Mix Examples - Pair of Linear Equations in Two Variables Questions in English

(C) Given pair of linear equations is:
$43x + 67y = -24 .....(i)$
$67x + 43y = 24 .....(ii)$
Adding equations $(i)$ and $(ii)$:
$(43 + 67)x + (67 + 43)y = -24 + 24$
$110x + 110y = 0$
$110(x + y) = 0$
$x + y = 0 \Rightarrow x = -y .....(iii)$
Subtracting equation $(i)$ from equation $(ii)$:
$(67 - 43)x + (43 - 67)y = 24 - (-24)$
$24x - 24y = 48$
$24(x - y) = 48$
$x - y = 2 .....(iv)$
Substituting $x = -y$ from $(iii)$ into $(iv)$:
$-y - y = 2$
$-2y = 2$
$y = -1$
Substituting $y = -1$ into $(iii)$:
$x = -(-1) = 1$
Thus,the solution is $x = 1$ and $y = -1$.

(A) Given pair of linear equations is:
$\frac{x}{a} + \frac{y}{b} = a + b \quad .....(i)$
$\frac{x}{a^{2}} + \frac{y}{b^{2}} = 2, \quad a, b \neq 0 \quad .....(ii)$
Multiply equation $(i)$ by $\frac{1}{a}$:
$\frac{x}{a^{2}} + \frac{y}{ab} = 1 + \frac{b}{a} \quad .....(iii)$
Subtract equation $(iii)$ from equation $(ii)$:
$\left( \frac{x}{a^{2}} + \frac{y}{b^{2}} \right) - \left( \frac{x}{a^{2}} + \frac{y}{ab} \right) = 2 - \left( 1 + \frac{b}{a} \right)$
$\frac{y}{b^{2}} - \frac{y}{ab} = 1 - \frac{b}{a}$
$y \left( \frac{a - b}{ab^{2}} \right) = \frac{a - b}{a}$
Since $a \neq b$ (assuming general case),we can cancel $(a - b)$:
$\frac{y}{ab^{2}} = \frac{1}{a}$
$y = b^{2}$
Substitute $y = b^{2}$ into equation $(ii)$:
$\frac{x}{a^{2}} + \frac{b^{2}}{b^{2}} = 2$
$\frac{x}{a^{2}} + 1 = 2$
$\frac{x}{a^{2}} = 1$
$x = a^{2}$
Thus,the solution is $x = a^{2}$ and $y = b^{2}$.

(A) Given pair of equations is:
$\frac{2xy}{x+y} = \frac{3}{2} \implies \frac{x+y}{2xy} = \frac{2}{3} \implies \frac{1}{y} + \frac{1}{x} = \frac{4}{3} \quad ...(i)$
$\frac{xy}{2x-y} = \frac{-3}{10} \implies \frac{2x-y}{xy} = \frac{-10}{3} \implies \frac{2}{y} - \frac{1}{x} = \frac{-10}{3} \quad ...(ii)$
Let $\frac{1}{x} = u$ and $\frac{1}{y} = v$. Then the equations become:
$v + u = \frac{4}{3} \quad ...(iii)$
$2v - u = \frac{-10}{3} \quad ...(iv)$
Adding $(iii)$ and $(iv)$:
$(v + u) + (2v - u) = \frac{4}{3} - \frac{10}{3}$
$3v = \frac{-6}{3} = -2 \implies v = \frac{-2}{3}$
Substituting $v = \frac{-2}{3}$ in $(iii)$:
$u + (\frac{-2}{3}) = \frac{4}{3} \implies u = \frac{4}{3} + \frac{2}{3} = \frac{6}{3} = 2$
Since $u = \frac{1}{x} = 2$,we get $x = \frac{1}{2}$.
Since $v = \frac{1}{y} = \frac{-2}{3}$,we get $y = \frac{-3}{2}$.
Thus,the solution is $x = \frac{1}{2}, y = \frac{-3}{2}$.

(B) Given pair of equations is:
$\frac{x}{10} + \frac{y}{5} = 1 \dots (i)$
$\frac{x}{8} + \frac{y}{6} = 15 \dots (ii)$
Multiplying equation $(i)$ by $10$,we get $x + 2y = 10 \dots (iii)$.
Multiplying equation $(ii)$ by $24$,we get $3x + 4y = 360 \dots (iv)$.
To eliminate $y$,multiply equation $(iii)$ by $2$:
$2x + 4y = 20 \dots (v)$
Subtracting equation $(v)$ from equation $(iv)$:
$(3x + 4y) - (2x + 4y) = 360 - 20$
$x = 340$
Substituting $x = 340$ into equation $(iii)$:
$340 + 2y = 10$
$2y = -330$
$y = -165$
Given the relation $y = \lambda x + 5$,substitute $x = 340$ and $y = -165$:
$-165 = \lambda(340) + 5$
$-170 = 340\lambda$
$\lambda = -\frac{170}{340} = -\frac{1}{2}$
Thus,the solution is $x = 340, y = -165$ and $\lambda = -\frac{1}{2}$.

(A) The given pair of equations is:
$3x + y + 4 = 0 .....(i)$
$6x - 2y + 4 = 0 .....(ii)$
Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
$a_1 = 3, b_1 = 1, c_1 = 4$
$a_2 = 6, b_2 = -2, c_2 = 4$
Calculating ratios:
$\frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2}$
$\frac{b_1}{b_2} = \frac{1}{-2} = -\frac{1}{2}$
Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$,the lines intersect at a unique point. Therefore,the system is consistent.
For equation $(i)$,$y = -3x - 4$:

$x$	$0$	$-1$	$-2$
$y$	$-4$	$-1$	$2$

For equation $(ii)$,$2y = 6x + 4 \Rightarrow y = 3x + 2$:

$x$	$-1$	$0$	$1$
$y$	$-1$	$2$	$5$

Plotting these lines on a graph,they intersect at the point $(-1, -1)$. Thus,the solution is $x = -1, y = -1$.

