# Pair of Linear Equations in Two Variables

## Class 10 NCERT

### NCERT

1   1 Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.

##### Solution :

Let the present age of Aftab be x$\\$ And, present age of his daughter = y$\\$ Seven years ago,$\\$ Age of Aftab = x - 7$\\$ Age of his daughter = y - 7$\\$ According to the question,$\\$ $(x-7) = 7(y-7)$ $\\$ $x-7 = 7y -49$ $\\$ $x-7y = -42 \qquad (1)$ $\\$ Three years hence, $\\$ Age of Aftab = x + 3 $\\$ Age of his daughter = y + 3 $\\$ According to the question, $\\$ $(x + 3) = 3(y+3)$ $\\$ $x+ 3 = 3y + 9$ $\\$ $x-3y = 6 \qquad (2)$ $\\$ Therefore, the algebraic representation is $\\$ $x-7y = -42$ $\\$ $x-3y = 6$ $\\$ for $x-7y = -42 ,$ $\\$ $x = -42 +7y,$ $\\$ The Solution table is

$\begin{array}{|c|c|c|c|} \hline x& -7 & 0 & 7 \\ \hline y& 5 & 6 & 7 \\ \hline \end{array}$$\\ For x-3y = 6 \\ x = 6 +3y \\ The solution table is \begin{array}{|c|c|c|c|} \hline x&6 &3 & 0 \\ \hline y&0 &-1 & -2 \\ \hline \end{array}$$\\$ The graphical representation is as follow

2   2. The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 2 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically.

##### Solution :

Let the cost of a bat be Rs x $\\$ And, cost of a ball = Rs y $\\$ According to the question, the algebraic representation is, $\\$ $3x + 6y = 3900$ $\\$ $x + 2y = 1300$ $\\$ For, $3x + 6y = 3900$ $\\$ $x = \frac{3900 - 6y}{3}$

$\begin{array}{|c|c|c|c|} \hline x & 300 & 100 & -100 \\ \hline y& 500 & 600 & 700 \\ \hline \end{array}$$\\ For x+2y = 1300 , \\ x = 1300 - 2y \\ The solution table is \\ \begin{array}{|c|c|c|c|} \hline x & 300 & 100 & -100 \\ \hline y& 500 & 600 & 700 \\ \hline \end{array}$$\\$ The graphical representation is as follows. $\\$

3   3. The cost of $2$ kg of apples and 1 kg of grapes on a day was found to be Rs $160$. After a month, the cost of $4$ kg of apples and $2$kg of grapes is Rs $300$. Represent the situation algebraically and geometrically.

##### Solution :

Let the cost of 1 kg of apples be Rs x $\\$ And, cost of 1 kg of grapes = Rs y $\\$ According to the question, the algebraic representation is $\\$ $2x +y = 160$ $\\$ $4x + 2y = 300$ $\\$ For, $2x +y = 160$ $\\$ $y= 160 -2x$

The solution table is $\\$ $\begin{array}{|c|c|c|c|} \hline x&50 &60 & 70 \\ \hline y&60 &40 & 20 \\ \hline \end{array}$$\\ \\ For 4x + 2y = 300 , \\ y = \frac{300 - 4x}{2} \\ \begin{array}{|c|c|c|c|} \hline x&70 &80 & 75 \\ \hline y&10 &-10 & 0 \\ \hline \end{array}$$\\$

The graphical representation is as follows.

4   Form the pair of linear equations in the following problems, and find their solutions graphically. $\\$ (i) $10$ students of Class X took part in a Mathematics quiz. If the number of girls is $4$ more than the number of boys, find the number of boys and girls who took part in the quiz. (ii) $5$ pencils and $7$ pens together cost Rs $50$, whereas $7$ pencils and 5 pens together cost Rs $46$. Find the cost of one pencil and that of one pen.

##### Solution :

(i) Let the number of girls be x and the number of boys be y $\\$ According to the question, the algebraic representation is $\\$ $x + y = 10$ $\\$ $x - y = 4$ $\\$ For $x + y = 10,$ $\\$ $x = 10 - y$

$\begin{array}{|c|c|c|c|} \hline x&5 &4 & 6 \\ \hline y&5 &6 & 4 \\ \hline \end{array}$$\\ \\ For x-y =4 \\ x = 4 + y \\ \begin{array}{|c|c|c|c|} \hline x&5 &4 & 3 \\ \hline y&1 &0 & -1 \\ \hline \end{array}$$\\$ Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines intersect each other at point (7, 3). $\\$ (ii) Let the cost of 1 pencil be Rs x and the cost of 1 pen be Rs y. $\\$ According to the question, the algebraic representation is $\\$ $5x + 7y = 50$ $\\$ $7x + 5y = 46$ For $5x + 7y = 50,$ $\\$ $x = \frac{50-7y}{5}$

$\begin{array}{|c|c|c|c|} \hline x&3 &10 & -4 \\ \hline y&5 & 0 & 10 \\ \hline \end{array}$$\\ 7x + 5y =46 \\ x = \frac{46 - 5y}{7} \begin{array}{|c|c|c|c|} \hline x&8 &3 & -2 \\ \hline y&-2 &5 & 12 \\ \hline \end{array}$$\\$

From the figure, it can be observed that these lines intersect each other at point (3, 5). $\\$ Therefore, the cost of a pencil and a pen are Rs 3 and Rs 5 respectively.

5   On comparing the ratios $\frac{a_1}{a_2}, \frac{b_1}{b_2}, \quad and \quad \frac{c_1}{c_2}$ , find out whether the lines representing the following pairs of linear equations at a point, are parallel or coincident: $\\$ (i) $5x -4y + 8 = 0 \qquad 7x + 6y =0$ $\\$ (ii) $9x +3y +12 =0 \qquad 18x + 6y + 24 =0$ $\\$ (iii) $6x-3y + 10 = 0 \qquad 2x - y + 9 =0$ $\\$

##### Solution :

(i) $5x - 4y + 8 = 0$ $\\$ $7x + 6y - 9 = 0$ $\\$ Comparing these equations with $a_1x + b_1y + c_1 = 0$ $\\$ and $a_2x + b_2y + c_2 = 0,$ we obtain $\\$ $a_1 = 5, \qquad b_1 =-4 \qquad c_1 =8$ $\\$ $a_2 = 7, \qquad b_2 =6 \qquad c_2 = -9$ $\\$ $\frac{a_1}{a_2} = \frac{5}{7}$ $\\$ $\frac{b_1}{b_2} = \frac{-4}{6} = \frac{-2}{3}$ $\\$ Since, $\frac{a_1}{a_2} \neq \frac{b_1}{b_2},$ $\\$ Hence, the lines representing the given pair of equations have a unique solution and the pair of lines intersects at exactly one point. $\\$ (ii) $9x + 3y + 12 = 0$ $\\$ $18x + 6y + 24 = 0$ $\\$ Comparing these equations with, $a_1x + b_2y +c_1 = 0$ and $a_2x +b_2y +c_2 = 0,$ we obtain $\\$ $a_1 = 9, \qquad b_1 =3 \qquad c_1 =12$ $\\$ $a_2 = 18, \qquad b_2 =6 \qquad c_2 = 24$ $\\$ $\frac{a_1}{a_2} = \frac{9}{18} = \frac{1}{2}$ $\\$ $\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$ $\\$ $\frac{c_1}{c_2} = \frac{12}{24} = \frac{1}{2}$ $\\$

Since, $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ $\\$ Hence, the lines representing the given pair of equations are coincident and there are infinite possible solutions for the given pair of equations. $\\$ (iii) $6x - 3y + 10 = 0$ $\\$ $2x - y +9 =0$ $\\$ Comparing these equations with $a_1x + b_1y +c_1 = 0$ and $a_2x + b_2y +c_2 = 0$ we obtain $\\$ $a_1 = 6, \qquad b_1 =-3, \qquad c_1=10$ $\\$ $a_2 = 2, \qquad b_2 =-1, \qquad c_2=9$ $\\$ $\frac{a_1}{a_2} =\frac{6}{2} = \frac{3}{1}$ $\\$ $\frac{b_1}{b_2} = \frac{-3}{-1} = \frac{3}{1}$ $\\$ $\frac{c_1}{c_2} = \frac{10}{9}$ $\\$ Since, $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ $\\$ Hence, the lines representing the given pair of equations are parallel to each other and hence, these lines will never intersect each other at any point or there is no possible solution for the given pair of equations.

