Time, Speed, Distance and concept of proportionality: Questions, Examples

Concept of Proportionality in Time, Speed, and Distance

The concept of proportionality plays a crucial role in solving advanced time, speed, and distance problems in mathematics and quantitative aptitude. Proportionality helps us understand how one variable changes when another variable changes, while the remaining quantity stays constant. In this chapter, we mainly deal with direct variation and inverse (indirect) variation, both of which are frequently tested in competitive exams like CAT, XAT, CUET, SSC, and Banking exams. Let us describe each concept in detail:

Direct Proportionality in Time, Speed, and Distance

This section focuses on direct proportionality in time, speed, and distance, where two quantities increase or decrease together while the third remains constant. It explains how distance varies directly with speed or time, supported by simple examples and concepts frequently tested in quantitative aptitude and competitive exams.

Direct Variation of Distance with Speed (Time Constant)

When time remains constant, the distance travelled is directly proportional to speed. This means that if the speed increases, the distance covered increases in the same ratio.

Distance ∝ Speed (when Time is constant)

Example:
A car covers a distance of 100 km at a speed of 25 km/hr in a fixed time.
If the speed is increased to 50 km/hr (which is double), the distance covered in the same time will also double.

New distance = 100 × 2 = 200 km

This clearly shows that increasing speed while keeping time constant results in more distance covered, a classic case of direct proportionality.

Direct Variation of Distance with Time (Speed Constant)

When speed is constant, the distance travelled is directly proportional to time. In simple terms, the longer you travel at the same speed, the more distance you cover.

Distance ∝ Time (when Speed is constant)

Example:
A car covers 100 km in 2 hours at a constant speed.
If the travel time is increased to 4 hours at the same speed, the distance covered will also double.

New distance = 100 × 2 = 200 km

This type of question is common in basic time–distance problems and helps build intuition for proportional reasoning.

Inverse Proportionality in Time, Speed, and Distance

This section covers the concept of inverse proportionality in time, speed, and distance, where one quantity increases as the other decreases while the distance remains constant. It explains how speed and time are inversely related, along with clear logic and examples commonly asked in aptitude and competitive exam questions.

Indirect (Inverse) Variation of Speed with Time (Distance Constant)

When the distance remains constant, speed is inversely proportional to time. This means that if you want to take less time to cover the same distance, you must travel at a higher speed.

Speed ∝ $\frac{1}{\text{Time}}$

Example:
A journey of a fixed distance takes 3 hours at a speed of 40 km/hr.
If the time is reduced to 2 hours, the required speed will be:

New speed = $\frac{40 × 3}{2}$ = 60 km/hr

This shows that reducing time for the same distance forces an increase in speed, which is a key idea behind inverse proportionality questions.

Understanding these direct and inverse proportionality concepts makes it much easier to solve tricky time, speed, and distance problems, especially those involving ratio-based shortcuts and exam-level applications. Once this logic is clear, many seemingly difficult questions become quick and manageable.

Tips and Tricks

• Understand the basic formulas of time, speed and distance and their interrelation.

• Always convert all units to a consistent set before solving a problem. For instance, if the distance is in kilometres and time is in hours, ensure the speed is in km/hr.

• The formula for average speed when travelling different distances at different speeds is:

Average speed = $\frac{\text{Total distance}}{\text{Total time}}$.

• When two objects are moving in the same direction, the relative speed is the difference in their speeds.

• When two objects are moving in opposite directions, the relative speed is the sum of their speeds.

• Time, speed and distance concept is widely used in the concept of Average speed and Relative speed as well as in application of Time, speed, and distance in problems on trains.

Practice Questions/Solved Examples

Q.1.A man travels $\frac{3}{4}$th of the distance of his journey by bus, $\frac{1}{6}$th by rickshaw, and the remaining 2 km on foot. The total distance travelled by the man is:

• 12 km

• 18 km

• 20 km

• 24 km

Hint: Let the total distance be $x$ km. Find the remaining distance and equate it with 2 km.

Solution:

Given:
A man travelled by bus $\frac{3}{4}$th of the total distance.
The man travelled by rickshaw $\frac{1}{6}$th of the total distance.
The man travelled on foot = 2 km

Let the total distance be $x$ km.

Remaining distance
$= x - \frac{3x}{4} - \frac{x}{6}$

According to the question,

$x - \frac{3x}{4} - \frac{x}{6} = 2$

$⇒ \frac{12x - 9x - 2x}{12} = 2$

$⇒ \frac{x}{12} = 2$

$\therefore x = 24$

Hence, the correct answer is option (4).

Q.2. To cover a certain distance at a speed of 60 km/hr, a train takes 15 hours. If it covers the same distance in 12 hours, what will its speed be?

• 65 km/hr

• 70 km/hr

• 75 km/hr

• 80 km/hr

Hint: Use the formula Distance = Speed × Time.