(D) The given pair of equations is:
$x - 2y = 6$ ..... $(i)$
$3x - 6y = 0$ ..... $(ii)$
Comparing these with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:
$a_1 = 1, b_1 = -2, c_1 = -6$
$a_2 = 3, b_2 = -6, c_2 = 0$
Now,calculating the ratios:
$\frac{a_1}{a_2} = \frac{1}{3}$
$\frac{b_1}{b_2} = \frac{-2}{-6} = \frac{1}{3}$
$\frac{c_1}{c_2} = \frac{-6}{0}$ (which is undefined)
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$,the lines represented by these equations are parallel to each other.
Because the lines are parallel,they do not intersect at any point.
Therefore,the given pair of linear equations is inconsistent and has no solution.

(CONSISTENT) Given pair of equations is:
$x+y=3 .....(i)$
$3x+3y=9 .....(ii)$
On comparing with $ax+by+c=0$,we get:
$a_1=1, b_1=1, c_1=-3$ [from Eq. $(i)$]
$a_2=3, b_2=3, c_2=-9$ [from Eq. $(ii)$]
$\frac{a_1}{a_2} = \frac{1}{3}, \frac{b_1}{b_2} = \frac{1}{3}, \frac{c_1}{c_2} = \frac{-3}{-9} = \frac{1}{3}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$,the given pair of lines is coincident. Therefore,these lines have infinitely many solutions. Hence,the given pair of linear equations is consistent.
Now,for $x+y=3 \Rightarrow y=3-x$:

$x$	$0$	$3$
$y$	$3$	$0$
Points	$A$	$B$

For $3x+3y=9 \Rightarrow y = \frac{9-3x}{3} = 3-x$:

$x$	$0$	$1$	$3$
$y$	$3$	$2$	$0$
Points	$C$	$D$	$E$

Plotting the points $A(0,3)$ and $B(3,0)$,we get the line $AB$. Similarly,plotting the points $C(0,3), D(1,2)$,and $E(3,0)$,we get the same line. We observe that the lines represented by Eqs. $(i)$ and $(ii)$ are coincident.

(A) The given pair of linear equations are $2x + y = 4$ and $2x - y = 4$.
Table for line $2x + y = 4$:

$x$	$0$	$2$
$y = 4 - 2x$	$4$	$0$
Points	$A(0, 4)$	$B(2, 0)$

Table for line $2x - y = 4$:

$x$	$0$	$2$
$y = 2x - 4$	$-4$	$0$
Points	$C(0, -4)$	$B(2, 0)$

By plotting these points on a graph,we observe that the two lines intersect at point $B(2, 0)$ and intersect the $y$-axis at points $A(0, 4)$ and $C(0, -4)$.
The triangle formed by these lines and the $y$-axis is $\triangle ABC$.
The vertices of the triangle are $A(0, 4)$,$B(2, 0)$,and $C(0, -4)$.
The base of the triangle lies on the $y$-axis,with length $AC = |4 - (-4)| = 8$ units.
The height of the triangle is the perpendicular distance from point $B(2, 0)$ to the $y$-axis,which is $2$ units.
Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 2 = 8$ sq units.
Thus,the vertices are $(0, 4), (2, 0), (0, -4)$ and the area is $8$ sq units.

(D) First,we solve the pair of linear equations:
$x + y = 2$ --- $(i)$
$2x - y = 1$ --- $(ii)$
Adding equations $(i)$ and $(ii)$,we get:
$(x + y) + (2x - y) = 2 + 1$
$3x = 3$
$x = 1$
Substituting $x = 1$ in equation $(i)$:
$1 + y = 2$
$y = 1$
The solution point is $(1, 1)$.
Any line passing through $(1, 1)$ can be written in the form $y - 1 = m(x - 1)$,where $m$ is the slope. For example,if $m = 1$,the equation is $y - 1 = x - 1$,which simplifies to $y = x$.
Since there are infinitely many possible values for the slope $m$,there are infinitely many such lines that can pass through the point $(1, 1)$.

(D) Given that $(x+1)$ is a factor of $f(x) = 2x^{3} + ax^{2} + 2bx + 1$,by the Factor Theorem,$f(-1) = 0$.
Substituting $x = -1$ into the polynomial:
$2(-1)^{3} + a(-1)^{2} + 2b(-1) + 1 = 0$
$-2 + a - 2b + 1 = 0$
$a - 2b = 1$ ... $(i)$
We are also given the equation $2a - 3b = 4$ ... $(ii)$.
From equation $(i)$,we have $a = 2b + 1$.
Substituting this into equation $(ii)$:
$2(2b + 1) - 3b = 4$
$4b + 2 - 3b = 4$
$b + 2 = 4$
$b = 2$
Now,substitute $b = 2$ into $a = 2b + 1$:
$a = 2(2) + 1 = 5$.
Thus,the values are $a = 5$ and $b = 2$.

(A) Given that $x, y$ and $40^{\circ}$ are the angles of a triangle.
Since the sum of all angles of a triangle is $180^{\circ}$,we have:
$x + y + 40^{\circ} = 180^{\circ}$
$\Rightarrow x + y = 140^{\circ} \quad \dots(i)$
Also,the difference between the two angles is given as:
$x - y = 30^{\circ} \quad \dots(ii)$
Adding equations $(i)$ and $(ii)$:
$(x + y) + (x - y) = 140^{\circ} + 30^{\circ}$
$2x = 170^{\circ}$
$x = 85^{\circ}$
Substituting $x = 85^{\circ}$ in equation $(i)$:
$85^{\circ} + y = 140^{\circ}$
$y = 140^{\circ} - 85^{\circ} = 55^{\circ}$
Thus,the values of $x$ and $y$ are $85^{\circ}$ and $55^{\circ}$ respectively.

