Let x be the time taken in minutes and y be the distance travelled in km.
\(\begin{array}{|l|l|l|l|l|} \hline \text { Time taken } \mathbf{x} \text { (in minutes) } & 60 & 120 & 180 & 240 \\ \hline \text { Distance } \mathbf{y} \text { (in km) } & 50 & 100 & 150 & 200 \\ \hline \end{array}\)
(i) Observe that as time increases, the distance travelled also increases. Therefore, the variation is a direct variation. It is of the form y = kx .
Constant of variation
\(k=\frac{y}{x}=\frac{50}{60}=\frac{100}{120}=\frac{150}{180}=\frac{200}{240}=\frac{5}{6}\)
Hence, the relation may be given as
\(y=k x \ \Rightarrow \ y=\frac{5}{6} x\)
(ii) From the graph, \(y=\frac{5 x}{6},\) if x = 90 , then \(y=\frac{5}{6} \times 90=75 \mathrm{~km}\)
The distance travelled for \(1 \frac{1}{2}\)hours (i.e.,) 90 minutes is 75 km.
(iii) From the graph, \(y=\frac{5 x}{6}, \text { if } y=300\) then \(x=\frac{6 y}{5}=\frac{6}{5} \times 300=360\) minutes (or) 6 hours.
The time taken to cover 300 km is 360 minutes, that is 6 hours.

Usual speed = x km/hr
After increasing speed = x + 25 km/hr
Distance = 150 km
\(\text { Time }=\frac{\text { Distance }}{\text { speed }}\)
\(\frac{150}{x}-\frac{150}{x+25}=\frac{1}{2} h r \ \left[\because 30 \mathrm{~min}=\frac{1}{2} h r\right]\)
\(150 x+\frac{150 \times 25}{x(x+25)}-150 x=\frac{1}{2}\)
\(150 \times 25 \times 2 =x(x+25) \)
\(7500 =x^{2}+25 x \)
\(x^{2}+25 x-750 =0 \)
\((x-75)(x+100) =0 \)
\(x =75 \text { (or) } x=-100 \)
x is +ve
x = 75km/hr
speed = 75 km/hr

Let speed be x
and time be y
Distance = speed x time
Therefore Distance = xy
\((x+10)(y-2)=x y \)
\(x y-2 x+10 y-20=x y \)
\(-2 x+10 y=20 \)
\(x-5 y=-10 \)
\(\text {From condition(2) }\)
\((x-10)(y+3)=x y \)
\(x y+3 x-10 y-30=x y \)
\(3 x-10 y=30 \)
\((2) \Rightarrow 3 x-10 y=30\)
\(\text { (1) } \times 3 \Rightarrow 3 x-15 y =-30 \)
\(5 y =60 \)
\(y =12 \)
\(\text { (1) } \Rightarrow x-5(12) =-10 \)
\(x-60 =-10 \)
\(x =50 \mathrm{~km} / \mathrm{hr} \)
\(\text { Distance } =x y \)
\(=50 \times 12 \)
\(=600 \mathrm{~km} \)

\(\frac{a^{2}-16}{a^{3}-8} \times \frac{2 a^{2}-3 a-2}{2 a^{2}+9 a+4} \div \frac{3 a^{2}-11 a-4}{a^{2}-2 a+4}\)
\(\frac{(a+4)(a-4)}{(a-2)\left(a^{2}+2 a+4\right)} \times \frac{(2 a+1)(a-2)}{(2 a+1)(a+4)} \times \frac{a^{2}-2 a+4}{(3 a+1)(a-4)}\)
\(=\frac{a^{2}-2 a+4}{(3 a+1)\left(a^{2}+2 a+4\right)}\)

\(\frac{1}{x+1}+\frac{2}{x+2} =\frac{4}{x+4} \)
\(\frac{x+2+2 x+2}{(x+1)(x+2)} =\frac{4}{x+4} \)
\((3 x+4)(x+4) =4(x+1)(x+2) \)
\(3 x^{2}+16 x+16 =4\left(x^{2}+3 x+2\right) \)
\(3 x^{2}+16 x+16 =4 x^{2}+12 x+8 \)
\(3 x^{2}+16 x+16-4 x^{2}-12 x-8=0 \)
\(-x^{2}+4 x+8=0 \)
\(\Rightarrow x^{2}-4 x-8=0 \)
\(\text { By formula method, }\)
\(x =\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \)
\(a =1, b=-4, c=-8 \)
\(x =\frac{4 \pm \sqrt{16+32}}{2}=\frac{4 \pm \sqrt{48}}{2} \)
\(=\frac{4 \pm 4 \sqrt{3}}{2} \)
\(x =2 \pm 2 \sqrt{3} \)

