Q: A train covered a certain distance at a uniform speed. If the train would have been \$6\$ km/h faster, it would have taken \$4\$ hours less than the scheduled time. And, if the train were slower by \$6\$ km/h, it would have taken \$6\$ hours more than the scheduled time. Find the length of the journey.

Let the actual speed of the train be \$x\$ km/hr and the actual time taken be \$y\$ hours. Then,Distance covered \$=(xy)km\$ ..(i) \$[\therefore\$ Distance \$=\$ Speed \$\times\$ Time\$]\$If the speed is increased by \$6\$ km/hr, then time of journey is reduced by \$4\$ hours i.e., when speed is \$(x+6)\$km/hr, time of journey is \$(y-4)\$ hours.\$\therefore\$ Distance covered \$=(x+6)(y-4)\$\$\Rightarrow xy=(x+6)(y-4)\$ [Using (i)]\$\Rightarrow -4x+6y-24=0\$\$\Rightarrow -2x+3y-12=0\$ ..(ii)When the speed is reduced by \$6\$ km/hr, then the time of journey is increased by \$6\$ hours i.e., when speed is \$(x-6)\$ km/hr, time of journey is \$(y-6)\$ hours.\$\therefore\$ Distance covered \$=(x-6)(y+6)\$\$\Rightarrow xy=(x-6)(y+6)\$ [Using (i)]\$\Rightarrow 6x-6y-36=0\$\$\Rightarrow x-y-6=0\$ (iii)Thus, we obtain the following system of equations:\$-2x+3y-12=0\$\$x-y-6=0\$By using cross-multiplication, we have,\$\dfrac{x}{3\times -6-(-1)\times -12}=\dfrac{-y}{-2\times -6-1\times -12}=\dfrac{1}{-2\times -1-1\times 3}\$\$\Rightarrow \dfrac{x}{-30}=\dfrac{-y}{24}=\dfrac{1}{-1}\$\$\Rightarrow x=30\$ and \$y=24\$Putting the values of x and y in equation (i), we obtainDistance \$=(30\times 24)\$km \$=720\$km.Hence, the length of the journey is \$720\$km.

Q: Draw the graphs of \$x-y+1=0\$ and \$3x+2y-12=0\$. Determine the coordinates of the vertices of the triangle formed by these lines and \$x\$-axis and shade the triangular area.

Given,\$x-y+1=0\$....(1)\$3x+2y-12=0\$....(2)put \$x=0\$ in (1)\$0-y+1=0\Rightarrow y=1\$\$(0,1)\$.....(a)put \$y=0\$ in (2)\$3x+2(0)-12=0\rightarrow x=4\$\$(4,0)\$.....(b)put \$x=0\$ in (2)\$3(0)+2y-12=0\rightarrow y=6\$\$(0,6)\$......(c)put \$y=0\$ in (1)\$x-0+1=0\Rightarrow x=-1\$\$(-1,0)\$...(d)From figure, the vertices of triangle are \$(-1,0),(2,3),(4,0)\$

Q: 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\$

The given pair of equations are:\$3x+y = 1...(1)\$\$\left ( 2k-1 \right )x +\left ( k-1 \right )y = 2k+1..(2)\$Now rearranging eq1 and eq2 will get\$3x+y-1=0...(3)\$\$\left ( 2k-1 \right )x +\left ( k-1 \right )y -\left( 2k+1 \right )= 0..(4)\$Now compare with\$a_{1} = 3, b_{1} = 1, c_{1} = -1\$\$a_{2} = 2k-1, b_{2} = k-1, c_{2} = -\left( 2k+1 \right )\$Now we get\$ \dfrac{a_{1}}{a_{2}} =\dfrac{3}{2k-1} , \dfrac{b_{1}}{b_{2}} = \dfrac{1}{k-1},\dfrac{c_{1}}{c_{2}} = \dfrac{-1}{-\left( 2k+1 \right )} \$Now will take\$ \dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} \$\$\Rightarrow \dfrac{3}{2k-1} = \dfrac{1}{k-1}\$\$\Rightarrow 3k-3= 2k-1\$\$\Rightarrow 3k-2k = -1+3\$\$\Rightarrow k = 2\$Hence \$ k =2 \$ is the value.

