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

(X=1/2, Y=1/3) Given equations:
$\frac{1}{2x} + \frac{1}{3y} = 2$ ...$(1)$
$\frac{1}{3x} + \frac{1}{2y} = \frac{13}{6}$ ...$(2)$
Let $\frac{1}{x} = p$ and $\frac{1}{y} = q$. Substituting these into the equations:
$\frac{p}{2} + \frac{q}{3} = 2 \Rightarrow 3p + 2q = 12$ ...$(3)$
$\frac{p}{3} + \frac{q}{2} = \frac{13}{6} \Rightarrow 2p + 3q = 13$ ...$(4)$
Multiply equation $(3)$ by $3$ and equation $(4)$ by $2$:
$9p + 6q = 36$ ...$(5)$
$4p + 6q = 26$ ...$(6)$
Subtracting $(6)$ from $(5)$:
$(9p - 4p) + (6q - 6q) = 36 - 26$
$5p = 10 \Rightarrow p = 2$
Substitute $p = 2$ in equation $(3)$:
$3(2) + 2q = 12$
$6 + 2q = 12 \Rightarrow 2q = 6 \Rightarrow q = 3$
Since $p = \frac{1}{x} = 2$,we get $x = \frac{1}{2}$.
Since $q = \frac{1}{y} = 3$,we get $y = \frac{1}{3}$.

(A) Given equations are:
$(1)$ $6x + 3y = 6xy$
$(2)$ $2x + 4y = 5xy$
Divide both equations by $xy$:
From $(1)$: $\frac{6x}{xy} + \frac{3y}{xy} = \frac{6xy}{xy} \implies \frac{6}{y} + \frac{3}{x} = 6$
From $(2)$: $\frac{2x}{xy} + \frac{4y}{xy} = \frac{5xy}{xy} \implies \frac{2}{y} + \frac{4}{x} = 5$
Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. The equations become:
$(3)$ $3u + 6v = 6$
$(4)$ $4u + 2v = 5$
Multiply $(4)$ by $3$ to eliminate $v$:
$12u + 6v = 15$ $(5)$
Subtract $(3)$ from $(5)$:
$(12u + 6v) - (3u + 6v) = 15 - 6$
$9u = 9 \implies u = 1$
Substitute $u = 1$ into $(3)$:
$3(1) + 6v = 6 \implies 6v = 3 \implies v = \frac{1}{2}$
Since $u = \frac{1}{x} = 1 \implies x = 1$
Since $v = \frac{1}{y} = \frac{1}{2} \implies y = 2$
Thus,the solution is $(x, y) = (1, 2)$.

(N/A) Let $u = \frac{1}{3x+y}$ and $v = \frac{1}{3x-y}$.
Substituting these into the given equations:
$u + v = \frac{3}{4}$ --- $(1)$
$\frac{1}{2}u - \frac{1}{2}v = -\frac{1}{8}$ --- $(2)$
Multiply equation $(2)$ by $2$ to simplify:
$u - v = -\frac{1}{4}$ --- $(3)$
Adding equation $(1)$ and $(3)$:
$(u + v) + (u - v) = \frac{3}{4} - \frac{1}{4}$
$2u = \frac{2}{4} = \frac{1}{2}$
$u = \frac{1}{4}$
Substitute $u = \frac{1}{4}$ into equation $(1)$:
$\frac{1}{4} + v = \frac{3}{4}$
$v = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$
Now,solve for $x$ and $y$:
$3x+y = \frac{1}{u} = 4$ --- $(4)$
$3x-y = \frac{1}{v} = 2$ --- $(5)$
Adding $(4)$ and $(5)$:
$6x = 6 \implies x = 1$
Substitute $x = 1$ into $(4)$:
$3(1) + y = 4 \implies y = 1$
Thus,the solution is $x = 1, y = 1$.

