Robert plans to reach point A at 2 PM. If he rides at 10 km/h he reaches at 2 PM; if he rides at 15 km/h he reaches at 12 noon. At what constant speed must he ride to reach at 1 PM?

Problem restatement
For the same distance, 10 km/h arrival is 2 PM, 15 km/h arrival is 12 noon (2 hours earlier). Find the speed to arrive 1 hour earlier than 2 PM, i.e., at 1 PM.

Given data

• Distance = D km (constant).
• t₁₀ = time at 10 km/h; t₁₅ = time at 15 km/h.
• t₁₀ − t₁₅ = 2 h.

Concept/Approach
Use D = v × t and relate the two times. Then compute D and the time needed for 1 PM, finally the required speed.

Step-by-step calculation
D = 10t₁₀ = 15t₁₅ ⇒ t₁₀ = 1.5t₁₅t₁₀ − t₁₅ = 2 ⇒ 1.5t₁₅ − t₁₅ = 2 ⇒ 0.5t₁₅ = 2 ⇒ t₁₅ = 4 ht₁₀ = 6 h, so D = 10 × 6 = 60 kmTo reach at 1 PM (1 hour earlier than 2 PM): time required = 6 − 1 = 5 hSpeed needed = D ÷ time = 60 ÷ 5 = 12 km/h

Verification/Alternative
Check endpoints: at 10 km/h → 6 h; at 15 km/h → 4 h (2 hours saved). Our target time 5 h is between them, giving 12 km/h.

Common pitfalls
Do not mix calendar clock times with durations; compute durations first.

Final Answer
12 km/h