Q: For what value of \$K\$ will the following pair of linear equations have infinitely many solutions?\$Kx+3y-(K-3)=0\$ \$12x+Ky-K=0\$

Given \$kx+3y-(k-3)=0\$Comparing with \$a_1x+b_1y+c1=0\$\$\therefore a_1=k,\ b_1=3,\ c=-(k-3)\$\$12x+ky-k=0\$Comparing with \$a_1x+b_1y+c1=0\$\$\therefore a_1=12,\ b_1=k,\ c=-k\$Since equation has infinite number of solutionsSo, \$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}\$\$\dfrac{k}{12}=\dfrac{3}{k}=\dfrac{k-3}{k}\$\$\dfrac{k}{12}=\dfrac{3}{k}\$\$k^2=12\times3\$\$k^2=36\$\$k=\pm 6\$\$\dfrac{3}{k}=\dfrac{k-3}{k}\$\$3k=k(k-3)\$\$3k=k^2-3k\$\$k^2-3k-3k=0\$\$k^2-6k=0\$\$k(k-6)=0\$\$k=0,6\$Therefore, \$k=6\$ satisfies both equationsHence, \$k=6\$

Q: Graphically, the pair of equations \$6x - 3y + 10 = 0, 2x - y + 9 = 0\$ represents two lines which are

The linear equations are\$ 6x-3y+10=0\$\$ 2x-y+9=0\$Here, \$a_{1}=6,b_{1}=-3,c_{1}=10\$\$a_{2}=2,b_{2}=-1,c_{2}=9\$\$\therefore \displaystyle \frac{a_{1}}{a_{2}}=\frac{6}{2}=3\$\$\displaystyle \frac{b_{1}}{b_{2}}=\frac{-3}{-1}=3\$\$\displaystyle \frac{c_{1}}{c_{2}}=\frac{10}{9} \$\$\Rightarrow \displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{2}}{b_{2}}\neq \frac{c_{1}}{c_{2}}\$\$\therefore\$ The lines represented by the given equations are parallel.

Q: The pair of equations \$x + 2y + 5 = 0\$ and \$- 3x - 6y + 1 = 0\$ have

The given equations are:\$x+2y+5= 0 \$\$-3x-6y+1 = 0\$From the given equations we have:\$ \displaystyle\frac{a_{1}}{a_{2}} = \frac{1}{-3}\$\$ \displaystyle \frac{b_{1}}{b_{2}} = \frac{2}{-6} = \frac{1}{-3} \$\$\displaystyle\frac{c_{1}}{c_{2}} =\displaystyle \frac{5}{1}\$\$\Rightarrow \displaystyle\frac{a_{1}}{a_{2}} = \displaystyle \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\$Hence the given pair of equations have no solution.

Q: If \$3x-2y=5\$ and \$3y-2x=3\$ then find the value of \$(x+y)\$.

\$3x-2y=5\$............ \$(1)\$\$3y-2x=3\$\$-2x+3y=3\$Multiply \$(1)\$ by \$2\$ & eq \$(2)\$ by \$3\$\$6x-4y=10.......... (3)\\ -6x+9y=19.......... (4)\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_\\ Adding\ eqn\ (3)\ and\ (4),\ we\ get\$\$6x-4y=10\\ -6x+9y=19\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad \quad \quad 5y=1\$\$y=\dfrac{1}{5}\$Putting the value in \$6x-\dfrac{4}{5}=10\$\$6x=10+\dfrac{4}{5}=\dfrac{54}{5}\$\$x=\dfrac{54}{5\times 6}=\dfrac{9}{5}\$\$x=\dfrac{9}{5}\$Now \$x+y=\dfrac{9}{5}+\dfrac{1}{5}=\dfrac{10}{5}\$=2\$\$x+y=2

Question 1

A train covered a certain distance at a uniform speed. If the train would have been $6$ km/h faster, it would have taken $4$ hours less than the scheduled time. And, if the train were slower by $6$ km/h, it would have taken $6$ hours more than the scheduled time. Find the length of the journey.