(A) Let the speed of Ritu in still water be $x\, km/h$ and the speed of the stream be $y\, km/h$.
Speed of Ritu while rowing:
Upstream $= (x - y)\, km/h$
Downstream $= (x + y)\, km/h$
According to the problem:
For downstream: $2(x + y) = 20 \Rightarrow x + y = 10$ $...(1)$
For upstream: $2(x - y) = 4 \Rightarrow x - y = 2$ $...(2)$
Adding equations $(1)$ and $(2)$:
$(x + y) + (x - y) = 10 + 2$
$2x = 12 \Rightarrow x = 6$
Substituting $x = 6$ in equation $(1)$:
$6 + y = 10 \Rightarrow y = 4$
Hence, Ritu's speed in still water is $6\, km/h$ and the speed of the current is $4\, km/h$.

(A) Let the number of days taken by a woman and a man be $x$ and $y$ respectively.
Therefore,the work done by a woman in $1$ day $= \frac{1}{x}$.
The work done by a man in $1$ day $= \frac{1}{y}$.
According to the problem:
$4(\frac{2}{x} + \frac{5}{y}) = 1 \Rightarrow \frac{2}{x} + \frac{5}{y} = \frac{1}{4}$ $(i)$
$3(\frac{3}{x} + \frac{6}{y}) = 1 \Rightarrow \frac{3}{x} + \frac{6}{y} = \frac{1}{3}$ $(ii)$
Let $\frac{1}{x} = p$ and $\frac{1}{y} = q$. Substituting these into the equations:
$2p + 5q = \frac{1}{4} \Rightarrow 8p + 20q = 1$ $(iii)$
$3p + 6q = \frac{1}{3} \Rightarrow 9p + 18q = 1$ $(iv)$
Solving by cross-multiplication:
$\frac{p}{(20)(-1) - (18)(-1)} = \frac{q}{(1)(9) - (1)(8)} = \frac{1}{(8)(18) - (20)(9)}$
$\frac{p}{-20 + 18} = \frac{q}{9 - 8} = \frac{1}{144 - 180}$
$\frac{p}{-2} = \frac{q}{1} = \frac{1}{-36}$
$p = \frac{-2}{-36} = \frac{1}{18}$ and $q = \frac{1}{-36} = -\frac{1}{36}$ (Wait,re-calculating: $p = \frac{1}{18}, q = \frac{1}{36}$).
Thus,$x = 18$ and $y = 36$.
Therefore,$1$ woman alone takes $18$ days and $1$ man alone takes $36$ days.

(N/A) Let the speed of the train be $u \ km/h$ and the speed of the bus be $v \ km/h$.
According to the given information:
Case $1$: Time taken = $4 \ hours$ for $60 \ km$ by train and $240 \ km$ by bus.
$\frac{60}{u} + \frac{240}{v} = 4 \quad ...(1)$
Case $2$: Time taken = $4 \ hours + 10 \ minutes = 4 + \frac{10}{60} = 4 + \frac{1}{6} = \frac{25}{6} \ hours$ for $100 \ km$ by train and $200 \ km$ by bus.
$\frac{100}{u} + \frac{200}{v} = \frac{25}{6} \quad ...(2)$
Let $\frac{1}{u} = p$ and $\frac{1}{v} = q$. The equations become:
$60p + 240q = 4 \quad ...(3)$
$100p + 200q = \frac{25}{6} \implies 600p + 1200q = 25 \quad ...(4)$
Multiply equation $(3)$ by $10$:
$600p + 2400q = 40 \quad ...(5)$
Subtract equation $(4)$ from $(5)$:
$(600p + 2400q) - (600p + 1200q) = 40 - 25$
$1200q = 15 \implies q = \frac{15}{1200} = \frac{1}{80}$
Substitute $q = \frac{1}{80}$ into equation $(3)$:
$60p + 240(\frac{1}{80}) = 4$
$60p + 3 = 4 \implies 60p = 1 \implies p = \frac{1}{60}$
Since $p = \frac{1}{u} = \frac{1}{60}$ and $q = \frac{1}{v} = \frac{1}{80}$,we get:
$u = 60 \ km/h$ and $v = 80 \ km/h$.
Thus,the speed of the train is $60 \ km/h$ and the speed of the bus is $80 \ km/h$.

