Solution

\[ \begin{align*} x + x + x + x &= 4x \end{align*} \]

Answer \( {\color{orange} (ii)} \ \color{red} 4x \)

Solution

\[ \begin{align*} & = (9z + 7) - 5(2z - 3) + 3(2z + 3) \\ &= 9z + 7 - 10z + 15 + 6z + 9 \\ &= 9z - 10z + 6z + 7 + 15 + 9 \\ &= 5z + 31 \end{align*} \]

Answer \( {\color{orange} (i)} \ \color{red} 5z + 31 \)

Solution

\[ \begin{align*} 4x - 3 &= 21 \\ 4x &= 21 + 3 \\ 4x &= 24 \\ x &= \frac{\cancel{24}^6}{\cancel4_1} \\ \\ x &= 6 \\ \\ \text{Now } \implies & 3x - 5 \\ &= 3(6) - 5 \\ &= 18 - 5 \\ &= 13 \end{align*} \]

Answer \( {\color{orange} (iii)} \ \color{red} 13 \)

Solution

\[ \begin{align*} 3x - 7 &= 7 - 4x \\ 3x + 4x &= 7 + 7 \\ 7x &= 14 \\ x &= \frac{\cancel{14}^2}{\cancel 7_1} \\ x &= 2 \end{align*} \]

Answer \( {\color{orange} (iii)} \ \color{red} x = 2 \)

Solution

\[ \begin{align*} 4x - 2 &= 3x + 3 \\ 4(5) - 2 &= 3(5) + 3 \\ 20 - 2 &= 15 + 3 \\ 18 &= 18 \end{align*} \]

Answer \( {\color{orange} (iii)} \ \color{red} 4x - 2 = 3x + 3 \)

Solution

\[ \begin{aligned} \text{Let the no. of boxes be } & x \\ \text{Buying 50 more boxes} &= x + 50 \\ \text{Half of these were used} &= \frac{x + 50}{2} \\ \text{Remaining boxes} &= 40 \\ \\ \color{magenta} (x + 50) - \left( \frac{x + 50}{2} \right) &= \color{magenta}40 \\ \\ x + 50 - \left( \frac{x + 50}{2} \right) &= 40 \\ \\ \frac{x {\color{green} \times 2}}{1 {\color{green} \times 2}} + \frac{50 {\color{green} \times 2}}{1 {\color{green} \times 2}} - \left( \frac{x + 50}{2} \right) &= 40 \\ \\ \frac{2x + 100 - x - 50}{2} &= 40 \\ \\ \frac{x+50}{2} &= 40 \\ \\ x+50 &= 2 \times 40 \\ x+50 &= 80 \\ x &= 80 - 50 \\ x &= 30 \end{aligned} \]

Answer The confectioner had \( \color{red} 30 \ boxes \) in the beginning.

Solution

\[ \begin{aligned} Kx + 8 &= x - 7 \\ K(3) + 8 &= 3 - 7 \\ 3K + 8 &= -4 \\ 3K &= -4 - 8 \\ 3K &= -12 \\ K &= \frac{-\cancel{12}^4}{\cancel3_1} \\ \\ K &= -4 \end{aligned} \]

Answer \( \color{red} K = -4 \)

Solution

\[ \begin{aligned} \text{Let the length } &\text{of wire be } x \\ \text{Side of the square} &= 4 \ cm \\ \text{Perimeter} &= 4 \times 4 \implies 16 \text{ cm} \\ \text{Remaining wire} &= 5 \ cm \\ \\ x & = \text{Perimeter} + \text{Remaining wire}\\ x &= 4 \times 4 + 5 \\ x &= 16 + 5 \\ x &= 21 \ cm \\ \end{aligned} \]

Answer Seema had \( \color{red} 21 \) cm of wire.

Solution

\[ \begin{aligned} 2A + B &= 6 \\ 2(2a - 3) + (5 - 3a) &= 6 \\ 4a - 6 + 5 - 3a &= 6 \\ 4a - 3a -6 + 5 &= 6 \\ a - 1 &= 6 \\ a &= 6 + 1 \\ a &= 7 \end{aligned} \]

Answer \( \color{red} a = 7 \)

Solution

\[ \begin{aligned} \color{magenta} LHS & = 1.2(3y - 4) - 0.3(2y - 6) \\ & = 1.2[3(1) - 4] - 0.3[2(1) - 6] \\ & = 1.2(3 - 4) - 0.3(2 - 6) \\ & = 1.2(-1) - 0.3(-4) \\ & = -1.2 + 1.2 \\ & = 0 \\ \\ \color{magenta} RHS & = 0 \\ \\ LHS & = RHS \end{aligned} \]

Answer \( \color{red} y = 1 \) is a solution of the equation.