(B) Let Salim's current age be $x$ years and his daughter's current age be $y$ years.
According to the first condition,two years ago:
$x - 2 = 3(y - 2)$
$x - 2 = 3y - 6$
$x - 3y = -4$ --- $(i)$
According to the second condition,six years later:
$x + 6 = 2(y + 6) + 4$
$x + 6 = 2y + 12 + 4$
$x - 2y = 10$ --- $(ii)$
Subtracting equation $(i)$ from equation $(ii)$:
$(x - 2y) - (x - 3y) = 10 - (-4)$
$x - 2y - x + 3y = 10 + 4$
$y = 14$
Substituting $y = 14$ into equation $(ii)$:
$x - 2(14) = 10$
$x - 28 = 10$
$x = 38$
Therefore,Salim's current age is $38$ years and his daughter's current age is $14$ years.

(C) Let the present age of the father be $x$ years and the sum of the present ages of his two children be $S$ years.
According to the first condition: $x = 2S$ ... $(i)$
After $20$ years,the father's age will be $(x + 20)$ years.
After $20$ years,each child's age increases by $20$ years,so the sum of their ages will be $(S + 20 + 20) = (S + 40)$ years.
According to the second condition: $x + 20 = S + 40$
Substituting $S = x/2$ from equation $(i)$ into the second condition:
$x + 20 = x/2 + 40$
$x - x/2 = 40 - 20$
$x/2 = 20$
$x = 40$
Thus,the father's present age is $40$ years.

(D) Let the two numbers be $5x$ and $6x$ based on the given ratio $5:6$.
According to the second condition,if $8$ is subtracted from each number,the ratio becomes $4:5$.
So,$\frac{5x - 8}{6x - 8} = \frac{4}{5}$.
Cross-multiplying,we get $5(5x - 8) = 4(6x - 8)$.
$25x - 40 = 24x - 32$.
$25x - 24x = 40 - 32$.
$x = 8$.
Therefore,the first number is $5x = 5 \times 8 = 40$.
The second number is $6x = 6 \times 8 = 48$.
Thus,the required numbers are $40$ and $48$.

(A) Let the number of students in halls $A$ and $B$ be $x$ and $y$,respectively.
According to the first condition,if $10$ students are sent from $A$ to $B$,the number of students becomes equal:
$x - 10 = y + 10$
$\Rightarrow x - y = 20$ ---$(i)$
According to the second condition,if $20$ students are sent from $B$ to $A$,the number of students in $A$ becomes double the number of students in $B$:
$x + 20 = 2(y - 20)$
$x + 20 = 2y - 40$
$\Rightarrow x - 2y = -60$ ---(ii)
Subtracting equation (ii) from equation $(i)$:
$(x - y) - (x - 2y) = 20 - (-60)$
$x - y - x + 2y = 80$
$y = 80$
Substituting $y = 80$ in equation $(i)$:
$x - 80 = 20$
$x = 100$
Thus,there are $100$ students in hall $A$ and $80$ students in hall $B$.

(B) Let the fixed charge for the first two days be $Rs. x$ and the additional charge for each day thereafter be $Rs. y$.
According to the first condition,Latika paid $Rs. 22$ for a book kept for six days (first $2$ days + $4$ extra days):
$x + 4y = 22$ --- $(i)$
According to the second condition,Anand paid $Rs. 16$ for a book kept for four days (first $2$ days + $2$ extra days):
$x + 2y = 16$ --- $(ii)$
Subtracting equation $(ii)$ from equation $(i)$:
$(x + 4y) - (x + 2y) = 22 - 16$
$2y = 6$
$y = 3$
Substituting $y = 3$ in equation $(ii)$:
$x + 2(3) = 16$
$x + 6 = 16$
$x = 10$
Thus,the fixed charge is $Rs. 10$ and the charge for each extra day is $Rs. 3$.

(C) Let $x$ be the number of correct answers and $(120-x)$ be the number of wrong answers.
According to the problem,the total marks obtained is $90$.
The equation is: $x \times 1 - (120 - x) \times \frac{1}{2} = 90$.
Expanding the equation: $x - 60 + \frac{x}{2} = 90$.
Adding $60$ to both sides: $x + \frac{x}{2} = 150$.
Combining terms: $\frac{3x}{2} = 150$.
Solving for $x$: $x = \frac{150 \times 2}{3} = 100$.
Therefore,Jayanti answered $100$ questions correctly.

(A) We know that,by the property of a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.
$\angle A + \angle C = (6x + 10)^{\circ} + (x + y)^{\circ} = 180^{\circ}$
$\Rightarrow 7x + y = 170 \quad \dots(i)$
Similarly,$\angle B + \angle D = (5x)^{\circ} + (3y - 10)^{\circ} = 180^{\circ}$
$\Rightarrow 5x + 3y = 190 \quad \dots(ii)$
Multiplying equation $(i)$ by $3$,we get $21x + 3y = 510 \quad \dots(iii)$
Subtracting equation $(ii)$ from $(iii)$:
$(21x + 3y) - (5x + 3y) = 510 - 190$
$16x = 320 \Rightarrow x = 20$
Substituting $x = 20$ in equation $(i)$:
$7(20) + y = 170 \Rightarrow 140 + y = 170 \Rightarrow y = 30$
Now,calculating the angles:
$\angle A = 6(20) + 10 = 130^{\circ}$
$\angle B = 5(20) = 100^{\circ}$
$\angle C = 20 + 30 = 50^{\circ}$
$\angle D = 3(30) - 10 = 80^{\circ}$
Thus,$x = 20$,$y = 30$,and the angles are $130^{\circ}, 100^{\circ}, 50^{\circ}, 80^{\circ}$.

