Speed, Time & Distance for Quantitative Aptitude

Average Speed

Average Speed Formula

Average speed is defined as the total distance traveled divided by the total time taken for the journey. It is a scalar quantity and provides the overall rate of motion, regardless of direction or variations in speed during the trip.

Average Speed = (Total distance traveled) / (Total time taken)
It is a measure of the total path length covered per unit of total time.

Special Cases for Average Speed

These formulas apply in specific, simplified scenarios where either the distance covered at each speed is the same, or the time spent traveling at each speed is the same.

Case 1 – Constant Distance

When the same distance is covered at two different speeds, the average speed is the harmonic mean of the two speeds. This accounts for the fact that more time is spent at the lower speed.

Average speed = 2xyx+y\frac{2xy}{x+y}x+y2xy
Where x and y are the two speeds at which the same distance has been covered.
This formula is derived because total time = (d/x + d/y), and total distance = 2d.

Case 2 – Constant Time

When traveling for equal durations at two different speeds, the average speed is simply the arithmetic mean of the two speeds.

Average speed = x+y2\frac{x + y}{2}2x+y
Where x and y are the two speeds at which we traveled for the same amount of time.
This formula works because total distance = (xt + yt), and total time = 2t.

Example Problems

Example 1: Constant Distance Trip

A person travels from one place to another at 30 km/hr and returns at 120 km/hr. If the total time taken is 5 hours, find the Distance.

Solution:

The distance for each leg of the trip is constant.
Speed ratio: 30:120 = 1:4.
Since time is inversely proportional to speed (for constant distance), the time ratio is the inverse: 4:1.
Total time = 5 hours. Therefore, time going = 4 hours, time returning = 1 hour.
Distance = Speed * Time = 30 km/hr * 4 hr = 120 km.

Example 2: Late Due to Reduced Speed

Traveling at 3/4th of the original speed, a train is 10 minutes late. Find the usual time taken by the train.

Solution:

Let usual speed = S1 and usual time = T1. Distance is constant.
New speed S2 = (3/4)S1.
For constant distance, time is inversely proportional to speed. Therefore, new time T2 = (4/3)T1.
Given: T2 – T1 = 10 minutes.
So, (4/3)T1 – T1 = 10 => (1/3)T1 = 10 => T1 = 30 minutes.

Inverse Proportionality of Speed & Time

When distance is constant, speed and time are inversely proportional. This means if speed increases, time decreases proportionally, and vice versa, provided the distance covered remains the same.

Speed () is inversely proportional to Time () when Distance () is constant: — Relationship between speed and time for a fixed distance.
If speeds are in the ratio m:nm:nm:n, then the times taken will be in the inverse ratio n:mn:mn:m.
There are two methods to solve related questions.

Inverse Proportionality and Constant Product Rule

Using Inverse Proportionality

Inverse proportionality describes a relationship where one quantity increases as another decreases proportionally. In speed-distance-time problems, time is inversely proportional to speed when distance is constant. This principle allows us to set up equations comparing scenarios with different speeds and times.

Key relationship: When distance is constant, T∝1ST \propto \frac{1}{S}T∝S1 or S∝1TS \propto \frac{1}{T}S∝T1
Constant product: S×T=constantS \times T = \text{constant}S×T=constant for a fixed distance
Application: Used to solve problems where speed changes cause time differences, especially in cases of delays or early arrivals

Using Constant Product Rule

The constant product rule is a direct application of inverse proportionality: if two variables are inversely proportional, their product remains constant. In motion problems, this often means speed × time for a given segment is constant, allowing comparisons between normal and altered conditions.

Rule: For a fixed distance , S1×T1=S2×T2=DS_1 \times T_1 = S_2 \times T_2 = DS1×T1=S2×T2=D
Derivation: Since D=S×TD = S \times TD=S×T, if DDD is unchanged, any change in must be compensated by an inverse change in TTT
Common use: Calculating delays or time differences when speed is reduced or increased over the same distance

Example

This example demonstrates applying inverse proportionality to a train delay problem with two scenarios. The train's speed drops to of its usual speed after an accident, causing different delays depending on where the accident occurs.

Problem setup: Accident causes speed reduction to 34\frac{3}{4}43 of usual speed
Two scenarios given:
1. Accident at 50 km → 45 minutes late
2. Accident at 60 km (10 km further) → 35 minutes late
Goal: Find the usual speed of the train

Solutions

The solution uses inverse proportionality to relate the time differences in the two scenarios. By comparing the additional time taken after the accident point in both cases, we can determine the usual speed.

Let usual speed = SSS km/hr
Reduced speed = 34S\frac{3}{4}S43S km/hr
Time difference after accident point: For the 10 km difference in accident location, the time delay difference is 10 minutes (45 min − 35 min = 10 min)
Equation setup: Time for 10 km at reduced speed − Time for 10 km at usual speed = 10 minutes
Formula: 10(3/4)S−10S=1060\frac{10}{(3/4)S} - \frac{10}{S} = \frac{10}{60}(3/4)S10−S10=6010 — [Time difference in hours for the 10 km segment where accident location differs]
Solve: 403S−10S=16\frac{40}{3S} - \frac{10}{S} = \frac{1}{6}3S40−S10=61 → 40−303S=16\frac{40 - 30}{3S} = \frac{1}{6}3S40−30=61 → 103S=16\frac{10}{3S} = \frac{1}{6}3S10=61 → 3S=603S = 603S=60 → S=20S = 20S=20 km/hr
Final answer: Usual speed = 20 km/hr

Using Inverse Proportionality Method

Case Analysis

There are two scenarios: an accident occurring at 50 km and at 60 km. The only difference between the two cases is the 10 km stretch between 50 km and 60 km. The observed 10-minute time difference between the cases is solely due to how this 10 km is traveled.

