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

Before the jet overtakes the cargo plane, both will be at the same distance from the airport.

Method I

Let t = the time the jet overtakes the cargo plane after taking off.

Since the cargo plane leaves 1 hour before the jet, at the time of the overtake, the cargo plane's time will be t + 1

Distance = Speed x Time

Speed of cargo plane = 180 km/h

Distance at point of overtaking = 180 x (t + 1)

Speed of jet = 450 km/h

Distance at point of overtaking = 450 x t

Since at the point of overtaking, the distance are the same,

180 x (t + 1) = 450 x t

180t + 180 = 450t

180 = 450t - 180t

180 = 270t

Divide both sides by 270

t = $\frac{180}{270}$ hours

If hour = 60 minutes

$\frac{180}{270}$ = 60 minutes x $\frac{180}{270}$ hour ÷ 1 hour

$\frac{180}{270}$ = 60 minutes x $\frac{180}{270}$ = 40 minutes

∴ the jet will overtake the cargo plane after 40 minutes of leaving the airport at 1.00 pm hence the time to overtake the cargo plane is 1:40 pm

Method II

Let t = the time travelled by the cargo plane before the jet overtakes it.

Time the jet travelled before overtaking the cargo plane = t - 1

Distance = Speed x Time

Speed of cargo plane = 180 km/h

Distance at point of overtaking = 180 x t

Speed of jet = 450 km/h

Distance at point of overtaking = 450 x (t - 1)

Since at the point of overtaking, the distance are the same,

180 x t = 450 x (t - 1)

180t = 450t - 450

450 = 450t - 180t

450 = 270t

Divide both sides by 270

t = $\frac{450}{270}$ = $\frac{45}{27}$hours

If 1 hour = 60 minutes

$\frac{45}{27}$ hour = 60 minutes x $\frac{45}{27}$ hour ÷ 1 hour

$\frac{45}{27}$ hour = 60 minutes x $\frac{45}{27}$ = 100 minutes

100 minutes = 1 hour 40 minutes

∴ the jet overtakes the cargo plane at 1:40 pm.

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

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

Let x = initial speed

Time taken to cover the 120 km by the initial speed

Time = $\frac{120}{\mathrm{x}}$

The speed was increased by 20 km/h

New speed = x + 20 km/h

Time taken to cover the 120 km by the new speed

Time = $\frac{120}{\mathrm{x\; +\; 20}}$

If you subtract the time taken by the new speed from the time taken by the initial speed, you will get 30 minutes which is half an hour.

$\frac{120}{\mathrm{x}}$ - $\frac{120}{\mathrm{x\; +\; 20}}$ = $\frac{1}{2}$

Multiply both sides by the L.C.M of the denominators 2, x and x + 20

2x(x + 20) x $\frac{120}{\mathrm{x}}$ - 2x(x + 20) x $\frac{120}{\mathrm{x\; +\; 20}}$ = 2x(x + 20) x $\frac{1}{2}$

2(x + 20) x 120 - 2x x 120 = 1 x x(x + 20)

240(x + 20) - 240x = x² + 20x

240x + 4800 - 240x = x² + 20x

4800 = x² + 20x

x² + 20x - 4800 = 0

80 - 60 = 20 and 80 x -60 = - 4800

x² + 80x - 60x - 4800 = 0

x(x + 80) - 60(x + 80) = 0

(x + 80)(x - 60) = 0

When x + 80 = 0 ⇒ x = - 80

When x - 60 = 0 ⇒ x = 60

The speed is positive hence initial speed = 60 km/h

The distance covered whether delayed or on time is the same.

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

Average speed for regular schedule = r

Time taken during regular schedule = $\frac{120}{\mathrm{r}}$

Average speed for delayed schedule = r - 5

Time taken during delayed schedule = $\frac{120}{\mathrm{r\; -\; 5}}$

The delayed schedule will take longer time than the regular schedule.

If you subtract the regular time from the time taken for the delayed trip, you will get 20 minutes.

Since the speed is in km/h, change the minutes to hours.

20 minutes = $\frac{20}{60}$ = $\frac{1}{3}$ hours

$\frac{120}{\mathrm{r\; -\; 5}}$ - $\frac{120}{\mathrm{r}}$ = $\frac{1}{3}$

Multiply both sides by the L.C.M of the denominators [r x (r - 5) x 3]

r x (r - 5) x 3 x $\frac{120}{\mathrm{r\; -\; 5}}$ - r x (r - 5) x 3 x $\frac{120}{\mathrm{r}}$ = r x (r - 5) x 3 x $\frac{1}{3}$

360r - 360(r - 5) = (r - 5) x r

360r - 360r + 1800 = r² - 5r

1800 = r² - 5r

r² - 5r - 1800 = 0

Solve for r

Split - 5 to be in the form two numbers when you multiply you will get -1800 and when you add you will get - 5

The numbers are -45 and 40

r² - 45r + 40r - 1800 = 0

r(r - 45) + 40(r - 45) = 0

(r - 45)(r + 40) = 0

r = 45 or - 40

But the speed is positive, hence the average speed r = 45 km/h

The sum of the distances covered by each train should give you the distance between them before the journey.

Let s = speed of train A

train B, travels $\frac{8}{7}$ as fast as train A

Speed of train B = $\frac{8}{7}$ x s

Time taken by both to cover the distance = 2 hours.

Distance covered by train A = s x 2 = 2s

Distance covered by train B = $\frac{8}{7}$ x s x 2 = $\frac{\mathrm{16s}}{7}$

Sum of both distance covered should give 450 km

2s + $\frac{\mathrm{16s}}{7}$ = 450

Get rid of the fraction by multiplying both sides by the denominator 7

7 x 2s + $\frac{\mathrm{16s}}{7}$ x 7 = 450 x 7

14s + 16s = 3150

30s = 3150

Divide both sides by 30

s = $\frac{3150}{30}$ = 105 km/h

∴ speed of train A = 105 km/h

Speed of train B = $\frac{8}{7}$ x s

Speed of train B = $\frac{8}{7}$ x 105 km/h = 120 km/h