Accepted Answer

Let the actual speed of the train be $x$ km/hr and the actual time taken be $y$ hours. Then,Distance covered $=(xy)km$ ..(i) $[\therefore$ Distance $=$ Speed $\times$ Time$]$If the speed is increased by $6$ km/hr, then time of journey is reduced by $4$ hours i.e., when speed is $(x+6)$km/hr, time of journey is $(y-4)$ hours.$\therefore$ Distance covered $=(x+6)(y-4)$$\Rightarrow xy=(x+6)(y-4)$ [Using (i)]$\Rightarrow -4x+6y-24=0$$\Rightarrow -2x+3y-12=0$ ..(ii)When the speed is reduced by $6$ km/hr, then the time of journey is increased by $6$ hours i.e., when speed is $(x-6)$ km/hr, time of journey is $(y-6)$ hours.$\therefore$ Distance covered $=(x-6)(y+6)$$\Rightarrow xy=(x-6)(y+6)$ [Using (i)]$\Rightarrow 6x-6y-36=0$$\Rightarrow x-y-6=0$ (iii)Thus, we obtain the following system of equations:$-2x+3y-12=0$$x-y-6=0$By using cross-multiplication, we have,$\dfrac{x}{3\times -6-(-1)\times -12}=\dfrac{-y}{-2\times -6-1\times -12}=\dfrac{1}{-2\times -1-1\times 3}$$\Rightarrow \dfrac{x}{-30}=\dfrac{-y}{24}=\dfrac{1}{-1}$$\Rightarrow x=30$ and $y=24$Putting the values of x and y in equation (i), we obtainDistance $=(30\times 24)$km $=720$km.Hence, the length of the journey is $720$km.

Question 2

The area of a rectangle gets reduced by $9$ square units if its length is reduced by $5$ units and the breadth is increased by $3$ units. If we increase the length by $3$ units and breadth by $2$ units, the area is increased by $67$ square units. Find the length and breadth of the rectangle.

Accepted Answer

Let the length and breadth of the rectangle be x and y units respectively. Then,Area $=xy$ sq. units.If length is reduced by $5$ units and the breadth is increases by $3$ units, then area is reduced by $9$ square units.$\therefore xy-9=(x-5)(y+3)$$\Rightarrow xy-9=xy+3x-5y-15$$\Rightarrow 3x-5y-6=0$ ...(i)When length is increased by $3$ units and breadth by $2$ units, the area is increased by $67$ sq. units.$\therefore xy+67=(x+3)(y+2)$$\Rightarrow xy+67=xy+2x+3y+6$$\Rightarrow 2x+3y-61=0$ ...(ii)Thus, we get the following system of linear equations:$3x-5y-6=0$$2x+3y-61=0$By using cross-multiplication, we have$\dfrac{x}{305+18}=\dfrac{-y}{-183+12}=\dfrac{1}{9+10}$$\Rightarrow x=\dfrac{323}{19}=17$ and $y=\dfrac{171}{19}=9$Hence, the length and breadth of the rectangle are $17$ units and $9$ units respectively.

Question 3

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

Accepted Answer

Given,$x-y+1=0$....(1)$3x+2y-12=0$....(2)put $x=0$ in (1)$0-y+1=0\Rightarrow y=1$$(0,1)$.....(a)put $y=0$ in (2)$3x+2(0)-12=0\rightarrow x=4$$(4,0)$.....(b)put $x=0$ in (2)$3(0)+2y-12=0\rightarrow y=6$$(0,6)$......(c)put $y=0$ in (1)$x-0+1=0\Rightarrow x=-1$$(-1,0)$...(d)From figure, the vertices of triangle are $(-1,0),(2,3),(4,0)$

Question 4

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$

Accepted Answer

&#10;&#9;&#10;&#9;&#10;&#9;&#10;&#10;&#10;The&#10;given pair of equations are:$3x+y&#10;= 1...(1)$$\left&#10;( 2k-1 \right )x +\left ( k-1 \right )y = 2k+1..(2)$Now&#10;rearranging eq1 and eq2 will get$3x+y-1=0...(3)$$\left&#10;( 2k-1 \right )x +\left ( k-1 \right )y -\left( 2k+1 \right )=&#10;0..(4)$Now&#10;compare with$a_{1}&#10;= 3, b_{1} = 1, c_{1} = -1$$a_{2}&#10;= 2k-1, b_{2} = k-1, c_{2} = -\left( 2k+1 \right )$Now&#10;we get$&#10;\dfrac{a_{1}}{a_{2}} =\dfrac{3}{2k-1} , \dfrac{b_{1}}{b_{2}} =&#10;\dfrac{1}{k-1},\dfrac{c_{1}}{c_{2}} = \dfrac{-1}{-\left( 2k+1 \right&#10;)} $Now&#10;will take$&#10;\dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} $$\Rightarrow&#10;\dfrac{3}{2k-1} = \dfrac{1}{k-1}$$\Rightarrow&#10;3k-3= 2k-1$$\Rightarrow&#10;3k-2k = -1+3$$\Rightarrow&#10;k = 2$Hence&#10;$ k =2 $ is the value.&#10;

