DAV Class 7 Maths Chapter 2 Worksheet 3

DAV Class 7 Maths Chapter 2 Worksheet 3

Operations on Rational Numbers Worksheet 3


1. Multiply and express the result as a rational number in the standard form.

(i) \( \displaystyle \frac{11}{7} \) by \( \displaystyle \frac{-3}{8} \)

Solution

\[ \begin{align*} &= \frac{11}{7} \times \frac{-3}{8} \\ \\ &= \frac{11 \times (-3)}{7 \times 8}\\ \\ &= \frac{-33}{56} \end{align*} \]

Answer \( \displaystyle \color{red} \frac{-33}{56} \)

(ii) \( \displaystyle \frac{-7}{4} \) by \( \displaystyle \frac{2}{3} \)

Solution

\[ \begin{align*} &= \frac{-7}{4} \times \frac{2}{3} \\ \\ &= \frac{(-7) \times \cancel2^1}{\cancel{4}_2 \times 3}\\ \\ &= \frac{(-7) \times 1}{2 \times 3}\\ \\ &= \frac{-7}{6} \end{align*} \]

Answer \( \displaystyle \color{red} \frac{-7}{6} \)

(iii) \( \displaystyle \frac{-3}{12} \) by \( -48 \)

Solution

\[ \begin{align*} &= \frac{-3}{12} \times \frac{-48}{1} \\ \\ &= \frac{(-3) \times (-\cancel{48}^4)}{\cancel{12}_1 \times 1}\\ \\ &= \frac{(-3) \times (-4)}{1}\\ \\ &= 12 \end{align*} \]

Answer \( \color{red} 12 \)

(iv) \( \displaystyle \frac{-14}{9} \) by \( \displaystyle \frac{-3}{7} \)

Solution

\[ \begin{align*} &= \frac{-14}{9} \times \frac{-3}{7} \\ \\ &= \frac{(-\cancel{14}^2) \times (-\cancel3^1)}{\cancel9_3 \times \cancel7_1}\\ \\ &= \frac{{-2} \times (-1)}{3 \times 1}\\ \\ &= \frac{2}{3} \end{align*} \]

Answer \( \displaystyle \color{red} \frac{2}{3} \)

(v) \( \displaystyle \frac{23}{5} \) by \( \displaystyle \frac{-25}{11} \)

Solution

\[ \begin{align*} &= \frac{23}{\cancel5_1} \times \frac{-\cancel{25}^5}{11} \\ \\ &= \frac{23 \times (-5)}{1 \times 11}\\ \\ &= \frac{-115}{11} \end{align*} \]

Answer \( \displaystyle \color{red} \frac{-115}{11} \)

(vi) \( 7 \) by \( \frac{-15}{63} \)

Solution

\[ \begin{align*} &= \frac{\cancel7^1}{1} \times \frac{-15}{\cancel{63}_9} \\ \\ &= \frac{-\cancel{15}^5}{\cancel9_3}\\ \\ &= \frac{-5}{3} \end{align*} \]

Answer \( \displaystyle \color{red} \frac{-5}{3} \)

2. For the following values of xand y, verify \( \color{red} x \times y = y \times x\)

(i) \( \displaystyle x = \frac{7}{9} , y = \frac{3}{2} \)

Answer

\[ \begin{array}{c|c} \begin{aligned} & \text{LHS} \\ & =\color{green} x \times y \\ \\ & = \frac{7}{\cancel9_3} \times \frac{\cancel3^1}{2}\\ \\ & = \color{green} \boxed{\frac{7}{6}} \end{aligned} & \begin{aligned} & \text{RHS} \\ & =\color{green} y \times x \\ \\ & = \frac{\cancel3^1}{2} \times \frac{7}{\cancel9_3}\\ \\ & = \color{green} \boxed{\frac{7}{6}} \end{aligned} \end{array} \] \[ \begin{align*} \text{Hence verified} \\ \color{red} x \times y = y \times x \end{align*} \]

(ii) \( \displaystyle x = \frac{-2}{7}, y = \frac{5}{8} \)

Answer

\[ \begin{array}{c|c} \begin{aligned} & \text{LHS} \\ & =\color{green} x \times y \\ \\ & = \frac{-\cancel2^1}{7} \times \frac{5}{\cancel8_4}\\ \\ & = \color{green} \boxed{\frac{-5}{28}} \end{aligned} & \begin{aligned} & \text{RHS} \\ & =\color{green} y \times x \\ \\ & = \frac{5}{\cancel8_4} \times \frac{-\cancel2^1}{7}\\ \\ & = \color{green} \boxed{\frac{-5}{28}} \end{aligned} \end{array} \] \[ \begin{align*} \text{Hence verified} \\ \color{red} x \times y = y \times x \end{align*} \]