(N/A) We know that the graph of $x=-2$ is a line parallel to the $y$-axis at a distance of $2$ units to the left of it.
The graph of $y=3$ is a line parallel to the $x$-axis at a distance of $3$ units above it.
The figure enclosed by the lines $x=-2$,$y=3$,the $x$-axis,and the $y$-axis is $OABC$,which is a rectangle.
$A$ is a point on the $y$-axis at a distance of $3$ units above the $x$-axis. So,the coordinates of $A$ are $(0, 3)$.
$C$ is a point on the $x$-axis at a distance of $2$ units to the left of the $y$-axis. So,the coordinates of $C$ are $(-2, 0)$.
$B$ is the intersection point of the lines $x=-2$ and $y=3$. So,the coordinates of $B$ are $(-2, 3)$.
The origin $O$ is $(0, 0)$.
Thus,the vertices of the rectangle $OABC$ are $O(0, 0)$,$A(0, 3)$,$B(-2, 3)$,and $C(-2, 0)$.
The length and breadth of this rectangle are $2$ units and $3$ units,respectively.
Since the area of a rectangle $=$ length $\times$ breadth,
the area of rectangle $OABC = 2 \times 3 = 6 \text{ sq. units}$.

(B) The vertices of a triangle are the intersection points of the lines forming its sides. We solve the equations in pairs to find these points.
Pair $1$: $5x - y = 5$ and $x + 2y = 1$.
From the first equation,$y = 5x - 5$. Substituting into the second: $x + 2(5x - 5) = 1 \implies x + 10x - 10 = 1 \implies 11x = 11 \implies x = 1$. Then $y = 5(1) - 5 = 0$. Vertex: $(1, 0)$.
Pair $2$: $x + 2y = 1$ and $6x + y = 17$.
From the second equation,$y = 17 - 6x$. Substituting into the first: $x + 2(17 - 6x) = 1 \implies x + 34 - 12x = 1 \implies -11x = -33 \implies x = 3$. Then $y = 17 - 6(3) = -1$. Vertex: $(3, -1)$.
Pair $3$: $5x - y = 5$ and $6x + y = 17$.
Adding the two equations: $(5x - y) + (6x + y) = 5 + 17 \implies 11x = 22 \implies x = 2$. Substituting $x = 2$ into $5x - y = 5$: $5(2) - y = 5 \implies 10 - y = 5 \implies y = 5$. Vertex: $(2, 5)$.
The vertices are $(1, 0), (3, -1)$,and $(2, 5)$.

(C) Let the cost price of the table be $Rs.\, x$ and the cost price of the chair be $Rs.\, y$.
The selling price of the table at $10\%$ profit is $x + 0.10x = 1.10x$.
The selling price of the chair at $25\%$ profit is $y + 0.25y = 1.25y$.
Given,$1.10x + 1.25y = 1050$. Multiplying by $100$,we get $110x + 125y = 105000$ ....$(1)$
If the table is sold at $25\%$ profit and the chair at $10\%$ profit,the selling price is $1.25x + 1.10y = 1065$.
Multiplying by $100$,we get $125x + 110y = 106500$ ....$(2)$
Adding $(1)$ and $(2)$:
$235x + 235y = 211500 \implies x + y = 900$ ....$(3)$
Subtracting $(2)$ from $(1)$:
$-15x + 15y = -1500 \implies x - y = 100$ ....$(4)$
Adding $(3)$ and $(4)$:
$2x = 1000 \implies x = 500$.
Substituting $x = 500$ in $(3)$:
$500 + y = 900 \implies y = 400$.
Thus,the cost price of the table is $Rs.\, 500$ and the cost price of the chair is $Rs.\, 400$.

(D) Let the time taken by the larger pipe to fill the pool be $x$ hours and the time taken by the smaller pipe be $y$ hours.
In $1$ hour,the larger pipe fills $1/x$ part of the pool and the smaller pipe fills $1/y$ part of the pool.
According to the first condition,if both pipes are used,they fill the pool in $12$ hours:
$1/x + 1/y = 1/12$ --- (Equation $1$)
According to the second condition,the larger pipe works for $4$ hours and the smaller pipe for $9$ hours to fill half the pool:
$4/x + 9/y = 1/2$ --- (Equation $2$)
Let $u = 1/x$ and $v = 1/y$. The equations become:
$u + v = 1/12$ --- (Equation $3$)
$4u + 9v = 1/2$ --- (Equation $4$)
Multiply Equation $3$ by $4$: $4u + 4v = 4/12 = 1/3$.
Subtract this from Equation $4$: $(4u + 9v) - (4u + 4v) = 1/2 - 1/3$.
$5v = 1/6$,so $v = 1/30$.
Substitute $v = 1/30$ into Equation $3$: $u + 1/30 = 1/12$.
$u = 1/12 - 1/30 = (5 - 2)/60 = 3/60 = 1/20$.
Since $u = 1/x = 1/20$,$x = 20$ hours.
Since $v = 1/y = 1/30$,$y = 30$ hours.
Thus,the larger pipe takes $20$ hours and the smaller pipe takes $30$ hours to fill the pool separately.

(4:1) Given equations are $2x + y = 6$ and $2x - y + 2 = 0$.
Table for equation $2x + y = 6$:

$x$	$0$	$3$
$y$	$6$	$0$
Points	$B$	$A$

Table for equation $2x - y + 2 = 0$:

$x$	$0$	$-1$
$y$	$2$	$0$
Points	$D$	$C$

Let $A_1$ and $A_2$ represent the areas of $\triangle ACE$ and $\triangle BDE$,respectively.
Now,$A_1 = \text{Area of } \triangle ACE = \frac{1}{2} \times AC \times PE = \frac{1}{2} \times 4 \times 4 = 8$.
And $A_2 = \text{Area of } \triangle BDE = \frac{1}{2} \times BD \times QE = \frac{1}{2} \times 4 \times 1 = 2$.
Therefore,$A_1 : A_2 = 8 : 2 = 4 : 1$.
Hence,the pair of equations intersect graphically at point $E(1, 4)$,i.e.,$x = 1$ and $y = 4$.

