\displaystyle \textbf{Question 1: }\text{Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel}
\displaystyle \text{and play Hoopla (a game in which you throw a ring on the items kept in the stall, and if}
\displaystyle \text{the ring covers any object completely you get it). The number of times she played Hoopla}
\displaystyle \text{is half the number of rides she had on the Giant Wheel. Each ride costs Rs. }3\text{, and a}
\displaystyle \text{game of Hoopla costs Rs. }4\text{. If she spent Rs. }20\text{ in the fair, represent this situation}
\displaystyle \text{algebraically and graphically.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of Giant Wheel rides be }x\text{ and the number of Hoopla games be }y.
\displaystyle \text{The number of Hoopla games is half the number of Giant Wheel rides.}
\displaystyle \therefore y=\frac{x}{2}
\displaystyle \Rightarrow x-2y=0\qquad\ldots\text{(i)}
\displaystyle \text{Cost of each Giant Wheel ride}= \text{Rs. }3
\displaystyle \text{Cost of each Hoopla game}= \text{Rs. }4
\displaystyle \text{Total amount spent}= \text{Rs. }20
\displaystyle \therefore 3x+4y=20\qquad\ldots\text{(ii)}
\displaystyle \text{Thus, the algebraic representation is}
\displaystyle x-2y=0\quad\text{and}\quad3x+4y=20.
\displaystyle \text{For the graphical representation, we find two solutions of each equation.}
\displaystyle \text{For }x-2y=0:
\displaystyle \begin{array}{c|cc}x&0&4\\ \hline y&0&2\end{array}
\displaystyle \text{Thus, the line }x-2y=0\text{ passes through }(0,0)\text{ and }(4,2).
\displaystyle \text{For }3x+4y=20:
\displaystyle \begin{array}{c|cc}x&0&4\\ \hline y&5&2\end{array}
\displaystyle \text{Thus, the line }3x+4y=20\text{ passes through }(0,5)\text{ and }(4,2).
\displaystyle \text{Plot these points and draw the two straight lines through them.}
\displaystyle \text{The two lines intersect at }(4,2).
\displaystyle \therefore x=4,\quad y=2.
\displaystyle \therefore \text{Akhila had }4\text{ Giant Wheel rides and played Hoopla }2\text{ times.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Aftab tells his daughter, ``Seven years ago, I was seven times as old as you}
\displaystyle \text{were then. Also, three years from now, I shall be three times as old as you will be.''}
\displaystyle \text{Is not this interesting? Represent this situation algebraically and graphically.}
\displaystyle \text{Answer:}
\displaystyle \text{Let Aftab's present age be }x\text{ years and his daughter's present age be }y\text{ years.}
\displaystyle \text{Seven years ago, their ages were }(x-7)\text{ years and }(y-7)\text{ years respectively.}
\displaystyle \therefore x-7=7(y-7)
\displaystyle \Rightarrow x-7=7y-49
\displaystyle \Rightarrow x-7y=-42\qquad\ldots\text{(i)}
\displaystyle \text{Three years from now, their ages will be }(x+3)\text{ years and }(y+3)\text{ years respectively.}
\displaystyle \therefore x+3=3(y+3)
\displaystyle \Rightarrow x+3=3y+9
\displaystyle \Rightarrow x-3y=6\qquad\ldots\text{(ii)}
\displaystyle \text{Thus, the algebraic representation is}
\displaystyle x-7y=-42\quad\text{and}\quad x-3y=6.
\displaystyle \text{For the graphical representation, we find two solutions of each equation.}
\displaystyle \text{For }x-7y=-42:
\displaystyle \begin{array}{c|cc}x&0&42\\ \hline y&6&12\end{array}
\displaystyle \text{Thus, the line }x-7y=-42\text{ passes through }(0,6)\text{ and }(42,12).
\displaystyle \text{For }x-3y=6:
\displaystyle \begin{array}{c|cc}x&6&42\\ \hline y&0&12\end{array}
\displaystyle \text{Thus, the line }x-3y=6\text{ passes through }(6,0)\text{ and }(42,12).
\displaystyle \text{Plot these points and draw the two straight lines through them.}