(iii) \( \displaystyle x = \frac{4}{9}, y = \frac{-5}{11} \)

Answer

\[ \begin{array}{c|c} \begin{aligned} & \text{LHS} \\ & =\color{green} x \times y \\ \\ & = \frac{4}{9} \times \frac{-5}{11}\\ \\ & = \color{green} \boxed{\frac{-20}{99}} \end{aligned} & \begin{aligned} & \text{RHS} \\ & =\color{green} y \times x \\ \\ & = \frac{-5}{11} \times \frac{4}{9}\\ \\ & = \color{green} \boxed{\frac{-20}{99}} \end{aligned} \end{array} \] \[ \begin{align*} \text{Hence verified} \\ \color{red} x \times y = y \times x \end{align*} \]

(iv) \( \displaystyle x = \frac{-17}{48}, y = \frac{-96}{51} \)

Answer

\[ \begin{array}{c|c} \begin{aligned} & \text{LHS} \\ & =\color{green} x \times y \\ \\ & = \frac{-\cancel{17}^1}{\cancel{48}_1} \times \frac{-\cancel{96}^2}{\cancel{51}_3}\\ \\ & = \frac{-1 \times -2}{3}\\ \\ & = \color{green} \boxed{\frac{2}{3}} \end{aligned} & \begin{aligned} & \text{RHS} \\ & =\color{green} y \times x \\ \\ & = \frac{-\cancel{96}^2}{\cancel{51}_3} \times \frac{-\cancel{17}^1}{\cancel{48}_1}\\ \\ & = \frac{-2 \times -1}{3}\\ \\ & = \color{green} \boxed{\frac{2}{3}} \end{aligned} \end{array} \] \[ \begin{align*} \text{Hence verified} \\ \color{red} x \times y = y \times x \end{align*} \]

3. For the following values of x, yand z, find the products \( (x \times y) \times z \) and \( x \times (y \times z)\) and observe the result \( \color{red}(x \times y) \times z = x \times (y \times z) \).

(i) \( \displaystyle x = \frac{3}{5} , y = \frac{-7}{3} , z = \frac{8}{11} \)

Answer

\[ \begin{aligned} \text{LHS} & =\color{green} (x \times y) \times z \\ \\ & = \left(\frac{\cancel3^1}{5} \times \frac{-7}{\cancel3_1}\right) \times \frac{8}{11}\\ \\ & = \frac{-7}{5} \times \frac{8}{11}\\ \\ & = \color{green} \boxed{\frac{-56}{55}} \\ \\ \text{RHS} & =\color{green} x \times (y \times z) \\ \\ & = \frac{3}{5} \times \left(\frac{-7}{3} \times \frac{8}{11}\right)\\ \\ & = \frac{\cancel3^1}{5} \times \frac{-56}{\cancel{33}_{11}} \\ \\ & = \color{green} \boxed{\frac{-56}{55}} \end{aligned} \] \[ \begin{align*} &\color{red} (x \times y) \times z = x \times (y \times z) \end{align*} \]

(ii) \( \displaystyle x = \frac{-7}{11}, y = \frac{4}{5}, z = \frac{3}{8} \)

Answer

\[ \begin{aligned} \text{LHS} & =\color{green} (x \times y) \times z \\ \\ & = \left(\frac{-7}{11} \times \frac{4}{5}\right) \times \frac{3}{8}\\ \\ & = \frac{-\cancel{28}^7}{55} \times \frac{3}{\cancel8_2}\\ \\ & = \color{green} \boxed{\frac{-21}{110}} \\ \\ \text{RHS} & =\color{green} x \times (y \times z) \\ \\ & = \frac{-7}{11} \times \left(\frac{\cancel4^1}{5} \times \frac{3}{\cancel8_2}\right)\\ \\ & = \frac{-7}{11} \times \frac{3}{10} \\ \\ & = \color{green} \boxed{\frac{-21}{110}} \end{aligned} \] \[ \begin{align*} &\color{red} (x \times y) \times z = x \times (y \times z) \end{align*} \]

(iii) \( \displaystyle x = \frac{-4}{7}, y = \frac{-3}{8}, z = \frac{16}{5} \)