(O(0,0), Q(4,4), D(6,2)) The given linear equations are:
$y=x \quad (i)$
$3y=x \quad (ii)$
$x+y=8 \quad (iii)$
To find the vertices,we determine the intersection points of these lines taken in pairs.
$1$. Intersection of $(i)$ and $(ii)$:
Substituting $y=x$ into $3y=x$,we get $3x=x$,which implies $2x=0$,so $x=0$. Thus,$y=0$. The intersection point is $O(0,0)$.
$2$. Intersection of $(i)$ and $(iii)$:
Substituting $y=x$ into $x+y=8$,we get $x+x=8$,so $2x=8$,which means $x=4$. Thus,$y=4$. The intersection point is $Q(4,4)$.
$3$. Intersection of $(ii)$ and $(iii)$:
From $(ii)$,$x=3y$. Substituting this into $(iii)$,we get $3y+y=8$,so $4y=8$,which means $y=2$. Then $x=3(2)=6$. The intersection point is $D(6,2)$.
Thus,the vertices of the triangle formed by the lines are $O(0,0)$,$Q(4,4)$,and $D(6,2)$.

(8) Given equations of lines are $2x-y-4=0$,$x=3$,and $x=5$.
For the line $2x-y-4=0$,we find two points:

$x$	$0$	$2$
$y=2x-4$	$-4$	$0$

Plot the points $P(0, -4)$ and $Q(2, 0)$ and draw the line passing through them. Also,draw the vertical lines $x=3$ and $x=5$.
The quadrilateral formed by the lines $x=3$,$x=5$,the $x$-axis,and the line $2x-y-4=0$ is a trapezium with parallel sides at $x=3$ and $x=5$.
At $x=3$,$y = 2(3)-4 = 2$. So,point $D$ is $(3, 2)$.
At $x=5$,$y = 2(5)-4 = 6$. So,point $C$ is $(5, 6)$.
The parallel sides of the trapezium are $AD = 2$ units and $BC = 6$ units. The distance between them is $AB = 5-3 = 2$ units.
Area of trapezium $= \frac{1}{2} \times (\text{sum of parallel sides}) \times (\text{height})$
$= \frac{1}{2} \times (AD + BC) \times AB$
$= \frac{1}{2} \times (2 + 6) \times 2$
$= 8 \text{ sq units}$.
Thus,the area of the quadrilateral is $8 \text{ sq units}$.

(D) Let the cost of a pen be $Rs.\, x$ and the cost of a pencil box be $Rs.\, y$.
According to the first condition,$4x + 4y = 100$,which simplifies to $x + y = 25$ (Equation $i$).
According to the second condition,$3x = y + 15$,which can be written as $3x - y = 15$ (Equation $ii$).
Adding Equation $(i)$ and Equation $(ii)$:
$(x + y) + (3x - y) = 25 + 15$
$4x = 40$
$x = 10$
Substituting $x = 10$ into Equation $(i)$:
$10 + y = 25$
$y = 15$
Therefore,the cost of a pen is $Rs.\, 10$ and the cost of a pencil box is $Rs.\, 15$.

(A) The given equations of the lines are:
$3x - y = 3 \quad ...(i)$
$2x - 3y = 2 \quad ...(ii)$
$x + 2y = 8 \quad ...(iii)$
Let the lines $(i), (ii),$ and $(iii)$ represent the sides $AB, BC,$ and $CA$ of $\triangle ABC$ respectively.
$1$. To find vertex $B$,solve lines $(i)$ and $(ii)$:
Multiply Eq. $(i)$ by $3$: $9x - 3y = 9$
Subtract Eq. $(ii)$ from this: $(9x - 3y) - (2x - 3y) = 9 - 2 \Rightarrow 7x = 7 \Rightarrow x = 1$.
Substitute $x = 1$ in Eq. $(i)$: $3(1) - y = 3 \Rightarrow y = 0$.
So,vertex $B$ is $(1, 0)$.
$2$. To find vertex $C$,solve lines $(ii)$ and $(iii)$:
Multiply Eq. $(iii)$ by $2$: $2x + 4y = 16$.
Subtract Eq. $(ii)$ from this: $(2x + 4y) - (2x - 3y) = 16 - 2 \Rightarrow 7y = 14 \Rightarrow y = 2$.
Substitute $y = 2$ in Eq. $(iii)$: $x + 2(2) = 8 \Rightarrow x = 4$.
So,vertex $C$ is $(4, 2)$.
$3$. To find vertex $A$,solve lines $(iii)$ and $(i)$:
Multiply Eq. $(i)$ by $2$: $6x - 2y = 6$.
Add Eq. $(iii)$: $(6x - 2y) + (x + 2y) = 6 + 8 \Rightarrow 7x = 14 \Rightarrow x = 2$.
Substitute $x = 2$ in Eq. $(i)$: $3(2) - y = 3 \Rightarrow 6 - y = 3 \Rightarrow y = 3$.
So,vertex $A$ is $(2, 3)$.
Thus,the vertices of the triangle are $A(2, 3), B(1, 0),$ and $C(4, 2)$.

(B) Let the speed of the rickshaw be $x \,km/h$ and the speed of the bus be $y \,km/h$.
Time taken to travel $2 \,km$ by rickshaw is $t_1 = \frac{2}{x} \,h$.
Time taken to travel the remaining distance $(14 - 2) = 12 \,km$ by bus is $t_2 = \frac{12}{y} \,h$.
According to the first condition: $\frac{2}{x} + \frac{12}{y} = \frac{1}{2} \quad ...(i)$
Time taken to travel $4 \,km$ by rickshaw is $t_3 = \frac{4}{x} \,h$.
Time taken to travel the remaining distance $(14 - 4) = 10 \,km$ by bus is $t_4 = \frac{10}{y} \,h$.
Since she takes $9 \,minutes$ longer,the total time is $\frac{1}{2} + \frac{9}{60} = \frac{1}{2} + \frac{3}{20} = \frac{13}{20} \,h$.
According to the second condition: $\frac{4}{x} + \frac{10}{y} = \frac{13}{20} \quad ...(ii)$
Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. The equations become:
$2u + 12v = \frac{1}{2} \quad ...(iii)$
$4u + 10v = \frac{13}{20} \quad ...(iv)$
Multiply equation $(iii)$ by $2$: $4u + 24v = 1 \quad ...(v)$
Subtract $(iv)$ from $(v)$:
$(4u + 24v) - (4u + 10v) = 1 - \frac{13}{20}$
$14v = \frac{7}{20} \Rightarrow v = \frac{7}{20 \times 14} = \frac{1}{40}$.
Thus,$y = 40 \,km/h$.
Substitute $v = \frac{1}{40}$ into $(iii)$:
$2u + 12(\frac{1}{40}) = \frac{1}{2} \Rightarrow 2u + \frac{3}{10} = \frac{1}{2}$
$2u = \frac{1}{2} - \frac{3}{10} = \frac{5-3}{10} = \frac{2}{10} = \frac{1}{5}$
$u = \frac{1}{10}$. Thus,$x = 10 \,km/h$.
The speed of the rickshaw is $10 \,km/h$ and the speed of the bus is $40 \,km/h$.

