Answer: Option C. -> 60∘,  120∘,  60∘,  120∘
:
C
Let the measure of the angle be $x$ and that of its adjacent angle be $y$.
$x=2y...\left(i\right)$ (given)
$x+y={180}^{\circ }...\left(ii\right)$
(sum of adjacent angles of a parallelogram is ${180}^{\circ }$)
On substituting (i) in (ii), we get
$2y+y={180}^{\circ }$
$\Rightarrow y={60}^{\circ }$
Since $x=2y$,$x={120}^{\circ }$.
We know thatopposite angles of parallelogram are equal.
$\Rightarrow$ Measure of the angles are${60}^{\circ }$, ${120}^{\circ }$, ${60}^{\circ }$ and ${120}^{\circ }$.

Answer: Option B. -> 26 m, 20 m
:
B
Let $l$and $b$ be the length and breadth of the room respectively.
Then, the perimeter of the room is$2(l+b)$ metres.
From the question,
$l=6+b$ ...(1)
$\frac{1}{2}\times 2(l+b)=46$
$\Rightarrow l+b=46$ ...(2)
Let's solve these two equations using substitution method.
On substituting the value of $l$ from (1) in (2), we get
$6+b+b=46$
$\Rightarrow$ $6+2b=46$
$\Rightarrow 2b=40$
$\Rightarrow b=20m$
$\Rightarrow l=20+6=26m$
Therefore, length and breadth are 26 m and 20 m long respectively.

Answer: Option C. -> (2, -1) 
:
C
The solutions of a linear equation must be able to satisfy the equation. Among the given options, point (2,-1) satisfies the given equation. Substituing the value of x and y in the equation we get:
3(2) + 4(-1) = 2.

Answer: Option B. -> False
:
B
$x+2y=5$ ...(1)
$7x+3y=13$ ...(2)
Let's solve these equations using elimination method.
Onmultiplying the first equation by 7, we get
$7x+14y=35$ ...(3)
On subtracting (2) from (3), we have
$7x+14y=35$
$-7x-3y=-13$
_______________
$11y=22$
$\Rightarrow y=2$
On subsituting value of y in$x+2y=5$, we get
$x+2\left(2\right)=5$
$\Rightarrow x=1$
$\Rightarrow x=1$and$y=2$
Hence, the solution of the given equations is (1, 2).

Answer: Option B. -> u=2231,v=1123
:
B
Divide the given equations by uv,
$8v−3u=5uv⇒\frac{8}{u}−\frac{3}{v}=5...\left(1\right)$
$6v−5u=−2uv⇒\frac{6}{u}−\frac{5}{v}=−2...\left(2\right)$
Assume $\frac{1}{u}=x$ and $\frac{1}{v}=y$
Put the values of $\frac{1}{u}$ and $\frac{1}{v}$ in (1) and (2)
$8x−3y=5...\left(3\right)$
$6x−5y=−2...\left(4\right)$
Solving equations (3) and (4) we get x and y
Multiply (3) with 5 and (4) with 3 to equate the coefficients of y.
$40x−15y=25$
$18x−15y=−6$

$⇒x=\frac{31}{22}⇒u=\frac{22}{31}$
Substituting x in (3)
$8×\frac{31}{22}−3y=5→3y=\frac{69}{11}$
$→y=\frac{23}{11}⇒v=\frac{11}{23}$
So, $u=\frac{22}{31}$ and $v=\frac{11}{23}$

Answer: Option B. -> 2, -1
:
B
Given
$2x-3y=7$...(i)
$5x+y=9$...(ii)
Rearranging (ii), we get$y=9-5x$...(iii)
Substituting (iii) in (i),we get
$2x-3(9-5x)=7$
$\Rightarrow$$17x=34$
$\Rightarrow x=2$.
Substituting the value of $x$ in (i), we get
$2\left(2\right)-3y=7$
$\Rightarrow y=-1$.

Answer: Option A. -> a unique
:
A
If the graph oflinear equations represented by the lines intersectat only one point, then thepoint isits only solution.
Here, the lines meet at onlyone point (1,-1). Therefore,the given pair of equations has a unique solution.

Answer: Option B. -> x + 3y = 16
:
B
The solution of the linear equation should satisfy the equation. The given values of x and y are 7 and 3 respectively.
Among the given equations, only the equation x + 3y = 16 is able to satisfy the equation.
$\Rightarrow 7+3\left(3\right)=16$
Hence $x=7andy=3$ is the solution ofx + 3y = 16.

Answer: Option B. -> (6, 5)
:
B
The solutionof a linear equation satisfiesthe equation.
$3\times 3+4\times 4=9+16=25\ne 38\to$ (3, 4) does not satisfy the equation.
$3\times 6+4\times 5=18+20=38\to$ (6, 5) satisfiesthe equation.
$3\times 2+4\times 19=6+76=82\ne 38\to$ (2, 19) does not satisfy the equation.
$3\times 3+4\times 12=9+48=57\ne 38\to$ (3, 12) does not satisfy the equation.

Answer: Option A. -> 4 km/hr
:
A
Let's assume that the speed of the boat in still water is x km/hr and speed of the stream is y km/hr.
So, the speed of the boat in upstream will be (x-y) km/hr.
Similarly, the speed of the boat downstream will be (x+y) km/hr.
We know $\text{time}=\left(\frac{\text{distance}}{\text{speed}}\right)$.
Using the above formula we can form the equations in two variables.
Taking the first case,
$\frac{30}{\text{x - y}}+\frac{44}{\text{x + y}}=14$.
Taking the second case,
$\frac{60}{\text{x - y}}+\frac{55}{\text{x + y}}=25$.
Now, we have the equations in two variables but the equations are not linear.
So, we will assume $\frac{1}{\text{x - y}}=\text{u}$ and $\frac{1}{\text{x + y}}=\text{v}$.
So on substituting $\text{u}$ and $\text{v}$ in the above two equations, we get
$30\text{u}+44\text{v}=14$ ...(1)
$60\text{u}+55\text{v}=25$ ...(2)
We can solve the above two equations using the elimination method.
$60\text{u}+88\text{v}=28$ ...(3)
(by multiplyingequation (1) by 2)
On subtracting equation (2) from (3),we get $\text{v}=\frac{1}{11}$
On substituting $\text{v}$ in equation (2)we get $\text{u}=\frac{1}{3}$
Now as we have assumed
$\frac{1}{\text{x - y}}=\text{u}$ and $\frac{1}{\text{x + y}}=\text{v}$
On substituting the values of $\text{u}$ and $\text{v}$,
we get a pair of linear equations in $\text{x and y}$
$\text{x - y}=3$...(4)
$\text{x + y}=11$...(5)
On adding (5) from (4), we have
$2x=14$
$x=7$
On subsituting the value of x in $x-y=3$,weget y = 4.
So, the speed of the boat in still water is 7 km/hr and the speed of the stream is 4 km/hr.