\(16 x^{2}-24 x^{3}+(a-1) x^{2}+(b+1) x+49\)

Case: I
a - 8 - 56 = 0
b + 1 + 42 = 0
a = 65, b = -43
Case: II
a - 8 - 56 = 0
b + 1 + 42 = 0
a + 48 = 0 , b - 41 = 0
a = -48, b = 41

\(2 x+y+4 z=15 \)   ...(1)
\(x-2 y+3 z=13 \)    ...(2)
\(3 x+y-z=2 \)        ...(3)

\(\text { Sub } x=2 \text { in }(5), \)
\(10(2)+y =19 \)
\(20+y =19 \)
\(y =19-20 \Rightarrow y=-1 \)
\(\text { Sub : } x=2 \text { and } y=-1 \text { in (1) }\)
\(2(2)-1+4 z =15 \)
\(3+4 z =15 \)
\(4 z =15-3 \)
\(4 z =12 \)
\(z =3 \)
\(x=2 ; y =1 ; z=3 \)

\(A B=\left[\begin{array}{ccc} 1 & 1 & -1 \\ 2 & 0 & 3 \\ 3 & -1 & 2 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 0 & 2 \\ 1 & 4 \end{array}\right]\)
\(=\left[\begin{array}{cc} 1+0-1 & 3+2-4 \\ 2+0+3 & 6+0+12 \\ 3+0+2 & 9-2+8 \end{array}\right]\)
\(=\left[\begin{array}{cc} 0 & 1 \\ 5 & 18 \\ 5 & 15 \end{array}\right]\)
LHS = (AB) C
\(=\left[\begin{array}{cc} 0 & 1 \\ 5 & 18 \\ 5 & 15 \end{array}\right]\left[\begin{array}{cccc} 1 & 2 & 3 & -4 \\ 2 & 0 & -2 & 1 \end{array}\right]\)
\(=\left[\begin{array}{cccc} 0+2 & 0+0 & 0-2 & 0+1 \\ 5+36 & 10+0 & 15-36 & -20+18 \\ 5+30 & 10+0 & 15-30 & -20+15 \end{array}\right]\)
\(=\left[\begin{array}{cccc} 2 & 0 & -2 & 1 \\ 41 & 10 & -21 & -2 \\ 35 & 10 & -15 & -5 \end{array}\right]\)
\(\text { Now, } B C=\left[\begin{array}{ll} 1 & 3 \\ 0 & 2 \\ 1 & 4 \end{array}\right]\left[\begin{array}{rrrr} 1 & 2 & 3 & -4 \\ 2 & 0 & -2 & 1 \end{array}\right]\)
\(=\left[\begin{array}{cccc} 1+6 & 2+0 & 3-6 & -4+3 \\ 0+4 & 0+0 & 0-4 & 0+2 \\ 1+8 & 2+0 & 3-8 & -4+4 \end{array}\right]\)
\(=\left[\begin{array}{cccc} 7 & 2 & -3 & -1 \\ 4 & 0 & -4 & 2 \\ 9 & 2 & -5 & 0 \end{array}\right]\)
\(A(B C)=\left[\begin{array}{ccc} 1 & 1 & -1 \\ 2 & 0 & 3 \\ 3 & -1 & 2 \end{array}\right]\left[\begin{array}{rrrr} 7 & 2 & -3 & -1 \\ 4 & 0 & -4 & 2 \\ 9 & 2 & -5 & 0 \end{array}\right]\)
\(=\left[\begin{array}{rrrr} 7+4-9 & 2+0-2 & -3-4+5 & -1+2+0 \\ 14+0+27 & 4+0+6 & -6+0-15 & -2+0+0 \\ 21-4+18 & 6+0+4 & -9+4-10 & -3-2+0 \end{array}\right]\)
\(=\left[\begin{array}{cccc} 2 & 0 & -2 & 1 \\ 41 & 10 & -21 & -2 \\ 35 & 10 & -15 & -5 \end{array}\right]\)
LHS = RHS
\(\therefore\) (AB)C = A(BC)
Hence proved

