must divide both original partitions evenly. - Use the smaller common multiple for efficient renaming, then add the common-sized parts.

Solution

1. Panel a) shows \(3\) of \(8\) parts shaded, so it represents \(\frac{3}{8}\). Panel b) shows \(5\) of \(12\) parts shaded, so it represents \(\frac{5}{12}\). 2. Both \(24\) and \(48\) are common multiples of \(8\) and \(12\), so both can provide equal-sized common units. Twenty-fourths are more efficient because \(24\) is the least common multiple. 3. \(\frac{3}{8}=\frac{9}{24}\) and \(\frac{5}{12}=\frac{10}{24}\), so the sum is \(\frac{19}{24}\).

Answer

a) \(\frac{3}{8}\) and \(\frac{5}{12}\) b) Twenty-fourths are more efficient because \(24\) is the least common multiple; forty-eighths also work because \(48\) is another common multiple. c) \(\frac{19}{24}\)
Add fractions with unlike denominators 33
Decide whether \(\frac{21}{40}+\frac{19}{42}\) is greater than or less than \(1\) without finding the exact sum. Explain your reasoning by comparing both fractions with \(\frac{1}{2}\).

Hints

- 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.

Solution

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 \(1\).

Answer

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.
Add fractions with unlike denominators 34
Let \(S=\frac{25}{49}+\frac{26}{53}\). Decide whether \(S>1\) or \(S<1\) without finding the exact sum. Explain your reasoning by finding and comparing each fraction's distance from \(\frac{1}{2}\).

Hints

- 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.

Solution

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 \(S>1\).

Answer

\(S>1\). 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 \(1\).
Add fractions with unlike denominators 35
Find two different ways to represent \(\frac{2}{5}\) using distinct unit fractions. 1) Write it as a sum of exactly two different unit fractions. 2) Write it as a sum of exactly three different unit fractions.

Hints

- Begin with a unit fraction smaller than \(\frac{2}{5}\). - For the first part, check whether the remainder is a unit fraction. - For the second part, try splitting the remaining amount into two different unit fractions.

Solution

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

Answer

1) \(\frac{2}{5}=\frac{1}{3}+\frac{1}{15}\) 2) \(\frac{2}{5}=\frac{1}{4}+\frac{1}{10}+\frac{1}{20}\)
Add fractions with unlike denominators 36
The fraction \(\frac{3}{10}\) has these two unit-fraction decompositions: Method A: \(\frac{1}{4}+\frac{1}{20}\) Method B: \(\frac{1}{5}+\frac{1}{10}\) a) Verify both decompositions. b) In the greedy method, you begin with the greatest unit fraction that does not exceed the original fraction. Which method follows this rule? Explain.

Hints

- Give each sum a common denominator to verify it. - Compare the two starting unit fractions. - Test whether any larger unit fraction could be the first term without exceeding the original amount.

Solution

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

Answer

a) Both methods equal \(\frac{3}{10}\). b) Method A: \(\frac{1}{4}\) is the largest unit fraction at most \(\frac{3}{10}\).
Add fractions with unlike denominators 37
In the greedy method for unit fractions, choose the greatest unit fraction that does not exceed the current fraction, subtract it, and repeat with the remainder. Use this method to write \(\frac{3}{14}\) as a sum of unit fractions. 1) Find the greatest unit fraction \(\frac{1}{k}\) that is no larger than \(\frac{3}{14}\). 2) Subtract that fraction and give the full decomposition.

Hints

- Compare the original fraction to nearby unit fractions. - The greatest allowed unit fraction must not make the subtraction negative. - After subtracting, see whether the remainder is already a unit fraction.

Solution

1. Since \(\frac{1}{4}>\frac{3}{14}\) but \(\frac{1}{5}<\frac{3}{14}\), choose \(\frac{1}{5}\). 2. Subtract: \(\frac{3}{14}-\frac{1}{5}=\frac{15}{70}-\frac{14}{70}=\frac{1}{70}\). 3. Therefore, \(\frac{3}{14}=\frac{1}{5}+\frac{1}{70}\).

Answer

1) \(\frac{1}{5}\) 2) \(\frac{3}{14}=\frac{1}{5}+\frac{1}{70}\)
Add fractions with unlike denominators 38
A unit fraction has numerator \(1\). Find two different ways to write \(\frac{1}{4}\) as a sum of two distinct unit fractions. Verify each sum.