\displaystyle \text{The two lines intersect at }(42,12).
\displaystyle \therefore x=42,\quad y=12.
\displaystyle \therefore \text{Aftab's present age is }42\text{ years and his daughter's present age is }12\text{ years.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The path of a train }A\text{ is given by the equation }3x+4y-12=0\text{ and the path}
\displaystyle \text{of another train }B\text{ is given by the equation }6x+8y-48=0.\text{ Represent this situation}
\displaystyle \text{graphically.}
\displaystyle \text{Answer:}
\displaystyle \text{The paths of the two trains are given by}
\displaystyle 3x+4y-12=0\qquad\ldots\text{(i)}
\displaystyle 6x+8y-48=0\qquad\ldots\text{(ii)}
\displaystyle \text{For }3x+4y-12=0:
\displaystyle 3x+4y=12
\displaystyle \begin{array}{c|cc}x&4&0\\ \hline y&0&3\end{array}
\displaystyle \text{Thus, the line }3x+4y-12=0\text{ passes through }(4,0)\text{ and }(0,3).
\displaystyle \text{For }6x+8y-48=0:
\displaystyle 3x+4y=24
\displaystyle \begin{array}{c|cc}x&8&0\\ \hline y&0&6\end{array}
\displaystyle \text{Thus, the line }6x+8y-48=0\text{ passes through }(8,0)\text{ and }(0,6).
\displaystyle \text{Plot these points and draw the two straight lines through them.}
\displaystyle \text{The two lines are parallel and do not intersect.}
\displaystyle \therefore \text{The paths of the two trains are parallel and they do not meet.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Gloria is walking along the path joining }(-2,3)\text{ and }(2,-2)\text{, while Suresh is}
\displaystyle \text{walking along the path joining }(0,5)\text{ and }(4,0).\text{ Represent this situation graphically.}
\displaystyle \text{Answer:}
\displaystyle \text{Gloria's path passes through }(-2,3)\text{ and }(2,-2).
\displaystyle \text{Using }\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1},
\displaystyle \frac{y-3}{-2-3}=\frac{x+2}{2+2}
\displaystyle \Rightarrow \frac{y-3}{-5}=\frac{x+2}{4}
\displaystyle \Rightarrow 4y-12=-5x-10
\displaystyle \Rightarrow 5x+4y=2\qquad\ldots\text{(i)}
\displaystyle \text{Thus, Gloria's path is represented by }5x+4y=2.
\displaystyle \text{Suresh's path passes through }(0,5)\text{ and }(4,0).
\displaystyle \frac{y-5}{0-5}=\frac{x-0}{4-0}
\displaystyle \Rightarrow \frac{y-5}{-5}=\frac{x}{4}
\displaystyle \Rightarrow 4y-20=-5x
\displaystyle \Rightarrow 5x+4y=20\qquad\ldots\text{(ii)}
\displaystyle \text{Thus, Suresh's path is represented by }5x+4y=20.
\displaystyle \text{Plot }(-2,3),(2,-2)\text{ and draw the straight line representing Gloria's path.}
\displaystyle \text{Plot }(0,5),(4,0)\text{ and draw the straight line representing Suresh's path.}
\displaystyle \text{The two lines are parallel and do not intersect.}
\displaystyle \therefore \text{Gloria and Suresh are walking along parallel paths.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{On comparing the ratios }\frac{a_1}{a_2},\frac{b_1}{b_2}\text{ and }\frac{c_1}{c_2}\text{, and without drawing them,}
\displaystyle \text{find out whether the lines representing the following pairs of linear equations intersect at a}
\displaystyle \text{point, are parallel or coincident:}
\displaystyle \text{(i) }5x-4y+8=0\qquad 7x+6y-9=0
\displaystyle \text{(ii) }9x+3y+12=0\qquad 18x+6y+24=0
\displaystyle \text{(iii) }6x-3y+10=0\qquad 2x-y+9=0
\displaystyle \text{Answer:}
\displaystyle \text{(i) }5x-4y+8=0\quad\text{and}\quad7x+6y-9=0
\displaystyle a_1=5,\ b_1=-4,\ c_1=8,\quad a_2=7,\ b_2=6,\ c_2=-9
\displaystyle \frac{a_1}{a_2}=\frac{5}{7},\qquad\frac{b_1}{b_2}=\frac{-4}{6}=-\frac{2}{3}
\displaystyle \therefore \frac{a_1}{a_2}\ne\frac{b_1}{b_2}
\displaystyle \therefore \text{The two lines intersect at a point.}