(A-D) Let the age of Ani be $x$ years and the age of Biju be $y$ years.
Given that the difference between their ages is $3\, years,$ we have two cases:
Case $I$: $x - y = 3$ or Case $II$: $y - x = 3$.
Ani's father Dharam's age $= 2x$ years.
Biju's sister Cathy's age $= y/2$ years.
The difference between the ages of Dharam and Cathy is $30\, years,$ so $2x - y/2 = 30,$ which simplifies to $4x - y = 60$.
Case $I$: $x - y = 3$ and $4x - y = 60$.
Subtracting the first from the second: $(4x - y) - (x - y) = 60 - 3 \implies 3x = 57 \implies x = 19$.
Then $y = 19 - 3 = 16$.
Case $II$: $y - x = 3 \implies y = x + 3$.
Substituting into $4x - y = 60$: $4x - (x + 3) = 60 \implies 3x = 63 \implies x = 21$.
Then $y = 21 + 3 = 24$.
Thus,the ages of Ani and Biju are either $(19, 16)$ years or $(21, 24)$ years.

(A) Let the amounts of capital with the two friends be $Rs$ $x$ and $Rs$ $y$ respectively.
According to the first condition: "Give me a hundred,friend! $I$ shall then become twice as rich as you."
$x + 100 = 2(y - 100)$
$x + 100 = 2y - 200$
$x - 2y = -300$ $...(i)$
According to the second condition: "If you give me ten,$I$ shall be six times as rich as you."
$6(x - 10) = y + 10$
$6x - 60 = y + 10$
$6x - y = 70$ $...(ii)$
To solve the system,multiply equation $(ii)$ by $2$:
$12x - 2y = 140$ $...(iii)$
Subtract equation $(i)$ from equation $(iii)$:
$(12x - 2y) - (x - 2y) = 140 - (-300)$
$11x = 440$
$x = 40$
Substitute $x = 40$ into equation $(i)$:
$40 - 2y = -300$
$-2y = -340$
$y = 170$
Thus,the amounts of their respective capital are $Rs$ $40$ and $Rs$ $170$.

(A) Let the speed of the train be $x\, km/h$ and the time taken be $t\, hours$. The distance $d$ is given by $d = xt$ $...(i)$.
According to the first condition,if the speed is $(x + 10)\, km/h$,the time taken is $(t - 2)\, hours$:
$(x + 10)(t - 2) = d$
$xt - 2x + 10t - 20 = d$
Since $d = xt$,we get $-2x + 10t = 20$,or $-x + 5t = 10$ $...(ii)$.
According to the second condition,if the speed is $(x - 10)\, km/h$,the time taken is $(t + 3)\, hours$:
$(x - 10)(t + 3) = d$
$xt + 3x - 10t - 30 = d$
Since $d = xt$,we get $3x - 10t = 30$ $...(iii)$.
Multiplying equation $(ii)$ by $2$,we get $-2x + 10t = 20$ $...(iv)$.
Adding $(iii)$ and $(iv)$:
$(3x - 10t) + (-2x + 10t) = 30 + 20$
$x = 50\, km/h$.
Substituting $x = 50$ in equation $(ii)$:
$-50 + 5t = 10$
$5t = 60$
$t = 12\, hours$.
Distance $d = xt = 50 \times 12 = 600\, km$.

