Class IX Mathematics worksheet on rational numbers with problems involving addition, subtraction, multiplication, and simplification.

Problem: Solve the given problems related to rational numbers.

#### 1. Reduce in the standard form:

We need to simplify the given fractions to their simplest forms.

(i) \(\frac{36}{-24}\)

- Simplify by finding the greatest common divisor (GCD) of 36 and 24, which is 12.
- Divide both the numerator and the denominator by 12:
\[
\frac{36}{-24} = \frac{36 \div 12}{-24 \div 12} = \frac{3}{-2} = -\frac{3}{2}
\]

(ii) \(\frac{-3}{-15}\)

- Simplify by finding the GCD of 3 and 15, which is 3.
- Divide both the numerator and the denominator by 3:
\[
\frac{-3}{-15} = \frac{-3 \div 3}{-15 \div 3} = \frac{-1}{-5} = \frac{1}{5}
\]

Solution for Question 1:
\[
\boxed{-\frac{3}{2}, \frac{1}{5}}
\]

---

#### 2. Find the sum:

We need to add the given rational numbers.

(i) \(\frac{5}{4} + \left(-\frac{11}{4}\right)\)

- Since the denominators are the same, we can directly add the numerators:
\[
\frac{5}{4} + \left(-\frac{11}{4}\right) = \frac{5 + (-11)}{4} = \frac{5 - 11}{4} = \frac{-6}{4}
\]
- Simplify \(\frac{-6}{4}\) by dividing both the numerator and the denominator by their GCD, which is 2:
\[
\frac{-6}{4} = \frac{-6 \div 2}{4 \div 2} = \frac{-3}{2}
\]

(ii) \(\frac{5}{3} + \frac{3}{5}\)

- The denominators are different, so we need a common denominator. The least common multiple (LCM) of 3 and 5 is 15.
- Rewrite each fraction with the common denominator:
\[
\frac{5}{3} = \frac{5 \times 5}{3 \times 5} = \frac{25}{15}, \quad \frac{3}{5} = \frac{3 \times 3}{5 \times 3} = \frac{9}{15}
\]
- Add the fractions:
\[
\frac{5}{3} + \frac{3}{5} = \frac{25}{15} + \frac{9}{15} = \frac{25 + 9}{15} = \frac{34}{15}
\]

(iii) \(\frac{-9}{10} + \frac{22}{15}\)

- The denominators are different, so we need a common denominator. The LCM of 10 and 15 is 30.
- Rewrite each fraction with the common denominator:
\[
\frac{-9}{10} = \frac{-9 \times 3}{10 \times 3} = \frac{-27}{30}, \quad \frac{22}{15} = \frac{22 \times 2}{15 \times 2} = \frac{44}{30}
\]
- Add the fractions:
\[
\frac{-9}{10} + \frac{22}{15} = \frac{-27}{30} + \frac{44}{30} = \frac{-27 + 44}{30} = \frac{17}{30}
\]

(iv) \(\frac{-3}{11} + \frac{5}{9}\)

- The denominators are different, so we need a common denominator. The LCM of 11 and 9 is 99.
- Rewrite each fraction with the common denominator:
\[
\frac{-3}{11} = \frac{-3 \times 9}{11 \times 9} = \frac{-27}{99}, \quad \frac{5}{9} = \frac{5 \times 11}{9 \times 11} = \frac{55}{99}
\]
- Add the fractions:
\[
\frac{-3}{11} + \frac{5}{9} = \frac{-27}{99} + \frac{55}{99} = \frac{-27 + 55}{99} = \frac{28}{99}
\]

(v) \(\frac{-8}{19} + \left(-\frac{2}{57}\right)\)

- The denominators are different, so we need a common denominator. The LCM of 19 and 57 is 57.
- Rewrite each fraction with the common denominator:
\[
\frac{-8}{19} = \frac{-8 \times 3}{19 \times 3} = \frac{-24}{57}, \quad \left(-\frac{2}{57}\right) = \frac{-2}{57}
\]
- Add the fractions:
\[
\frac{-8}{19} + \left(-\frac{2}{57}\right) = \frac{-24}{57} + \frac{-2}{57} = \frac{-24 - 2}{57} = \frac{-26}{57}
\]