Answer

\[ \begin{aligned} \text{LHS} & =\color{green} (x \times y) \times z \\ \\ & = \left(\frac{-4}{7} \times \frac{-3}{8}\right) \times \frac{16}{5}\\ \\ & = \left(\frac{-\cancel4^1}{7} \times \frac{-3}{\cancel8_2}\right) \times \frac{16}{5}\\ \\ & = \left(\frac{-1}{7} \times \frac{-3}{2}\right) \times \frac{16}{5}\\ \\ & = \frac{3}{\cancel{14}_7} \times \frac{\cancel{16}^8}{5} \\ \\ & = \color{green} \boxed{\frac{24}{35}} \\ \\ \text{RHS} & =\color{green} x \times (y \times z) \\ \\ & = \frac{-4}{7} \times \left(\frac{-3}{8} \times \frac{16}{5}\right)\\ \\ & = \frac{-4}{7} \times \left(\frac{-3}{\cancel{8}_1} \times \frac{\cancel{16}^2}{5}\right)\\ \\ & = \frac{-4}{7} \times \left(\frac{-3}{1} \times \frac{2}{5}\right)\\ \\ & = \frac{-4}{7} \times \frac{-6}{5} \\ \\ & = \color{green} \boxed{\frac{24}{35}} \end{aligned} \] \[ \begin{align*} &\color{red} (x \times y) \times z = x \times (y \times z) \end{align*} \]

(iv) \( \displaystyle x = -3, y = \frac{-4}{9}, z = \frac{-7}{3} \)

Answer

\[ \begin{aligned} \text{LHS}& =\color{green} (x \times y) \times z \\ \\ & = \left(\frac{-3}{1} \times \frac{-4}{9}\right) \times \frac{-7}{3}\\ \\ & = \left(\frac{-\cancel3^1}{1} \times \frac{-4}{\cancel9_3}\right) \times \frac{-7}{3}\\ \\ & = \frac{4}{3} \times \frac{-7}{3}\\ \\ & = \color{green} \boxed{\frac{-28}{9}} \\ \\ \text{RHS} & =\color{green} x \times (y \times z) \\ \\ & = \frac{-3}{1} \times \left(\frac{-4}{9} \times \frac{-7}{3}\right)\\ \\ & = \frac{-\cancel3^1}{1} \times \frac{28}{\cancel{27}_9} \\ \\ & = \frac{-1 \times 28}{1 \times 9} \\ \\ & = \color{green} \boxed{\frac{-28}{9}} \end{aligned} \] \[ \begin{align*} &\color{red} (x \times y) \times z = x \times (y \times z) \end{align*} \]

4. Verify the property \( \color{red} x \times (y + z) = x \times y + x \times z \) by taking -

(i) \( \displaystyle x = \frac{1}{3} , y = \frac{1}{5} , z = \frac{1}{7} \)

Answer

\[ \begin{array}{c|c} \begin{aligned} &\text{LHS}\\ & =\color{green} x \times (y + z) \\ \\ \\ & = \frac{1}{3} \times \left(\frac{1}{5} + \frac{1}{7}\right) \\ \\ \\ & = \frac{1}{3} \times \left(\frac{7 + 5}{35}\right) \\ \\ \\ & = \frac{1}{\cancel3_1} \times \frac{\cancel{12}^4}{35} \\ \\ \\ & = \color{green} \boxed{\frac{4}{35}} \\ \\ \end{aligned} & \begin{aligned} &\text{RHS} \\ & =\color{green} x \times y + x \times z \\ \\ & = \frac{1}{3} \times \frac{1}{5} + \frac{1}{3} \times \frac{1}{7}\\ \\ & = \frac{1}{15} + \frac{1}{21}\\ \\ &\text{LCM} = 105 \\ \\ & = \frac{7 + 5}{105} \\ \\ & = \frac{\cancel{12}^4}{\cancel{105}_{35}} \\ \\ & = \color{green} \boxed{\frac{4}{35}} \\ \\ \end{aligned} \end{array} \] \[ \begin{align*} \text{Hence} & \text{ Verified} \\ \color{red} x \times (y + z) &= \color{red} x \times y + x \times z \end{align*} \]

(ii) \( \displaystyle x = \frac{-3}{7} , y = \frac{2}{5} , z = \frac{-4}{9} \)

