# Challenging Speed, Time, Distance Questions (Hints & Answers)

## ✩ You must already know:

1. Speed, Time, Distance Formula :
2. The Pythagoras Theorem

## Questions

Try to use a minimum number of hints to reach the solution. All the best!

If you find these challenging, learn to solve the speed, time and distance questions here.

### 1 Question

A train goes from station A to station B at a speed of 200 Km/h. While returning, the train has a better engine. It is faster by 100 Km/h than the old engine. What is the train’s average speed for the round trip?

Let the distance between stations = d Km.

Speed from A to B = 200 Km/h

Speed from B to A = 200 Km/h + 100 Km/h = 300 Km/h

Can you find the time taken from A to B and B to A? What about total time?

### 2 Question

A bus left town A for town B. Having traveled 300 km, it stopped for 30 minutes due to road blockage. It traveled 60% of the total distance by this time. After it started again, the driver increased the speed by 20 km/h and reached town B at the scheduled time. What was the original speed of the bus?

Can you find the relation between the original speed and time till bus hits the road block?

### 3 Question

A motor-boat goes down the river to a port at a distance of 18 km. Then it travels upstream for 16 km. It takes seven hours to complete the trip. The speed of the boat in still water is 5 Km/h. Find the speed of the water flowing in the river.

Still water neither propels nor resists the boat in any direction. However, traveling downstream, the boat is going with the flow of water. It propels the boat with it. Its speed gets added to the speed of the boat.

Let the speed of water be x Km/h.

Therefore downstream boat speed = 5 + x Km/h.

Can you now find the time taken to travel downstream?

### 4 Question

Two friends, Maya and Sofia, live 12 kilometers apart. They start walking towards each other at the same time. Maya flies a drone towards Sofia at the start, which returns to Maya upon reaching Sofia. Find the total distance traveled by the drone if its speed is 15 Km/hr while the friends walk at 5 Km/h.

In how much time does the drone meet Sofia? What is the distance traveled by the drone during this time?

### 5 Question

A boat at point A travels to point B on the other side of the river, directly opposite to point A. The river is 1.4 Km wide and flows at 12 m/s. The speed of the boat is 37 m/sec in still water. Find the shortest time it takes to complete the trip.

Direction Of Boat:

If the boat travels on the straight line AB, by the time it will reach the opposite side, the river will carry it forward to some point D. Therefore, it has to travel on the line AC, to reach point B, directly opposite to point A.

Let the time taken by the boat to complete the trip = t

Can you find distances AC and CB in terms of t?

Let the distance between stations = d Km.

Speed from A to B = 200 Km/h

Speed from B to A = 200 Km/h + 100 Km/h = 300 Km/h

From the speed formula:

Time = Distance/Speed

Time from A to B = tAB = d/200

Time from B to A =tBA = d/300

Total Time = tAB + tBA = d/200 + d/300

hours

Let us calculate the average speed now.

Total distance = AB + BA = d + d = 2d

AverageSpeed = TotalDistance/TotalTime

Km/h

Average Speed = 240 Km/h

The bus stops at point C for 1/2 hour due to blockage.

Let the initial speed be x Km/h

Time = Distance/Speed

Time to travel the distance AC = tAC = 300/x hours

Let’s first calculate CB

60% of AB = 300 Km

AB × 0.60 = 300

AB = 300/0.60 = 500 Km

CB = Remaining distance = 500 − 300 = 200 Km

Speed of bus after point C = (x + 20) Km / h

Time to travel the distance CB = tCB = 200/(x + 20) hours

Had there been no blockage, the bus would have traveled at xKm/h. Its travel time would have been:

tusual = AB/x = 500/x hours

The bus reached town B on time, despite blockage, therefore:

Usual time = Time to travel the distance AC + Delay + Time to travel the distance CB

Multiplying both sides by 2x(x + 20):

500 × 2(x + 20) = 300 × 2(x + 20) + x(x + 20) + 200 × 2x

1000x + 20000 = 600x + 12000 + x2 + 20x + 400x

1000x + 20000 = x2 + 1020x + 12000

0 = x2 + 1020x − 1000x + 12000 − 20000

x2 + 20x − 8000 = 0

The speed is calculated by solving the above quadratic equation. Factorizing the quadratic expression we get:

(x + 100)(x − 80) = 0

x = − 100, or x = 80.

Speed is not a negative quantity, so x = 80Km/h.

Still water neither propels nor resists the boat in any direction. However, traveling downstream, the boat is going with the flow of water. It propels the boat with it. Its speed gets added to the speed of the boat.

Let the speed of water be x Km/h.

Therefore downstream boat speed = 5 + x Km/h.

Downstream Distance = 18 Km

Downstream Speed = (5 + x) Km/h

Downstream Time =

hours

While traveling upstream, the boat is going against the flow of water. So water carries it in the opposite direction. The reduction in its speed is equal to the speed of the water. Therefore the speed of the boat upstream is 5 − x Km/h.

Upstream Distance = 16 Km

Upstream Speed = (5 − x) Km/h

Upstream Time =

hours

Total time = tdown + tup = 7 Hours

90 − 18x + 80 + 16x = 7(5 + x)(5 − x)

−2x + 170 = 7(25 − x2)

−2x + 170 = 175 − 7x2

7x2 − 2x − 5 = 0

(7x + 5)(x − 1) = 0

(7x + 5) = 0 Or (x − 1) = 0 Therefore Or x = 1

As the speed of water cannot be negative, x = 1 is the only acceptable solution.

Speed of Water = 1 Km/h

Let’s say the Maya starts from A and Sofia from B

When the drone meets Sofia at point C, Maya reaches point D

Drone flies back to meet the Maya at point F

Distance flown by drone = AC + CF

Speed of Drone = Km/hr

Speed of Maya/Sofia = Km/hr

Let’s say drone and Sofia travel toward each other for time to meet at point C.

Distance traveled by drone = AC =

At the same time, Sofia reaches point C.

Distance traveled by Sofia = CB =

Total distance traveled by both = AC + CB = AB = 12 Km

Therefore AC = Km

By the time drone reaches point C, the Maya reaches point D.

The distance remaining between drone and Maya:

Km

Both meet at point F, after time .

Equation 1

Distance traveled by the drone in time :

CF = FC = Equation 2

Distance traveled by Maya in time :

DF = Equation 3

Putting values of FC and DF (from equation 2 and 3) in Equation 1:

Using equation 2, CF = Km

Let us finally calculate the total distance traveled by the drone:

= AC + CF = 9 + 4.5 = 13.5 Km

Direction Of Boat:

If the boat travels on the straight line AB, by the time it will reach the opposite side, the river will carry it forward to some point D. Therefore, it has to travel on the line AC, to reach point B, directly opposite to point A.

Let the time taken by the boat to complete the trip = t

Distance Calculation:

The distance river carries the boat during this time = CB = SpeedOfRiver × t

CB = 12t

The distance traveled by boat = AC = SpeedOfBoat × TimeTaken

AC = 37t

As unit of speed is m/s we convert the width of river (= AB) to meters.

AB = 1.4 Km = 1400 m

From the Pythagoras theorem we have:

AC2 = CB2 + AB2 = (12t)2 + (1400)2

(37t)2 = (12t)2 + (1400)2

1359t2 − 144t2 = (1400)2

1225t2 = (1400)2

seconds

The students familiar with vectors can try to solve the problem using the velocities of boat and river.

The Speed Formula (with Practice Questions) ➤

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