(C) Let the speed of the stream be $v \, km/h$.
Given that,the speed of the person in still water is $5 \, km/h$.
The speed of the person downstream is $(5 + v) \, km/h$.
The speed of the person upstream is $(5 - v) \, km/h$.
Time taken to cover $40 \, km$ downstream is $t_1 = \frac{40}{5 + v} \, h$.
Time taken to cover $40 \, km$ upstream is $t_2 = \frac{40}{5 - v} \, h$.
According to the problem,the time taken upstream is thrice the time taken downstream,so $t_2 = 3 \times t_1$.
Substituting the values: $\frac{40}{5 - v} = 3 \times \frac{40}{5 + v}$.
Dividing both sides by $40$: $\frac{1}{5 - v} = \frac{3}{5 + v}$.
Cross-multiplying: $5 + v = 3(5 - v)$.
$5 + v = 15 - 3v$.
$4v = 10$.
$v = 2.5 \, km/h$.
Thus,the speed of the stream is $2.5 \, km/h$.

(D) Let the speed of the motorboat in still water be $u \, km/h$ and the speed of the stream be $v \, km/h$.
Then,the speed of the motorboat downstream is $(u+v) \, km/h$ and the speed of the motorboat upstream is $(u-v) \, km/h$.
According to the first condition,the time taken to travel $30 \, km$ upstream and $28 \, km$ downstream is $7 \, hours$:
$\frac{30}{u-v} + \frac{28}{u+v} = 7 \quad ...(i)$
According to the second condition,the time taken to travel $21 \, km$ upstream and return $21 \, km$ downstream is $5 \, hours$:
$\frac{21}{u-v} + \frac{21}{u+v} = 5 \quad ...(ii)$
Let $x = \frac{1}{u+v}$ and $y = \frac{1}{u-v}$.
Substituting these into equations $(i)$ and $(ii)$:
$30y + 28x = 7 \quad ...(iii)$
$21y + 21x = 5 \Rightarrow x + y = \frac{5}{21} \quad ...(iv)$
From $(iv)$,$x = \frac{5}{21} - y$. Substituting this into $(iii)$:
$30y + 28(\frac{5}{21} - y) = 7$
$30y + \frac{20}{3} - 28y = 7$
$2y = 7 - \frac{20}{3} = \frac{1}{3} \Rightarrow y = \frac{1}{6}$
Now,$x = \frac{5}{21} - \frac{1}{6} = \frac{10-7}{42} = \frac{3}{42} = \frac{1}{14}$.
Since $y = \frac{1}{u-v} = \frac{1}{6}$,we have $u-v = 6 \quad ...(v)$
Since $x = \frac{1}{u+v} = \frac{1}{14}$,we have $u+v = 14 \quad ...(vi)$
Adding $(v)$ and $(vi)$:
$2u = 20 \Rightarrow u = 10 \, km/h$.
Substituting $u=10$ in $(vi)$:
$10 + v = 14 \Rightarrow v = 4 \, km/h$.
Thus,the speed of the boat in still water is $10 \, km/h$ and the speed of the stream is $4 \, km/h$.

(A) Let the two-digit number be $10x + y$,where $x$ is the tens digit and $y$ is the units digit.
Case $I$: Multiplying the sum of the digits by $8$ and then subtracting $5$ gives the number:
$8(x + y) - 5 = 10x + y$
$8x + 8y - 5 = 10x + y$
$2x - 7y = -5$ ... $(i)$
Case $II$: Multiplying the difference of the digits by $16$ and then adding $3$ gives the number:
$16(x - y) + 3 = 10x + y$
$16x - 16y + 3 = 10x + y$
$6x - 17y = -3$ ... $(ii)$
To solve the system,multiply equation $(i)$ by $3$:
$6x - 21y = -15$ ... $(iii)$
Subtract equation $(iii)$ from equation $(ii)$:
$(6x - 17y) - (6x - 21y) = -3 - (-15)$
$4y = 12$
$y = 3$
Substitute $y = 3$ into equation $(i)$:
$2x - 7(3) = -5$
$2x - 21 = -5$
$2x = 16$
$x = 8$
The number is $10x + y = 10(8) + 3 = 83$.

(B) Let the full first class fare be $x$ and the reservation charges per ticket be $y$.
Case $I$: The cost of one reserved first class ticket from station $A$ to $B$ is $Rs. 2530$.
$x + y = 2530$ ... $(i)$
Case $II$: The cost of one reserved first class ticket and one reserved first class half ticket from station $A$ to $B$ is $Rs. 3810$.
The cost of a half ticket is $\frac{x}{2} + y$.
So,$(x + y) + (\frac{x}{2} + y) = 3810$
$\frac{3x}{2} + 2y = 3810$
$3x + 4y = 7620$ ... $(ii)$
Multiply equation $(i)$ by $4$:
$4x + 4y = 10120$ ... $(iii)$
Subtract equation $(ii)$ from equation $(iii)$:
$(4x + 4y) - (3x + 4y) = 10120 - 7620$
$x = 2500$
Substitute $x = 2500$ into equation $(i)$:
$2500 + y = 2530$
$y = 30$
Thus,the full first class fare is $Rs. 2500$ and the reservation charge is $Rs. 30$.