Answer

\[ \begin{aligned} \text{LHS} & =\color{green} x \times (y + z) \\ \\ \\ & = \frac{-3}{7} \times \left(\frac{2}{5} + \frac{-4}{9}\right) \\ \\ & = \frac{-3}{7} \times \left(\frac{18 - 20}{45}\right) \\ \\ & = \frac{-\cancel3^1}{7} \times \frac{-2}{\cancel{45}_{15}} \\ \\ & = \frac{-1}{7} \times \frac{-2}{15} \\ \\ & = \color{green} \boxed{\frac{2}{105}} \\ \\ \text{RHS} & =\color{green} x \times y + x \times z \\ \\ & = \left(\frac{-3}{7} \times \frac{2}{5} \right) + \left(\frac{-\cancel3^1}{7} \times \frac{-4}{\cancel9_3} \right) \\ \\ & = \frac{-6}{35} + \frac{4}{21}\\ \\ &\text{LCM} = 105 \\ \\ & = \frac{-18 + 20}{105} \\ \\ & = \color{green} \boxed{\frac{2}{105}} \\ \\ \end{aligned} \] \[ \begin{align*} \text{Hence} & \text{ Verified} \\ \color{red} x \times (y + z) &= \color{red} x \times y + x \times z \end{align*} \]

5. Show that \( \displaystyle \color{red} \frac{-4}{3} \times \left(\frac{2}{5} + \frac{-7}{10}\right) = \left(\frac{-4}{3} \times \frac{2}{5}\right) + \left(\frac{-4}{3} \times \frac{-7}{10}\right) \)

Answer

\[ \begin{aligned} \text{LHS}& =\color{green} \frac{-4}{3} \times \left(\frac{2}{5} + \frac{-7}{10}\right) \\ \\ & = \frac{-4}{3} \times \left(\frac{4 - 7}{10}\right) \\ \\ & = \frac{\cancel{-4}^{-2}}{\cancel3_1} \times \frac{\cancel{-3}^{-1}}{\cancel{10}_5} \\ \\ & = \frac{-2 \times (-1)}{1 \times 5} \\ \\ & = \color{green} \boxed{\frac{2}{5}} \\ \\ \text{RHS} & =\color{green} \left(\frac{-4}{3} \times \frac{2}{5}\right) + \left(\frac{-4}{3} \times \frac{-7}{10}\right) \\ \\ & =\left(\frac{-4}{3} \times \frac{2}{5}\right) + \left( \frac{\cancel{-4}^{-2}}{3} \times \frac{-7}{\cancel{10}_5} \right) \\ \\ & = \frac{-8}{15} + \frac{14}{15} \\ \\ & = \frac{-8 + 14}{15} \\ \\ & = \frac{\cancel6^2}{\cancel{15}_5} \\ \\ & = \color{green} \boxed{\frac{2}{5}} \\ \end{aligned} \] \[ \begin{align*} \text{Hence Verified} \\ \end{align*} \] \( \displaystyle \color{red} \frac{-4}{3} \times \left(\frac{2}{5} + \frac{-7}{10}\right) = \left(\frac{-4}{3} \times \frac{2}{5}\right) + \left(\frac{-4}{3} \times \frac{-7}{10}\right) \)

6 . Show that \( \displaystyle \color{red} \frac{3}{5} \times \left(\frac{-1}{7} - \frac{5}{14}\right) = \left(\frac{3}{5} \times \frac{-1}{7}\right) - \left(\frac{3}{5} \times \frac{5}{14}\right) \)

Answer

\[ \begin{aligned} \text{LHS} & =\color{green} \frac{3}{5} \times \left(\frac{-1}{7} - \frac{5}{14}\right) \\ \\ & = \frac{3}{5} \times \left(\frac{-2 - 5}{14}\right) \\ \\ & = \frac{3}{5} \times \frac{\cancel{-7}^{-1}}{\cancel{14}_2} \\ \\ & = \frac{3 \times (-1)}{5 \times 2} \\ \\ & = \color{green} \boxed{\frac{-3}{10}} \\ \\ \text{RHS} & =\color{green} \left(\frac{3}{5} \times \frac{-1}{7}\right) - \left(\frac{3}{5} \times \frac{5}{14}\right) \\ \\ & = \left(\frac{3}{5} \times \frac{-1}{7}\right) - \left(\frac{3}{\cancel5_1} \times \frac{\cancel5^1}{14}\right) \\ \\ & = \frac{-3}{35} - \frac{3}{14} \\ \\ LCM &= 70 \\ \\ & = \frac{-3 {\color{green} \times 2}}{35 {\color{green} \times 2}} - \frac{3 {\color{green} \times 5}}{14 {\color{green} \times 5}} \\ \\ & = \frac{-6 - 15}{70} \\ \\ & = \frac{\cancel{-21}^{-3}}{\cancel{70}_{10}} \\ \\ & = \color{green} \boxed{\frac{-3}{10}} \\ \end{aligned} \] \[ \begin{align*} &\text{Hence Verified} \\ \end{align*} \] \( \displaystyle \color{red} \frac{3}{5} \times \left(\frac{-1}{7} - \frac{5}{14}\right) = \left(\frac{3}{5} \times \frac{-1}{7}\right) - \left(\frac{3}{5} \times \frac{5}{14}\right) \)