6   On comparing the ratios $\frac{a_1}{a_2}, \frac{b_1}{b_2} and \frac{c_1}{c_2}$, $\\$find out whether the following pair of linear equations are consistent, or inconsistent. (i)$3x +2y =5, \qquad 2x-3y =7$ $\\$ (ii)$2x -3y =8, \qquad 4x-6y =9$ $\\$ (iii)$\frac{3}{2}x + \frac{5}{3}y =7, \qquad 9x-10y =14$ $\\$ (iv)$5x -3y =11, \qquad -10x+6y =-22$ $\\$ (v)$\frac{4}{3}x + 2y =8, \qquad 2x+3y = 12$ $\\$

##### Solution :

(i) $3x + 2y = 5$ $\\$ $2x - 3y = 7$ $\\$ $\frac{a_1}{a_2} = \frac{3}{2}, \quad \frac{b_1}{b_2} = \frac{-2}{3}, \quad \frac{c_1}{c_2} = \frac{5}{7}$ $\\$ $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ $\\$ These linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent. $\\$ (ii)$2x - 3y = 8$ $\\$ $4x - 6y =9$ $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{8}{9}$ $\\$ Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

(iii) $\ frac{4}{3}x + 2y = 8$ $\\$ $2x - 3y =12$ $\\$ $\frac{a_1}{a_2} = \frac{\frac{4}{3}}{2} = \frac{2}{3}, \quad \frac{2}{3}, \quad \frac{c_1}{c_2} = \frac{8}{12} = \frac{2}{3}$ $\\$ Since, $\frac{a_1}{a_2} = \frac{b_1}{b_2},$ Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent. $\\$

Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent. $\\$ (iii) $\ frac{3}{2}x + \frac{5}{3}y = 7$ $\\$ $9x - 10y =14$ $\\$ $\frac{a_1}{a_2} = \frac{\frac{3}{2}}{9} = \frac{1}{6}, \quad \frac{\frac{5}{3}}{-10} = \frac{-1}{6}, \quad \frac{c_1}{c_2} = \frac{7}{14} = \frac{1}{2}$ $\\$ Since, $\frac{a_1}{a_2} \neq \frac{b_1}{b_2},$ Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent. $\\$ (iv) $5x - 3 y = 11$ $- 10x + 6y = - 22$ $\frac{a_1}{a_2} = \frac{5}{-10} = \frac{-1}{2}, \quad \frac{-3}{6} = \frac{-1}{2}, \quad \frac{c_1}{c_2} = \frac{11}{-22} = \frac{-1}{2}$ $\\$ Since, $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2},$ Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent. $\\$

7   Which of the following pairs of linear equations are consistent/ inconsistent? If consistent, obtain the solution graphically: $\\$ (i) $x+y =5, \qquad 2x+2y =10$ $\\$ (ii) $x-y =8, \qquad 3x-3y =16$ $\\$ (iii) $2x+y -6 =0, \qquad 4x-2y -4 =0$ $\\$ (iv) $2x-2y -2 =0, \qquad 4x-4y - 5 =0$ $\\$

##### Solution :

(i) $x + y = 5$ $\\$ $2x + 2y = 10$ $\\$ $\frac{a_1}{a_2} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{5}{10} =\frac{1}{2}$ $\\$ Since, $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2},$ $\\$ Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent. $\\$ $x + y =5$ $\\$ $x = 5 - y$ $\\$

$\begin{array}{|c|c|c|c|} \hline x&4 &3 & 2 \\ \hline y&1 &2 & 3 \\ \hline \end{array}$$\\ \\ And, 2x + 2y = 10 \\ x = \frac{10 - 2y}{2} \\ \begin{array}{|c|c|c|c|} \hline x&4 &3 & 2 \\ \hline y&1 &2 & 3 \\ \hline \end{array}$$\\$ $\\$ Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines are overlapping each other. Therefore, infinite solutions are possible for the given pair of equations. $\\$ (ii) $x - y = 8$ $\\$ $3x - 3y = 16$ $\\$ $\frac{a_1}{a_2} = \frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{-1}{-3}, \quad \frac{c_1}{c_2} = \frac{8}{12} = \frac{1}{2}$ $\\$ Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{_2}$ $\\$ Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent. $\\$ (iii) $2x + y - 6 = 0$ $\\$ $4x - 2y - 4 = 0$ $\\$ $\frac{a_1}{a_2} = \frac{2}{4}=\frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{-1}{2}, \quad \frac{c_1}{c_2} = \frac{-6}{-4} = \frac{3}{2}$ $\\$ Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ $\\$ Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent. $\\$ $2x + y - 6 = 0$ $\\$ $y = 6 - 2x$

$\begin{array}{|c|c|c|c|} \hline x& 0 & 1 & 2 \\ \hline y& 6 & 4 & 2 \\ \hline \end{array}$$\\ \\ And 4x- 2y - 4 = 0 \\ y = \frac{4x -4}{2} \begin{array}{|c|c|c|c|} \hline x& 1 & 2 & 3 \\ \hline y& 6 & 4 & 2 \\ \hline \end{array}$$\\$ $\\$ Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines are intersecting each other at the only point i.e., (2, 2) and it is the solution for the given pair of equations. $\\$ (iv) $2x - 2y- 2 = 0$ $\\$ $4x - 4y - 5 = 0$ $\\$ $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{-2}{-4} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{2}{5}$ $\\$ Since, $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2},$ $\\$ Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.

8   Half the perimeter of a rectangular garden, whose length is $4 m$ more than its width, is $36 m$. Find the dimensions of the garden.

##### Solution :

Let the width of the garden be x and length be y. $\\$ According to the question, $\\$ $y - x = 4 (1)$ $\\$ $y + x = 36 (2)$ $\\$ $y - x = 4$ $\\$ $y = x + 4$ $\\$

$\begin{array}{|c|c|c|c|} \hline x & 0 & 8 & 12 \\ \hline y & 4 & 12 & 16 \\ \hline \end{array}$$\\ y +x = 36 \begin{array}{|c|c|c|c|} \hline x & 0 & 36 & 16 \\ \hline y & 36 & 0 & 20 \\ \hline \end{array}$$\\$ Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines are intersecting each other at only point i.e., (16, 20). Therefore, the length and width of the given garden is 20 m and 16 m respectively.

9   Given the linear equation $2x + 3y - 8 = 0$, write another linear equations in two variables such that the geometrical representation of the pair so formed is: $\\$ (i) intersecting lines $\\$ (ii) parallel lines $\\$ (iii) coincident lines

##### Solution :

(i)Intersecting lines: $\\$ For this condition, $\\$ $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ $\\$ The second line such that it is intersecting the given line is $\\$ $2x + 4y -6 =0 \quad as\quad \frac{a_1}{a_2} = \frac{2}{2} =1, \quad \frac{b_1}{b_2} = \frac{3}{4} \quad and \quad \frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ $\\$ (ii) Parallel lines: $\\$ For this condition, $\\$ $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ $\\$ Hence, the second line can be $\\$ $4x + 6y - 8 = 0$

as $\frac{a_1}{a_2} = \frac{2}{3} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{3}{6}= \frac{1}{2} \quad ]frac{c_1}{c_2} = \frac{-8}{-8} =1$ $\\$ and clearly, $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ $\\$ (iii)Coincident lines: $\\$ For coincident lines, $\\$ $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ Hence, the second line can be $\\$ $6x + 9y - 24 = 0$ $\\$ as $\frac{a_1}{a_2} = \frac{2}{6} = \frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{3}{9} = \frac{1}{3}, \quad \frac{c_1}{c_2} = \frac{-8}{-24} = \frac{1}{3}$ $\\$ And Clearly, $\frac{a_1}{a_2}= \frac{b_1}{b_2} = \frac{c_1}{c_2}$

10   Draw the graphs of the equations $x - y + 1 = 0$ and $3x + 2y - 12 = 0$. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

##### Solution :

$x - y + 1 = 0$ $\\$ $x = y - 1$

From the figure, it can be observed that these lines are intersecting each other at point (2, 3) and x-axis at (-1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), (-1, 0), and (4, 0).

$\begin {array}{|c|c|} \hline x & 0 & 1 & 2 \\ \hline y & 1 & 2 & 3 \\ \hline \end {array}$ $\\$ $3x + 2y - 12 = 0$ $\\$ $x = \frac{12 - 2y}{3}$ $\\$ $\begin {array}{|c|c|} \hline x & 4 & 2 & 0 \\ \hline y & 0 & 3 & 6 \\ \hline \end {array}$ $\\$ Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines are intersecting each other at point (2, 3) and x-axis at (-1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), (-1, 0), and (4, 0).