(C) Let the cost price of the saree be $Rs. x$ and the list price of the sweater be $Rs. y$.
Case $I$: Sells a saree at $8 \%$ profit and a sweater at $10 \%$ discount,total sum $= Rs. 1008$.
$1.08x + 0.90y = 1008$ --- $(i)$
Case $II$: Sells a saree at $10 \%$ profit and a sweater at $8 \%$ discount,total sum $= Rs. 1028$.
$1.10x + 0.92y = 1028$ --- $(ii)$
Multiply $(i)$ by $92$ and $(ii)$ by $90$ to eliminate $y$:
$99.36x + 82.8y = 92736$ --- $(iii)$
$99x + 82.8y = 92520$ --- $(iv)$
Subtract $(iv)$ from $(iii)$:
$0.36x = 216$
$x = 600$
Substitute $x = 600$ into $(i)$:
$1.08(600) + 0.9y = 1008$
$648 + 0.9y = 1008$
$0.9y = 360$
$y = 400$
Thus,the cost price of the saree is $Rs. 600$ and the list price of the sweater is $Rs. 400$.

(A) Let the amount invested in schemes $A$ and $B$ be $Rs.\, x$ and $Rs.\, y,$ respectively.
$Case \, I$: Interest at $8 \%$ per annum on scheme $A$ + Interest at $9 \%$ per annum on scheme $B = 1860$.
$\Rightarrow \frac{8x}{100} + \frac{9y}{100} = 1860 \Rightarrow 8x + 9y = 186000 \quad \dots(i)$
$Case \, II$: If amounts are interchanged,interest is $1860 + 20 = 1880$.
$\Rightarrow \frac{9x}{100} + \frac{8y}{100} = 1880 \Rightarrow 9x + 8y = 188000 \quad \dots(ii)$
Multiply $(i)$ by $8$ and $(ii)$ by $9$:
$64x + 72y = 1488000 \quad \dots(iii)$
$81x + 72y = 1692000 \quad \dots(iv)$
Subtract $(iii)$ from $(iv)$:
$(81x - 64x) = 1692000 - 1488000$
$17x = 204000 \Rightarrow x = 12000$
Substitute $x = 12000$ in $(i)$:
$8(12000) + 9y = 186000$
$96000 + 9y = 186000$
$9y = 90000 \Rightarrow y = 10000$
Thus,she invested $Rs.\, 12000$ in scheme $A$ and $Rs.\, 10000$ in scheme $B$.

(A) Let the number of bananas in lots $A$ and $B$ be $x$ and $y$,respectively.
Case $I$: Cost of the first lot at the rate of $Rs. 2$ for $3$ bananas $+$ cost of the second lot at the rate of $Rs. 1$ per banana $=$ amount received.
$\frac{2}{3}x + y = 400$
$2x + 3y = 1200 \dots (i)$
Case $II$: Cost of the first lot at the rate of $Rs. 1$ per banana $+$ cost of the second lot at the rate of $Rs. 4$ for $5$ bananas $=$ amount received.
$x + \frac{4}{5}y = 460$
$5x + 4y = 2300 \dots (ii)$
On multiplying Eq. $(i)$ by $4$ and Eq. $(ii)$ by $3$,we get:
$8x + 12y = 4800$
$15x + 12y = 6900$
Subtracting the first from the second:
$(15x - 8x) + (12y - 12y) = 6900 - 4800$
$7x = 2100$
$x = 300$
Now,substitute the value of $x$ in Eq. $(i)$:
$2(300) + 3y = 1200$
$600 + 3y = 1200$
$3y = 600$
$y = 200$
Total number of bananas $= x + y = 300 + 200 = 500$.

(N/A) Let the cost of $1\,kg$ of apples be Rs. $x$ and the cost of $1\,kg$ of grapes be Rs. $y$.
According to the first condition,the total cost of $2\,kg$ of apples and $1\,kg$ of grapes is Rs. $160$.
Therefore,the equation is: $2x + y = 160$ $... (1)$
According to the second condition,the total cost of $4\,kg$ of apples and $2\,kg$ of grapes is Rs. $300$.
Therefore,the equation is: $4x + 2y = 300$ $... (2)$
Thus,the pair of linear equations in two variables representing the given situation is $2x + y = 160$ and $4x + 2y = 300$.

Let the cost of one textbook of Mathematics $IX$ be Rs. $x$ and the cost of one textbook of Mathematics $X$ be Rs. $y$.
Kushan purchased $2$ textbooks of Mathematics $IX$ and $3$ textbooks of Mathematics $X$. The total cost is represented by the expression $(2x + 3y)$.
Given that the total cost is Rs. $250$,we have:
$2x + 3y = 250$ ........ $(1)$
Rajan purchased $4$ textbooks of Mathematics $IX$ and $6$ textbooks of Mathematics $X$. The total cost is represented by the expression $(4x + 6y)$.
Given that the total cost is Rs. $500$,we have:
$4x + 6y = 500$ ........ $(2)$
Thus,the equations $(1)$ and $(2)$ represent the given situation as a pair of linear equations in two variables.

(N/A) Let Devarshi's marks be $x$ and Maharshi's marks be $y$ in the annual mathematics examination of standard $10$.
According to the problem,Devarshi got thrice the marks obtained by Maharshi:
$x = 3y$
$\therefore x - 3y = 0$ .......... $(1)$
The sum of the marks obtained by Devarshi and Maharshi is $150$:
$x + y = 150$ .......... $(2)$
Thus,the pair of linear equations in two variables representing the given situation is:
$x - 3y = 0$
$x + y = 150$

(N/A) Suppose the present age of the first girl is $x$ years and the present age of the second girl is $y$ years.
The ratio of their present ages is $x:y = 5:7$.
This implies $\frac{x}{y} = \frac{5}{7}$,which simplifies to $7x = 5y$ or $7x - 5y = 0$ ......... $(1)$
Eight years ago,the age of the first girl was $(x - 8)$ years and the age of the second girl was $(y - 8)$ years.
The ratio of their ages eight years ago was $(x - 8) : (y - 8) = 7 : 13$.
This implies $\frac{x - 8}{y - 8} = \frac{7}{13}$.
Cross-multiplying gives $13(x - 8) = 7(y - 8)$.
$13x - 104 = 7y - 56$.
$13x - 7y = 104 - 56$.
$13x - 7y = 48$ ......... $(2)$
Thus,the pair of linear equations representing the situation is $7x - 5y = 0$ and $13x - 7y = 48$.

