Rational Number- Grade VIII Part -I

Rational Numbers Worksheet

1. Question: Subtract \( \frac{-11}{5} \) from additive inverse of sum of \( \frac{-5}{8} \) and \( \frac{5}{6} \).

   Solution: \( -\left(\frac{-5}{8} + \frac{5}{6}\right) - \frac{-11}{5} = -\left(\frac{-15 + 20}{24}\right) + \frac{11}{5} = -\left(\frac{5}{24}\right) + \frac{11}{5} = -\frac{5}{24} + \frac{132}{24} = \frac{127}{24} \).

2. Question: Subtract the sum of \( \frac{-4}{7} \) and \( \frac{5}{14} \) from the sum of \( \frac{9}{14} \) and \( \frac{23}{28} \).

   Solution: \( \left(\frac{9}{14} + \frac{23}{28}\right) - \left(\frac{-4}{7} + \frac{5}{14}\right) = \frac{18 + 46}{28} - \left(\frac{-8 + 5}{14}\right) = \frac{64}{28} - \left(\frac{-3}{14}\right) = \frac{64}{28} + \frac{3}{14} = \frac{71}{14} \).

3. Question: What number should be added to \( \frac{-7}{8} \) so as to get \( \frac{5}{9} \)?

   Solution: Let the number be \( x \). Therefore, \( x + \frac{-7}{8} = \frac{5}{9} \). Solving for \( x \), \( x = \frac{5}{9} + \frac{7}{8} = \frac{40 + 63}{72} = \frac{103}{72} \).

4. Question: Verify the distributive law of multiplication over subtraction for the rational numbers \( x = \frac{-3}{4} \), \( y = \frac{5}{7} \), \( z = \frac{-4}{5} \).

   Solution: \( x(y - z) = x \cdot y - x \cdot z = \frac{-3}{4} \cdot \frac{5}{7} - \frac{-3}{4} \cdot \frac{-4}{5} = \frac{-15}{28} + \frac{12}{20} = \frac{-15}{28} + \frac{42}{28} = \frac{27}{28} \).

5. Question: Write 6 rational numbers between \( \frac{3}{-7} \) and \( \frac{2}{5} \).

   Solution: Six rational numbers between \( \frac{3}{-7} \) and \( \frac{2}{5} \) are \( \frac{-2}{7} \), \( \frac{-1}{7} \), \( 0 \), \( \frac{1}{5} \), \( \frac{2}{7} \), and \( \frac{3}{7} \).

6. Question: Write in standard form (a) \( \frac{12}{-240} \) (b) \( \frac{-19}{362} \).

   Solution: (a) \( \frac{12}{-240} = \frac{-1}{20} \) (b) \( \frac{-19}{362} \) is already in standard form.

7. Question: Using appropriate property find \( \frac{-2}{3} \times \frac{3}{5} + \frac{5}{2} \times \frac{4}{7} + \frac{3}{5} \times \frac{1}{6} \).

   Solution: \( \frac{-2}{3} \times \frac{3}{5} + \frac{5}{2} \times \frac{4}{7} + \frac{3}{5} \times \frac{1}{6} = \frac{-2}{5} + \frac{10}{7} + \frac{1}{10} = \frac{-14}{35} + \frac{50}{35} + \frac{7}{35} = \frac{43}{35} \).

8. Question: What should be subtracted from the product of \( \frac{3}{7} \) and \( \frac{2}{5} \) to get \( \frac{4}{35} \)?

   Solution: Let the number be \( x \). Therefore, \( \frac{3}{7} \times \frac{2}{5} - x = \frac{4}{35} \). Solving for \( x \), \( x = \frac{3}{7} \times \frac{2}{5} - \frac{4}{35} = \frac{6}{35} - \frac{4}{35} = \frac{2}{35} \).

9. Question: One sixth of the students of a class joined the sports club. Three fifth of these students opted to play table tennis. If 6 students play table tennis, how many students are there in the class?

   Solution: Let the total number of students be \( x \). Therefore, \( \frac{1}{6}x \times \frac{3}{5} = 6 \). Solving for \( x \), \( x = \frac{6 \times 6 \times 5}{3} = 60 \). So, there are 60 students in the class.

10. Question: Simplify \( \left(\frac{10}{21} \times \frac{-7}{5}\right) - \left(\frac{2}{3} \times \frac{9}{-16}\right) + \left(\frac{-4}{15} \times \frac{-6}{16}\right) \).

    Solution: \( \left(\frac{10}{21} \times \frac{-7}{5}\right) - \left(\frac{2}{3} \times \frac{9}{-16}\right) + \left(\frac{-4}{15} \times \frac{-6}{16}\right) = \frac{-70}{105} + \frac{3}{8} + \frac{1}{10} = \frac{-56}{105} + \frac{42}{105} + \frac{21}{105} = \frac{7}{105} \).

11. Question: Is \( \frac{4}{11} \) the multiplication inverse of \( -3 \frac{3}{4} \)? Give reasons.

    Solution: \( -3 \frac{3}{4} \) is equal to \( -\frac{15}{4} \) and its multiplicative inverse is \( -\frac{4}{15} \). Since \( -\frac{4}{15} \) is not equal to \( \frac{4}{11} \), \( \frac{4}{11} \) is not the multiplicative inverse of \( -3 \frac{3}{4} \).