Average Speed — Total by Total

🎯 Learning Objective

Compute average speed as total distance over total time and apply 2xy/(x+y) for equal distances.

💡 Concept

• Average speed = TOTAL distance ÷ TOTAL time — never the average of speeds
• Equal DISTANCES at x and y → average = 2xy/(x+y) (harmonic mean)
• Equal TIMES at x and y → average = (x+y)/2
• 2xy/(x+y) always lands closer to the SLOWER speed

🧮 Key Formulas

Avg speed = Total D ÷ Total T

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Equal distances: 2xy/(x+y)

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Equal times: (x+y)/2

✏️ Easy Example

Q. A man drives to office at 30 km/h and returns by the same route at 20 km/h. Find his average speed.

1. Equal distances → 2xy/(x+y)
2. = (2 × 30 × 20)/(30 + 20)
3. = 1200/50

Answer: 24 km/h

🇮🇳 Real-Life Example

The Mumbai local feels fast, but your door-to-door average counts the auto ride, the ticket line and the platform wait — total by total is why the 'fast train' commute still averages barely 35 km/h.

📝 Exam-Level Example

Q. A car covers 120 km at 60 km/h and the next 180 km at 45 km/h. Find its average speed.

1. T₁ = 120/60 = 2 h; T₂ = 180/45 = 4 h
2. Total = 300 km in 6 h
3. 300 ÷ 6 = 50

Answer: 50 km/h

📝 Exam-Level Example

Q. A bike covers half a distance at 40 km/h and the other half at 60 km/h. Find the average speed.

1. 2xy/(x+y) = (2 × 40 × 60)/100
2. = 4800/100

Answer: 48 km/h

🪄 Memory Trick

(x+y)/2 is a planted TRAP option every single time. Equal distances → it is 2xy/(x+y), full stop.

⚠️ Common Mistakes

• ❌ Averaging the speeds directly (30 and 20 → 25 ✗, correct is 24)
• ❌ Using 2xy/(x+y) when TIMES are equal instead of distances

🏆 Exam Tips

• ✅ When distances differ, compute each leg's time separately
• ✅ Answer must sit between the two speeds, nearer the slower one

📌 Summary

• Average = total D ÷ total T
• Equal distances → 2xy/(x+y)
• Equal times → (x+y)/2
• Result leans towards the slower speed