a train is delayed 2 hours more than the usual time so that the loco pilot increase his speed by 30% to reach its destination which is 780 km find the increase speed of the train

Let's denote the usual speed of the train as \( S \) km/h and the usual time taken to cover the distance of 780 km as \( T \) hours.

Given that:
1. The train is delayed by 2 hours, so the total time taken is \( T + 2 \).
2. The train's speed is increased by 30%, so the new speed becomes \( 1.3S \) km/h.

### Step 1: Calculate the usual time \( T \).
The usual time \( T \) is the time taken to cover 780 km at the usual speed \( S \), which can be expressed as:
\[
T = \frac{780}{S}
\]

### Step 2: Calculate the new time with the increased speed.
With the increased speed, the time taken to cover the same distance is reduced to:
\[
T + 2 = \frac{780}{1.3S}
\]

### Step 3: Set up the equation and solve for \( S \).
We now have the equation:
\[
T + 2 = \frac{780}{1.3S}
\]
Substitute \( T = \frac{780}{S} \) into this equation:
\[
\frac{780}{S} + 2 = \frac{780}{1.3S}
\]
Multiply the entire equation by \( S \) to eliminate the denominators:
\[
780 + 2S = \frac{780S}{1.3}
\]
Multiply both sides by 1.3 to clear the fraction:
\[
1.3(780 + 2S) = 780S
\]
Expand the left side:
\[
1.3 \times 780 + 1.3 \times 2S = 780S
\]
\[
1014 + 2.6S = 780S
\]
Bring all terms involving \( S \) to one side:
\[
1014 = 780S - 2.6S
\]
\[
1014 = 777.4S
\]
Solve for \( S \):
\[
S = \frac{1014}{777.4} \approx 1.304 \text{ km/h}
\]

### Step 4: Calculate the increased speed.
The new speed is \( 1.3S \):
\[
1.3S = 1.3 \times 130.4 \approx 169.52 \text{ km/h}
\]

Thus, the increased speed of the train is approximately **169.52 km/h**.

Do you want details or have any questions?

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Here are 5 related questions to explore:
1. How would the time change if the train's speed increased by 50% instead of 30%?
2. What would be the effect of decreasing the distance to 500 km while keeping the same conditions?
3. How can we calculate the time saved due to the increase in speed?
4. What if the train was delayed by 3 hours instead of 2 hours? How would the result change?
5. How would the calculation change if the train had an initial speed of 100 km/h?

**Tip**: Increasing speed to make up for delays works well, but doubling speed doesn't mean halving the time, due to real-world factors like acceleration limits and fuel consumption.

A train is delayed 2 hours more than the usual time so that the loco pilot increases his speed by 30% to reach its destination which is 780 km. Find the increased speed of the train.

Discover how to calculate the increased speed of a train after a 2-hour delay, where the loco pilot boosts the speed by 30% to cover 780 km. This problem involves algebra, speed-distance-time relationships, and percentage increase, suitable for Grades 9-12.

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