Class 8 Mathematics worksheet on rational numbers with 15 questions covering addition, subtraction, multiplication, division, and properties of rational numbers.

Problem: Solve the given problems related to rational numbers.

#### 1. The sum of a rational number and its additive inverse is equal to ______.

- Solution:
The additive inverse of a rational number \( x \) is \( -x \).
The sum of a rational number and its additive inverse is:
\[
x + (-x) = 0
\]
Therefore, the answer is:
\[
\boxed{0}
\]

#### 2. Write the multiplicative inverse of \( -\frac{2}{3} + \frac{1}{10} \).

- Solution:
First, simplify the expression \( -\frac{2}{3} + \frac{1}{10} \):
\[
-\frac{2}{3} + \frac{1}{10} = \frac{-20}{30} + \frac{3}{30} = \frac{-20 + 3}{30} = \frac{-17}{30}
\]
The multiplicative inverse of a number \( x \) is \( \frac{1}{x} \). Therefore, the multiplicative inverse of \( \frac{-17}{30} \) is:
\[
\frac{1}{\frac{-17}{30}} = \frac{30}{-17} = -\frac{30}{17}
\]
Therefore, the answer is:
\[
\boxed{-\frac{30}{17}}
\]

#### 3. Check whether \( 2 \frac{1}{3} \) is the multiplicative inverse of \( 2.2 \).

- Solution:
Convert \( 2 \frac{1}{3} \) to an improper fraction:
\[
2 \frac{1}{3} = \frac{7}{3}
\]
Convert \( 2.2 \) to a fraction:
\[
2.2 = \frac{22}{10} = \frac{11}{5}
\]
The multiplicative inverse of \( \frac{11}{5} \) is \( \frac{5}{11} \).
Now, check if \( \frac{7}{3} \) is equal to \( \frac{5}{11} \):
\[
\frac{7}{3} \neq \frac{5}{11}
\]
Therefore, \( 2 \frac{1}{3} \) is not the multiplicative inverse of \( 2.2 \).
The answer is:
\[
\boxed{\text{No}}
\]

#### 4. Find the product of \( \frac{13}{15} \) and the additive inverse of \( -\frac{5}{26} \).

- Solution:
The additive inverse of \( -\frac{5}{26} \) is \( \frac{5}{26} \).
Now, find the product of \( \frac{13}{15} \) and \( \frac{5}{26} \):
\[
\frac{13}{15} \times \frac{5}{26} = \frac{13 \times 5}{15 \times 26} = \frac{65}{390}
\]
Simplify the fraction:
\[
\frac{65}{390} = \frac{1}{6}
\]
Therefore, the answer is:
\[
\boxed{\frac{1}{6}}
\]

#### 5. Show \( \left| \frac{7}{8} - \frac{-4}{15} \right| \) on the number line.

- Solution:
First, simplify the expression inside the absolute value:
\[
\frac{7}{8} - \frac{-4}{15} = \frac{7}{8} + \frac{4}{15}
\]
Find a common denominator for \( \frac{7}{8} \) and \( \frac{4}{15} \):
\[
\text{LCM}(8, 15) = 120
\]
Convert the fractions:
\[
\frac{7}{8} = \frac{7 \times 15}{8 \times 15} = \frac{105}{120}, \quad \frac{4}{15} = \frac{4 \times 8}{15 \times 8} = \frac{32}{120}
\]
Add the fractions:
\[
\frac{105}{120} + \frac{32}{120} = \frac{105 + 32}{120} = \frac{137}{120}
\]
The absolute value is:
\[
\left| \frac{137}{120} \right| = \frac{137}{120}
\]
To show this on the number line, mark the point \( \frac{137}{120} \), which is slightly more than 1 (since \( \frac{137}{120} \approx 1.1417 \)).