Solution

\begin{align*} \frac{x}{2} + \frac{x}{3} - \frac{x}{4} &= 7 \\ \\ \frac{x {\color{green} \times 6}}{2 {\color{green} \times 6}} + \frac{x {\color{green} \times 4}}{3 {\color{green} \times 4}} - \frac{x {\color{green} \times 3}}{4 {\color{green} \times 3}} &= 7 \\ \\ \frac{6x + 4x - 3x}{12} &= 7 \\ \\ \frac{7x}{12} &= 7 \\ x &= \frac{\cancel 7^1 \times 12}{\cancel 7_1} \\ \color{red} x & \, \color{red} = 12 \end{align*}

Solution

\begin{align*} \frac{2}{3}(y - 5) - \frac{1}{4}(y - 2) &= \frac{9}{2} \\ \\ \frac{2y - 10}{3} - \frac{y - 2}{4} &= \frac{9}{2} \\ \\ \frac{(2y - 10) {\color{green} \times 6}}{3 {\color{green} \times 4}} - \frac{(y - 2) {\color{green} \times 3}}{4 {\color{green} \times 3}} &= \frac{9}{2} \\ \\ \\ \frac{8y - 40 - (3y - 6)}{12} &= \frac{9}{2} \\ \\ \frac{8y - 40 - 3y + 6}{12} &= \frac{9}{2} \\ \\ \frac{8y - 3y - 40 + 6}{12} &= \frac{9}{2} \\ \\ \frac{5y - 34}{12} &= \frac{9}{2} \\ \\ (5y - 34) \times 2 &= 9 \times 12 \\ 10y - 68 &= 108 \\ 10y &= 108 + 68 \\ 10y &= 176 \\ y &= \frac{\cancel{176}^{88}}{\cancel{10}_5} \\ \\ \color{red} y & \, \color{red} = \frac{88}{5} \end{align*}

Solution

\begin{align*} 13(z - 4) - 3(z - 9) - 5(z + 4) &= 0 \\ 13z - 52 - 3z + 27 - 5z - 20 &= 0 \\ 13z - 3z - 5z - 52 + 27 - 20 &= 0 \\ 5z - 45 &= 0 \\ 5z &= 45 \\ z &= \frac{\cancel{45}^9}{\cancel 5_1} \\ \color{red} z & \, \color{red} = 9 \end{align*}

Solution

\begin{align*} (x + 2)(x + 3) + (x - 3)(x - 2) - 2x(x + 1) &= 0 \\ x(x + 3)+ 2(x + 3) + x(x - 2) -3(x - 2) - 2x(x + 1) &= 0 \\ x^2 + 3x + 2x + 6 + x^2 - 2x -3x + 6 - 2x^2 - 2x &= 0 \\ {\color{magenta} x^2 + x^2 - 2x^2} + {\color{green} 3x + 2x - 2x -3x - 2x} + {\color{blue}6 + 6} &= 0 \\ {\color{magenta} 2x^2 - 2x^2} + {\color{green} 5x - 7x} + {\color{blue}12} &= 0 \\ 0 - 2x + 12 &= 0 \\ -2x &= -12 \\ x &= \frac{\cancel{-12}^6}{\cancel{-2}_1} \\ \color{red} x & \, \color{red} = 6 \end{align*}

Solution

\begin{align*} \frac{2x + 14}{3x + 6} &= 4 \\ \\ 2x + 14 &= 4(3x + 6) \\ 2x + 14 &= 12x + 24 \\ 2x - 12x &= 24 - 14 \\ -10x &= 10 \\ -x &= \frac{\cancel{10}^1}{\cancel{10}_1} \\ -x &= 1 \\ \color{red} x & \, \color{red} = -1 \end{align*}

Solution

\begin{align*} a - \frac{a - 1}{2} &= 1 - \frac{a - 2}{3} \\ \\ a - \frac{(a -1)}{2} + \frac{(a - 2)}{3} &= 1 \\ \\ \frac{a {\color{green} \times 6}}{1 {\color{green} \times 6}} - \frac{(a -1) {\color{green} \times 3}}{2 {\color{green} \times 3}} + \frac{(a - 2) {\color{green} \times 2}}{3 {\color{green} \times 2}} &= 1 \\ \\ \frac{6a - 3(a-1) +2(a-2) }{6} &= 1 \\ \\ 6a - 3a + 3 + 2a - 4 &= 6 \\ 6a - 3a + 2a + 3 - 4 &= 6 \\ 5a - 1 &= 6 \\ 5a &= 6 +1 \\ 5a &= 7 \\ \color{red} a & \, \color{red} = \frac{7}{5} \end{align*}

Solution

\begin{align*} \frac{x + 2}{6} - \left[ \frac{11 - x}{3} - \frac{1}{4} \right] &= \frac{3x - 14}{12} \\ \\ \frac{x + 2}{6} - \frac{11 - x}{3} + \frac{1}{4} &= \frac{3x - 14}{12} \\ \\ \frac{(x + 2) {\color{green} \times 2}}{6 {\color{green} \times 2}} - \frac{(11 - x) {\color{green} \times 4}}{3 {\color{green} \times 4}} + \frac{1 {\color{green} \times 3}}{4 {\color{green} \times 3}} &= \frac{3x - 14}{12} \\ \\ \frac{2(x + 2) -4(11-x) +3}{12} &= \frac{3x - 14}{12} \\ \\ 2x + 4 -44+4x +3 &= \frac{3x - 14}{\cancel{12}} \times \cancel{12} \\ \\ 6x - 37 &= 3x - 14 \\ 6x - 3x &= -14 + 37 \\ 3x &= 23 \\ \color{red} x & \, \color{red} = \frac{23}{3} \end{align*}