11   Solve the following pair of linear equations by the substitution method. $\\$ (i) $x+y = 14, \qquad x-y =4$ $\\$ (ii) $s-t = 3 \qquad \frac{s}{3} + \frac{t}{2} = 6$ $\\$ (iii) $3x -y = 3 \qquad 9x-3y =9$ $\\$ (iv) $0.2x + 0.3y=1.3 \qquad 0.4x + 0.5y =2.3$ $\\$ (v) $\sqrt{4}x + \sqrt{3}y =0 \qquad \sqrt{3}x - \sqrt{8}y = 0$ $\\$ (vi) $\frac{3x}{2} - \frac{5y}{3} = -2 \qquad \frac{x}{3} + \frac{y}{2} = \frac{13}{6}$

##### Solution :

(i) $x + y = 14 \qquad ..............(1)$ $\\$ $x - y = 4 \qquad ............(2)$ $\\$ From (1), we obtain $\\$ $x = 14 - y \qquad ............(3)$ $\\$ Substituting this value in equation (2), we obtain $\\$ $(14 -y) -y = 4$ $\\$ $14- 2y 4$ $\\$ $10 = 2y$ $\\$ $y = 5 \qquad ........(4)$ $\\$ Substituting this in equation (3), we obtain $\\$ $x = 9$ $\\$ $\therefore x = 9, y =5$ $\\$ (ii) $s-t =3 \qquad .......(1)$ $\\$ $\frac{s}{3} + \frac{t}{2} =6 \qquad .......(2)$ $\\$ From (1), we obtain $\\$ $s =t +3 \qquad .........(3)$ $\\$ Substituting this value in equation (2), we obtain $\\$ $\frac{t+3}{3} + \frac{t}{2} = 6$ $\\$ $2t +6 +3t = 36$ $\\$ $5t =30$ $\\$ $t = 6 \qquad ...(4)$ $\\$ Substituting in equation (3), we obtain

$s = 9$ $\\$ $\therefore s =9, \quad t =6$ $\\$ (iii) $3x - y = 3 \qquad ........... (1)$ $\\$ $9x - 3y = 9 \qquad ....... (2)$ $\\$ From (1), we obtain $\\$ $y = 3x - 3 \qquad ......(3)$ $\\$ Substituting this value in equation (2), we obtain $\\$ $9x -3(3x -3) =9$ $\\$ $9x -9x + 9 = 9$ $\\$ $9 =9$ $\\$ This is always true. $\\$ Hence, the given pair of equations has infinite possible solutions and the relation between these variables can be given by $\\$ y = 3x - 3 $\\$ Therefore, one of its possible solutions is x = 1, y = 0. $\\$

(iv) $0.2x + 0.3y = 1.3 \qquad ...(1)$ $\\$ $0.4x + 0.5y = 2.3 \qquad ...(2)$ $\\$ From equation (1), we obrain, $\\$ $x = \frac{1.3 - 0.3y}{0.2} \qquad ..(3)$ $\\$ Substituting this value in equation (2), we obtain $\\$ $0.4 \Bigg( \frac{1.3 - 0.3y}{0.2} \Bigg) + 0.5y = 2.3$ $\\$ $2.6 - 0.6 + 0.5y = 2.3$ $\\$ $0.3 = 0.1y$ $\\$ $y = 3 \qquad ....(4)$ $\\$ Substituting this value in equation (3), we obtain $\\$ $x = \frac{1.3 - 0.3 \times 3}{0.2}$ $= \frac{1.3 -0.9 }{0.2} = \frac{0.4}{0.2} = 2$ $\\$ $\therefore x = 2, y =3$ $\\$ (v) $\sqrt{2}x + \sqrt{3}y = 0 \qquad ..(1)$ $\\$ $\sqrt{3}x - \sqrt{8}y = 0 \qquad ...(2)$ $\\$ From equation (1), we obtain $\\$ $x = \frac{-\sqrt{3}y}{\sqrt{2} } \qquad ...(3)$ $\\$ Substituting this value in equation (2), we obtain $\\$ $\sqrt{3} \Bigg( -\frac{\sqrt{3}y}{\sqrt{2}} \Bigg) - \sqrt{8}y = 0$ $\\$ $-\frac{3y}{\sqrt{2} } - 2\sqrt{2}y = 0$ $\\$ $y \Bigg( -\frac{3}{\sqrt{2}} - 2 \sqrt{2} \Bigg) =0$ $\\$ $y =0 \qquad ....(4)$ $\\$ Substituting this value in equation (3), we obtain

$x = 0$ $\\$ $\therefore x = 0,\quad y = 0$ $\\$ (vi) $\frac{3}{2}x - \frac{5}{3}y = -2 \qquad ..(1)$ $\\$ $\frac{x}{3} + \frac{y}{2} = \frac{13}{6} \qquad..{2}$ $\\$ From equation (1), we obtain $\\$ $9x -10y = -12$ $\\$ $x =\frac{-12 + 10y}{9} \qquad ..(3)$ $\\$ Substituting this value in equation (2), we obtain $\\$ $\frac{\frac{-12 +10y}{9}}{3} + \frac{y}{2} = \frac{13}{6}$ $\\$ $\frac{-12 + 10y}{27} + \frac{y}{2} = \frac{13}{6}$ $\\$ $\frac{-24 + 20y + 27y}{54} = \frac{13}{6}$ $\\$ $47y = 117 + 24$ $\\$ $47y = 141$ $\\$ $y = 3 \qquad ....(4)$ $\\$ Substituting this value in equation (3), we obtain $\\$ $x = \frac{-12 + 10 \times 3}{9} = \frac{18}{9} = 2$ $\\$ Hence, x = 2, y = 3

12   Solve $2x + 3y = 11$ and $2x - 4y = - 24$ and hence find the value of ‘m’ for which $y = mx + 3$.

##### Solution :

$2x + 3y =11 \qquad (1)$ $\\$ $2x - 4 y = -24 \qquad (1)$ $\\$ From equation (1), we obtain $\\$ $x = \frac{11 -3y}{2} \qquad(3)$ $\\$ Substituting this value in equation (2), we obtain $\\$ $2 \Bigg( \frac{11 -3y}{2} \Bigg) -4y =-24$ $\\$ $11 -3y -4y = -24$ $\\$ $-7y = -35$ $\\$ $y = 5 \qquad (5)$ $\\$ Putting this value in equation (3), we obtain $\\$ $x = \frac{11 - 3 \times 5}{2} = -\frac{4}{2} = -2$ $\\$ Hence,$x = -2, y = 5$ $\\$ Also, $y = mx + 3$ $\\$ $5 = -2m +3$ $\\$ $-2m =2$ $\\$ $m =-1$ $\\$

13   Form the pair of linear equations for the following problems and find their solution by substitution method. $\\$ (i) The difference between two numbers is $26$ and one number is three times the other. Find them. $\\$ (ii) The larger of two supplementary angles exceeds the smaller by $18$ degrees. Find them. $\\$ (iii) The coach of a cricket team buys $7$ bats and $6$ balls for Rs $3800$. Later, she buys $3$ bats and $5$ balls for Rs $1750$ Find the cost of each bat and each ball.$\\$ (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of $10$ km, the charge paid is Rs $105$ and for a journey of 15 km, the charge paid is Rs $155$. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of $25$ km.$\\$ (v) A fraction becomes $\frac{9}{11}$, if 2 is added to both the numerator and the denominator. If, $3$ is added to both the numerator and the denominator it becomes $\frac{5}{6}$. Find the fraction. $\\$ (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

##### Solution :

(i) Let the first number be x and the other number be y such that y > x. According to the given information, $\\$ $y = 3x \qquad (1)$ $\\$ $y-x =26 \qquad (2)$ $\\$ On substituting the value of y from equation (1) into equation (2), we obtain$\\$ $3x -x =26$ $\\$ $x = 13$ $\\$ Substituting this in equation (1), we obtain $\\$ $y =39$ $\\$ Hence, the numbers are 13 and 39. $\\$

(ii) Let the larger angle be x and smaller angle be y. $\\$ We know that the sum of the measures of angles of a supplementary pair is always $180^{\circ}.$ $\\$ According to the given information, $\\$ $x +y =180^{\circ} \qquad (1)$ $\\$ $x -y =180^{\circ} \qquad (2)$ $\\$ From (1), we obtain $x = 180^{\circ} - y \qquad (3)$ $\\$ Substituting this in equation (2), we obtain $\\$ $180^{\circ} -y -y =18^{\circ}$ $\\$ $162^{\circ} = 2y$ $\\$ $81^{\circ} = y \qquad(4)$ $\\$ Putting this in equation (3), we obtain$\\$ $x = 180^{\circ} - 81^{\circ} = 99^{\circ}$ $\\$ Hence, the angles are $81^{\circ}$ and $99^{\circ}$ $\\$

(iii) Let the cost of a bat and a ball be x and y respectively. $\\$ According to the given information, $\\$ $7x +6y =3800 \qquad (1)$ $\\$ $3x + 5y =1750 \qquad (2)$ $\\$ From (1), we obtain $\\$ $y = \frac{3800 - 7x}{6} \qquad(3)$ $\\$ Substituting this value in equation (2), we obtain $\\$ $3x + 5 \Bigg( \frac{3800 - 7x}{6} \Bigg) = 1750$ $\\$ $3x + \frac{9500}{3} - \frac{35x}{6} = 1750$ $\\$ $3x - \frac{35x}{3} - \frac{35x}{6} = 1750$ $\\$ $3x - \frac{35x}{6} =1750 - \frac{9500}{3}$ $\\$ $\frac{18x - 35x}{6} = \frac{5250 - 9500}{3}$ $\\$ $-\frac{17x}{6} = \frac{-4250}{3}$ $\\$ $-17x = -8500$ $\\$ $x =500 \qquad (4)$ $\\$ Substituting this in equation (3), we obtain $\\$ $y = \frac{3800 - 7 \times 500}{6}$ $\\$ $= \frac{300}{6} = 50$ $\\$ Hence, the cost of a bat is Rs 500 and that of a ball is Rs 50. $\\$

