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?
6 : 4 : 3
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.
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 \(= 1/2 : 1/3 : 1/4\)
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:
So, the ratio of the time taken by the three cars is 6 : 4 : 3.
The steps to find the time ratio from the speed ratio for the same distance are:
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.
| 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) |
| 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 \(6 : 4 : 3\). |
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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