MPT Q6 - Ontario Mathematics Proficiency Test

A train traveling at a steady 120 kilometers per hour covers a total distance of 390 kilometers after 3 hours and 15 minutes, leaving 96 kilometers still to travel to reach its destination. The key move is converting 15 minutes into a fraction of an hour, then using the distance formula d = v × t to compute how far the train has already gone.

The solution starts with the standard speed relationship: speed equals distance divided by time, so distance equals speed times time. Because the train’s speed is given in kilometers per hour, the elapsed time must also be expressed in hours. Three hours and 15 minutes becomes 3 + 15/60 hours, since 60 minutes make one hour. That fraction simplifies to 0.25, so the total time is 3.25 hours.

With t = 3.25 hours and v = 120 km/h, the distance traveled is d = 120 × 3.25. Breaking 3.25 into 3 + 0.25 makes the arithmetic straightforward: 120 × 3 = 360, and 120 × 0.25 = 30. Adding them gives 390 kilometers. A quick check confirms the same result: 120 × 3.25 equals 390.

The question then asks how much farther the train must go. The destination distance is 486 kilometers, so the remaining distance is 486 − 390. Subtracting yields 96 kilometers. The final answer matches the multiple-choice option identified as choice A.

Beyond the calculation, the approach emphasizes unit consistency and intuition-building. Since the train’s speed stays constant, the distance traveled increases linearly with time. The method suggests using a Desmos visualization to model distance versus time as a straight line with slope 120 (kilometers per hour) and an intercept of 0, then reading off the distance at t = 3.25. That graphical check aligns with the computed 390 kilometers and shows the remaining gap of 96 kilometers to the 486-kilometer destination.

In short: convert minutes to hours, multiply by 120 km/h to get the distance covered (390 km), then subtract from the total required distance (486 km) to find what’s left (96 km).