(iv)Let the fixed charge be Rs x and per km charge be Rs y. $\\$ According to the given information, $\\$ $x +10y =105 \qquad (1)$ $\\$ $x + 15y =155 \qquad (2)$ $\\$ From (3), we obtain $x = 105 -10y \qquad (3)$ $\\$ Substituting this in equation (2), we obtain $\\$ $105 - 10y + 15y =155$ $\\$ $5y = 50$ $\\$ $y =10 \qquad (4)$ $\\$ Putting this in equation (3), we obtain $\\$ $x = 105 - 10 \times 10$ $\\$ $x =5$ $\\$ Hence, fixed charge = Rs$5$ $\\$ And per km charge = Rs $10$ $\\$ Charge for $25 km = x +25y$ $\\$ $= 5 + 250 = Rs 255$ $\\$

(v) Let the fraction be $\frac{x}{y}$ $\\$ According to the given information, $\\$ $\frac{x +2}{y +2} = \frac{9}{11}$ $\\$ $11x + 22 = 9y +18$ $\\$ $11x -9y = -4 \qquad (1)$ $\\$ $\frac{ x +3}{y+3} = \frac{5}{6}$ $\\$ $6x +18 =5y +15$ $\\$ $6x -5y =-3 \qquad (2)$ $\\$ From equation (1), we obtain $x = \frac{-4 + 9y}{11} \qquad (3)$ $\\$ $6\Bigg( \frac{-4 + 9y}{11} \Bigg) -5y = -3$ $\\$ $-24 + 54 y - 55y = -33$ $\\$ $-y = -9$ $\\$ $y =9$ $\\$ Substituting this in equation (3), we obtain $\\$ $x = \frac{-4 + 81}{11} = 7$ $\\$ Hence, the fraction is $\frac{7}{9}.$

(vi) Let the age of Jacob be x and the age of his son be y. $\\$ According to the given information, $\\$ $(x + 5) = 3(y + 5)$ $\\$ $x-3y =10 \qquad (1)$ $\\$ $(x-5)=7(y-5)$ $\\$ $x -7y = -30$\qquad (2) $\\$ From (1), we obtain $x = 3y +10 \qquad (3)$ $\\$ Substituting this value in equation (2), we obtain $\\$ $3y +10 -7y =-30$ $\\$ $-4y =-40$ $\\$ $y =10 \qquad (4)$ $\\$ Substituting this value in equation (3), we obtain $\\$ $x =3 \times 10 +10 = 40$ $\\$ Hence, the present age of Jacob is 40 years whereas the present age of his son is 10 years.

14   Solve the following pair of linear equations by the elimination method and the substitution method: $\\$ (i) $x+y = 5 \qquad 2x -3y =4$ $\\$ (ii) $3x+4y = 10 \qquad 2x - 2y =2$ $\\$ (iii) $3x - 5y -4 =0 \qquad 9x = 2y +7$ $\\$ (iv) $\frac{x}{2} + \frac{2y}{3} = -1 \qquad x -\frac{y}{3} = 3$ $\\$

##### Solution :

(i) By elimination method $\\$ $x+y =5 \qquad (1)$ $\\$ $2x -3y =4 \qquad (2)$ $\\$ Multiplying equation (1) by (2), we obtain $\\$ $2x + 2y = 10 \qquad (3)$ $\\$ Subtracting equation (2) from equation (3), we obtain $\\$ $5y =6$ $\\$ $y = \frac{6}{5} \qquad (4)$ $\\$ Substituting the value in equation (1), we obtain $x = 5 - \frac{6}{5} = \frac{19}{5}$ $\\$ $\therefore x = \frac{19}{5}, \quad y = \frac{6}{5}$ $\\$ By substitution method $\\$ From equation (1), we obtain $\\$ $x = 5-y \qquad (5)$ $\\$ Putting this value in equation (2), we obtain $\\$ $2(5 -y) -3y =4$ $\\$ $-5y = -6$ $\\$ $y = \frac{6}{5}$ $\\$ Substituting the value in equation (5), we obtain $\\$ $x = 5 - \frac{ 6}{5} = \frac{19}{5}$ $\\$ $\therefore x = \frac{19}{y}, \quad y =\frac{6}{5}$ $\\$

(ii) By elimination method $\\$ $3x + 4y =10 \qquad (1)$ $\\$ $2x -2y =2 \qquad (2)$ $\\$ Multiplying equation (2) by 2, we obtain $\\$ $4x -4y =4 \qquad (3)$ $\\$ Adding equation (1) and (3), we obtain $\\$ $7x = 14$ $\\$ $x = 2 \qquad (4)$ $\\$ Substituting in equation (1), we obtain $\\$ $6 + 4y =10$ $\\$ $4y =4$ $\\$ $y =1$ $\\$ Hence, x = 2, y = 1 $\\$ By substitution method $\\$ From equation (2), we obtain $\\$ $x =1+y \qquad (5)$ $\\$ Putting this value in equation (1), we obtain $\\$ $3(1+y) + 4y =10$ $\\$ $7y =7$ $\\$ $y =1$ $\\$ Substituting the value in equation (5), we obtain $\\$ $x = 1 + 1 = 2$ $\\$ $\therefore x =2, y =1$ $\\$

(iii) By elimination method $3x -5y -4 = o \qquad(1)$ $\\$ $9x = 2y+ 7$ $\\$ $9x -2y -7 =0 \qquad (2)$ $\\$ Multiplying equation (1) by 3, we obtain $\\$ $9x -15y -12 =0 \qquad (3)$ $\\$ Subtracting equation (3) from equation (2), we obtain $\\$ $13y = -5$ $\\$ $y = \frac{-5}{13} \qquad (4)$ Substituting in equation (1), we obtain $\\$ $3x + \frac{25}{13} -4 =0$ $\\$ $3x = \frac{27}{13}$ $\\$ $x = \frac{9}{13}$ $\\$ $\therefore x = \frac{9}{13}, \quad y = \frac{-5}{13}$ $\\$ By substitution method $\\$ From equation (1), we obtain $\\$ $x = \frac{5y +4}{3} \qquad (5)$ $\\$ Putting this value in equation (2), we obtain $\\$ $9 \Bigg( \frac{5y + 4}{3} \Bigg) -2y -7 =0$ $\\$ $13y = -5$ $\\$ $y = - \frac{5}{13}$ $\\$ Substituting the value in equation (5), we obtain $\\$ $x = \frac{ 5 \Bigg( \frac{-5}{13}\Bigg) +4 }{3}$ $\\$ $x = \frac{9}{13}$ $\\$ $\therefore x = \frac{9}{13}, \quad y =\frac{-5}{13}$

(iv)By elimination method $\\$ $\frac{x}{2} + \frac{2y}{3} = -1$ $\\$ $3x +4y =-6 \qquad (1)$ $\\$ $x -\frac{y}{3} = 3$ $\\$ $3x -y =9 \qquad (2)$ $\\$ Subtracting equation (2) from equation (1), we obtain $\\$ $5y = -15$ $\\$ $y = -3 \qquad (3)$ $\\$ Substituting this value in equation (1), we obtain $\\$ $3x -12 = -6$ $\\$ $3x =6$ $\\$ $x =2$ $\\$ Hence, x = 2, y = -3 $\\$ By substitution method $\\$ From equation (2), we obtain $\\$ $x = \frac{y+9}{3} \qquad (5)$ $\\$ Putting this value in equation (1), we obtain $\\$ $3\Bigg( \frac{y+9}{3} \Bigg) +4y = -6$ $\\$ $5y = -5$ $\\$ $y = -3$ $\\$ Substituting the value in equation (5), we obtain $\\$ $x = \frac{-3 + 9 }{3} = 2$ $\therefore x= 2, y=-3$

15   Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method: $\\$ (i) If we add $1$ to the numerator and subtract 1 from the denominator, a fraction reduces to $1$. It becomes $\frac{1}{2}$ if we only add 1 to the denominator. What is the fraction? $\\$ (ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu? $\\$ (iii) The sum of the digits of a two-digit number is $9$. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number. $\\$ (iv) Meena went to bank to withdraw $Rs 2000$. She asked the cashier to give her $Rs 50$ and $Rs 100$ notes only. Meena got $25$ notes in all. Find how many notes of $Rs 50$ and $Rs 100$ she received. $\\$ (v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid $Rs 27$ for a book kept for seven days, while Susy paid $Rs 21$ for the book she kept for five days. Find the fixed charge and the charge for each extra day.

