Rajeev travels 80 km from his home to college. One day he leaves 1 hour later than usual, so he increases his speed by 4 km/h and still reaches the college at the normal time. What is his increased speed (in km/h)?

Introduction / Context: This problem tests your understanding of the relationship between distance, speed, and time, and how a change in departure time can be compensated for by increasing speed. Rajeev travels a fixed distance every day, and on one particular day he leaves late but still manages to arrive at his college at the usual time by walking faster.

Given Data / Assumptions:

- Distance between home and college = 80 km. - Usual speed of Rajeev = v km/h (unknown). - On the late day, his speed = v + 4 km/h. - He leaves 1 hour later than normal but still reaches at the normal arrival time. - Travel is along a straight route at constant speed.

Concept / Approach: The fundamental formula is distance = speed * time. Since the distance remains the same on both days, we can express the usual and increased travel times in terms of v. Because he departs 1 hour late but arrives at the normal time, the travel time on the late day must be exactly 1 hour less than his usual travel time. This gives a simple equation in v that we can solve to get the usual and then the increased speed.

Step-by-Step Solution: Let the usual speed of Rajeev be v km/h. Usual time to travel 80 km = 80 / v hours. On the late day, his speed is v + 4 km/h. So time on the late day = 80 / (v + 4) hours. He leaves 1 hour late but still arrives at the same time, so: 80 / (v + 4) = 80 / v - 1. Rearrange: 80 / v - 80 / (v + 4) = 1. Multiply through by v (v + 4): 80 (v + 4) - 80 v = v (v + 4). 80 v + 320 - 80 v = v^2 + 4 v. 320 = v^2 + 4 v. Rearrange: v^2 + 4 v - 320 = 0. Solve the quadratic: v = 16 km/h (reject the negative root). Thus the increased speed = v + 4 = 20 km/h.

Verification / Alternative check: At the usual speed of 16 km/h, time taken = 80 / 16 = 5 hours. At the increased speed of 20 km/h, time taken = 80 / 20 = 4 hours. The difference in travel time is 5 - 4 = 1 hour, which matches the given condition about leaving 1 hour late and still arriving on time. So our calculation is consistent.

Why Other Options Are Wrong: 28 km/h: This would not give a travel time 1 hour less than the usual time for a consistent distance of 80 km. 30 km/h and 40 km/h: These speeds would give even shorter travel times, and no usual speed would satisfy the given 1 hour difference with these as v + 4.

Common Pitfalls: A common mistake is to mix up which time is greater or to add 1 hour instead of subtracting it in the equation. Another frequent error is to plug in options directly without forming a proper time equation, which can be confusing when numbers are larger. Always relate changes in departure time to changes in travel time using the basic distance formula.

Final Answer: The increased speed of Rajeev on that day is 20 km/h.