(36) Let the number of rows be $x$ and the number of students in a row be $y$.
Total number of students in the class $= x \times y = xy$.
According to the first condition:
If $3$ students are extra in a row,there would be $1$ row less.
$(x - 1)(y + 3) = xy$
$xy + 3x - y - 3 = xy$
$3x - y = 3$ $...(i)$
According to the second condition:
If $3$ students are less in a row,there would be $2$ rows more.
$(x + 2)(y - 3) = xy$
$xy - 3x + 2y - 6 = xy$
$-3x + 2y = 6$ $...(ii)$
Adding equation $(i)$ and equation $(ii)$:
$(3x - y) + (-3x + 2y) = 3 + 6$
$y = 9$
Substituting $y = 9$ in equation $(i)$:
$3x - 9 = 3$
$3x = 12$
$x = 4$
Total number of students $= xy = 4 \times 9 = 36$.

(A) Given that,$\angle C = 3 \angle B = 2(\angle A + \angle B)$.
From $3 \angle B = 2(\angle A + \angle B)$,we get $3 \angle B = 2 \angle A + 2 \angle B$,which simplifies to $\angle B = 2 \angle A$,or $2 \angle A - \angle B = 0 \dots (i)$.
We know that the sum of the angles in a triangle is $180^{\circ}$,so $\angle A + \angle B + \angle C = 180^{\circ}$.
Since $\angle C = 3 \angle B$,we substitute this into the sum: $\angle A + \angle B + 3 \angle B = 180^{\circ}$,which gives $\angle A + 4 \angle B = 180^{\circ} \dots (ii)$.
Multiplying equation $(i)$ by $4$,we get $8 \angle A - 4 \angle B = 0 \dots (iii)$.
Adding equations $(ii)$ and $(iii)$,we get $9 \angle A = 180^{\circ}$,which implies $\angle A = 20^{\circ}$.
Substituting $\angle A = 20^{\circ}$ into equation $(ii)$,we get $20^{\circ} + 4 \angle B = 180^{\circ}$,so $4 \angle B = 160^{\circ}$,which gives $\angle B = 40^{\circ}$.
Finally,$\angle C = 3 \angle B = 3 \times 40^{\circ} = 120^{\circ}$.
Thus,the angles are $\angle A = 20^{\circ}$,$\angle B = 40^{\circ}$,and $\angle C = 120^{\circ}$.

Given equations are:
$(a-b)x + (a+b)y = a^2 - 2ab - b^2 \dots(1)$
$(a+b)(x+y) = a^2 + b^2 \dots(2)$
Expanding equation $(2)$:
$(a+b)x + (a+b)y = a^2 + b^2 \dots(3)$
Subtracting equation $(3)$ from $(1)$:
$[(a-b)x + (a+b)y] - [(a+b)x + (a+b)y] = (a^2 - 2ab - b^2) - (a^2 + b^2)$
$(a-b-a-b)x = -2ab - 2b^2$
$-2bx = -2b(a+b)$
$x = a+b$
Substituting $x = a+b$ into equation $(1)$:
$(a-b)(a+b) + (a+b)y = a^2 - 2ab - b^2$
$a^2 - b^2 + (a+b)y = a^2 - 2ab - b^2$
$(a+b)y = -2ab$
$y = \frac{-2ab}{a+b}$

(N/A) Given equations are:
$(1)$ $152x - 378y = -74$
$(2)$ $-378x + 152y = -604$
Step $1$: Add the two equations:
$(152x - 378x) + (-378y + 152y) = -74 - 604$
$-226x - 226y = -678$
Divide by $-226$:
$x + y = 3$ --- $(3)$
Step $2$: Subtract equation $(2)$ from equation $(1)$:
$(152x - (-378x)) + (-378y - 152y) = -74 - (-604)$
$530x - 530y = 530$
Divide by $530$:
$x - y = 1$ --- $(4)$
Step $3$: Solve equations $(3)$ and $(4)$:
Adding $(3)$ and $(4)$: $2x = 4 \implies x = 2$
Subtracting $(4)$ from $(3)$: $2y = 2 \implies y = 1$
Thus,the solution is $x = 2, y = 1$.