#### 6. Write the additive inverse of \( \frac{4}{8} + \frac{7}{8} + \frac{3}{8} \).

- Solution:
First, simplify the expression \( \frac{4}{8} + \frac{7}{8} + \frac{3}{8} \):
\[
\frac{4}{8} + \frac{7}{8} + \frac{3}{8} = \frac{4 + 7 + 3}{8} = \frac{14}{8} = \frac{7}{4}
\]
The additive inverse of \( \frac{7}{4} \) is \( -\frac{7}{4} \).
Therefore, the answer is:
\[
\boxed{-\frac{7}{4}}
\]

#### 7. Write four rational numbers between \( -\frac{2}{3} \) and \( -\frac{3}{4} \).

- Solution:
Convert \( -\frac{2}{3} \) and \( -\frac{3}{4} \) to decimals for easier comparison:
\[
-\frac{2}{3} \approx -0.6667, \quad -\frac{3}{4} = -0.75
\]
Four rational numbers between \( -0.6667 \) and \( -0.75 \) are:
\[
-\frac{13}{20}, -\frac{14}{20}, -\frac{15}{20}, -\frac{16}{20}
\]
Simplify these fractions:
\[
-\frac{13}{20}, -\frac{7}{10}, -\frac{3}{4}, -\frac{4}{5}
\]
Therefore, the answer is:
\[
\boxed{-\frac{13}{20}, -\frac{7}{10}, -\frac{3}{4}, -\frac{4}{5}}
\]

#### 8. Divide the sum of \( 5 \frac{5}{13} \) and \( 2 \frac{3}{5} \) by their difference.

- Solution:
Convert the mixed numbers to improper fractions:
\[
5 \frac{5}{13} = \frac{70}{13}, \quad 2 \frac{3}{5} = \frac{13}{5}
\]
Find the sum:
\[
\frac{70}{13} + \frac{13}{5} = \frac{70 \times 5}{13 \times 5} + \frac{13 \times 13}{5 \times 13} = \frac{350}{65} + \frac{169}{65} = \frac{350 + 169}{65} = \frac{519}{65}
\]
Find the difference:
\[
\frac{70}{13} - \frac{13}{5} = \frac{70 \times 5}{13 \times 5} - \frac{13 \times 13}{5 \times 13} = \frac{350}{65} - \frac{169}{65} = \frac{350 - 169}{65} = \frac{181}{65}
\]
Divide the sum by the difference:
\[
\frac{\frac{519}{65}}{\frac{181}{65}} = \frac{519}{65} \times \frac{65}{181} = \frac{519}{181}
\]
Therefore, the answer is:
\[
\boxed{\frac{519}{181}}
\]

#### 9. By what number should \( -5 \frac{1}{2} \) be divided to get \( \frac{11}{8} \)?

- Solution:
Convert \( -5 \frac{1}{2} \) to an improper fraction:
\[
-5 \frac{1}{2} = -\frac{11}{2}
\]
Let the unknown number be \( x \). We need to solve:
\[
-\frac{11}{2} \div x = \frac{11}{8}
\]
This can be rewritten as:
\[
-\frac{11}{2} \times \frac{1}{x} = \frac{11}{8}
\]
Simplify:
\[
\frac{-11}{2x} = \frac{11}{8}
\]
Cross-multiply:
\[
-11 \times 8 = 11 \times 2x \implies -88 = 22x \implies x = \frac{-88}{22} = -4
\]
Therefore, the answer is:
\[
\boxed{-4}
\]

#### 10. What should be subtracted from \( \frac{7}{12} - \frac{2}{3} \) to get \( -\frac{1}{6} \)?

- Solution:
First, simplify \( \frac{7}{12} - \frac{2}{3} \):
\[
\frac{7}{12} - \frac{2}{3} = \frac{7}{12} - \frac{8}{12} = \frac{7 - 8}{12} = \frac{-1}{12}
\]
Let the number to be subtracted be \( x \). We need to solve:
\[
\frac{-1}{12} - x = -\frac{1}{6}
\]
Rearrange to solve for \( x \):
\[
-x = -\frac{1}{6} - \left( \frac{-1}{12} \right) = -\frac{1}{6} + \frac{1}{12}
\]
Find a common denominator:
\[
-\frac{1}{6} + \frac{1}{12} = -\frac{2}{12} + \frac{1}{12} = \frac{-2 + 1}{12} = \frac{-1}{12}
\]
Therefore:
\[
-x = \frac{-1}{12} \implies x = \frac{1}{12}
\]
Therefore, the answer is:
\[
\boxed{\frac{1}{12}}
\]

