Deconstructing the QuestionLet \(D\) be the distance to work.
Let \(T\) be the exact time needed to arrive on time.
Scenario 1: Speed = 40 mph. Arrives 3 mins late (\(T + 3\) min).
Scenario 2: Speed = 60 mph. Arrives 3 mins early (\(T - 3\) min).
The difference in travel time between the two speeds is:
\(\Delta t = (T + 3) - (T - 3) = 6\) minutes = \(\frac{1}{10}\) hours.
Step 1: Set up the Distance EquationTime = Distance / Speed.
\(t_{slow} - t_{fast} = \frac{1}{10}\)
\(\frac{D}{40} - \frac{D}{60} = \frac{1}{10}\)
Multiply by 120 (LCM of 40 and 60) to clear denominators:
\(3D - 2D = 12\)
\(D = 12\) miles.
Step 2: Find the Correct TimeUsing the 40 mph scenario:
Time taken = \(\frac{12}{40} = \frac{3}{10}\) hours = 18 minutes.
Since he is 3 minutes late, the required time is:
\(T_{target} = 18 - 3 = 15\) minutes = \(\frac{1}{4}\) hours.
Step 3: Calculate Required Speed\(Speed = \frac{Distance}{Time}\)
\(Speed = \frac{12}{1/4} = 48\) mph.
Answer: B