##### Solution :

(i)Let the fraction be $\frac{x}{y}$ $\\$ According to the given information, $\\$ $\frac{x+1}{y-1} =1 \qquad \Rightarrow x-y = -2 \qquad (1)$ $\\$ $\frac{x}{y+1} =\frac{1}{2} \qquad \Rightarrow 2x-y = 1 \qquad (2)$ $\\$ Subtracting equation (1) from equation (2), we obtain $\\$ $x = 3 (3)$ $\\$ Substituting this value in equation (1), we obtain $\\$ $3-y =-2$ $\\$ $-y =-5$ $\\$ $y =5$ $\\$ Hence, the fraction is $\frac{3}{5}$ $\\$ (ii)Let present age of Nuri = x $\\$ and present age of Sonu = y $\\$ According to the given information, $\\$ $(x -5) =3 (y -5 )$ $\\$ $x-3y =-10 \qquad (1)$ $\\$ $( x + 10 )= 2(y + 10)$ $\\$ $x - 2y =10 \qquad (2)$ $\\$ Subtracting equation (1) from equation (2), we obtain $\\$ $y = 20 \qquad (3)$ $\\$ Substituting it in equation (1), we obtain$\\$ $x -60 = -10$ $\\$ $x =50$ $\\$ Hence, age of Nuri = 50 years$\\$ And, age of Sonu = 20 years $\\$

(iii)Let the unit digit and tens digits of the number be x and y respectively. Then, number $= 10y + x$ $\\$ Number after reversing the digits = 10x + y  $\\$ According to the given information, $\\$ $x + y = 9 \qquad (1)$ $\\$ $9(10y + x) = 2(10x + y)$ $\\$ $88y -11x = 0$ $\\$ $-x + 8y =0 \qquad (2)$ $\\$ Adding equation (1) and (2), we obtain $\\$ $9y = 9$ $\\$ $y = 1 \qquad (3)$ $\\$ Substituting the value in equation (1), we obtain $\\$ $x = 8$ $\\$ Hence, the number is 10y + x = 10 × 1 + 8 = 18 $\\$ (iv)Let the number of Rs 50 notes and Rs 100 notes be x and y respectively. $\\$ According to the given information, $\\$ $x +y =25$ $\\$ $50x +100y = 2000 \qquad (2)$ $\\$ Multiplying equation (1) by 50, we obtain $\\$ $50x +50y =1250$ $\\$ Subtracting equation (3) from equation (2), we obtain $\\$ $50y = 750$ $\\$ $y =15$ $\\$ Substituting in equation (1), we have $x = 10$ $\\$ Hence, Meena has 10 notes of $Rs 50$ and 15 notes of $Rs 100.$ $\\$ (v)Let the fixed charge for first three days and each day charge thereafter be Rs x and Rs y respectively. $\\$ According to the given information, $\\$ $x +4y =27 \qquad (1)$ $\\$ $x +2y =21 \qquad (2)$ $\\$ Subtracting equation (2) from equation (1), we obtain $\\$ $2y =6$ $\\$ $y =3 \qquad (3)$ $\\$ Substituting in equation (1), we obtain $\\$ $x +12 =27$ $\\$ $x =15$ $\\$ Hence, fixed charge = Rs 15 $\\$ And Charge per day = Rs 3 $\\$

16   Which of the following pairs of linear equations has unique solution, no solution or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method. $\\$ (i) $x -3y -3 =0 \qquad 3x -9y -2 =0$ $\\$ (ii) $2x + y = 5 \qquad 3x + 2y = 8$ $\\$ (iii) $3x - 5y = 20 \qquad x - 3y -7 = 0$ $\\$ (iv) $x - 3y -7 = 0 \qquad 3x - 3y -15 = 0$ $\\$

##### Solution :

( i ) $x-3y -3 =0$ $\\$ $3x -9y -2 =0$ $\\$ $\frac{a_1}{a_2} = \frac{1}{3}, \frac{b_1}{b_2} = \frac{-3}{-9} = \frac{1}{3}, \frac{c_1}{c_2} = \frac{-3}{-2} = \frac{3}{2}$ $\\$ $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ $\\$ Therefore, the given sets of lines are parallel to each other. Therefore, they will not intersect each other and thus, there will not be any solution for these equations. $\\$ (ii) $2x + y = 5$ $\\$ $3x + 2y =8$ $\\$ $\frac{a_1}{a_2} = \frac{2}{3}, \quad \frac{b_1}{b_1} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{-5}{-8}$ $\\$ $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ $\\$ Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations. $\\$ By cross-multiplication method,$\\$ $\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1 } = \frac{1}{a_1b_2 - a_2b_1 }$ $\\$ $\frac{x}{-8-(-10)} = \frac{y}{-15 +16} = \frac{1}{4-3}$ $\\$ $\frac{x}{2} = \frac{y}{1} = 1$ $\\$ $\frac{x}{2} =1. \quad \frac{y}{1} = 1$ $\\$ $\therefore x = 2, \quad y = 1$ $\\$ (iii)$3x -5y =20$ $\\$ $6x -10y =40$ $\\$ $\frac{a_1}{a_2} = \frac{3}{6} =\frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{-5}{-10}, \quad \frac{c_1}{c_2} = \frac{-20}{-40} =\frac{1}{2}$ $\\$ $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ $\\$ Therefore, the given sets of lines will be overlapping each other i.e., the lines will be coincident to each other and thus, there are infinite solutions possible for these equations. $\\$

(iv) $x -3y -7 =0$ $\\$ $3x - 3y -15 = 0$ $\\$ $\frac{a_1}{a_2} = \frac{1}{3}, \quad \frac{b_1}{b_2} = \frac{-3}{-3} =1, \quad \frac{c_1}{c_2} = \frac{-7}{-15} = \frac{7}{15}$ $\\$ $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ $\\$ Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations. $\\$ By cross-multiplication, $\\$ $\frac{x}{45 - (21)} = \frac{y}{-21 -(-15)} = \frac{1}{-3 - (-9)}$ $\\$ $\frac{x}{24} = \frac{y}{-6} = \frac{1}{6}$ $\\$ $\frac{x}{24} = \frac{1}{6} \quad and \quad \frac{y}{-6} =\frac{1}{6}$ $\\$ $x =4 \quad and \quad y =-1$ $\\$ $\therefore x =4, \quad y = -1$ $\\$

17   (i) For which values of a and b will the following pair of linear equations have an infinite number of solutions? $\\$ $2x +3y =7$ $\\$ $(a-b)x + (a+b)y = 3a +b -2$ $\\$ (ii) For which value of k will the following pair of linear equations have no solution? $\\$ $3x +y =1$ $\\$ $(2k -1)x + (k-1)y = 2k +1$ $\\$

##### Solution :

(i) $2x +3y =7$ $\\$ $(a-b)x + (a+b)y - (3a +b -2) = 0$ $\\$ $\frac{a_1}{a_2} = \frac{2}{a-b}, \quad \frac{b_1}{b_2} = \frac{3}{a+b}, \quad \frac{c_1}{c_2} = \frac{-7}{-(3a +b -2)} = \frac{7}{(3a +b -2)}$ $\\$ For infinitely many solutions, $\\$ $\frac{a_1}{a_2}= \frac{b_1}{b_2} = \frac{c_1}{c_2}$ $\\$ $\frac{2}{a-b} = \frac{7}{3a +b -2}$ $\\$ $6a + 2b - 4 = 7a -7b$ $\\$ $a-9b = -4 \qquad (1)$ $\\$ $\frac{2}{a-b} = \frac{3}{a+b}$ $\\$ $2a + 2b = 3a -3b$ $\\$ $a-5b =0 \qquad (2)$ $\\$ Subtracting (1) from (2), we obtain $\\$ $4b =4$ $\\$ $b =1$ $\\$ Substituting this in equation (2), we obtain $\\$ $a -5 \times 1 =0$ $\\$ $a =5$ $\\$ Hence, a = 5 and b = 1 are the values for which the given equations give infinitely many solutions. $\\$ (ii)$3x +y -1 = 0$ $\\$ $(2k -1)x + (k-1)y -2k -1 =0$ $\\$ $\frac{a_1}{a_2} = \frac{3}{2k-1}, \quad \frac{b_1}{b_2} = \frac{1}{k-1}, \quad \frac{c_1}{c_2} = \frac{-1}{-2k-1} = \frac{1}{2k +1}$ $\\$ For no solution, $\\$ $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ $\\$ $\frac{3}{2k-1} = \frac{1}{k-1} \neq \frac{1}{2k-1}$ $\\$ $\frac{3}{2k-1} = \frac{1}{ k-1}$ $\\$ $3k -3 = 2k-1$ $\\$ $k =2$ $\\$ Hence, for k = 2, the given equation has no solution.

