A train at 1/3 of its original speed reaches its destination by 4 hours late.Find the original timing

Questions : A train at 1/3 of its original speed reaches its destination by 4 hours late.Find the original timing.

Answer:  

 Let the original speed of the train be S and the original time taken to reach the destination be T hours.

When the train runs at 1/3 of its original speed, the time taken becomes 3T (since time and speed are inversely proportional).

According to the problem, the train is 4 hours late, meaning:

$$3T = T + 4$$

Solving for T:

$$3T - T = 4$$ $$2T = 4$$ $$T = 2 \text{ hours}$$

Thus, the original timing (time taken without delay) is 2 hours.

Alternative : 

Here's how to solve this problem:

Understanding the Relationship

• Speed and Time: When speed decreases, time increases, and vice versa. They are inversely proportional.
• The Key: The distance remains the same whether the train travels at its original speed or the reduced speed.

Let's use variables:

• Let 's' be the original speed of the train.
• Let 't' be the original time taken to reach the destination.
• Let 'd' be the distance to the destination.

Formulating Equations

• Original Journey: d = s * t
• Delayed Journey: The train travels at (1/3)s speed and takes t + 4 hours. So, d = (1/3)s * (t + 4)

Solving for 't'

Since the distance is the same in both cases, we can equate the two equations:

s * t = (1/3)s * (t + 4)

We can cancel 's' from both sides:

t = (1/3)(t + 4)

Now, solve for 't':

3t = t + 4

2t = 4

t = 2

Answer:

The original timing of the journey was 2 hours.

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