Let the speed of 1st car and 2nd car be u km/h and v km/h.
Respective speed of both cars while they are travelling in same direction = (u - v) km/h

Respective speed of both cars while they are travelling in opposite directions i.e., travelling towards each other = (u + v) km/h

According to the question,

5(u - v) = 100

⇒ u - v = 20 ... (i)

1(u + v) = 100 ... (ii)

Adding both the equations, we get

2u = 120

u = 60 km/h ... (iii)

Putting this value in equation (ii), we obtain
v = 40 km/h
Hence, speed of one car = 60 km/h and speed of other car = 40 km/h

Let length and breadth of rectangle be x unit and y unit respectively.
Area = xy
According to the question,
(x - 5) (y + 3) = xy - 9
⇒ 3x - 5y - 6 = 0 ... (i)
(x + 3) (y + 2) = xy + 67
⇒ 2x - 3y - 61 = 0 ... (ii)
By cross multiplication, we get
x/305-(-18) = y/-12-(-183) = 1/9-(-10)
x/323 = y/171 = 1/19
x = 17, y = 9
Hence, the length of the rectangle = 17 units and breadth of the rectangle = 9 units.

(i) 1/2x + 1/3y = 2
1/3x + 1/2y = 13/6
Let 1/x = p and 1/y = q, then the equations changes as below:
p/2 + q/3 = 2
⇒ 3p + 2q -12 = 0 ... (i)
p/3 + q/2 = 13/6
⇒ 2p + 3q -13 = 0 ... (ii)

By cross-multiplication method, we get

p/-26-(-36) = q/-24-(-39) = 1/9-4
p/10 = q/15 = 1/5
p/10 = 1/5 and q/15 = 1/5
p = 2 and q = 3
1/x = 2 and 1/y = 3
Hence, x = 1/2 and y = 1/3

(ii) 2/√x +3/√y = 2
4/√x - 9/√y = -1
Let 1/√x = p and 1/√y = q, then the equations changes as below:
2p + 3q = 2 ... (i)
4p - 9q = -1 ... (ii)
Multiplying equation (i) by 3, we get
6p + 9q = 6 ... (iii)
Adding equation (ii) and (iii), we get

10p = 5

p = 1/2 ... (iv)

Putting in equation (i), we get

2 × 1/2 + 3q = 2
3q = 1
q = 1/3

p = 1/√x = 1/2
√x = 2
x = 4
and
q = 1/√y = 1/3
√y = 3
y = 9
Hence, x = 4, y = 9

(iii) 4/x + 3y = 14
3/x - 4y = 23
Putting 1/x = p in the given equations, we get
4p + 3y = 14
⇒ 4p + 3y - 14 = 0
3p - 4y = 23
⇒ 3p - 4y -23 = 0
By cross-multiplication, we get
p/-69-56 = y/-42-(-92) = 1/-16-9
⇒ -p/125 = y/50 = -1/25
Now,
-p/125 = -1/25 and y/50 = -1/25
⇒ p = 5 and y = -2
Also, p = 1/x = 5
⇒ x = 1/5
So, x = 1/5 and y = -2 is the solution.

(iv) 5/x-1 + 1/y-2 = 2
6/x-1 - 3/y-2 = 1
Putting 1/x-1 = p and 1/y-2 = q in the given equations, we obtain
5p + q = 2 ... (i)
6p - 3q = 1 ... (ii)
Now, by multiplying equation (i) by 3 we get
15p + 3q = 6 ... (iii)
Now, adding equation (ii) and (iii)
21p = 7
⇒ p = 1/3
Putting this value in equation (ii) we get,
 6×1/3 - 3q =1
 ⇒ 2-3q = 1
 ⇒ -3q = 1-2
 ⇒ -3q = -1
 ⇒ q = 1/3
Now,
p = 1/x-1 = 1/3
 ⇒1/x-1 = 1/3
 ⇒ 3 = x - 1
 ⇒ x = 4
Also,
q = 1/y-2 = 1/3
 ⇒ 1/y-2 = 1/3
 ⇒ 3 = y-2
 ⇒ y = 5
Hence, x = 4 and y = 5 is the solution.

(v) 7x-2y/xy = 5
 ⇒ 7x/xy - 2y/xy = 5
 ⇒ 7/y - 2/x = 5 ... (i)
8x+7y/xy = 15
 ⇒ 8x/xy + 7y/xy = 15
 ⇒ 8/y + 7/x = 15 ... (ii)
Putting 1/x = p and 1/y = q in (i) and (ii) we get,
7q - 2p = 5 ... (iii)
8q + 7p = 15 ... (iv)
Multiplying equation (iii) by 7 and multiplying equation (iv) by 2 we get,
49q - 14p = 35 ... (v)
16q + 14p = 30 ... (vi)
Now, adding equation (v) and (vi) we get,
49q - 14p + 16q + 14p = 35 + 30
⇒ 65q = 65
⇒ q = 1
Putting the value of q in equation (iv)
8 + 7p = 15
⇒ 7p = 7
⇒ p = 1
Now,
p = 1/x = 1
⇒ 1/x = 1
⇒ x = 1
also, q = 1 = 1/y
⇒ 1/y = 1
⇒ y = 1
Hence, x =1 and y = 1 is the solution.