18   Solve the following pair of linear equations by the substitution and cross-multiplication methods: $\\$ $8x + 5y =9$ $\\$ $3x +2y =4$

##### Solution :

$8x + 5y =9 \qquad (i)$ $\\$ $3x +2y =4 \qquad (ii)$ $\\$ From equation (ii), we obtain $\\$ $x = \frac{4-2y}{3} \qquad (3)$ $\\$ Substituting this value in equation (i), we obtain $\\$ $8 \Bigg( \frac{4-2y}{3} \Bigg) + 5y =9$ $\\$ $32 -16y + 15y =27$ $\\$ $-y = -5$ $\\$ $y =5 \qquad(iv)$ $\\$ Substituting this value in equation (ii), we obtain $\\$ $3x + 10 =4$ $\\$ $x =-2$ $\\$ Hence, $x =-2, \quad y=5$ Again, by cross-multiplication method, we obtain $\\$ $8x + 5y -9 = 0$ $\\$ $3x + 2y -4 = 0$ $\\$ $\frac{x}{-20 -(-18)} = \frac{y}{-27 -(-32)} = \frac{1}{16- 15}$ $\\$ $\frac{x}{-2} =1, \quad and \quad \frac{y}{5} =1$ $\\$ $x =-2, \quad and \quad y =5$

19   Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method: $\\$ (i)A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for $20$ days she has to pay $Rs 1000$ as hostel charges whereas a student B, who takes food for $26$ days, pays $Rs 1180$ as hostel charges. Find the fixed charges and the cost of food per day. $\\$ (ii)A fraction becomes $\frac{1}{3}$ when $1$ is subtracted from the numerator and it becomes when 8 is added to its denominator. Find the fraction. $\\$ (iii)Yash scored $40$ marks in a test, getting $3$ marks for each right answer and losing $1$ mark for each wrong answer. Had $4$ marks been awarded for each correct answer and $2$ marks been deducted for each incorrect answer, then Yash would have scored $50$ marks. How many questions were there in the test? $\\$ (iv) Places A and B are $100 km$ apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in $5 hours$. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars? $\\$ (v)The area of a rectangle gets reduced by $9$ square units, if its length is reduced by $5$ units and breadth is increased by $3$ units. If we increase the length by $3$ units and the breadth by $2$ units, the area increases by 67 square units. Find the dimensions of the rectangle.

##### Solution :

(i) Let x be the fixed charge of the food and y be the charge for food per day. According to the given information, $\\$ $x +20y =1000$ $\\$ $x +26y = 1180$ $\\$ Subtracting equation (1) from equation (2), we obtain $\\$ $6y =180$ $\\$ $y =30$ $\\$ Substituting this value in equation (1), we obtain, $\\$ $x+20 \times 30 =1000$ $\\$ $x =1000 - 600$ $\\$ $x =400$ $\\$ Hence, fixed charge = $Rs 400$ $\\$ And charge per day = $Rs 30$ $\\$ (ii)Let the fraction be $\frac{x}{y}$ $\\$ According to the given information, $\\$ $\frac{x-1}{y} = \frac{1}{3} \qquad \Rightarrow \qquad 3x -y =3 \qquad (1)$ $\frac{x}{y + 8} = \frac{1}{4} \qquad \Rightarrow \qquad 4x -y =8 \qquad (2)$ Subtracting equation (1) from equation (2), we obtain $\\$ $x =5 \qquad (3)$ $\\$ Putting this value in equation (1), we obtain $\\$ $15 -y =3$ $\\$ $y =12$ $\\$ Hence, the fraction is $\frac{5}{12}$

(iii)Let the number of right answers and wrong answers be $x$ and $y$ respectively. $\\$ According to the given information, $\\$ $3x - y= 40 \qquad (1)$ $\\$ $4x -2y = 50$ $\\$ $\Rightarrow 2x - y =25 \qquad (2)$ $\\$ Subtracting equation (2) from equation (1), we obtain $\\$ $x = 15 \qquad (3)$ $\\$ Substituting this in equation (2), we obtain $\\$ $30 - y =25$ $\\$ $y =5$ $\\$ Therefore, number of right answers $= 15$ $\\$ And number of wrong answers $= 5$ $\\$ Total number of questions $= 20$ $\\$ (iv)Let the speed of $1^{st}$ car and $2 ^{nd}$ car be $u$ km/h and $v$ km/h. $\\$ Respective speed of both cars while they are travelling in same direction $= ( u-v)$ km/h $\\$ Respective speed of both cars while they are travelling in opposite directions i.e., travelling towards each other $= (u-v)$ km/h $\\$ According to the given information, $\\$ $5 (u-v) =100$ $\\$ $\Rightarrow u-v =10 \qquad ....(1)$ $\\$ $1(u + v) = 100 \qquad ...(2)$ $\\$ Adding both the equations, we obtain $\\$ $2u =120$ $\\$ $u = 60$ km/h $\qquad ...(3)$ $\\$ Substituting this value in equation (2), we obtain $\\$ $v = 40 km/h$ $\\$ Hence, speed of one car $= 60 km/h$ and speed of other car $= 40$km/h $\\$ (v) Let length and breadth of rectangle be x unit and y unit respectively. $\\$ Area $= xy$ $\\$ According to the question, $\\$ $(x-5)(x+3) = xy -9$ $\\$ $\Rightarrow 3x - 5y -6 =0 \qquad (1)$ $\\$ $(x + 3) (y +2) = xy +67$ $\\$ $\Rightarrow 2x + 3y -61 =0 \qquad (2)$ $\\$ By cross-multiplication method, we obtain $\\$ $\frac{x}{305 - (-18) }= \frac{y}{-12 - (-183)} = \frac{1}{9-(-10)}$ $\\$ $\frac{x}{323} = \frac{y}{171} = \frac{1}{19}$ $\\$ $x =17, \quad y =9$ $\\$ Hence, the length and breadth of the rectangle are 17 units and 9 units respectively.

20   Solve the following pairs of equations by reducing them to a pair of linear equations: $\\$ (i)$\frac{1}{2x} + \frac{1}{3y} =2 \qquad \frac{1}{3x} + \frac{1}{2y} = \frac{13}{6}$ $\\$ (ii)$\frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} =2 \qquad \frac{4}{\sqrt{x}} - \frac{9}{\sqrt{y}} = -1$ $\\$ (iii)$\frac{4}{x} + 3y = 14 \qquad \frac{3}{x} - 4y = 23$ $\\$ (iv)$\frac{1}{x-1} + \frac{1}{y-2} = 2 \qquad \frac{6}{x -1} - \frac{3}{y-2} = 1$ $\\$ (v)$\frac{7x -2y}{xy} =5 \qquad \frac{8x +7y}{x y} =15$ $\\$ (vi)$6x + 3y =6xy \qquad 2x + 4y = 5xy$ $\\$ (vii)$\frac{10}{x +y} + \frac{2}{x- y} =4 \qquad \frac{15 }{x +y} - \frac{5}{x -y} = -2$ $\\$ (viii)$\frac{1}{3x +y} + \frac{1}{3x-y} = \frac{3}{4} \qquad \frac{1}{2(3x +y)} + \frac{1}{2(3x -y)} = \frac{-1}{8}$ $\\$

##### Solution :

(i) $\frac{1}{2x} + \frac{1}{3y} = 2$ $\\$ $\frac{1}{3x} + \frac{1}{2y} = \frac{13}{6}$ $\\$ Let $\frac{1}{x} = p$ and $\frac{1}{y} =q,$ then the equation changes as follow. $\frac{p}{2} + \frac{q}{3} = 2 \qquad \Rightarrow \qquad 3p +2q -12 =0 \qquad (1)$ $\frac{p}{3} + \frac{q}{2} = \frac{13}{6} \qquad \Rightarrow \qquad 2p + 3q -13 =0 \qquad (2)$ $\\$ Using cross-multiplication method, we obtain $\\$ $\frac{p}{-26 -(-36)} = \frac{q}{-24 - (-39)} = \frac{1}{9-4}$ $\\$ $\frac{p}{10} = \frac{q}{15} =\frac{1}{5}$ $\\$ $\frac{p}{10} = \frac{1}{5} \quad and \quad \frac{q}{15 } = \frac{1}{5}$ $\\$ $p = 2 \quad and \quad q =3$ $\\$ $\frac{1}{x} = 2 \quad and \quad \frac{1}{y}=3$ $\\$ $x =\frac{1}{2} \quad and \quad y =\frac{1}{3}$ $\\$ (ii) $\frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2$ $\\$ $\frac{4}{\sqrt{x}} - \frac{9}{\sqrt{y}} =-1$ $\\$ Putting $\frac{1}{\sqrt{x}} = p$ and $\frac{1}{\sqrt{y}} = q$ in the given equations, we obtain $2p + 3q =2 \qquad (1)$ $\\$ $4p - 9q = -1 \qquad (2)$ $\\$ Multiplying equation (1) by 3, we obtain $\\$ $6p + 9q = 6 \qquad (3)$ $\\$ Adding equation (2) and (3), we obtain $\\$ $10p =5$ $\\$ $p =\frac{1}{2} \qquad (4)$ $\\$ Putting in equation (1), we obtain $\\$ $2 \times \frac{1}{2} + 3q =2$ $\\$ $3q =1$ $\\$ $q =\frac{1}{3}$ $\\$ $p = \frac{1}{\sqrt{x}} = \frac{1}{2}$ $\\$ $\sqrt{x} = 2$ $\\$ $x =4$ and $q = \frac{1}{\sqrt{y}} = \frac{1}{3}$ $\\$ $\sqrt{y} = 3$ $\\$ $y= 9$ $\\$ Hence, $x = 4, y =9$ $\\$