\(\mathrm{A}^{2}=\mathrm{A} \cdot \mathrm{A}\)
\(=\left[\begin{array}{rr} 1 & 0 \\ -1 & 7 \end{array}\right]\left[\begin{array}{rr} 1 & 0 \\ -1 & 7 \end{array}\right] \)
\(=\left[\begin{array}{rr} 1+0 & 0+0 \\ -1-7 & 0+49 \end{array}\right] \)
\(=\left[\begin{array}{rr} 1 & 0 \\ -8 & 49 \end{array}\right] \)
\(8 A=8\left[\begin{array}{rr} 1 & 0 \\ -1 & 7 \end{array}\right]=\left[\begin{array}{rr} 8 & 0 \\ -8 & 56 \end{array}\right]\)
\(7 I=7\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right]\)
\(\therefore A^{2}-8 A+7 I=\left[\begin{array}{rr} 1 & 0 \\ -8 & 49 \end{array}\right]-\left[\begin{array}{rr} 8 & 0 \\ -8 & 56 \end{array}\right]+\left[\begin{array}{cc} 7 & 0 \\ 0 & 7 \end{array}\right]\)
\(=\left[\begin{array}{rr} -7+7 & 0 \\ 0+0 & -7+7 \end{array}\right]\)
\(=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=0\)
Hence proved

\(A+B=\left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right]+\left[\begin{array}{ll} 2 & 1 \\ 1 & 0 \end{array}\right]=\left[\begin{array}{ll} 3 & 0 \\ 3 & 3 \end{array}\right]\)
\((A+B)^{2}=(A+B)(A+B)\)
\(=\left[\begin{array}{ll} 3 & 0 \\ 3 & 3 \end{array}\right]\left[\begin{array}{ll} 3 & 0 \\ 3 & 3 \end{array}\right]\)
\(=\left[\begin{array}{ll} 9+0 & 0+0 \\ 9+9 & 0+9 \end{array}\right]=\left[\begin{array}{rr} 9 & 0 \\ 18 & 9 \end{array}\right]\)
\(\text { Now } A^{2}=A \cdot A=\left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right]\left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right]\)
\(=\left[\begin{array}{ll} 1-2 & -1-3 \\ 2+6 & -2+9 \end{array}\right]=\left[\begin{array}{rr} -1 & -4 \\ 8 & 7 \end{array}\right]\)
\(A B=\left[\begin{array}{cc} 1 & -1 \\ 2 & 3 \end{array}\right]\left[\begin{array}{ll} 2 & 1 \\ 1 & 0 \end{array}\right]\)
\(=\left[\begin{array}{ll} 2-1 & 1+0 \\ 4+3 & 2+0 \end{array}\right]=\left[\begin{array}{ll} 1 & 1 \\ 7 & 2 \end{array}\right]\)
\(B^{2}=B \cdot B\)
\(=\left[\begin{array}{ll} 2 & 1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{ll} 2 & 1 \\ 1 & 0 \end{array}\right] \)
\(=\left[\begin{array}{cc} 4+1 & 2+0 \\ 2+0 & 1+0 \end{array}\right] \)
\(=\left[\begin{array}{ll} 5 & 2 \\ 2 & 1 \end{array}\right] \)
\(\text { Now } A^{2}+2 A B+B^{2}\)
\(=\left[\begin{array}{rr} -1 & -4 \\ 8 & 7 \end{array}\right]+\left[\begin{array}{rr} 2 & 2 \\ 14 & 4 \end{array}\right]+\left[\begin{array}{ll} 5 & 2 \\ 2 & 1 \end{array}\right]\)
\(=\left[\begin{array}{rr} 6 & 0 \\ 24 & 12 \end{array}\right]\)
\(\text { from }(1),(2),(\mathrm{A}+\mathrm{B})^{2} \neq \mathrm{A}^{2}+2 \mathrm{AB}+\mathrm{B}^{2}\)

Let the usual speed of the plane be x km/hr and usual time to cover 1500 km \(=\frac{1500}{x} \mathrm{hrs}\)
Given that speed increased by 250 km/hr.
Time taken to cover the distance with new speed = \(\frac{1500}{x+150}\)
\(\therefore \frac{1500}{x+250}-\frac{1500}{x}=\frac{1}{2}\)
On Simplifying we get, x² + 250x - 750000 = 0
x(x+ 1000) -750 (x+ 1000) = 0
(x - 750) (x + 1000) = 0
x = 750, x = - 1000 is not possible.
\(\therefore\) The usual speed of the plane is 750 km/hr.