\displaystyle \text{(ii) }9x+3y+12=0\quad\text{and}\quad18x+6y+24=0
\displaystyle a_1=9,\ b_1=3,\ c_1=12,\quad a_2=18,\ b_2=6,\ c_2=24
\displaystyle \frac{a_1}{a_2}=\frac{9}{18}=\frac{1}{2}
\displaystyle \frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}
\displaystyle \frac{c_1}{c_2}=\frac{12}{24}=\frac{1}{2}
\displaystyle \therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}
\displaystyle \therefore \text{The two lines are coincident.}

\displaystyle \text{(iii) }6x-3y+10=0\quad\text{and}\quad2x-y+9=0
\displaystyle a_1=6,\ b_1=-3,\ c_1=10,\quad a_2=2,\ b_2=-1,\ c_2=9
\displaystyle \frac{a_1}{a_2}=\frac{6}{2}=3,\qquad\frac{b_1}{b_2}=\frac{-3}{-1}=3
\displaystyle \frac{c_1}{c_2}=\frac{10}{9}
\displaystyle \therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}
\displaystyle \therefore \text{The two lines are parallel.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Given the linear equation }2x+3y-8=0\text{, write another linear equation in two}
\displaystyle \text{variables such that the geometrical representation of the pair so formed is:}
\displaystyle \text{(i) intersecting lines}\qquad\text{(ii) parallel lines}\qquad\text{(iii) coincident lines.}
\displaystyle \text{Answer:}
\displaystyle \text{Given equation is }2x+3y-8=0.
\displaystyle \therefore a_1=2,\quad b_1=3,\quad c_1=-8.
\displaystyle \text{(i) For intersecting lines, we must have }\frac{a_1}{a_2}\ne\frac{b_1}{b_2}.
\displaystyle \text{Let another equation be }x+y-2=0.
\displaystyle \frac{a_1}{a_2}=\frac{2}{1}=2,\qquad\frac{b_1}{b_2}=\frac{3}{1}=3.
\displaystyle \therefore \frac{a_1}{a_2}\ne\frac{b_1}{b_2}.
\displaystyle \therefore \text{One such equation is }x+y-2=0.

\displaystyle \text{(ii) For parallel lines, we must have }\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}.
\displaystyle \text{Let another equation be }4x+6y-10=0.
\displaystyle \frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\qquad\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}.
\displaystyle \frac{c_1}{c_2}=\frac{-8}{-10}=\frac{4}{5}.
\displaystyle \therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}.
\displaystyle \therefore \text{One such equation is }4x+6y-10=0.

\displaystyle \text{(iii) For coincident lines, we must have }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.
\displaystyle \text{Multiplying }2x+3y-8=0\text{ by }2,
\displaystyle 4x+6y-16=0.
\displaystyle \frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2},\qquad\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}.
\displaystyle \frac{c_1}{c_2}=\frac{-8}{-16}=\frac{1}{2}.
\displaystyle \therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.
\displaystyle \therefore \text{One such equation is }4x+6y-16=0.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The cost of }2\text{ kg of apples and }1\text{ kg of grapes on a day was found to be}
\displaystyle \text{Rs. }160\text{. After a month, the cost of }4\text{ kg of apples and }2\text{ kg of grapes is Rs. }300\text{. Represent}
\displaystyle \text{the situation algebraically and geometrically.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the cost of }1\text{ kg of apples be Rs. }x\text{ and the cost of }1\text{ kg of grapes be Rs. }y.
\displaystyle \text{On a day, the cost of }2\text{ kg of apples and }1\text{ kg of grapes was Rs. }160.
\displaystyle \therefore 2x+y=160\qquad\ldots\text{(i)}
\displaystyle \text{After a month, the cost of }4\text{ kg of apples and }2\text{ kg of grapes was Rs. }300.
\displaystyle \therefore 4x+2y=300\qquad\ldots\text{(ii)}
\displaystyle \text{Thus, the algebraic representation is}
\displaystyle 2x+y=160\quad\text{and}\quad4x+2y=300.
\displaystyle \text{For the geometrical representation, we find two solutions of each equation.}
\displaystyle \text{For }2x+y=160:
\displaystyle \begin{array}{c|cc}x&0&80\\ \hline y&160&0\end{array}
\displaystyle \text{Thus, the line }2x+y=160\text{ passes through }(0,160)\text{ and }(80,0).
\displaystyle \text{For }4x+2y=300:
\displaystyle \begin{array}{c|cc}x&0&75\\ \hline y&150&0\end{array}
\displaystyle \text{Thus, the line }4x+2y=300\text{ passes through }(0,150)\text{ and }(75,0).
\displaystyle \text{Plot these points and draw the two straight lines through them.}
\displaystyle \text{The two lines are parallel and do not intersect.}
\displaystyle \therefore \text{Geometrically, the given situation is represented by two parallel lines.}
\displaystyle \\

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