(vi) \(\frac{-2}{3} + \frac{4}{5}\)

- The denominators are different, so we need a common denominator. The LCM of 3 and 5 is 15.
- Rewrite each fraction with the common denominator:
\[
\frac{-2}{3} = \frac{-2 \times 5}{3 \times 5} = \frac{-10}{15}, \quad \frac{4}{5} = \frac{4 \times 3}{5 \times 3} = \frac{12}{15}
\]
- Add the fractions:
\[
\frac{-2}{3} + \frac{4}{5} = \frac{-10}{15} + \frac{12}{15} = \frac{-10 + 12}{15} = \frac{2}{15}
\]

Solution for Question 2:
\[
\boxed{-\frac{3}{2}, \frac{34}{15}, \frac{17}{30}, \frac{28}{99}, -\frac{26}{57}, \frac{2}{15}}
\]

---

#### 3. Find:

We need to perform the operations as specified.

(i) \(\frac{7}{24} - \frac{17}{36}\)

- The denominators are different, so we need a common denominator. The LCM of 24 and 36 is 72.
- Rewrite each fraction with the common denominator:
\[
\frac{7}{24} = \frac{7 \times 3}{24 \times 3} = \frac{21}{72}, \quad \frac{17}{36} = \frac{17 \times 2}{36 \times 2} = \frac{34}{72}
\]
- Subtract the fractions:
\[
\frac{7}{24} - \frac{17}{36} = \frac{21}{72} - \frac{34}{72} = \frac{21 - 34}{72} = \frac{-13}{72}
\]

(ii) \(\frac{5}{3} \times \left(-\frac{6}{21}\right)\)

- Multiply the numerators and the denominators:
\[
\frac{5}{3} \times \left(-\frac{6}{21}\right) = \frac{5 \times (-6)}{3 \times 21} = \frac{-30}{63}
\]
- Simplify \(\frac{-30}{63}\) by dividing both the numerator and the denominator by their GCD, which is 3:
\[
\frac{-30}{63} = \frac{-30 \div 3}{63 \div 3} = \frac{-10}{21}
\]

(iii) \(\frac{-6}{13} \div \left(-\frac{7}{15}\right)\)

- Dividing by a fraction is equivalent to multiplying by its reciprocal:
\[
\frac{-6}{13} \div \left(-\frac{7}{15}\right) = \frac{-6}{13} \times \left(-\frac{15}{7}\right) = \frac{-6 \times (-15)}{13 \times 7} = \frac{90}{91}
\]

(iv) \(\frac{-3}{8} \div \frac{7}{11}\)

- Dividing by a fraction is equivalent to multiplying by its reciprocal:
\[
\frac{-3}{8} \div \frac{7}{11} = \frac{-3}{8} \times \frac{11}{7} = \frac{-3 \times 11}{8 \times 7} = \frac{-33}{56}
\]

(v) \(\frac{-9}{5} \div \frac{3}{5}\)

- Dividing by a fraction is equivalent to multiplying by its reciprocal:
\[
\frac{-9}{5} \div \frac{3}{5} = \frac{-9}{5} \times \frac{5}{3} = \frac{-9 \times 5}{5 \times 3} = \frac{-45}{15}
\]
- Simplify \(\frac{-45}{15}\) by dividing both the numerator and the denominator by their GCD, which is 15:
\[
\frac{-45}{15} = \frac{-45 \div 15}{15 \div 15} = -3
\]

Solution for Question 3:
\[
\boxed{\frac{-13}{72}, \frac{-10}{21}, \frac{90}{91}, \frac{-33}{56}, -3}
\]

---

#### 4. Find the product:

We need to multiply the given rational numbers.