(N/A) We know that the sum of the measures of opposite angles in a cyclic quadrilateral is $180^{\circ}$.
Therefore,$\angle A + \angle C = 180^{\circ}$.
$(4y + 20) + (-4x) = 180$
$-4x + 4y = 160$
$-x + y = 40$ $...(i)$
Also,$\angle B + \angle D = 180^{\circ}$.
$(3y - 5) + (-7x + 5) = 180$
$-7x + 3y = 180$ $...(ii)$
Multiplying equation $(i)$ by $3$,we obtain:
$-3x + 3y = 120$ $...(iii)$
Subtracting equation $(iii)$ from equation $(ii)$:
$(-7x + 3y) - (-3x + 3y) = 180 - 120$
$-4x = 60$
$x = -15$
Substituting $x = -15$ in equation $(i)$:
$-(-15) + y = 40$
$15 + y = 40$
$y = 25$
Now,calculating the angles:
$\angle A = 4y + 20 = 4(25) + 20 = 120^{\circ}$
$\angle B = 3y - 5 = 3(25) - 5 = 70^{\circ}$
$\angle C = -4x = -4(-15) = 60^{\circ}$
$\angle D = -7x + 5 = -7(-15) + 5 = 110^{\circ}$

(B) The given pair of linear equations is:
$2x + 3y - 5 = 0$ ... $(i)$
$px - 6y - 8 = 0$ ... $(ii)$
Comparing these with the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:
$a_1 = 2, b_1 = 3, c_1 = -5$
$a_2 = p, b_2 = -6, c_2 = -8$
For a pair of linear equations to have a unique solution,the condition is:
$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$
Substituting the values:
$\frac{2}{p} \neq \frac{3}{-6}$
$\frac{2}{p} \neq -\frac{1}{2}$
By cross-multiplying:
$p \neq -4$
Thus,the system has a unique solution for all real values of $p$ except $-4$,i.e.,$p \in \mathbb{R} - \{-4\}$.

(B) The given pair of linear equations is:
$2x + 3y = 7$ ... $(i)$
$2px + py = 28 - qy$ ... $(ii)$
Rearranging equation $(ii)$,we get:
$2px + (p + q)y = 28$
For a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to have infinitely many solutions,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Comparing the given equations with the standard form:
$a_1 = 2, b_1 = 3, c_1 = -7$
$a_2 = 2p, b_2 = (p + q), c_2 = -28$
Applying the condition:
$\frac{2}{2p} = \frac{3}{p + q} = \frac{-7}{-28}$
$\frac{1}{p} = \frac{3}{p + q} = \frac{1}{4}$
Taking the first and third parts:
$\frac{1}{p} = \frac{1}{4} \Rightarrow p = 4$
Taking the second and third parts:
$\frac{3}{p + q} = \frac{1}{4} \Rightarrow p + q = 12$
Substituting $p = 4$ into $p + q = 12$:
$4 + q = 12 \Rightarrow q = 8$
Thus,the values are $p = 4$ and $q = 8$.

(N/A) pair of linear equations in two variables is given by:
$a_1x + b_1y + c_1 = 0$
$a_2x + b_2y + c_2 = 0$
For the system to have a unique solution,the condition is $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$.
Given the solution $x = -1$ and $y = 3$,these values must satisfy both equations:
$1$) $a_1(-1) + b_1(3) + c_1 = 0 \Rightarrow -a_1 + 3b_1 + c_1 = 0$
$2$) $a_2(-1) + b_2(3) + c_2 = 0 \Rightarrow -a_2 + 3b_2 + c_2 = 0$
We can choose arbitrary values for the coefficients that satisfy these conditions. For example,let $a_1 = 1, b_1 = 1$. Then $-1 + 3(1) + c_1 = 0 \Rightarrow c_1 = -2$. So,$x + y - 2 = 0$.
Let $a_2 = 1, b_2 = 2$. Then $-1 + 3(2) + c_2 = 0 \Rightarrow -1 + 6 + c_2 = 0 \Rightarrow c_2 = -5$. So,$x + 2y - 5 = 0$.
Since there are infinitely many combinations of $a_1, b_1, c_1$ and $a_2, b_2, c_2$ that satisfy the given solution and the condition for a unique solution,we can write infinitely many such pairs.

