Distance covered the first time = 80 km

Time covered the first time = 2 hours

Speed = $\frac{\mathrm{Distance}}{\mathrm{Time}}$

Distance = Speed x Time

Distance covered second time = 2.5 hours x 50 km/h = 125 km

Total distance covered = 80 km + 125 km = 205 km

Total time taken = 2 hours + 2.5 hours = 4.5 hours

Average Speed = $\frac{\mathrm{Total\; Distance}}{\mathrm{Total\; Time}}$

Average Speed = $\frac{\mathrm{205\; km}}{4.5}$

4.5 = $\frac{45}{10}$ = $\frac{9}{2}$

Average Speed = 205 ÷ $\frac{9}{2}$ km/h

Reciprocate the fraction at the right of the ÷ and change ÷ to multiplication (x)

Average Speed = 205 x $\frac{2}{9}$ km/h

Average Speed = $\frac{\mathrm{205\; x\; 2}}{9}$ km/h

Average Speed = $\frac{410}{9}$ km/h

Average Speed = 45$\frac{5}{9}$ km/h

Speed = $\frac{\mathrm{Distance}}{\mathrm{Time}}$

Speed x Time = Distance

Time = $\frac{\mathrm{Distance}}{\mathrm{Speed}}$

30kmh^-1 = 30000m^-1

Time = $\frac{200}{30000}$

Note: 200 divides itself once and 30000, 150 times.

Time = $\frac{1}{150}$ hour

1 hour = 3600seconds
$\frac{1}{150}$ hour = $\frac{1}{150}$ x 3600 seconds

Time = 24 seconds

Initial speed = X km/h

Time taken at the initial speed = 2 hours

Distance covered at the initial speed = X km/h x 2h

Distance covered at the initial speed = 2X km

Increased his speed by 2km/h means if you add 2 km/h to his current speed(initial speed), you will get the new speed.

New speed = (X + 2) km/h

Time taken at the new speed = 3 hours.

Distance covered at the new speed = (X + 2) km/h x 3h

Distance covered at the new speed = 3X km + 6 km

Total distance covered is 36 km

Distance covered at the initial speed + Distance covered at the new speed = 36 km

2X km + 3X km + 6 km = 36 km

2X + 3X + 6 = 36

5X = 36 - 6

5X = 30

Divide both sides by 5

X = $\frac{30}{5}$ = 6 km/h

The sum of the distances covered by both buses should give 240 km

Let t that travelled by both when they are 240 km apart.

Distance covered by the first bus = 72 km/h x t

Distance covered by the second bus = 48 km/h x t

72t + 48t = 240

120t = 240

t = $\frac{240}{120}$ = 2 hours

Since they started the journey at 9.00 am, in 2 hours time, the time will be 11.00 a.m and both buses will be 240 km apart.

The regular time to reach destination = $\frac{\mathrm{40\; km}}{\mathrm{50\; km/h}}$ = 0.8 hours

18 minutes = $\frac{18}{60}$ hours = 0.3 hours

Time left to get to destination = 0.8 hours - 0.3 hours = 0.5 hours.

The ferry must cover the 40 km in 0.5 hours.

Speed = $\frac{\mathrm{Distance}}{\mathrm{Time}}$ = $\frac{\mathrm{40\; km}}{\mathrm{0.5\; h}}$ = 80 km/h

Speed = 60 miles per hour

3600 s = 1 hour

Notes:
1. 60 minutes = 1 hour
2. 60 seconds = 1 minutes
3. 60 x 60 seconds = 3600 seconds = 1 hour

If 3600 seconds = 1 hour

15 seconds = $\frac{\mathrm{15\; seconds\; x\; 1\; hour}}{\mathrm{3600\; seconds}}$ = 0.00417 hour

Distance = Length of train

Distance = Speed x Time

Distance = 60 miles per hour x 0.00417 hour

Note: per hour cancels hour

Length of train ≈ 0.25 miles

If 1 mile = 1760 yards

0.25 mile = $\frac{\mathrm{0.25\; mile\; x\; 1760\; yards}}{\mathrm{1\; mile}}$ = 440.352 yards ≈ 440 yards

Note: the mile cancel each other

Distance covered the first time = 80 km

Time covered the first time = 2 hours

Speed = $\frac{\mathrm{Distance}}{\mathrm{Time}}$

Distance = Speed x Time

Distance covered second time = 2.5 hours x 50 km/h = 125 km

Total distance covered = 80 km + 125 km = 205 km

Total time taken = 2 hours + 2.5 hours = 4.5 hours

Average Speed = $\frac{\mathrm{Total\; Distance}}{\mathrm{Total\; Time}}$

Average Speed = $\frac{\mathrm{205\; km}}{4.5}$

4.5 = $\frac{45}{10}$ = $\frac{9}{2}$

Average Speed = 205 ÷ $\frac{9}{2}$ km/h

Reciprocate the fraction at the right of the ÷ and change ÷ to multiplication (x)

Average Speed = 205 x $\frac{2}{9}$ km/h

Average Speed = $\frac{\mathrm{205\; x\; 2}}{9}$ km/h

Average Speed = $\frac{410}{9}$ km/h

Average Speed = 45$\frac{5}{9}$ km/h

Speed = $\frac{\mathrm{Distance}}{\mathrm{Time}}$

Speed x Time = Distance

Time = $\frac{\mathrm{Distance}}{\mathrm{Speed}}$

30kmh^-1 = 30000m^-1

Time = $\frac{200}{30000}$

Note: 200 divides itself once and 30000, 150 times.

Time = $\frac{1}{150}$ hour

1 hour = 3600seconds
$\frac{1}{150}$ hour = $\frac{1}{150}$ x 3600 seconds

Time = 24 seconds

Initial speed = X km/h

Time taken at the initial speed = 2 hours

Distance covered at the initial speed = X km/h x 2h

Distance covered at the initial speed = 2X km

Increased his speed by 2km/h means if you add 2 km/h to his current speed(initial speed), you will get the new speed.

New speed = (X + 2) km/h

Time taken at the new speed = 3 hours.

Distance covered at the new speed = (X + 2) km/h x 3h

Distance covered at the new speed = 3X km + 6 km

Total distance covered is 36 km

Distance covered at the initial speed + Distance covered at the new speed = 36 km

2X km + 3X km + 6 km = 36 km

2X + 3X + 6 = 36

5X = 36 - 6

5X = 30

Divide both sides by 5

X = $\frac{30}{5}$ = 6 km/h

The sum of the distances covered by both buses should give 240 km

Let t that travelled by both when they are 240 km apart.

Distance covered by the first bus = 72 km/h x t

Distance covered by the second bus = 48 km/h x t

72t + 48t = 240

120t = 240

t = $\frac{240}{120}$ = 2 hours

Since they started the journey at 9.00 am, in 2 hours time, the time will be 11.00 a.m and both buses will be 240 km apart.

The regular time to reach destination = $\frac{\mathrm{40\; km}}{\mathrm{50\; km/h}}$ = 0.8 hours

18 minutes = $\frac{18}{60}$ hours = 0.3 hours

Time left to get to destination = 0.8 hours - 0.3 hours = 0.5 hours.

The ferry must cover the 40 km in 0.5 hours.

Speed = $\frac{\mathrm{Distance}}{\mathrm{Time}}$ = $\frac{\mathrm{40\; km}}{\mathrm{0.5\; h}}$ = 80 km/h