(N/A) Let the length of the rectangular garden be $x \, m$ and the width be $y \, m$.
According to the problem,the length is $4 \, m$ more than its width:
$x = y + 4$
$x - y = 4$ ...... $(1)$
The perimeter of a rectangle is $2(x + y)$.
Therefore,half the perimeter is $\frac{2(x + y)}{2} = x + y$.
Given that the half perimeter is $36 \, m$:
$x + y = 36$ ...... $(2)$
Thus,the pair of linear equations in two variables representing the given situation is:
$x - y = 4$
$x + y = 36$

(N/A) Let the cost of one pant be $x$ and the cost of one shirt be $y$.
According to the first condition,the total cost of $4$ pants and $3$ shirts is Rs. $5000$,which can be represented as $4x + 3y = 5000$.
According to the second condition,the cost of one pant and one shirt is Rs. $750$,which can be represented as $x + y = 750$.
Thus,the pair of linear equations is $4x + 3y = 5000$ and $x + y = 750$.

(A) Let the weight of the father be $x \, kg$ and the weight of the son be $y \, kg$.
According to the problem,the total weight is $x + y = 70$ (Equation $1$).
The weight of the son is one-sixth of the father's weight,so $y = \frac{1}{6}x$,which implies $x = 6y$ or $x - 6y = 0$ (Equation $2$).
Substitute $x = 6y$ into Equation $1$:
$6y + y = 70$
$7y = 70$
$y = 10 \, kg$.
Now,substitute $y = 10$ into $x = 6y$:
$x = 6 \times 10 = 60 \, kg$.
Therefore,the father's weight is $60 \, kg$ and the son's weight is $10 \, kg$.

(B) Let the larger number be $x$ and the smaller number be $y$.
According to the first condition,the sum of the two numbers is $25$,so:
$x+y=25$ --- $(1)$
According to the second condition,five times the smaller number $(5y)$ is $5$ more than three times the larger number $(3x)$:
$5y = 3x + 5$
Rearranging the terms to standard form $(ax+by+c=0)$:
$3x - 5y + 5 = 0$ --- $(2)$
Thus,the pair of linear equations is $x+y=25$ and $3x-5y+5=0$.

(A) Let the age of the son be $x$ and the age of the father be $y$.
According to the problem,the reciprocals of their ages are $\frac{1}{x}$ and $\frac{1}{y}$.
The sum of the reciprocals is given as $\frac{1}{x} + \frac{1}{y} = \frac{5}{40}$.
The difference of the reciprocals is given as $\frac{1}{x} - \frac{1}{y} = \frac{3}{40}$.
Thus,the pair of linear equations in two variables is $\frac{1}{x} + \frac{1}{y} = \frac{5}{40}$ and $\frac{1}{x} - \frac{1}{y} = \frac{3}{40}$.

(A) Let the length of the rectangle be $x$ and the width be $y$.
According to the problem,the length is $3$ less than twice the width,so $x = 2y - 3$,which can be written as $x - 2y + 3 = 0$.
The perimeter of a rectangle is given by $2(x + y) = 100$.
Dividing both sides by $2$,we get $x + y = 50$.
Thus,the pair of linear equations is $x - 2y + 3 = 0$ and $x + y = 50$.

(C) Let the present age of the mother be $x$ years and the present age of the son be $y$ years.
Case $1$: Five years ago,the mother's age was $(x - 5)$ and the son's age was $(y - 5)$.
According to the problem: $(x - 5) = 5(y - 5)$
$x - 5 = 5y - 25$
$x - 5y + 20 = 0$
Case $2$: After ten years,the mother's age will be $(x + 10)$ and the son's age will be $(y + 10)$.
According to the problem: $(x + 10) = 2(y + 10)$
$x + 10 = 2y + 20$
$x - 2y - 10 = 0$
Thus,the pair of linear equations is $x - 5y + 20 = 0$ and $x - 2y - 10 = 0$.

(A) Let the cost of one mobile be $x$ and the cost of one memory card be $y$.
According to the first condition,the cost of $2$ mobiles and $3$ memory cards is Rs. $6500$,which can be written as: $2x + 3y = 6500$.
According to the second condition,the cost of one mobile and one memory card is Rs. $2950$,which can be written as: $x + y = 2950$.
Thus,the pair of linear equations is: $2x + 3y = 6500$ and $x + y = 2950$.

(D) Let the marks obtained by Raj be $x$ and the marks obtained by Nil be $y$.
According to the first condition,Raj got $\frac{3}{4}$ times the marks of Nil,so $x = \frac{3}{4}y$,which simplifies to $4x = 3y$ or $4x - 3y = 0$.
According to the second condition,the sum of their marks is $140$,so $x + y = 140$.
Thus,the pair of linear equations is $4x - 3y = 0$ and $x + y = 140$.

(N/A) Let the cost of $1 \,kg$ of mangoes be $x$ and the cost of $1 \,kg$ of pomegranate be $y$.
According to the first condition: $3x + y = 200$.
According to the second condition: $x + y = 80$.
Thus,the pair of linear equations is $3x + y = 200$ and $x + y = 80$.

(A) Let the cost of one pen be $x$ and the cost of one pencil be $y$.
According to the first condition,the total cost of $7$ pens and $5$ pencils is Rs. $50$,which can be written as: $7x + 5y = 50$.
According to the second condition,the total cost of $5$ pens and $7$ pencils is Rs. $46$,which can be written as: $5x + 7y = 46$.
Thus,the pair of linear equations is:
$7x + 5y = 50$
$5x + 7y = 46$