#### 11. Simplify: \( \frac{7}{4} - \frac{3}{12} \times \frac{-24}{30} + 1 < 1 \)

- Solution:
Simplify step by step:
\[
\frac{7}{4} - \frac{3}{12} \times \frac{-24}{30} + 1
\]
Simplify \( \frac{3}{12} \):
\[
\frac{3}{12} = \frac{1}{4}
\]
Simplify \( \frac{-24}{30} \):
\[
\frac{-24}{30} = \frac{-4}{5}
\]
Now, calculate \( \frac{1}{4} \times \frac{-4}{5} \):
\[
\frac{1}{4} \times \frac{-4}{5} = \frac{1 \times -4}{4 \times 5} = \frac{-4}{20} = \frac{-1}{5}
\]
Substitute back:
\[
\frac{7}{4} - \left( \frac{-1}{5} \right) + 1 = \frac{7}{4} + \frac{1}{5} + 1
\]
Convert to a common denominator:
\[
\frac{7}{4} = \frac{35}{20}, \quad \frac{1}{5} = \frac{4}{20}, \quad 1 = \frac{20}{20}
\]
Add the fractions:
\[
\frac{35}{20} + \frac{4}{20} + \frac{20}{20} = \frac{35 + 4 + 20}{20} = \frac{59}{20}
\]
Compare with 1:
\[
\frac{59}{20} > 1
\]
Therefore, the inequality is not satisfied. The answer is:
\[
\boxed{\text{False}}
\]

#### 12. Simplify by using suitable property:
- I. \( \left( \frac{-5}{7} \times \frac{8}{15} \right) \times \frac{15}{8} \)
- II. \( \left( \frac{-5}{9} \times \frac{4}{15} \right) \times \frac{3}{5} \times \frac{-5}{9} \)
- III. \( -3 \frac{3}{4} \times \frac{-3}{2} + \frac{-3}{2} \times \frac{-7}{4} \)
- IV. \( \frac{2}{3} \times \frac{3}{7} - \frac{3}{14} \times \frac{3}{3} \)

- Solution for I:
\[
\left( \frac{-5}{7} \times \frac{8}{15} \right) \times \frac{15}{8} = \frac{-5}{7} \times \left( \frac{8}{15} \times \frac{15}{8} \right) = \frac{-5}{7} \times 1 = \frac{-5}{7}
\]
Therefore, the answer is:
\[
\boxed{\frac{-5}{7}}
\]

- Solution for II:
\[
\left( \frac{-5}{9} \times \frac{4}{15} \right) \times \frac{3}{5} \times \frac{-5}{9} = \frac{-5}{9} \times \frac{4}{15} \times \frac{3}{5} \times \frac{-5}{9}
\]
Simplify step by step:
\[
\frac{-5 \times 4 \times 3 \times -5}{9 \times 15 \times 5 \times 9} = \frac{300}{6075} = \frac{4}{81}
\]
Therefore, the answer is:
\[
\boxed{\frac{4}{81}}
\]

- Solution for III:
Convert \( -3 \frac{3}{4} \) to an improper fraction:
\[
-3 \frac{3}{4} = -\frac{15}{4}
\]
Use the distributive property:
\[
-\frac{15}{4} \times \frac{-3}{2} + \frac{-3}{2} \times \frac{-7}{4} = \frac{-3}{2} \left( -\frac{15}{4} + \frac{-7}{4} \right) = \frac{-3}{2} \left( \frac{-15 - 7}{4} \right) = \frac{-3}{2} \left( \frac{-22}{4} \right) = \frac{-3}{2} \times \frac{-11}{2} = \frac{33}{4}
\]
Therefore, the answer is:
\[
\boxed{\frac{33}{4}}
\]