Hints

- Try a unit fraction smaller than one fourth. - Subtract it from one fourth and see whether the remainder is also a unit fraction. - The two denominators in each sum must differ.

Solution

1. Starting with \(\frac{1}{5}\), the remainder is \(\frac{1}{4}-\frac{1}{5}=\frac{1}{20}\). So \(\frac{1}{4}=\frac{1}{5}+\frac{1}{20}\). 2. Starting with \(\frac{1}{6}\), the remainder is \(\frac{1}{4}-\frac{1}{6}=\frac{1}{12}\). So \(\frac{1}{4}=\frac{1}{6}+\frac{1}{12}\). 3. Check: \(\frac{4}{20}+\frac{1}{20}=\frac{5}{20}=\frac{1}{4}\) and \(\frac{2}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\).

Answer

\(\frac{1}{4}=\frac{1}{5}+\frac{1}{20}\) and \(\frac{1}{4}=\frac{1}{6}+\frac{1}{12}\)
Add fractions with unlike denominators 39
The greedy method writes a fraction as a sum of unit fractions by repeatedly subtracting the greatest unit fraction that is less than or equal to the current remainder. Apply the greedy method to \(\frac{4}{5}\) until you obtain a sum of three different unit fractions. Show your steps.

Hints

- At each stage, start with the greatest unit fraction that does not exceed what remains. - Use a common denominator for each subtraction. - You can stop once the remaining amount is a unit fraction.

Solution

1. The greatest unit fraction not exceeding \(\frac{4}{5}\) is \(\frac{1}{2}\). 2. Subtract: \(\frac{4}{5}-\frac{1}{2}=\frac{8}{10}-\frac{5}{10}=\frac{3}{10}\). 3. The greatest unit fraction not exceeding \(\frac{3}{10}\) is \(\frac{1}{4}\), because \(\frac{1}{3}>\frac{3}{10}\). 4. Subtract: \(\frac{3}{10}-\frac{1}{4}=\frac{6}{20}-\frac{5}{20}=\frac{1}{20}\). 5. Therefore, \(\frac{4}{5}=\frac{1}{2}+\frac{1}{4}+\frac{1}{20}\).

Answer

\(\frac{4}{5}=\frac{1}{2}+\frac{1}{4}+\frac{1}{20}\)
Add fractions with unlike denominators 40
A unit fraction has numerator \(1\). Write \(\frac{3}{4}\) as a sum of exactly three different unit fractions.

Hints

- Start with a relatively large unit fraction below \(\frac{3}{4}\). - Subtract it to find the leftover fraction. - Break that leftover fraction into two different unit fractions.

Solution

1. Start with \(\frac{1}{2}\): \(\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\). 2. Break up the remainder: \(\frac{1}{4}=\frac{1}{5}+\frac{1}{20}\). 3. So \(\frac{3}{4}=\frac{1}{2}+\frac{1}{5}+\frac{1}{20}\). The three denominators differ. 4. Another valid answer is \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}\).

Answer

One answer is \(\frac{3}{4}=\frac{1}{2}+\frac{1}{5}+\frac{1}{20}\).
Add fractions with unlike denominators 41
The bar shows one whole made from three colored unit-fraction pieces. A split replaces one unit fraction by two equal unit fractions, so the value stays the same. a) Starting from the decomposition shown, perform four splits and give one possible final decomposition. b) How many terms must every four-split result have? Explain.
Figure for problem 540737

Hints

- First identify the three fractions from the colored bar. - A split keeps the value but changes one term into two. - Track the number of terms after each split separately from their sizes.

Solution

1. The starting sum is \(\frac{1}{2}+\frac{1}{4}+\frac{1}{4}=1\), with three terms. 2. Split the half into two quarters, giving four quarter terms. 3. Split three of those quarters, one after another, into two eighths each. 4. The result is \(\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=1\). 5. Every split replaces one term by two, increasing the count by one. Three original terms plus four splits make seven terms.

Answer

a) For example, \(\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=1\). b) \(7\) terms, because each split adds one term.

All problems may be used, copied and printed free of charge for school and tutoring, including paid tutoring. Commercial adaptations as well as publication or redistribution on the internet are not permitted.

Building free math worksheets