Solution

\begin{align*} \frac{2}{3x} - 1 &= \frac{1}{12} \\ \\ \frac{2}{3x} &= \frac{1}{12} + 1 \\ \\ \frac{2}{3x} &= \frac{1}{12} + \frac{1 {\color{green} \times 12}}{1 {\color{green} \times 12}} \\ \\ \frac{2}{3x} &= \frac{1 + 12}{12} \\ \\ \frac{2}{3x} &= \frac{13}{12} \\ \\ 2 \times 12 &= 13 \times 3x \\ 24 &= 39x \\ x &= \frac{\cancel{24}^8}{\cancel{39}^{13}} \\ \\ \color{red} x & \, \color{red} = \frac{8}{13} \end{align*}

Solution

\begin{align*} \text{Let the denominator be} &= x \\ \text{Numerator} &= x - 6 \\ \\ \color{magenta} \frac{x - 6 + 3}{x} &= \color{magenta} \frac{2}{3} \\ \\ \frac{x - 3}{x} &= \frac{2}{3} \\ \\ 3(x - 3) &= 2x \\ 3x - 9 &= 2x \\ 3x - 2x &= 9 \\ x &= 9 \\ \\ \text{Numerator} &= x - 6 \\ &= 9 - 6 \\ \color{green} \text{Numerator} &=\color{green} 3 \\ \color{green} \text{Denominator} &= \color{green}9 \\ \\ \text{Original fraction} &= \frac{\cancel3^1}{\cancel9_3} \\ \\ &= \frac{1}{3} \end{align*}

Answer Original fraction \( = \color{red} \boxed{\frac{1}{3}}\)

Solution

\begin{align*} \text{Let the total journey be} &= x \text{ km} \\ \\ \text{Distance travelled by train} &= \frac{2}{5}x \text{ km} \\ \\ \text{Distance travelled by taxi} &= \frac{1}{3}x \text{ km} \\ \\ \text{Distance travelled by bus} &= \frac{1}{6}x \text{ km} \\ \\ \text{Distance travelled by foot} &= 10 \text{ km} \\ \\ \\ \color{magenta}\frac{2}{5}x + \frac{1}{3}x + \frac{1}{6}x + 10 &= \color{magenta}x \\ \\ \frac{2}{5}x + \frac{1}{3}x + \frac{1}{6}x - x &= -10 \\ \\ \frac{2x {\color{green} \times 6}}{5 {\color{green} \times 6}} + \frac{x {\color{green} \times 10}}{3 {\color{green} \times 10}} + \frac{x {\color{green} \times 5}}{6 {\color{green} \times 5}} - \frac{x {\color{green} \times 30}}{1 {\color{green} \times 30}} &= -10 \\ \\ \frac{12x + 10x + 5x -30x}{30} &= -10 \\ \\ \frac{-3x}{30} &= -10 \\ \\ -3x &= -10 \times 30 \\ \\ x &= \frac{\cancel{-300}^{100}}{\cancel{-3}_1} \\ \\ x &= 100 \\ \end{align*}

Answer The total journey is \(\color{red} 100 \ km \).

Solution

\begin{align*} \text{Let cost of a chair} & = x \\ \text{Cost of a table} & = x + 200 \\\\ \color{magenta} 2(x + 200) + 3x &= \color{magenta} 1400 \\ 2x + 400 + 3x &= 1400 \\ 5x + 400 &= 1400 \\ 5x &= 1400 - 400 \\ 5x &= 1000 \\ x &= \frac{\cancel{1000}^{200}}{\cancel5_1} \\ x &= 200 \\ \\ \text{Cost of a chair} &= \text{₹} 200 \\ \\ \text{Cost of a table} &= x + 200 \\ &= 200 + 200 \\ \text{Cost of a table}&= \text{₹} 400 \end{align*}

Answer The cost of a chair is \(\color{red} \text{₹} 200 \) and the cost of a table is \(\color{red} \text{₹}400 \).

Solution

\begin{align*} \text{Let Sunil's present age be} & = x \\ \text{ Then Anu's present age} & = x + 4 \\ \\ \text{Eight } & \text{years ago} \\ \text{Sunil's age} &= x - 8 \\ \text{Anu's age} &= x + 4 - 8 \\ & \implies x - 4 \\ \\ \text{Anu was three} & \text{ times Sunil's age} \\ \color{magenta} x - 4 &= \color{magenta} 3(x - 8) \\ x - 4 &= 3x - 24 \\ x - 3x &= 4 - 24 \\ -2x &= -20 \\ x &= \frac{\cancel{-20}^{10}}{\cancel{-2}_1} \\ x &= 10 \\ \\ \text{Sunil's present age} &= 10 \\ \text{Anu's present age} &= x + 4 \\ & = 10 + 4 \\ &= 14 \end{align*}

Answer Sunil is \(\color{red} 10 \) years old and Anu is \(\color{red} 14 \) years old.