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Question

The speeds of three cars are in the ratio 2 : 3 : 4. What is the ratio between the time taken by these cars to travel the same distance?

This question was previously asked in
CDS I 2016 English Previous Year Paper (14-Feb-2016)
The correct answer is

6 : 4 : 3

Understanding Speed, Time, and Distance Ratios

The problem provides the ratio of speeds of three cars and asks for the ratio of the time taken by these cars to cover the same distance. To solve this, we need to understand the relationship between speed, time, and distance.

The fundamental relationship is: Distance = Speed × Time.

When the distance is constant, Speed and Time are inversely proportional to each other. This means if speed increases, the time taken to cover the same distance decreases, and vice versa.

Calculating Time Ratio from Speed Ratio

Given the ratio of speeds of the three cars is 2 : 3 : 4.

Let the speeds of the three cars be \(S_1\), \(S_2\), and \(S_3\). So, \(S_1 : S_2 : S_3 = 2 : 3 : 4\).

Let the time taken by the three cars to travel the same distance be \(T_1\), \(T_2\), and \(T_3\).

Since the distance is the same for all three cars, and Speed is inversely proportional to Time (for constant distance), the ratio of the times taken will be the inverse ratio of their speeds.

The inverse ratio of speeds \(S_1 : S_2 : S_3\) is \(1/S_1 : 1/S_2 : 1/S_3\).

Substituting the given speed ratio:

Time Ratio \(= 1/2 : 1/3 : 1/4\)

Simplifying the Inverse Ratio

To express this ratio of fractions as a ratio of whole numbers, we need to find a common multiple of the denominators (2, 3, and 4). The least common multiple (LCM) of 2, 3, and 4 is 12.

Multiply each part of the ratio by the LCM, 12:

  • First term: \((1/2) \times 12 = 12/2 = 6\)
  • Second term: \((1/3) \times 12 = 12/3 = 4\)
  • Third term: \((1/4) \times 12 = 12/4 = 3\)

So, the ratio of the time taken by the three cars is 6 : 4 : 3.

Summary of Calculation

The steps to find the time ratio from the speed ratio for the same distance are:

  1. Identify the speed ratio.
  2. Take the inverse of each term in the ratio to get the initial time ratio (as fractions).
  3. Find the LCM of the denominators of the fractions.
  4. Multiply each term in the fractional time ratio by the LCM to get a ratio of whole numbers.

Applying these steps:

Speed Ratio = 2 : 3 : 4

Initial Time Ratio (inverse) = \(1/2 : 1/3 : 1/4\)

LCM of (2, 3, 4) = 12

Final Time Ratio = \((1/2)\times12 : (1/3)\times12 : (1/4)\times12 = 6 : 4 : 3\)

The ratio between the time taken by these cars to travel the same distance is 6 : 4 : 3.

Relationship between Speed, Time, and Distance
Condition Relationship
Distance is constant Speed \(\propto 1/\text{Time}\) (Inversely Proportional)
Speed is constant Distance \(\propto \text{Time}\) (Directly Proportional)
Time is constant Distance \(\propto \text{Speed}\) (Directly Proportional)

Revision Table: Speed and Time Ratio Conversion

Converting Speed Ratio to Time Ratio for Same Distance
Concept Details
Given Speed ratio \(S_1 : S_2 : S_3 = a : b : c\)
Condition Same Distance Covered
Relationship Time \(\propto 1/\text{Speed}\)
Time Ratio (Initial) \(1/a : 1/b : 1/c\)
LCM of Denominators Find LCM of a, b, c
Time Ratio (Simplified) Multiply \(1/a, 1/b, 1/c\) by LCM
Example (2:3:4) Inverse is \(1/2 : 1/3 : 1/4\). LCM is 12.
Ratio is \(6 : 4 : 3\).

Additional Information: Inverse Proportionality Explained

Inverse proportionality is a relationship between two quantities where if one quantity increases, the other quantity decreases proportionally. For example, if you double your speed, the time taken to cover the same distance is halved. If you triple your speed, the time taken is reduced to one-third.

Mathematically, if Quantity A is inversely proportional to Quantity B, we can write this as \(A \propto 1/B\). This means that the product of A and B is constant (\(A \times B = k\), where k is a constant).

In the context of speed and time for a fixed distance (D), Speed (S) and Time (T) are related by \(S \times T = D\). If D is constant, then \(S \times T\) is constant. This confirms that Speed and Time are inversely proportional when the distance is fixed.

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Important Questions from Relative Speed

  1. Two trains start from places A and B. respectively, and travel towards each other at the speeds of 60 km/h and 50 km/h. respectively. By the time they meet, the faster train has travelled 110 km more than the slower train. What is the distance between A and B?

  2. A train can cross a tunnel of length 600 m in 54 seconds, and it can cross a 350 m long bridge in 36 seconds. Which of the following statements is/are correct?

    (i) The speed of the train is 60 km/h.

    (ii) The length of the train is 150 m

  3. Raghu and Raman start together to walk a certain equal distance at a speed of 10 km/h and 8 km/h, respectively. Raghu arrives 30 minutes before Raman arrives. Find the distance between the start and the end point. 

  4. A and B started simultaneously and proceeded towards each other from places X and Y, respectively. After meeting each other at a certain point on the way, A and B took 3.2 hours and 1.8 hours, to reach Y and X, respectively. If the speed of B was 12 km/h, then the speed (in km/h) of A was:

  5. Anil started his journey in the morning. Till 10 a.m., he covered \(\frac{1}{2}\) of his journey, and on the same day till 1 a.m., he covered \(\frac{4}{5}\) of his journey. At what time did he start his journey?

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