Let the larger tap fill the tank in 'x' hours.
\(\therefore\) Smaller tap fill the tank in (x + 10) hours (given)
Portion of tank filled by larger tap in I hour \(= \frac{1}{x}\)
Portion of tank filled by smaller tap in t hour \(=\frac{1}{x+10}\)
\(\text { Given, Both the taps can fill the tank in } 9 \frac{3}{8}=\frac{75}{8} \text {Hours}\)
[Time and work done are reciprocals of each other]
\(\frac{1}{x}+\frac{1}{x+10}=\frac{8}{75} \)
\(\frac{x+10+x}{x(x+10)}=\frac{8}{75} \)
\(75(2 x+10) =8\left(x^{2}+10 x\right) \)
\(8 x^{2}-70 x-750 =0 \)
\(4 x^{2}-35 x-375 =0 \)
\(4 x^{2}-60 x+25 x-375 =0 \)
\(4 x(x-15)+25(x-15) =0 \)
\((4 x+25)(x-15) =0 \)
\(\mathrm{x}=15, \ \mathrm{x}=\frac{-25}{7} \text { is not possible. }\)
\(\therefore\) Time required to fill the tank by the 1^st tap is 15 hrs and by the 2^nd tap is 25 hours.

Let the sides of two squares be 'x' and 'y' respectively
Sum of areas x² + y² = 468
Difference of perimeters 4x - 4y = 24 [\(\because\) x > y ]
x - y = 6
y = x - 6
Substituting in (1)
\(x^{2}+(x-6)^{2} =468 \)
\(x^{2}+x^{2}-12 x+36-468 =0 \)
\(2 x^{2}-12 x-432 =0 \)
\(x^{2}-6 x-216 =0 \)
\((x-18)(x+12) =0 \)
\(x= 18\)
When x = 18:y = 18 - 6 = 12
\(\therefore\)The sides of two squares are 18 m and 12 m respectively.

Standard form of the given equations
\(x-3 z=-5 \)          ...(1)
\(2 x-y+2 z=16 \)   ...(2)
\(7 x-3 y-5 z=19\) ...(3)
Consider (2) and (3)

\(z=\frac{24}{8}=3\)
Substituting, \(z =3 \text { in }(1) \)
\(x =3(3)-5 \)
\(=9-5=4 \)
\(\text {Substituting,} \space x=4, z=3 \text { in (2) } \)
\(2(4)-y+2(3)=16\)
\(8+6-16=y\)
\(\Rightarrow \mathrm{y}=14-16=-2\)
\(\therefore \text { solution is } x=4, y=-2, z=3\)

\(4 x-2 y+3 z =1 \)
\(x+3 y-4 z =-7 \)
\(3 x+y+2 z =5 \)
\(\text { Consider (2) and (3) }\)
\((3) \times(3) \Rightarrow \)
\(\begin{aligned} &\begin{array}{r} 9 x+3 y+6 z=15 \\ x+3 y-4 z=-7 \\ \hline \end{array}\\ &\text { (4) }-(2) \Rightarrow 8 x+10 z=22 \end{aligned}\)
Now consider (1) and (3)
\(\begin{aligned} &(3) \times 2 \Rightarrow 6 x+2 y+4 z=10 \\ &4 x-2 y+3 z=1 \\ &\hline (6) +(1) \Rightarrow 10 x+7 z=11 \end{aligned}\)
Consider (5) and (7)
\(\begin{array}{r} (5) \times 7 \Rightarrow 56 x+70 z=154 \\ (7) \times 10 \Rightarrow 100 x+70 z=110 \\ \hline (8)-(9) \Rightarrow -44 x=44 \end{array}\)
\(x=\frac{44}{44}=-1\)
Substituting, x = -1 is in (7)
10(-1) + 7z = 11
-10  + 7z = 11
7z = 11 + 10
\(z=\frac{21}{7}=3\)
Substituting, x = - 1,z = 3 is in (3)
3(-1) + y + 2(3) = 5
-3 + y + 6 = 5
y = 5 - 3 = 2
\(\therefore\) solution x = - 1,y = 2,z = 3