2 Marks Questions

Question 12 Marks

Seema 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.

Answer

Assume the speed of current $=y \ km / h$
Speed of boat in still water $=x \ km / h$
So, Speed of boat in downstream $=x+y \ km / h$
And, Speed of boat in upstream $=x-y \ km / h$
According to the question,
$\frac{20}{x+y}=2\left(\because \text { Time }=\frac{\text { Distance }}{\text { speed }}\right)$
$\Rightarrow x+y=10 \ldots . .(i)$
Also, $\Rightarrow x+y=10$
$ \ldots \ldots(i)$
$\Rightarrow x-y=2 \ldots \ldots(ii)$
Adding equations $(i)$ and $(ii)$
$\Rightarrow x+y+x-y=10+2$
$\Rightarrow 2 x=12$
Divide the above equation by $2 ,$
$\Rightarrow x=6$
Substituting value of $x$ in Eqn. $(i)$
$\Rightarrow 6+y=10$
Subtract $6$ from both sides of the equation,
$\Rightarrow y=4$
So, speed of rowing in still water is $6 \ km / hr$ and speed of current is $4 \ km / hr$.

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Question 22 Marks

For what value of $c,$ the pair of equations $x-4 y=2$ and $3 x+c y=10$ has no solution?

Answer

Given equations are
$x-4 y=2 $ and $ 3 x+c y=10$
Comparing the equations with $a x+b y=c$, we get
$a_1=1, a_2=3, b_1=-4 $ on,
$\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
$\Rightarrow \frac{1}{3}=\frac{-4}{c}$
$\Rightarrow c=-4 \times 3$
$\Rightarrow c=-12$
Hence, the value of $c$ for equations to have no solution is $-12 .$

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Question 32 Marks

For what value of $k , 2 x+3 y=4$ and $(k+2) x=3 k+2$ will have infinitely many solutions.

Answer

The given equations are $2 x+3 y=4 \ldots \ldots (i)$ and
$(k+2) x=3 k+2 \ldots \ldots .(i i)$
For infinitely many solutions, $\frac{a_1}{a_2}=\frac{c_1}{c_2}$
Here, $a_1=2, a_2=k+2, c_1=4$ and $c_2=3 k+2$.
$\frac{2}{k+2}=\frac{4}{3 k+2}$
$\Rightarrow 2(3 k+2)=4(k+2)$
Using the distributive property $a(b+c)=a b+a c$
$\Rightarrow 6 k+4=4 k+8$
$\Rightarrow 2 k=4$
Divide the above equation by $2,$
$k=2$
Hence, for $k=2,2 x+3 y=4$ and $(k+2) x=3 k+2$ will have infinitely many solutions.

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Question 42 Marks

Find the value $(s)$ of $k$ for which the pair of equations $\left\{ kx +2 y =3, 3 x +6 y =10\right.\}$ has a unique solution.

Answer

Given:
$\Rightarrow kx+2 y=3$
$kx+2 y-3=0 ..... (i)$
Comparing with $a _1 x + b _1 y + c _1=0$
$\therefore a_1=k, b_1=2$ and $ c_1=-3$
Also, $3 x+6 y=10$
$3 x+6 y-10=0 ..... (ii)$
Comparing with $a _2 x + b _2 y + c _2=0$
$a_2=3, b_2=6 $ and $ c_2=-10$
As per question
Equations has unique solution
$\therefore \frac{a_1}{a_2} \neq \frac{b_1}{b_2}$
$\frac{k}{3} \neq \frac{2}{6}$
$\Rightarrow \frac{k}{3} \neq \frac{1}{3}$
$\Rightarrow k \neq 1$
So, for all values of $k$ except $1$ .

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Question 52 Marks

Find the solution of the pair of equations: $\frac{3}{x}+\frac{8}{y}=-1 ; \frac{1}{x}-\frac{2}{y}=2, x, y \neq 0$

Answer

$\frac{3}{x}+\frac{8}{y}=-1......(i)$
$\frac{1}{x}-\frac{2}.{y}=2.....(ii)$
On multiplying equation $(ii)$ by $4$ and adding equation $(i)$
$\frac{3}{x}+\frac{8}{y}=-1$
$\frac{+\frac{4}{x}-\frac{8}{y}=8}{\frac{3}{x}+\frac{4}{x}=-1+8}=\frac{7}{x}=7$
$\Rightarrow x=\frac{7}{7}=1$
Putting the value of $x$ in equation $(ii),$ we get
$\frac{1}{1}-\frac{2}{y}=2$
$\frac{-2}{y}=2-1 $
$\Rightarrow \frac{-2}{y}=1 $
$\Rightarrow y=-2$
Hence, $x=1$ and $y=-2$

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