$\frac{p}{-69 -56} = \frac{y}{-42-(-92) } = \frac{1}{-16-9}$ $\\$ $\frac{-p}{-125} = \frac{y}{50} = \frac{-1}{25}$ $\\$ $\frac{p}{-125 } = \frac{-1}{25} \quad$ and $\quad \frac{y}{50} = \frac{-1}{25}$ $\\$ $p =5 \quad$ and $\quad y =-2$ $\\$ $p = \frac{1}{x} = 5$ $\\$ $x = \frac{1}{5}$ $\\$ $y =-2$ $\\$

(iii) $\frac{4}{x} + 3y =14$ $\\$ $\frac{3}{x} -4y =23$ $\\$ Substituting $\frac{1}{x} = p$ in the given equations, we obtain$\\$ $4p + 3y =14 \qquad \Rightarrow \qquad 4p + 3y -14 =0 \qquad (1)$ $\\$ $3p -4y =23 \qquad \Rightarrow \qquad 3p -4y -23 = 0 \qquad (2)$ $\\$ By cross-multiplication, we obtain $\\$ $\frac{p}{-69 -56} = \frac{y}{-42-(-92) } = \frac{1}{-16-9}$ $\\$ $\frac{-p}{-125} = \frac{y}{50} = \frac{-1}{25}$ $\\$ $\frac{p}{-125 } = \frac{-1}{25} \quad$ and $\quad \frac{y}{50} = \frac{-1}{25}$ $\\$ $p =5 \quad$ and $\quad y =-2$ $\\$ $p = \frac{1}{x} = 5$ $\\$ $x = \frac{1}{5}$ $\\$ $y =-2$ $\\$

(iv) $\dfrac {5}{x-1} + \dfrac {1}{y -2} = 2$ $\\$ $\dfrac {6}{x-1} - \dfrac {3}{y -2} = 1$ $\\$ Putting $\dfrac {1}{x-1} = P and \dfrac {1}{y -2} = q$ in the given equation, we obtain $\\$ $5p + q = 2 \quad \quad \quad (1)$ $\\$ $6p -3q = 1 \quad \quad \quad (2)$ $\\$ Multiplying equation (1) by 3, we obtain $\\$ $15p + 3q = 6 \quad \quad \quad (3)$ $\\$ Adding (2) and (3), we obtain $\\$ $21p = 7$ $\\$ $p = \dfrac{1}{3}$ $\\$ Putting this value in equation (1), we obtain $\\$ $5 * \dfrac {1}{3} + q = 2$ $\\$ $q = 2 - \dfrac {5}{3} = \dfrac{1}{3}$ $\\$ $p = \dfrac{1}{1 - x} = \dfrac {1}{3} $$\\ \Rightarrow x - 1 = 3 \\ \Rightarrow x = 4 \\ q = \dfrac{1}{y - 2} = \dfrac {1}{3}$$\\$ $y - 2 = 3$ $\\$ $y = 5$ $\\$ $\therefore x =4 , y = 5$ $\\$

(v)$\dfrac {7x-2y}{xy} = 5$ $\\$ $\dfrac {7}{y} - \dfrac {2}{x} = 5 \quad \quad \quad (1)$ $\\$ $\dfrac {8x + 7y}{xy} = 15$ $\\$ $\dfrac {8}{y} + \dfrac {7}{x} = 15 \quad \quad \quad (2)$ $\\$ Putting $\dfrac{1}{x} = p$ and $\dfrac{1}{y} = q$ in the given equation, we obtain $\\$ $-2p + 7q = 5 \quad \quad \quad \Rightarrow -2p + 7q - 5 = 0 \quad \quad (3)$ $\\$ $7p + 8q = 15 \quad \quad \quad \Rightarrow 7p + 8q - 15 = 0 \quad \quad (4)$ $\\$ By cross-multiplication method, we obtain $\\$ $\dfrac {p}{- 105 - (-40)} = \dfrac {q}{- 35 - 30} = \dfrac {1}{- 16 - 49}$ $\\$ $\dfrac {p}{- 65} = \dfrac {q}{- 65} = \dfrac {1}{- 65}$ $\\$ $\dfrac {p}{- 65} = \dfrac {1}{- 65} \ and \ \dfrac {q}{- 65} = \dfrac {1}{- 65}$ $\\$ p $=1 \$ and $\ q =1$ $\\$ $p = \dfrac{1}{x} = 1 \ \ \ q = \dfrac{1}{y} = 1$ $\\$ $x = 1 \ \ \ \ y =1$

(vi) $6x + 3y = 6xy$ $\\$ $\Rightarrow \dfrac{6}{y} + \dfrac{3}{x} = 6 \quad \quad (1)$ $\\$ $2x + 4y = 5xy$ $\\$ $\dfrac{2}{y} + \dfrac{4}{x} = 5 \quad \quad \quad (2)$ $\\$ Putting $\dfrac{1}{x} = p$ and $\dfrac{1}{y} = q$ in the given equation, we obtain $\\$ $3p + 6q - 6 = 0$ $\\$ $4p + 2q - 5 = 0$ $\\$ By cross-multiplication method, we obtain $\\$ $\dfrac{p}{-30-(-12)} = \dfrac{q}{-24-(-15)} = \dfrac{1}{6 - 24}$ $\\$ $\dfrac{p}{-18} - \dfrac{1}{-18} = \dfrac{1}{ - 18}$ $\\$ $\dfrac{p}{-18} = \dfrac{1}{ - 18}$ and $\dfrac{q}{-9} = \dfrac{1}{ - 18}$ $\\$ $p = 1$ and $q = \dfrac {1}{2}$ $\\$ $p = \dfrac{1}{x} = 1 \quad \quad q = \dfrac{1}{y} = \dfrac{1}{2}$ $\\$ $x = 1 \quad \quad y = 2$ $\\$

(vii) $\dfrac{10}{x + y} + \dfrac{2}{x - y} = 4$ $\\$ $\dfrac{15}{x + y} - \dfrac{5}{x - y} = - 2$ $\\$ putting $\dfrac{1}{x - 1} = p$ and $\dfrac{1}{x - y} = q$ in the given equations, we obtain $\\$ $10p + 2q = 4 \quad \ \ \ \Rightarrow \quad 10p + 2q - 4 =0 \qquad (1)$ $\\$ $15p - 5q = -2 \quad \Rightarrow \quad 15p - 5q + 2 =0 \qquad (2)$ $\\$ Using cross-multiplication method, we obtain $\\$ $\frac{p}{4 - 20} = \frac{q}{ - 60 - (20)} = \frac{1}{ - 50 -30}$ $\\$ $\frac{p}{- 16} = \frac{q}{ - 80} = \frac{1}{ -80}$ $\\$ $\frac{p}{- 16} = \frac{1}{ - 80}$ and $\frac{q}{ -80} = \frac{1}{ -80}$ $\\$ $P = \frac{1}{5}$ and $q = 1$ $\\$ $P = \frac{1}{x+y} = \frac{1}{5}$ and $q = \frac{1}{x-y} = 1$ $\\$ $x + y = 5 \quad \quad (3)$ $\\$ and $x - y = 1 \quad \quad (4)$ $\\$ Adding equation (3) and (4), we obtain $\\$ $2x = 6$ $\\$ $x = 3 \quad \quad \quad (5)$ $\\$ Substituting in equation (3), we obtain $\\$ $y = 2$ $\\$ Hence, $x = 3, y =2$ $\\$

(viii) $\dfrac{1}{3x + y} + \dfrac{1}{3x - y} = \dfrac{3}{4}$ $\\$ $\dfrac{1}{2(3x +y)} + \dfrac{1}{2(3x -y)} = \dfrac{-1}{8}$ $\\$ Putting $\dfrac {1}{3x + y} = p$ and $\dfrac {1}{3x - y} = q$ in these equations, we obtain $\\$ $p + q = \dfrac{3}{4} \quad \quad \quad (1)$ $\\$ $\dfrac{p}{2} - \dfrac{q}{2} = \dfrac{-1}{8}$ $\\$ $p - q = \dfrac{-1}{4} \quad \quad \quad (2)$ $\\$ Adding $(1)$ and $(2),$ we obtain $\\$ $2p = \dfrac {3}{4} - \dfrac {1}{4}$ $\\$ $2p = \dfrac {1}{2}$ $\\$ $p = \dfrac {1}{4}$ $\\$ Substituting in $(2)$, we obtain $\\$ $\dfrac {1}{4} - q = \dfrac {-1}{4}$ $\\$ $q = \dfrac {1}{4} + \dfrac {1}{4} = \dfrac {1}{2}$ $\\$ $p = \dfrac {1}{3x + y} = \dfrac {1}{4}$ $\\$ $3x + y = 4 \quad \quad \quad (3)$ $\\$ $q = \dfrac {1}{3x - y} = \dfrac {1}{2}$ $\\$ $3x - y = 4 \quad \quad \quad (4)$ $\\$ Adding equations (3) and (4), we obtain $\\$ $6x = 6$ $\\$ $x = 1 \quad \quad \quad \quad \quad\quad \quad (5)$ $\\$ Substituting in $(3)$, we obtain $\\$ $3 (1) + y = 4$ $\\$ $y = 1$ $\\$ Hence, $x = 1, y = 1$ $\\$