7. Simplify and express the result in standard form.

(i) \( \displaystyle -4 \times \left(\frac{7}{3} - \frac{9}{10}\right) \)

Solution

\[ \begin{aligned} & = \frac{-4}{1} \times \left(\frac{7}{3} - \frac{9}{10}\right) \\ \\ & = \frac{-4}{1} \times \left(\frac{70-27}{30}\right) \\ \\ & = \frac{\cancel{-4}^{-2}}{1} \times \frac{43}{\cancel{30}_{15}} \\ \\ & = \frac{-2 \times 43}{15} \\ \\ & = \color{green} \boxed{\frac{-86}{15}} \\ \\ \end{aligned} \]

Answer \( \displaystyle \color{red}{\frac{-86}{15}} \)

(ii) \( \displaystyle \frac{7}{3} \times \left(\frac{9}{8} + 3\right) \)

Solution

\[ \begin{aligned} & = \frac{7}{3} \times \left(\frac{9}{8} + \frac{3}{1}\right) \\ \\ & = \frac{7}{3} \times \left(\frac{9+24}{8}\right) \\ \\ & = \frac{7}{\cancel3_1} \times \frac{\cancel{33}^{11}}{8} \\ \\ & = \color{green} \boxed{\frac{77}{8}} \\ \\ \end{aligned} \]

Answer \( \displaystyle \color{red}{\frac{77}{8}} \)

(iii) \( \displaystyle \left(\frac{-4}{3} + \frac{5}{7}\right) \times \frac{7}{9} \)

Solution

\[ \begin{aligned} & = \left(\frac{-4}{3} + \frac{5}{7}\right) \times \frac{7}{9} \\ \\ & = \left(\frac{-28 + 15}{21}\right) \times \frac{7}{9} \\ \\ & = \frac{-13}{\cancel{21}_3} \times \frac{\cancel7^1}{9} \\ \\ & = \color{green} \boxed{\frac{-13}{27}} \\ \\ \end{aligned} \]

Answer \( \displaystyle \color{red}{\frac{-13}{27}} \)

(iv) \( \displaystyle \left(\frac{5}{4} - \frac{6}{20}\right) \times \frac{8}{11} \)

Solution

\[ \begin{aligned} & = \left(\frac{5}{4} - \frac{6}{20}\right) \times \frac{8}{11} \\ \\ & = \left(\frac{25 - 6}{20}\right) \times \frac{8}{11} \\ \\ & = \frac{19}{\cancel{20}_5} \times \frac{\cancel8^2}{11} \\ \\ & = \frac{19 \times 2}{5 \times 11} \\ \\ & = \color{green} \boxed{\frac{38}{55}} \\ \\ \end{aligned} \]

Answer \( \displaystyle \color{red}{\frac{38}{55}} \)

8. Fill in the blanks

(i) \( \frac{-4}{7} \times \) \( 1 \) \( = \frac{-4}{7} \)

(ii) \( \frac{3}{8} \times \) \( -1 \) \( = \frac{-3}{8} \)

(iii) \(\left(\frac{-1}{3} \times \frac{4}{5}\right) \times \frac{6}{7} = \) \( \frac{-1}{3} \) \( \times \left(\frac{4}{5} \times \frac{6}{7} \right)\)

(iv) \( \frac{5}{3} \times \left(\frac{-7}{8} \times \frac{11}{3}\right) = (\frac{5}{3} \times \) \( \frac{-7}{8} \) \( )\times \frac{11}{3} \)

(v) \( \frac{3}{7} \times \frac{-6}{11} = \frac{-6}{11} \times \) \( \frac{3}{7} \)

(vi) \( \frac{4}{3} \times \) \( 0 \) \( = 0 \)

(vii) For any rational number \( x \), \( x \times 5 = x + x ... = \) \( 5 \) times.

(viii) \( \frac{2}{3} \times \left(\frac{7}{5} - \frac{2}{9}\right) = \frac{2}{3} \times \frac{7}{5} - \) \( \frac{2}{3} \times \frac{2}{9} \)

(ix) \( \frac{-5}{7} \times \frac{1}{3} + \frac{-5}{7} \times \frac{1}{6} = \frac{-5}{7} \times (\frac{1}{3} + \) \( \frac{1}{6} \) )

(x) \( \frac{4}{7} \times \frac{2}{3} - \frac{4}{5} \times \frac{5}{6} = \frac{4}{7} \times \) \( \left(\frac{2}{3} - \frac{5}{6}\right) \)