Question 5

For what value of $K$ will the following pair of linear equations have infinitely many solutions?$Kx+3y-(K-3)=0$ $12x+Ky-K=0$

Accepted Answer

Given $kx+3y-(k-3)=0$Comparing with $a_1x+b_1y+c1=0$$\therefore a_1=k,\ b_1=3,\ c=-(k-3)$$12x+ky-k=0$Comparing with $a_1x+b_1y+c1=0$$\therefore a_1=12,\ b_1=k,\ c=-k$Since equation has infinite number of solutionsSo, $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$$\dfrac{k}{12}=\dfrac{3}{k}=\dfrac{k-3}{k}$$\dfrac{k}{12}=\dfrac{3}{k}$$k^2=12\times3$$k^2=36$$k=\pm 6$$\dfrac{3}{k}=\dfrac{k-3}{k}$$3k=k(k-3)$$3k=k^2-3k$$k^2-3k-3k=0$$k^2-6k=0$$k(k-6)=0$$k=0,6$Therefore, $k=6$ satisfies both equationsHence, $k=6$

Question 6

Graphically, the pair of equations $6x - 3y + 10 = 0, 2x - y + 9 = 0$ represents two lines which are

Accepted Answer

The linear equations are$ 6x-3y+10=0$$ 2x-y+9=0$Here, $a_{1}=6,b_{1}=-3,c_{1}=10$$a_{2}=2,b_{2}=-1,c_{2}=9$$\therefore \displaystyle \frac{a_{1}}{a_{2}}=\frac{6}{2}=3$$\displaystyle \frac{b_{1}}{b_{2}}=\frac{-3}{-1}=3$$\displaystyle \frac{c_{1}}{c_{2}}=\frac{10}{9} $$\Rightarrow \displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{2}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$$\therefore$ The lines represented by the given equations are parallel.

Question 7

A boat covers $32$ km upstream and $36$ km downstream in $7$ hours. Also, it covers $40$ km upstream and $48$km downstream in $9$ hours. Find the speed of the boat in still water and that of the stream.

Accepted Answer

Let the speed of the boat in still water be x km/hr and the speed of the stream but y km/hr. Then,Speed upstream $=(x-y)$km/hrSpeed downstream $=(x+y)$ km/hrNow, Time taken to cover $32$km upstream $=\dfrac{32}{x-y}$ hrsTime taken to cover $36$ km downstream $=\dfrac{36}{x+y}$ hrsBut, total time of journey is $7$ hours.$\therefore \dfrac{32}{x-y}+\dfrac{36}{x+y}=7$ ..(i)Time taken to cover $40$km upstream $=\dfrac{40}{x-y}$Time taken to cover $48$ km downstream $=\dfrac{48}{x+y}$In this case, total time of journey is given to be $9$ hours.$\therefore \dfrac{40}{x-y}+\dfrac{48}{x+y}=9$ (ii)Putting $\dfrac{1}{x-y}=u$ and $\dfrac{1}{x+y}=v$ in equations (i) and (ii), we get$32u+36v=7\Rightarrow 32u-36v-7=0$ ..(iii)$40u+48v=9\Rightarrow 40u-48v-9=0$ ..(iv)Solving these equations by cross-multiplication, we get$\dfrac{u}{36\times -9-48\times -7}=\dfrac{-v}{32\times -9-40\times -7}=\dfrac{1}{32\times 48-40\times 36}$$\Rightarrow \dfrac{u}{-324+336}=\dfrac{-v}{-288+280}=\dfrac{1}{1536-1440}$$\Rightarrow \dfrac{u}{12}=\dfrac{v}{8}=\dfrac{1}{96}$$\Rightarrow u=\dfrac{12}{96}$ and $v=\dfrac{8}{96}$$\Rightarrow u=\dfrac{1}{8}$ and $v=\dfrac{1}{12}$Now, $u=\dfrac{1}{8}\Rightarrow \dfrac{1}{x-y}=\dfrac{1}{8}\Rightarrow x-y=8$ ..(v)and, $v=\dfrac{1}{12}\Rightarrow \dfrac{1}{x+y}=\dfrac{1}{12}\Rightarrow x+y=12$ ..(vi)Solving equations (v) and (vi), we get $x=10$ and $y=2$Hence, Speed of the boat in still water $=10$ km/hrand Speed of the stream $=2$km/hr.

