When the speed of a train is increased by 20%, it takes 20 minutes less to cover the same distance. What is the time taken to cover the same distance with the original speed?
120 minutes
Understanding the Train Speed and Time Problem
This problem involves the relationship between speed, time, and distance. The fundamental formula connecting these three quantities is:
\[\text{Distance} = \text{Speed} \times \text{Time}\]
In this specific scenario, the distance covered by the train remains constant, but the speed and time change.
Setting Up the Problem Variables
Let's define the variables to represent the original situation and the new situation:
- Let the original speed of the train be \(S_1\).
- Let the original time taken to cover the distance be \(T_1\) minutes.
- Let the distance covered be \(D\).
From the fundamental formula, we have:
\[D = S_1 \times T_1 \quad \text{(Equation 1)}\]
Analyzing the Changed Scenario
The problem states that the speed of the train is increased by 20%. The new speed, \(S_2\), is:
\[S_2 = S_1 + 20\% \text{ of } S_1\]
\[S_2 = S_1 + 0.20 S_1\]
\[S_2 = 1.20 S_1\]
The problem also states that the time taken to cover the same distance is 20 minutes less than the original time. The new time, \(T_2\), is:
\[T_2 = T_1 - 20 \text{ minutes}\]
Since the distance covered is the same in this new scenario, we can write:
\[D = S_2 \times T_2 \quad \text{(Equation 2)}\]
Solving for the Original Time
Now we have two equations for the distance \(D\). We can set them equal to each other:
\[S_1 \times T_1 = S_2 \times T_2\]
Substitute the expressions for \(S_2\) and \(T_2\) in terms of \(S_1\) and \(T_1\):
\[S_1 \times T_1 = (1.20 S_1) \times (T_1 - 20)\]
Since \(S_1\) represents the original speed, it must be greater than zero. We can divide both sides of the equation by \(S_1\):
\[T_1 = 1.20 \times (T_1 - 20)\]
Now, we solve this equation for \(T_1\):
\[T_1 = 1.20 T_1 - 1.20 \times 20\]
\[T_1 = 1.20 T_1 - 24\]
Subtract \(T_1\) from both sides:
\[0 = 1.20 T_1 - T_1 - 24\]
\[0 = 0.20 T_1 - 24\]
Add 24 to both sides:
\[24 = 0.20 T_1\]
To find \(T_1\), divide 24 by 0.20:
\[T_1 = \frac{24}{0.20}\]
\[T_1 = \frac{24}{\frac{20}{100}}\]
\[T_1 = 24 \times \frac{100}{20}\]
\[T_1 = 24 \times 5\]
\[T_1 = 120\]
So, the original time taken to cover the same distance was 120 minutes.
Verification
Let's check the answer. If the original time is 120 minutes, the new time is \(120 - 20 = 100\) minutes. The new speed is 1.2 times the original speed. For a constant distance, Speed and Time are inversely proportional. The ratio of new time to original time is \(100/120 = 5/6\). The ratio of original speed to new speed should be the inverse, which is \(6/5 = 1.2\). This matches our calculation that the new speed is 1.2 times the original speed (\(1.2 S_1\)). Thus, the original time of 120 minutes is correct.
Final Answer
The time taken to cover the same distance with the original speed is 120 minutes.
Revision Table: Speed, Time, Distance Concepts
| Concept | Formula | Relationship (Constant Distance) |
|---|---|---|
| Distance (D) | \(D = S \times T\) | \(D = \text{Constant}\) |
| Speed (S) | \(S = D / T\) | \(S \propto 1/T\) (Inversely proportional) |
| Time (T) | \(T = D / S\) | \(T \propto 1/S\) (Inversely proportional) |
Additional Information: Inverse Proportion Explained
When the distance is constant, the speed and the time taken to cover that distance are inversely proportional. This means that if you increase the speed, the time taken decreases, and if you decrease the speed, the time taken increases. The product of speed and time remains constant, equaling the distance.
In this problem, the speed increased by 20%, which means the new speed is 120% or 1.2 times the original speed. Since Speed \(\times\) Time = Constant Distance, we have:
\[S_{original} \times T_{original} = S_{new} \times T_{new}\]
\[S_1 \times T_1 = 1.2 S_1 \times (T_1 - 20)\]
As shown in the solution steps, this equation simplifies to \(T_1 = 1.2(T_1 - 20)\), leading to \(T_1 = 120\) minutes.
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Important Questions from Relative Speed
In a circular race of 2500 m, a man and a woman start from a point towards opposite directions with speeds of 37 km/h and 35 km/h, respectively. After how much time from the start of the race will they meet for the first time?
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A train of length 600 meters passes another train of length 1000 meters moving in the opposite direction in 96 seconds. If the speed of both the trains is the same, then what is the speed of each train?
The distance between Delhi and Patna is $480$ km. A train starts from Delhi and travels towards Patna at a speed of $60$ kmph. Another train starts from Patna towards Delhi at a speed of $80$ kmph, but starts $1$ hour after the first train. In how much time, after the first train started, will they meet each other?
Ram traveling at the speed of 3 km/h reaches 15 minutes late. Had he walked at 4 km/h, he would have reached 15 minutes earlier. How much distance does Ram have to cover?
