A man runs at three fourths of his usual speed and as a result reaches his destination 15 minutes late. What is his usual travel time in hours for this journey?

Introduction / Context:
This question highlights the relationship between speed and time when the distance remains fixed. Reducing speed increases travel time in a predictable way. You are given the amount of delay caused by running at three fourths of the usual speed and asked to determine the original, or usual, travel time.

Given Data / Assumptions:
The man usually travels a fixed distance in some time T hours.
On a particular day, he runs at three fourths of his usual speed.

Given Data / Assumptions:
Running at this reduced speed makes him 15 minutes late compared to his usual time.
Fifteen minutes is equal to 15 / 60 = 1 / 4 hours.
We must find the usual travel time T in hours.

Concept / Approach:
For a given distance, time is inversely proportional to speed. If the usual speed is v and usual time is T, then distance D = v * T. At reduced speed 3v / 4, the new time becomes D divided by (3v / 4). This new time is given as T plus 1 / 4 hour. Setting up an equation using these relationships allows us to solve for T directly.

Step-by-Step Solution:
Step 1: Let the usual speed be v km/h and the usual time be T hours.Step 2: Distance D = v * T.Step 3: On the slower day, speed = 3v / 4.Step 4: New time taken = D / (3v / 4) hours.Step 5: Simplify D / (3v / 4) = D * (4 / 3v) = (v * T) * 4 / 3v = 4T / 3.Step 6: The new time is 15 minutes, or 1 / 4 hour, more than T, so 4T / 3 = T + 1 / 4.Step 7: Multiply both sides by 12 to clear denominators: 12 * (4T / 3) = 12 * (T + 1 / 4).Step 8: Left side becomes 16T, right side becomes 12T + 3.Step 9: So 16T = 12T + 3.Step 10: Subtract 12T from both sides: 4T = 3.Step 11: Therefore, T = 3 / 4 hours.

Verification / Alternative check:
If usual time T is 3 / 4 hours, then usual speed v satisfies D = v * 3 / 4. At reduced speed 3v / 4, time taken is D divided by (3v / 4) = 4D / 3v. Substituting D = v * 3 / 4 gives 4 * (v * 3 / 4) / 3v = 1 hour. So the new time is 1 hour, which is indeed 15 minutes more than 3 / 4 hour. This confirms that our calculation for T is correct.

Why Other Options Are Wrong:
2/3 hours and 1/3 hours do not yield exactly 15 minutes of delay when you apply the same reduction in speed.
1/4 hour is far too short to be a realistic usual travel time given such a delay.
1 hour as the usual time would produce a different relationship between reduced time and delay, not matching the 15 minutes given in the question.

Common Pitfalls:
Many students misinterpret three fourths of the speed as meaning the time becomes three fourths of the original, when in fact time increases by a factor of four thirds. Another mistake is to convert 15 minutes incorrectly or to mix hours and minutes in the same equation without converting units properly. Working consistently in hours and carefully applying inverse proportionality avoids these issues.

Final Answer:
The man usually takes 3/4 hours to cover the distance.