By property of rectangle, we know that its opposite sides are of equal lengths.
i.e. DC=AB \(\Rightarrow\) x+3y=13   ...(i)
and AD=BC \(\Rightarrow\) 3x+y=7   .....(ii)
On multiplying Eq. (ii) by 3 and then subtracting Eq. (i), we get

\(\Rightarrow\) x=1
On putting x=1 in Eq. (i), we get
3y=12 \(\Rightarrow\) y=4
Hence, x=1 and y=4.

Given that, x, y ane 40⁰ are the angles of a triangle.
\(\therefore\) x+y+40⁰=180⁰  [\(\because\) sum of all the angles of a triangle is 180⁰ ]
\(\Rightarrow\)x+y=140⁰     ...(i)
Also, x-y=30⁰    ...(ii)
On adding Eqs. (i) and (ii), we get
2x=170⁰ \(\Rightarrow\)  x=85⁰
On putting  x=85⁰ in Eqs. (i), we get
85⁰+y=140⁰  \(\Rightarrow\)   y=55⁰
Hence, the required values of x and y are 85⁰ and 55⁰ respectively.

Let the number of students in halls A and B be x and y, respectively.
According to the question,
x-10=y+10 \(\Rightarrow\) x-y=20   ..(i)
and (x+20)=2(y-20) \(\Rightarrow\) x-2y=-60   ...(ii)
On subtracting Eq. (ii) from Eq. (i), we get
-y+2y=60+20 \(\Rightarrow\) y=80
On putting y=80 in Eq. (i), we get
x-80=20 \(\Rightarrow\) x=20+80 \(\Rightarrow\) x=100
Hence, there are 100 students in hall A and 80 students in hall B.                                                                                                                                                                        

Let the speed of steam be x km/h, then speed of person going upstream=(x+5) km/h and speed of person going downstream=(5-x)h
According to question
\(\frac { 4 }{ 5-x } =3\left( \frac { 4 }{ 5-x } \right) \)
2.4 km/h 
Lete the speed of boat in still water be x km/h and that of stream be y km/h, then according to question
\(\frac { 30 }{ x-y } +\frac { 28 }{ x+y } =7\quad \quad ...(i)\)
and \(\frac { 21 }{ x+y } +\frac { 21 }{ x-y } =5\quad \quad ...(ii)\)  
Solve Eq. (i) and Eq. (ii) to get the speed of boat and speed of stream.
Speed of motorboat in still water=10 km/h
Speed of stream=4 km/h

Let the two numbers be x and y, then \(\frac { x }{ y } =\frac { 5 }{ 6 } \quad \Rightarrow \quad y=\frac { 6x }{ 5 } \quad ..(i)\)
Also, \(\frac { x-8 }{ y-8 } =\frac { 4 }{ 5 } \quad \Rightarrow \quad 5x-4y=8\quad \quad ..(ii)\)
Draw graphs of Eq. (i) and Eq. (ii) to get the required numbers.
Numbe0 are 49 and 48.