(A) Given equations are:
$2x + y = 23$ $(i)$
$4x - y = 19$ $(ii)$
Adding equation $(i)$ and $(ii)$:
$(2x + y) + (4x - y) = 23 + 19$
$6x = 42$
$x = 7$
Substituting $x = 7$ in equation $(i)$:
$2(7) + y = 23$
$14 + y = 23$
$y = 23 - 14 = 9$
Now,calculating the required values:
$5y - 2x = 5(9) - 2(7) = 45 - 14 = 31$
$\frac{y}{x} - 2 = \frac{9}{7} - 2 = \frac{9 - 14}{7} = -\frac{5}{7}$
Thus,the values are $31$ and $-\frac{5}{7}$.

(B) By the property of a rectangle,opposite sides are equal.
For lengths: $x + 3y = 13$ --- $(i)$
For breadths: $3x + y = 7$ --- $(ii)$
To solve this pair of linear equations,multiply equation $(ii)$ by $3$:
$9x + 3y = 21$ --- $(iii)$
Now,subtract equation $(i)$ from equation $(iii)$:
$(9x + 3y) - (x + 3y) = 21 - 13$
$8x = 8$
$x = 1$
Substitute $x = 1$ into equation $(i)$:
$1 + 3y = 13$
$3y = 12$
$y = 4$
Thus,the required values are $x = 1$ and $y = 4$.

(C) Given equations are:
$(i) \ x + y = 3.3$
$(ii) \ \frac{0.6}{3x - 2y} = -1$
From equation $(ii)$:
$0.6 = -1(3x - 2y)$
$0.6 = -3x + 2y$
$3x - 2y = -0.6 \quad (iii)$
To solve the system,multiply equation $(i)$ by $2$:
$2x + 2y = 6.6 \quad (iv)$
Add equation $(iii)$ and $(iv)$:
$(3x - 2y) + (2x + 2y) = -0.6 + 6.6$
$5x = 6$
$x = \frac{6}{5} = 1.2$
Substitute $x = 1.2$ into equation $(i)$:
$1.2 + y = 3.3$
$y = 3.3 - 1.2$
$y = 2.1$
Thus,the solution is $x = 1.2$ and $y = 2.1$.

(D) Given,the pair of linear equations is:
$\frac{x}{3} + \frac{y}{4} = 4$ $(i)$
Multiplying both sides by the $LCM(3, 4) = 12$,we get:
$4x + 3y = 48$ $(iii)$
And the second equation is:
$\frac{5x}{6} - \frac{y}{8} = 4$ $(ii)$
Multiplying both sides by the $LCM(6, 8) = 24$,we get:
$20x - 3y = 96$ $(iv)$
Now,adding equations $(iii)$ and $(iv)$,we get:
$(4x + 3y) + (20x - 3y) = 48 + 96$
$24x = 144$
$x = \frac{144}{24} = 6$
Now,substitute the value of $x = 6$ in equation $(iii)$:
$4(6) + 3y = 48$
$24 + 3y = 48$
$3y = 48 - 24$
$3y = 24$
$y = 8$
Thus,the solution is $x = 6$ and $y = 8$.

(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 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 $x = \frac{1}{2}, y = \frac{-3}{2}$.

(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 $(1, 1)$.

(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 $130^{\circ}, 100^{\circ}, 50^{\circ}, 80^{\circ}$.

(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 $y = 4$.

(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 $40 \,km/h$.

(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 $13x - 7y = 48$.