(i) \(\frac{9}{2} \times \left(-\frac{7}{4}\right)\)

- Multiply the numerators and the denominators:
\[
\frac{9}{2} \times \left(-\frac{7}{4}\right) = \frac{9 \times (-7)}{2 \times 4} = \frac{-63}{8}
\]

(ii) \(\frac{3}{10} \times (-9)\)

- Rewrite \(-9\) as \(-\frac{9}{1}\):
\[
\frac{3}{10} \times (-9) = \frac{3}{10} \times \left(-\frac{9}{1}\right) = \frac{3 \times (-9)}{10 \times 1} = \frac{-27}{10}
\]

(iii) \(\frac{-6}{5} \times \frac{9}{11}\)

- Multiply the numerators and the denominators:
\[
\frac{-6}{5} \times \frac{9}{11} = \frac{-6 \times 9}{5 \times 11} = \frac{-54}{55}
\]

(iv) \(\frac{3}{5} \times \left(-\frac{2}{3}\right)\)

- Multiply the numerators and the denominators:
\[
\frac{3}{5} \times \left(-\frac{2}{3}\right) = \frac{3 \times (-2)}{5 \times 3} = \frac{-6}{15}
\]
- Simplify \(\frac{-6}{15}\) by dividing both the numerator and the denominator by their GCD, which is 3:
\[
\frac{-6}{15} = \frac{-6 \div 3}{15 \div 3} = \frac{-2}{5}
\]

(v) \(\frac{3}{17} \times \frac{2}{5}\)

- Multiply the numerators and the denominators:
\[
\frac{3}{17} \times \frac{2}{5} = \frac{3 \times 2}{17 \times 5} = \frac{6}{85}
\]

(vi) \(\frac{3}{-5} \times \frac{-5}{3}\)

- Multiply the numerators and the denominators:
\[
\frac{3}{-5} \times \frac{-5}{3} = \frac{3 \times (-5)}{-5 \times 3} = \frac{-15}{-15} = 1
\]

Solution for Question 4:
\[
\boxed{\frac{-63}{8}, \frac{-27}{10}, \frac{-54}{55}, \frac{-2}{5}, \frac{6}{85}, 1}
\]

---

#### 5. Find the value of:

We need to perform the operations as specified.

(i) \((-4) \div \frac{2}{3}\)

- Dividing by a fraction is equivalent to multiplying by its reciprocal:
\[
(-4) \div \frac{2}{3} = (-4) \times \frac{3}{2} = \frac{-4 \times 3}{2} = \frac{-12}{2} = -6
\]

(ii) \(\frac{-3}{5} + 2\)

- Rewrite 2 as \(\frac{2}{1}\):
\[
\frac{-3}{5} + 2 = \frac{-3}{5} + \frac{2}{1} = \frac{-3}{5} + \frac{2 \times 5}{1 \times 5} = \frac{-3}{5} + \frac{10}{5} = \frac{-3 + 10}{5} = \frac{7}{5}
\]

(iii) \(\frac{-4}{5} \div (-3)\)

- Dividing by a number is equivalent to multiplying by its reciprocal:
\[
\frac{-4}{5} \div (-3) = \frac{-4}{5} \times \left(-\frac{1}{3}\right) = \frac{-4 \times (-1)}{5 \times 3} = \frac{4}{15}
\]

(iv) \(\frac{-1}{8} \div \frac{3}{4}\)

- Dividing by a fraction is equivalent to multiplying by its reciprocal:
\[
\frac{-1}{8} \div \frac{3}{4} = \frac{-1}{8} \times \frac{4}{3} = \frac{-1 \times 4}{8 \times 3} = \frac{-4}{24}
\]
- Simplify \(\frac{-4}{24}\) by dividing both the numerator and the denominator by their GCD, which is 4:
\[
\frac{-4}{24} = \frac{-4 \div 4}{24 \div 4} = \frac{-1}{6}
\]

(v) \(\frac{-2}{13} + \frac{1}{7}\)

- The denominators are different, so we need a common denominator. The LCM of 13 and 7 is 91.
- Rewrite each fraction with the common denominator:
\[
\frac{-2}{13} = \frac{-2 \times 7}{13 \times 7} = \frac{-14}{91}, \quad \frac{1}{7} = \frac{1 \times 13}{7 \times 13} = \frac{13}{91}
\]
- Add the fractions:
\[
\frac{-2}{13} + \frac{1}{7} = \frac{-14}{91} + \frac{13}{91} = \frac{-14 + 13}{91} = \frac{-1}{91}
\]

(vi) \(\frac{-7}{12} + \left(-\frac{2}{13}\right)\)