21   Formulate the following problems as a pair of equations, and hence find their solutions: $\\$ (i) Ritu can row downstream $20\ km$ in $2$ hours, and upstream $4\ km$ in $2$ hours. Find her speed of rowing in still water and the speed of the current. $\\$ (ii) $2$ women and $5$ men can together finish an embroidery work in $4$ days, while $3$ women and $6$ men can finish it in $3$ days. Find the time taken by $1$ woman alone to finish the work, and also that taken by $1$ man alone. $\\$ (iii) Roohi travels $300\ km$ to her home partly by train and partly by bus. She takes $4$ hours if she travels $60\ km$ by train and the remaining by bus. If she travels $100\ km$ by train and the remaining by bus, she takes $10$ minutes longer. Find the speed of the train and the bus separately.$\\$

##### Solution :

(i) Let the speed of Ritu in still water and the speed of stream be $x \ km/h$ and $y \ km/h$ respectively. $\\$ Speed of Ritu while rowing $\\$ Upstream $= (x – y) km/h$ $\\$ Downstream $= (x + y) km/h$ $\\$ According to the question, $\\$ $2(x +y) = 20$ $\\$ $\Rightarrow \ \ x + y = 10 \quad \quad \quad (1)$ $\\$ $2(x - y) = 4$ $\\$ $\Rightarrow \ \ x - y = 2 \quad \quad \quad (2)$ $\\$ Adding equation $(1)$ and $(2),$we obtain $\\$ $2x = 12 \ \ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ x = 6$ $\\$ $y = 4$ $\\$ Hence, Ritu’s speed in still water is $6 km/h$ and the speed of the current is $4 km/h.$ $\\$ (ii) Let the number of days taken by a woman and a man be x and y respectively. $\\$ Therefore, work done by a woman in $1$ day $= \dfrac{1}{x}$ $\\$ work done by a man in $1$ day $= \dfrac{1}{y}$ $\\$ According to the question, $\\$ $4 \left(\dfrac {2}{x} + \dfrac {5}{Y} \right) = 1$ $\\$ $\dfrac{2}{x} + \dfrac{5}{y} = \dfrac{1}{4}$ $\\$ $3 \left(\dfrac {3}{x} + \dfrac {6}{Y} \right) = 1$ $\\$ $\dfrac{3}{x} + \dfrac{6}{y} = \dfrac{1}{3}$ $\\$ Putting $\dfrac{1}{x} = p$ and $\dfrac{1}{y} = q$ in these equations, we obtain $\\$ $2p + 5q = \dfrac{1}{4}$ $\\$ $\Rightarrow \ 8p + 20q = 1$ $\\$ $3p + 6q = \dfrac{1}{3}$ $\\$ $\Rightarrow \ 9p + 18q = 1$ $\\$ By cross-multiplication, we obtain $\\$ $\dfrac {p}{-20-(-18)} = \dfrac {q}{-9-(-8)} = \dfrac {1}{144 - 180)}$ $\\$ $\dfrac {p}{-2} = \dfrac {q}{-1} = \dfrac {1}{-36}$ $\\$ $\dfrac {p}{-2} = \dfrac {-1}{36}$and$\dfrac {q}{-1} = \dfrac {-1}{36}$ $\\$ $x = 18 \ \ \ \ \ \ \$ $y = 36$ $\\$ Hence, number of days taken by a woman $= 18$ $\\$ Number of days taken by a man $= 36$ $\\$

(iii) Let the speed of train and bus be u km/h and v km/h respectively. According to the given information, $\\$ $\dfrac {60}{u} + \dfrac {240}{v} = 4 \quad \quad \quad (1)$ $\\$ $\dfrac {100}{u} + \dfrac {200}{v} = \dfrac{25}{6} \quad \quad (2)$ $\\$ Putting $\dfrac{1}{u} = p \$ and $\dfrac{1}{v} = q$ q in these equations, we obtain $\\$ $60p + 240q = 4 \quad \quad \quad \quad (3) $$\\ 100p + 200q = \dfrac{25}{6}$$\\$ $600p + 1200q = 25 \quad \quad \quad (4) $$\\ Multiplying equation (3) by 10, we obtain \\ 600p + 2400q = 40 \quad \quad \quad (5)$$\\$ Subtracting equation $(4)$ from $(5)$, we obtain $\\$ $1200q = 15 $$\\ q = \dfrac{15}{1200} = \dfrac{1}{80} \quad \quad \quad (6) \\ Substituting in equation (3), we obtain \\ 60p + 3 = 4 \\ 60p = 1 \\ p = \dfrac{1}{60} \\ p = \dfrac{1}{u} = \dfrac{1}{60} annd q = \dfrac{1}{v} = \dfrac{1}{80} \\ u = 60 km/h and v = 80 km/h \\ Hence, speed of train = 60 km/h \\ Hence, speed of bus = 80 km/h \\ 22 The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differs by 30 years. Find the ages of Ani and Biju. ##### Solution : The difference between the ages of Biju and Ani is 3 years. Either Biju is 3 years older than Ani or Ani is 3 years older than Biju. However, it is obvious that in both cases, Ani’s father’s age will be 30 years more than that of Cathy’s age.\\ Let the age of Ani and Biju be x and y years respectively. Therefore, age of Ani’s father, Dharam = 2 × x = 2x years\\ And age of Biju’s sister Cathy=\dfrac{y}{2} years\\ By using the information given in the question,\\ Case (I) When Ani is older than Biju by 3 years,\\ x -y = 3 (i)\\ 4x - y = 60 (ii)\\ Subtracting (i) from (ii), we obtain 3x = 60 -3 = 57$$\\$ $x=\dfrac{57}{3} =19$$\\ Therefore, age of Ani = 19 years And age of Biju = 19 - 3 = 16 years \\ Case (II) When Biju is older than Ani,\\ y -x = 3 (i)\\ 2x-\dfrac{y}{2}=30\\ 4x - y = 60 (ii)\\ Adding (i) and (ii), we obtain 3x = 63\\ x = 21$$\\$ Therefore, age of Ani =$21$ years And age of Biju = $21 + 3 = 24$ years

23   One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what isthe amount of their (respective) capital? [From the Bijaganita of Bhaskara II] [Hint$\colon x + 100 = 2 (y - 100), y + 10 = 6(x- 10)$]

##### Solution :

Let those friends were having Rs x and y with them. Using the information given in the question, we obtain$\\$ $x + 100 = 2(y - 100) \\ x + 100 = 2y -200 x- 2y = -300 (i)$$\\ And, 6(x -10) = (y + 10)\\ 6x - 60 = y + 10\\ 6x - y = 70 (ii) \\ Multiplying equation (ii) by 2, we obtain\\ 12x - 2y = 140 (iii) \\ Subtracting equation (i) from equation (iii), we obtain\\ 11x = 140 + 300\\ 11x = 440\\ x = 40$$\\$ Using this in equation (i), we obtain$\\$ $40 -2y = -300\\ 40 + 300 = 2y\\ 2y = 340\\ y = 170$$\\ Therefore, those friends had Rs 40 and Rs 170 with them respectively. 24 A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And if the train were slower by 10$$ km/h$; it would have taken $3$ hours more than the scheduled time. Find the distance covered by the train.

##### Solution :

Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel was d km. We know that,$\\$ speed=$\dfrac{Distance \ \ travelled}{Time \ \ taken \ \ to \ \ travel \ \ that \ \ distance}$$\\ x=\dfrac{d}{t}$$\\$ Or $d=xt \dots$(i)$\\$ Using the information given in the question, we obtain$\\$ $(x + 10) =\dfrac{d}{t-2}\\(x+10)(t-2)=d$$\\ By using equation (i), we obtain\\ -2x + 10t = 20 \ \ (ii) \\ (x-10)=\dfrac{d}{t+3}\\ (x-10)(t+3)=d\\ xt-10t=30 \ \ \ (iii)$$\\$ Adding equations (ii) and (iii), we obtain$\\$ $x = 50$ Using equation (ii), we obtain$\\$ $(-2) × (50) + 10t = 20\\ -100 + 10t = 20\\ 10t = 120\\ t = 12$hours$\\$ From equation (i), we obtain$\\$ Distance to travel = $d = xt$$\\ = 50 × 12\\ = 600 km$$\\$ Hence, the distance covered by the train is$600 km.$

25   The students of a class are made to stand in rows. If $3$ students are extra in a row, there would be $1$ row less. If $3$ students are less in a row, there would be $2$ rows more. Find the number of students in the class.