Solution:

Given:
Speed = 60 km/hr
Time = 15 hours

Distance = Speed × Time
$= 60 × 15 = 900$ km

Required speed to cover this distance in 12 hours
$= \frac{900}{12} = 75$ km/hr

Hence, the correct answer is option (3).

Q.3. Sound travels at 330 metres per second. The distance (in kilometres) of a thundercloud when its sound follows the flash after 10 seconds is:

• 0.33 km

• 3.3 km

• 33 km

• 33.3 km

Hint: Use the formula Distance = Speed × Time.

Solution:

Given:
Speed of sound = 330 m/s

Distance = Speed × Time
$= 330 × 10 = 3300$ metres

$= 3.3$ km

Hence, the correct answer is option (2).

Q.4. At an average speed of 80 km/hr, the Shatabdi Express reaches Ranchi from Kolkata in 7 hours. The distance between Kolkata and Ranchi is:

• 506 km

• 560 km

• 506 m

• 650 m

Hint: Use Distance = Average Speed × Time.

Solution:

Given:
Average speed = 80 km/hr
Time = 7 hours

Distance = $80 × 7 = 560$ km

Hence, the correct answer is option (2).

Q.5. A truck covers a distance of 550 metres in one minute whereas a bus covers a distance of 33 km in $\frac{3}{4}$ hour. Then the ratio of their speeds is:

• $1:2$

• $2:3$

• $3:4$

• $4:5$

Hint: Use Speed = $\frac{\text{Distance}}{\text{Time}}$.

Solution:

Speed of truck
$= \frac{550}{60} = \frac{55}{6}$ m/s

Speed of bus
$= \frac{33 × 1000}{\frac{3}{4} × 3600} = \frac{110}{9}$ m/s

Required ratio
$= \frac{\frac{55}{6}}{\frac{110}{9}}$

$= \frac{55}{6} × \frac{9}{110}$

$= \frac{3}{4}$

$= 3:4$

Hence, the correct answer is option (3).

Q.6. Rameshwar goes on a trip on his motorcycle and rides for 405 kilometres. If he rides for 6 hours at a speed of 45 km/hr, find at what speed he travels for the remaining 3 hours of the journey.

• 55 km/hr

• 52 km/hr

• 45 km/hr

• 40 km/hr

Hint: Use Speed = $\frac{\text{Distance}}{\text{Time}}$.

Solution:

Total distance = 405 km

Distance covered in 6 hours
$= 6 × 45 = 270$ km

Remaining distance
$= 405 - 270 = 135$ km

Speed for remaining 3 hours
$= \frac{135}{3} = 45$ km/hr

Hence, the correct answer is option (3).

Q.7. If a train maintains an average speed of 42 km/hr, it arrives at its destination on time. If the speed is 40 km/hr, it arrives 15 minutes late. Find the distance of the journey.

• 210 km

• 209 km

• 205 km

• 200 km

Hint: Use Distance = Speed × Time.

Solution:

Let time be $x$ hours at 42 km/hr.
At 40 km/hr, time becomes $x + \frac{1}{4}$ hours.

$42x = 40\left(x + \frac{1}{4}\right)$

$⇒ 2x = 10$

$\therefore x = 5$

Distance
$= 42 × 5 = 210$ km

Hence, the correct answer is option (1).

Q.8. The speeds of three cars are in the ratio $1:3:5$. The ratio of the time taken to cover the same distance is:

• $3:5:15$

• $15:3:5$

• $15:5:3$

• $5:3:1$

Hint: Time is inversely proportional to speed.

Solution:

Time ∝ $\frac{1}{\text{Speed}}$

Required ratio
$= 1 : \frac{1}{3} : \frac{1}{5}$

LCM of 1, 3, and 5 = 15

$= 15 : 5 : 3$

Hence, the correct answer is option (3).

Q.9. Anuj drives at 70 km/hr with 3 halts of 10 minutes each, while Anup drives at 80 km/hr with 4 halts of 15 minutes each. If Anuj takes 8.5 hours, find the time taken by Anup.

• 8 hours

• 7 hours

• 4 hours

• 6 hours

Hint: Use Speed = $\frac{\text{Distance}}{\text{Time}}$.

Solution:

Anuj’s halt time
$= 30$ minutes = 0.5 hours

Actual travel time
$= 8.5 - 0.5 = 8$ hours

Distance
$= 70 × 8 = 560$ km

Anup’s halt time
$= 60$ minutes = 1 hour

Travel time
$= \frac{560}{80} = 7$ hours

Total time
$= 7 + 1 = 8$ hours

Hence, the correct answer is option (1).

Q.10. Raghav covers a distance of 3300 m in 45 minutes. What is his speed (in km/hr)?

• 5

• 4.4

• 3.75

• 2.25

Hint: Use Speed = $\frac{\text{Distance}}{\text{Time}}$.

Solution:

Distance
$= \frac{3300}{1000} = 3.3$ km

Time
$= \frac{45}{60} = 0.75$ hours

Speed
$= \frac{3.3}{0.75} = 4.4$ km/hr

Hence, the correct answer is option (2).