- The denominators are different, so we need a common denominator. The LCM of 12 and 13 is 156.
- Rewrite each fraction with the common denominator:
\[
\frac{-7}{12} = \frac{-7 \times 13}{12 \times 13} = \frac{-91}{156}, \quad \left(-\frac{2}{13}\right) = \frac{-2 \times 12}{13 \times 12} = \frac{-24}{156}
\]
- Add the fractions:
\[
\frac{-7}{12} + \left(-\frac{2}{13}\right) = \frac{-91}{156} + \frac{-24}{156} = \frac{-91 - 24}{156} = \frac{-115}{156}
\]

Solution for Question 5:
\[
\boxed{-6, \frac{7}{5}, \frac{4}{15}, \frac{-1}{6}, \frac{-1}{91}, \frac{-115}{156}}
\]

---

#### 6. Find the value of:

We need to perform the operations as specified.

(i) \(-\frac{2}{3} \times \frac{3}{5} \times \frac{5}{7} \times \frac{1}{3} \times \frac{1}{6}\)

- Multiply the numerators and the denominators:
\[
-\frac{2}{3} \times \frac{3}{5} \times \frac{5}{7} \times \frac{1}{3} \times \frac{1}{6} = \frac{-2 \times 3 \times 5 \times 1 \times 1}{3 \times 5 \times 7 \times 3 \times 6}
\]
- Simplify step-by-step:
\[
\frac{-2 \times 3 \times 5 \times 1 \times 1}{3 \times 5 \times 7 \times 3 \times 6} = \frac{-2 \times 1 \times 1 \times 1 \times 1}{1 \times 1 \times 7 \times 3 \times 6} = \frac{-2}{126}
\]
- Simplify \(\frac{-2}{126}\) by dividing both the numerator and the denominator by their GCD, which is 2:
\[
\frac{-2}{126} = \frac{-2 \div 2}{126 \div 2} = \frac{-1}{63}
\]

(ii) \(\frac{2}{3} \times \left(-\frac{3}{7}\right) - \frac{1}{6} \times \frac{2}{7} + \frac{2}{14}\)

- Perform each multiplication separately:
\[
\frac{2}{3} \times \left(-\frac{3}{7}\right) = \frac{2 \times (-3)}{3 \times 7} = \frac{-6}{21} = \frac{-2}{7}
\]
\[
\frac{1}{6} \times \frac{2}{7} = \frac{1 \times 2}{6 \times 7} = \frac{2}{42} = \frac{1}{21}
\]
\[
\frac{2}{14} = \frac{1}{7}
\]
- Substitute back into the expression:
\[
\frac{2}{3} \times \left(-\frac{3}{7}\right) - \frac{1}{6} \times \frac{2}{7} + \frac{2}{14} = \frac{-2}{7} - \frac{1}{21} + \frac{1}{7}
\]
- Combine the fractions. The LCM of 7 and 21 is 21:
\[
\frac{-2}{7} = \frac{-2 \times 3}{7 \times 3} = \frac{-6}{21}, \quad \frac{1}{7} = \frac{1 \times 3}{7 \times 3} = \frac{3}{21}
\]
- Add the fractions:
\[
\frac{-2}{7} - \frac{1}{21} + \frac{1}{7} = \frac{-6}{21} - \frac{1}{21} + \frac{3}{21} = \frac{-6 - 1 + 3}{21} = \frac{-4}{21}
\]

Solution for Question 6:
\[
\boxed{-\frac{1}{63}, -\frac{4}{21}}
\]

---

Final Answer:

\[
\boxed{-\frac{3}{2}, \frac{1}{5}, -\frac{3}{2}, \frac{34}{15}, \frac{17}{30}, \frac{28}{99}, -\frac{26}{57}, \frac{2}{15}, \frac{-13}{72}, \frac{-10}{21}, \frac{90}{91}, \frac{-33}{56}, -3, \frac{-63}{8}, \frac{-27}{10}, \frac{-54}{55}, \frac{-2}{5}, \frac{6}{85}, 1, -6, \frac{7}{5}, \frac{4}{15}, \frac{-1}{6}, \frac{-1}{91}, \frac{-115}{156}, -\frac{1}{63}, -\frac{4}{21}}
\]