- Solution for IV:
Simplify each term:
\[
\frac{2}{3} \times \frac{3}{7} - \frac{3}{14} \times \frac{3}{3} = \frac{2 \times 3}{3 \times 7} - \frac{3 \times 3}{14 \times 3} = \frac{6}{21} - \frac{9}{42} = \frac{2}{7} - \frac{3}{14}
\]
Find a common denominator:
\[
\frac{2}{7} = \frac{4}{14}, \quad \frac{3}{14} = \frac{3}{14}
\]
Subtract the fractions:
\[
\frac{4}{14} - \frac{3}{14} = \frac{1}{14}
\]
Therefore, the answer is:
\[
\boxed{\frac{1}{14}}
\]

#### 13. The sum of two rational numbers is \( -3 \). If one of them is \( -\frac{10}{3} \), find the other number.

- Solution:
Let the other number be \( x \). We know:
\[
x + \left( -\frac{10}{3} \right) = -3
\]
Simplify:
\[
x - \frac{10}{3} = -3
\]
Add \( \frac{10}{3} \) to both sides:
\[
x = -3 + \frac{10}{3}
\]
Convert \( -3 \) to a fraction:
\[
-3 = \frac{-9}{3}
\]
Add the fractions:
\[
x = \frac{-9}{3} + \frac{10}{3} = \frac{-9 + 10}{3} = \frac{1}{3}
\]
Therefore, the answer is:
\[
\boxed{\frac{1}{3}}
\]

#### 14. The product of two rational numbers is \( -\frac{28}{27} \). If one of them is \( -\frac{4}{9} \), find the other number.

- Solution:
Let the other number be \( x \). We know:
\[
x \times \left( -\frac{4}{9} \right) = -\frac{28}{27}
\]
Solve for \( x \):
\[
x = \frac{-\frac{28}{27}}{-\frac{4}{9}} = \frac{-28}{27} \times \frac{9}{-4} = \frac{28 \times 9}{27 \times 4} = \frac{252}{108} = \frac{7}{3}
\]
Therefore, the answer is:
\[
\boxed{\frac{7}{3}}
\]

#### 15. Find the area of a square park with each side \( 8 \frac{1}{2} \) m.

- Solution:
Convert \( 8 \frac{1}{2} \) to an improper fraction:
\[
8 \frac{1}{2} = \frac{17}{2}
\]
The area of a square is given by \( \text{side}^2 \):
\[
\left( \frac{17}{2} \right)^2 = \frac{17^2}{2^2} = \frac{289}{4}
\]
Convert to a mixed number:
\[
\frac{289}{4} = 72 \frac{1}{4}
\]
Therefore, the answer is:
\[
\boxed{72 \frac{1}{4}}
\]

Final Answers:

1. \(\boxed{0}\)
2. \(\boxed{-\frac{30}{17}}\)
3. \(\boxed{\text{No}}\)
4. \(\boxed{\frac{1}{6}}\)
5. \(\boxed{\frac{137}{120}}\) (on the number line)
6. \(\boxed{-\frac{7}{4}}\)
7. \(\boxed{-\frac{13}{20}, -\frac{7}{10}, -\frac{3}{4}, -\frac{4}{5}}\)
8. \(\boxed{\frac{519}{181}}\)
9. \(\boxed{-4}\)
10. \(\boxed{\frac{1}{12}}\)
11. \(\boxed{\text{False}}\)
12. I. \(\boxed{\frac{-5}{7}}\), II. \(\boxed{\frac{4}{81}}\), III. \(\boxed{\frac{33}{4}}\), IV. \(\boxed{\frac{1}{14}}\)
13. \(\boxed{\frac{1}{3}}\)
14. \(\boxed{\frac{7}{3}}\)
15. \(\boxed{72 \frac{1}{4}}\)