Time, Speed & Distance – MBA CUET PG – Notes & Practice Questions

Core Formula

This is the foundation of all TSD problems:

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

From this:

$$\text{Distance} = \text{Speed} \times \text{Time}$$

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

Unit Conversions

km/hr ↔ m/s

$$1 \text{ km/hr} = \frac{5}{18} \text{ m/s}$$

$$1 \text{ m/s} = \frac{18}{5} \text{ km/hr}$$

Shortcut:
km/hr → m/s: multiply by 5/18
m/s → km/hr: multiply by 18/5

Relative Speed

Same Direction:

If speeds are $u$ and $v$:

$$\text{Relative Speed} = |u-v|$$

Opposite Direction:

$$\text{Relative Speed} = u + v$$

Used in:

• Trains crossing

• Boats & streams

• Chasing problems

Distance Between Two Moving Objects

If two objects start at same point or different points:

$$\text{Distance Covered} = \text{Relative Speed} \times \text{Time}$$

If meeting after time $t$:

$$(u+v)t = \text{Initial Gap}$$

If one catches another:

$$(u – v)t = \text{Initial Gap}$$

Average Speed

General Formula:

$$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$$

Special Case: Equal Distances:

If a distance is covered at speed $u$ and same distance at speed $v$:

$$\text{Average Speed} = \frac{2uv}{u+v}$$

Important: Never use simple average for different distances.

Speed Increase/Decrease

If speed increases by $x\%$:
Time decreases by:

$$\frac{x}{100+x}\times 100\%$$

If speed decreases by $x\%$:
Time increases by:

$$\frac{x}{100-x}\times 100\%$$

Time Difference Concept

If two people run the same distance at speeds $u$ and $v$:

$$\text{Time Difference} = d\left(\frac{1}{u} – \frac{1}{v}\right)$$

Used for races.

Train Problems

Length Calculation:

$$\text{Distance Covered} = \text{Length of Train} + \text{Length of Object Crossed}$$

Train Passing Standing Object

$$t = \frac{\text{Length of Train}}{\text{Speed}}$$

Two Trains Crossing (Opposite Direction)

$$t = \frac{\text{Sum of Lengths}}{u+v}$$

Two Trains Overtaking (Same Direction)

$$t = \frac{\text{Sum of Lengths}}{|u-v|}$$

Boat & Stream

Let boat speed in still water = $b$
Stream speed = $s$

Downstream (with current):

$$\text{Speed}_d = b + s$$

Upstream (against current):

$$\text{Speed}_u = b – s$$

Useful Relations:

$$b = \frac{\text{Speed}_u + \text{Speed}_d}{2}$$

$$s = \frac{\text{Speed}_d – \text{Speed}_u}{2}$$

Circular Track Problems

If two people run on a circular track in opposite directions:

$$\text{Meeting Time} = \frac{\text{Track Length}}{u+v}$$

​

Same direction:

$$\text{Meeting Time} = \frac{\text{Track Length}}{|u-v|}$$

​

Ratio Method

If distances same:

Time ∝ 1/Speed
Speed ∝ 1/Time

$$\frac{T_1}{T_2} = \frac{S_2}{S_1}$$​​

If times same:

Distance ∝ Speed

If speeds same:

Time ∝ Distance

LCM Trick

Used when multiple people cover same distance at different speeds → assign distance = LCM of speeds.

Example:
Speeds = 6, 10
LCM = 30 → time = 30/6 = 5 h, 30/10 = 3 h.

Speeds become “efficiencies” like in work-time problems.

Shortcut for Trains with Percent Increase in Speed

If speed of train increases by x%, crossing time becomes:

$$T_{\text{new}} = T_{\text{old}} \times \frac{100}{100+x}$$​

If speed decreases:

$$T_{\text{new}} = T_{\text{old}} \times \frac{100}{100-x}$$​

Chase Problems

If A chases B with gap $d$ and speed difference $u – v$:

$$t = \frac{d}{u – v}$$​

If A gains x meters per second:

$$t = \frac{d}{x}$$​

Speed Conversion in Percentage Terms

If A is x% faster than B:

$$\frac{S_A}{S_B} = 1 + \frac{x}{100}$$

​

Time ratio:

$$\frac{T_A}{T_B} = \frac{1}{1 + \frac{x}{100}}$$

Meeting at Specified Points : Circular Tracks

Distance traveled before meeting = LCM of simplifiable lap distances.

Example:
Two runners on 100 m track
Speeds 6 m/s and 4 m/s
Relative = 10 m/s
Time to meet = 100/10 = 10 s.

Race / Head Start Problems

If B gets head start D and A’s speed is \(S_A\), B = \(S_B\):

Catch-up time:

$$t = \frac{D}{S_A – S_B}$$​

If catching happens before finish → A wins.

Speed of Moving Platform / River

Train relative to platform:

$$\text{Relative speed} = \text{Train speed}$$

Train relative to man running on platform:

• Same direction: $u – v$

• Opposite direction: $u + v$

Boat relative to flow works the same way.

Distance Reduction Trick

If time reduces from $t_1$ to $t_2$ at same distance:

$$\frac{S_2}{S_1} = \frac{t_1}{t_2}$$​​

Useful Fractions for Fast Mental TSD

• 1/6 = 0.166 → often used in train + stream

• 1/5 = 0.2 → clean distance splits

• 3/2 × speed = 150% faster

• 4/3 × speed = 133% faster

Observations for Exams

• Any TSD question involving crossing, meeting, chasing, streams, races, or trains is solved by relative speed.

• Any TSD question involving two different speeds in equal distance uses harmonic mean average speed formula.

• Any TSD question involving percent change in speed is solved using inverse percent change for time.

• If question gives ratios, always use LCM method.

Practice Questions

1. A person walks 30 km at 5 km/h and returns at 6 km/h. Find average speed.
A) 5.4
B) 5.5
C) 5.6
D) 5.7

2. Two trains 180 m and 240 m long move in opposite directions at 54 km/h and 36 km/h. Time to cross each other?
A) 12 s
B) 14 s
C) 15 s
D) 18 s

3. A runner increases speed by 25% and takes 16 min less to cover a distance. Original time = ?
A) 60
B) 64
C) 70
D) 80

4. A man travels 40 km at x km/h and returns at (x + 10) km/h. Average speed = 40. Find x.
A) 20
B) 30
C) 40
D) 50

5. A train 150 m long crosses a platform 450 m long in 30 s. Speed (km/h)?
A) 60
B) 72
C) 84
D) 96