Time Difference Derivation

In Case 1, the 10 km is covered at (3/4) of the usual speed. In Case 2, the same 10 km is covered at the usual speed. Let the usual time to cover 10 km be t.

The time taken in Case 1 at reduced speed is 43t\frac{4}{3}t34t (since speed is inversely proportional to time).
The time difference is given as 10 minutes.
Equation: 43t−t=10\frac{4}{3}t - t = 1034t−t=10 minutes
Solving: 13t=10\frac{1}{3}t = 1031t=10 → minutes

Calculating Usual Speed

Distance, d=10d = 10d=10 km
Usual time, t=30t = 30t=30 minutes = 0.50.50.5 hours
Usual Speed = dt=10 km0.5 hr=20\frac{d}{t} = \frac{10 \text{ km}}{0.5 \text{ hr}} = 20td=0.5 hr10 km=20 km/hr

Using Constant Product Rule Method

Inverse Proportionality and Time Calculation

When speed decreases by a fraction, time increases inversely. A 1/4 decrease in speed causes a 1/3 increase in time because speed and time are inversely proportional for a fixed distance.

Rule: For inversely proportional parameters, a 1/x1/x1/x increase in one parameter results in a 1/(x+1)1/(x+1)1/(x+1) decrease in the other.
The delay from this speed decrease is given as 10 minutes.
Therefore, (1/3)T=10(1/3)T = 10(1/3)T=10, solving gives minutes or 0.5 hours.

Distance Calculation

The distance is computed using the original time and the context of the problem.

Distance = 10 km — derived from the scenario's given values or standard speed-time relationship.

Meeting Point Questions

Meeting Point Basics

When two people travel from points A and B towards each other and meet at point P, the total distance covered by both is AB. The time taken by both to meet is the same. Since time is constant, distances AP and BP are proportional to their speeds.

Total distance covered on meeting = AB
Time to meet is the same for both travelers
Distances AP and BP are in the ratio of their speeds
If distance between A and B is d:
- First meeting: combined distance = d
- Second meeting: combined distance = 3d
- Third meeting: combined distance = 5d
- Pattern: successive meetings add 2d to combined distance

Inverse Proportionality of Speed and Time

When distance is constant, time taken is inversely proportional to speed. If speeds are in ratio a:b, times taken are in ratio b:a.

Constant distance → time ∝ 1/speed
Speed ratio 2:3 → time ratio 3:2
Example: Amit (60 kmph) vs Aman (90 kmph) → speed ratio 60:90 = 2:3 → time ratio 3:2. If Amit takes 9 hours, Aman takes 6 hours

Sample Problem Solutions

Q1: Home to Office Timing Problem

A man travels from home to office at 4 km/hr (arrives 20 min late) or 6 km/hr (arrives 10 min early). Find distance.

Solution:

Let distance = d, on-time duration = x minutes
Case 1 (late): time = (x+20) minutes = (x+20)/60 hours
Case 2 (early): time = (x-10) minutes = (x-10)/60 hours
Distance constant → speed ratio inverse of time ratio
Speed ratio = 4:6 = 2:3 → time ratio = 3:2
Equation: (x+20)/(x-10) = 3/2
Solve: 2x+40 = 3x-30 → x = 70 minutes
Using case 1: 4 = d/((70+20)/60) = d/(90/60) → d = 4 × 1.5 = 6 km

Q2: Boat in River Speed Problem

Ram rows in still water at 10 kmph. Upstream takes 2 hours, downstream takes 1.5 hours. Find river speed.

Solution:

Let river speed = y kmph
Upstream speed = 10 - y, downstream speed = 10 + y
Distance constant → speed ratio inverse of time ratio
Time ratio (downstream:upstream) = 1.5:2 = 3:4
Speed ratio (downstream:upstream) = 4:3
Equation: (10+y)/(10-y) = 4/3
Solve: 30+3y = 40-4y → 7y = 10 → y = 10/7 kmph

Q3: Meeting Multiple Times Problem

Ram (2 m/s) and Shyam (1 m/s) start 30 m apart. Find Ram's distance when they meet third time.

Solution:

Room width d = 30 m
Third meeting → combined distance = 5d = 5×30 = 150 m
Speed ratio Ram:Shyam = 2:1 → distance ratio = 2:1 (time constant)
Ram's distance = (2/3) × 150 = 100 m

Q4: Two-Part Journey Speed Problem

Cover 6 km in 84 minutes. Cover first 2/3 at 4 km/hr, find speed for remainder.

Solution:

Total distance = 6 km, total time = 84 minutes = 1.4 hours
First part: (2/3)×6 = 4 km at 4 km/hr
Time for first part = 4/4 = 1 hour = 60 minutes
Remaining distance = 2 km, remaining time = 84-60 = 24 minutes = 0.4 hours
Required speed = 2 km / 0.4 hr = 5 kmph

Q5: Average Speed with Two Speeds

Ramesh travels to office at 30 kmph and returns at 45 kmph. Find average speed.

Solution:

When distance is same for both segments, average speed = 2ab/(a+b)
Here a = 30 kmph, b = 45 kmph
Average speed = (2×30×45)/(30+45) = 2700/75 = 36 kmph

Time Distance Speed