Adding and subtracting fractions word problems for Grade 4 students, with problems involving class attendance, construction paper, and pencils.

Problem Analysis:

The worksheet contains three word problems involving fractions. Let's solve each problem step by step.

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Problem 1:

Question: On Friday, many of the students were missing in her class. $\frac{1}{4}$ of the class went to a basketball tournament and $\frac{1}{5}$ of the class called in sick. What fraction of the class was in school?

#### Solution:
1. Identify the fractions of students who are absent:
- Students who went to the basketball tournament: $\frac{1}{4}$
- Students who called in sick: $\frac{1}{5}$

2. Find the total fraction of students who are absent:
To add these fractions, we need a common denominator. The least common denominator (LCD) of 4 and 5 is 20.
\[
\frac{1}{4} = \frac{1 \times 5}{4 \times 5} = \frac{5}{20}
\]
\[
\frac{1}{5} = \frac{1 \times 4}{5 \times 4} = \frac{4}{20}
\]
Now, add the fractions:
\[
\frac{1}{4} + \frac{1}{5} = \frac{5}{20} + \frac{4}{20} = \frac{9}{20}
\]

3. Determine the fraction of students who are in school:
Since the total fraction of the class is 1 (or $\frac{20}{20}$), the fraction of students in school is:
\[
1 - \frac{9}{20} = \frac{20}{20} - \frac{9}{20} = \frac{11}{20}
\]

#### Final Answer for Problem 1:
\[
\boxed{\frac{11}{20}}
\]

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Problem 2:

Question: Ms. Bloom has $\frac{15}{8}$ of a pack of construction paper at her desk. She takes another 2 new packs of construction paper to the class for an art project. The students use $1 \frac{3}{4}$ packs for the art project. How many packs of construction papers are left?

#### Solution:
1. Convert mixed numbers to improper fractions:
- $1 \frac{3}{4} = \frac{7}{4}$ (since $1 \frac{3}{4} = \frac{4}{4} + \frac{3}{4} = \frac{7}{4}$)

2. Calculate the total number of packs Ms. Bloom initially had:
- She starts with $\frac{15}{8}$ packs.
- She adds 2 new packs: $2 = \frac{16}{8}$ (since $2 = \frac{16}{8}$).
- Total packs before the project:
\[
\frac{15}{8} + \frac{16}{8} = \frac{31}{8}
\]

3. Subtract the packs used by the students:
- The students used $\frac{7}{4}$ packs.
- Convert $\frac{7}{4}$ to eighths for consistency:
\[
\frac{7}{4} = \frac{7 \times 2}{4 \times 2} = \frac{14}{8}
\]
- Subtract the used packs from the total:
\[
\frac{31}{8} - \frac{14}{8} = \frac{17}{8}
\]

4. Convert the result back to a mixed number (if needed):
- $\frac{17}{8} = 2 \frac{1}{8}$ (since $17 \div 8 = 2$ remainder $1$).

#### Final Answer for Problem 2:
\[
\boxed{2 \frac{1}{8}}
\]

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Problem 3:

Question: Ms. Bloom had $4 \frac{7}{12}$ boxes of pencils but $2 \frac{1}{4}$ boxes of the pencils were broken. After she threw out the broken pencils, how many boxes of pencils were left?

#### Solution:
1. Convert mixed numbers to improper fractions:
- $4 \frac{7}{12} = \frac{55}{12}$ (since $4 \frac{7}{12} = \frac{48}{12} + \frac{7}{12} = \frac{55}{12}$)
- $2 \frac{1}{4} = \frac{9}{4}$ (since $2 \frac{1}{4} = \frac{8}{4} + \frac{1}{4} = \frac{9}{4}$)

2. Find a common denominator to subtract the fractions:
- The denominators are 12 and 4. The LCD is 12.
- Convert $\frac{9}{4}$ to twelfths:
\[
\frac{9}{4} = \frac{9 \times 3}{4 \times 3} = \frac{27}{12}
\]

3. Subtract the broken boxes from the total boxes:
\[
\frac{55}{12} - \frac{27}{12} = \frac{28}{12}
\]

4. Simplify the result:
- $\frac{28}{12}$ can be simplified by dividing both numerator and denominator by their greatest common divisor (GCD), which is 4:
\[
\frac{28}{12} = \frac{28 \div 4}{12 \div 4} = \frac{7}{3}
\]

5. Convert the improper fraction to a mixed number:
- $\frac{7}{3} = 2 \frac{1}{3}$ (since $7 \div 3 = 2$ remainder $1$).

#### Final Answer for Problem 3:
\[
\boxed{2 \frac{1}{3}}
\]

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Final Answers:

1. $\boxed{\frac{11}{20}}$
2. $\boxed{2 \frac{1}{8}}$
3. $\boxed{2 \frac{1}{3}}$