Question 8

It can take $12$ hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for four hours and the pipe of smaller diameter for $9$ hours, only half of the pool can be filled. How long would it take for each pipe to fill the pool separately?

Accepted Answer

Let there are two pipes A and B and diameter of A is larger than B.Now suppose that Pipe A take $x$ hours and Pipe B takes $y$ hours to fill the pool separately .In 1 hour pipe A can fill the pool $=\dfrac{1}{x}$In 1 hour pipe B can fill the pool$=\dfrac{1}{y}$If both pipe are together then they take $12$ hour to fill the pool.If the are togther then in 1 hour they fill the pool.$\dfrac{1}{x} $+ $\dfrac{1}{y}$=$\dfrac{1}{12}----(1)$Now In $4$ hour pipe A can fill the pool is$=\dfrac{4}{x}$In $9$ hour pipe B can fill the pool is$=\dfrac{9}{y}$ than they fill half of the pool,so$\dfrac{4}{x}$+$\dfrac{9}{y}$ $=\dfrac{1}{2}-----(2)$ Now Let $\dfrac{1}{x}=P$ &$\dfrac{1}{y}=Q$  and puuting in equation 1 &2, than$P+Q$=$\dfrac{1}{12}$ $\Rightarrow12P+12Q=1----(3)$And $4P+9Q$ $=\dfrac{1}{2}$$\Rightarrow8P+18Q=1-----(4)$Now solving equation 3 & 4Equation (3) multiplying by 2 and (4) multiplying by 3 and substract.$\Rightarrow  -30Q=-1$$\Rightarrow Q$$=\dfrac{1}{30}$now put value of $Q$ in equation $3$We get $P$$=\dfrac{1}{20}$So we found$P$$=\dfrac{1}{20}$   & $ Q$$=\dfrac{1}{30}$So $x=20$ and $y=30$Hence the A pipe would take 20 hours and the pipe B would take 30 hours separately  .

Question 9

The pair of equations $x + 2y + 5 = 0$ and $- 3x - 6y + 1 = 0$ have

Accepted Answer

The given equations are:$x+2y+5= 0 $$-3x-6y+1 = 0$From the given equations we have:$ \displaystyle\frac{a_{1}}{a_{2}} = \frac{1}{-3}$$ \displaystyle \frac{b_{1}}{b_{2}} = \frac{2}{-6} = \frac{1}{-3} $$\displaystyle\frac{c_{1}}{c_{2}} =\displaystyle \frac{5}{1}$$\Rightarrow \displaystyle\frac{a_{1}}{a_{2}} = \displaystyle \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$Hence the given pair of equations have no solution.

Question 10

If $3x-2y=5$ and $3y-2x=3$ then find the value of $(x+y)$.

Accepted Answer

$3x-2y=5$............ $(1)$$3y-2x=3$$-2x+3y=3$Multiply $(1)$ by $2$ & eq $(2)$ by $3$$6x-4y=10.......... (3)\ -6x+9y=19.......... (4)\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_\ Adding\ eqn\ (3)\ and\ (4),\ we\ get$$6x-4y=10\ -6x+9y=19\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \ \quad \quad \quad \quad 5y=1$$y=\dfrac{1}{5}$Putting the value in $6x-\dfrac{4}{5}=10$$6x=10+\dfrac{4}{5}=\dfrac{54}{5}$$x=\dfrac{54}{5\times 6}=\dfrac{9}{5}$$x=\dfrac{9}{5}$Now $x+y=\dfrac{9}{5}+\dfrac{1}{5}=\dfrac{10}{5}$=2x+y=2