(vi) 6x + 3y = 6xy
⇒ 6x/xy + 3y/xy = 6
⇒ 6/y + 3/x = 6 ... (i)
2x + 4y = 5xy
⇒ 2x/xy + 4y/xy = 5
⇒ 2/y + 4/x = 5 ... (ii)
Putting 1/x = p and 1/y = q in (i) and (ii) we get,
6q + 3p - 6 = 0
2q + 4p - 5 = 0
By cross multiplication method, we get
p/-30-(-12) = q/-24-(-15) = 1/6-24
p/-18 = q/-9 = 1/-18
p/-18 = 1/-18 and q/-9 = 1/-18
p = 1 and q = 1/2
p = 1/x = 1 and q = 1/y = 1/2
x = 1, y = 2
Hence, x = 1 and y = 2

(vii) 10/x+y + 2/x-y = 4
15/x+y - 5/x-y = -2
Putting 1/x+y = p and 1/x-y = q in the given equations, we get:
10p + 2q = 4
⇒ 10p + 2q - 4 = 0 ... (i)
15p - 5q = -2
⇒ 15p - 5q + 2 = 0 ... (ii)
Using cross multiplication, we get
p/4-20 = q/-60-(-20) = 1/-50-30
p/-16 = q/-80 = 1/-80
p/-16 = 1/-80 and q/-80 = 1/-80
p = 1/5 and q = 1
p = 1/x+y = 1/5 and q = 1/x-y = 1
x + y = 5 ... (iii)
and x - y = 1 ... (iv)
Adding equation (iii) and (iv), we get
2x = 6
x = 3 .... (v)
Putting value of x in equation (iii), we get
y = 2
Hence, x = 3 and y = 2

(viii) 1/3x+y + 1/3x-y = 3/4
1/2(3x-y) - 1/2(3x-y) = -1/8
Putting 1/3x+y = p and 1/3x-y = q in the given equations, we get
p + q = 3/4 ... (i)
p/2 - q/2 = -1/8
p - q = -1/4 ... (ii)
Adding (i) and (ii), we get
2p = 3/4 - 1/4
2p = 1/2
p = 1/4
Putting the value in equation (ii), we get
1/4 - q = -1/4
q = 1/4 + 1/4 = 1/2
p = 1/3x+y = 1/4
3x + y = 4 ... (iii)
q = 1/3x-y = 1/2
3x - y = 2 ... (iv)
Adding equations (iii) and (iv), we get
6x = 6
x = 1 ... (v)
Putting the value in equation (iii), we get
3(1) + y = 4
y = 1
Hence, x = 1 and y = 1

Let the speed of Ritu in still water and the speed of stream be x km/h
and y km/h respectively.
Speed of Ritu while rowing
Upstream = (x - y) km/h

Downstream = (x + y) km/h

According to question,

2(x + y) = 20

⇒ x + y = 10 ... (i)
2(x - y) = 4
⇒ x - y = 2 ... (ii)
Adding equation (i) and (ii), we get

Putting this equation in (i), we get

y = 4

Hence, Ritu's speed in still water is 6 km/h and the speed of the current is 4 km/h.

Let the number of days taken by a woman and a man be x and y respectively.
Therefore, work done by a woman in 1 day = 1/x

According to the question,

4(2/x + 5/y) = 1

2/x + 5/y = 1/4

3(3/x + 6/y) = 1

3/x + 6/y = 1/3

Putting 1/x = p and 1/y = q in these equations, we get

2p + 5q = 1/4

By cross multiplication, we get

p/-20-(-18) = q/-9-(-18) = 1/144-180

p/-2 = q/-1 = 1/-36

p/-2 = -1/36 and q/-1 = 1/-36

p = 1/18 and q = 1/36

p = 1/x = 1/18 and q = 1/y = 1/36

x = 18 and y = 36

Hence, number of days taken by a woman = 18 and number of days taken by a man = 36

Let the speed of train and bus be u km/h and v km/h respectively.
According to the given information,

60/u + 240/v = 4 ... (i)
100/u + 200/v = 25/6 ... (ii)
Putting 1/u = p and 1/v = q in the equations, we get
60p + 240q = 4 ... (iii)
100p + 200q = 25/6
600p + 1200q = 25 ... (iv)
Multiplying equation (iii) by 10, we get
600p + 2400q = 40 .... (v)
Subtracting equation (iv) from (v), we get1200q = 15
q = 15/200 = 1/80 ... (vi)
Putting equation (iii), we get
60p + 3 = 4
60p = 1
p = 1/60
p = 1/u = 1/60 and q = 1/v = 1/80
u = 60 and v = 80
Hence, speed of train = 60 km/h and speed of bus = 80 km/h.

Let the age of Ani and Biju be x and y years respectively.

Therefore, age of Ani’s father, Dharam = 2 × x = 2x years

And age of Biju’s sister Cathy  = y / 2 years

Case (i) :When Ani is older than Biju by 3 years,

x − y = 3 …...............(i)

2x – y / 2 = 30

4x − y = 60 …...............(ii)

Subtracting (i) from (ii), we obtain

3x = 60 − 3 = 57

x = 57 / 3 = 19

Therefore, age of Ani = 19 years

And age of Biju = 19 − 3 = 16 years

Case (ii) :When Biju is older than Ani,

y − x = 3 …........ (I)

2x – y / 2 = 30

4x − y = 60 …...............(ii)

Adding (i) and (ii), we obtain

3x = 63

x = 21

Therefore, age of Ani = 21 years

And age of Biju = 21 + 3 = 24 years.

Let those friends were having Rs x and y with them.

Using the information given in the question, we obtain

x + 100 = 2(y − 100)

x + 100 = 2y − 200

x − 2y = −300 (i)

And, 6(x − 10) = (y + 10)

6x − 60 = y + 10

6x − y = 70 (ii)

Multiplying equation (ii) by 2, we obtain

12x − 2y = 140 (iii)

Subtracting equation (i) from equation (iii), we obtain

11x = 140 + 300

11x = 440

x = 40

Using this in equation (i), we obtain

40 − 2y = −300

40 + 300 = 2y

2y = 340

y = 170

Therefore, those friends had Rs 40 and Rs 170 with them respectively.