Speed, Time & Distance — Advanced Exam Problems

🎯 Learning Objective

Solve the toughest RRB speed patterns — fraction-of-speed lateness, late-early pairs, post-meeting speeds and stoppage time.

💡 Concept

• Usual-speed pattern: at a/b of speed, time stretches to b/a of usual — the late minutes equal the stretch
• Late-early double condition: one distance, two speeds — D/s₁ − D/s₂ = (late + early) time gap
• Two starters meeting: after the meeting, S₁/S₂ = √(t₂/t₁) of the leftover times
• Bus-stoppage classic: minutes halted per hour = (speed lost/running speed) × 60
• Every hard STD question is one distance equation — name D, write both times, subtract

🧮 Key Formulas

At (a/b) of usual speed, late by t → usual time = t × a/(b − a)

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Late-early pair: D/s₁ − D/s₂ = total time gap

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After meeting: S₁/S₂ = √(t₂/t₁)

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Stoppage: minutes lost per hour = (speed drop/running speed) × 60

✏️ Easy Example

Q. Walking at 3/4 of his usual speed, a man reaches office 20 minutes late. Find his usual time.

1. Speed becomes 3/4, so time becomes 4/3 of usual — same distance makes speed and time inverse
2. Extra time = 4/3 − 1 = 1/3 of the usual time
3. 1/3 of usual = 20 minutes → usual = 20 × 3 = 60 minutes

Answer: 60 minutes (1 hour)

🇮🇳 Real-Life Example

Every 'should I leave 10 minutes earlier or ride faster?' debate before office is exactly this speed-time trade-off.

📝 Exam-Level Example

Q. A student cycling at 10 km/h reaches school 8 minutes late, but at 15 km/h he reaches 4 minutes early. Find the distance to school.

1. The two trips differ by 8 + 4 = 12 minutes = 1/5 hour — late and early ADD because they sit on opposite sides of the bell
2. Time at 10 km/h minus time at 15 km/h = D/10 − D/15
3. D/10 − D/15 = D(3 − 2)/30 = D/30 — take LCM 30 to subtract the fractions
4. Set the gap equal: D/30 = 1/5
5. D = 30/5 = 6; check: 36 min vs 24 min, gap 12 ✓

Answer: 6 km

📝 Exam-Level Example

Q. Two trains start at the same time from stations A and B towards each other. After meeting, they take 9 hours and 4 hours respectively to finish their journeys. If the first train runs at 40 km/h, find the speed of the second train.

1. Post-meeting rule: S₁/S₂ = √(t₂/t₁) — the faster train has LESS distance left, so the leftover times flip as a square
2. Here t₁ = 9 h (first train after meeting) and t₂ = 4 h (second train)
3. S₁/S₂ = √(4/9) = 2/3 — the first train is the slower one
4. 40/S₂ = 2/3 → S₂ = 40 × 3/2
5. S₂ = 60 km/h

Answer: 60 km/h

📝 Exam-Level Example

Q. Excluding stoppages, the speed of a bus is 54 km/h; including stoppages it is 45 km/h. For how many minutes does the bus stop per hour?

1. Due to stoppages the bus effectively covers 45 km instead of 54 km each hour — a loss of 9 km
2. Those missing 9 km would have been covered at the RUNNING speed of 54 km/h — that is the time eaten by halts
3. Time lost = distance lost ÷ running speed = 9/54 hour = 1/6 hour
4. 1/6 × 60 = 10 minutes of stoppage per hour

Answer: 10 minutes per hour

🪄 Memory Trick

Usual-time one-liner: at a/b of speed and t minutes late, usual time = t × a/(b − a). At 3/4 speed, 20 late → 20 × 3/1 = 60.

⚠️ Common Mistakes

• ❌ Subtracting late and early times instead of adding them (they sit on opposite sides of on-time)
• ❌ Using √(t₁/t₂) instead of √(t₂/t₁) — the ratio flips; slower train keeps the bigger leftover time
• ❌ Dividing the lost distance by the WITH-stoppage speed instead of the running speed

🏆 Exam Tips

• ✅ Same distance → time ratio is the inverse of the speed ratio; write both ratios side by side
• ✅ Convert every time gap to hours before mixing it with km/h
• ✅ Before meeting, distances covered are in the ratio of speeds — a free extra equation

📌 Summary

• Fraction-of-speed → inverse fraction of time; the gap is the clue
• Late + early both add into the time-gap equation
• After-meeting times: speeds vary as √ of the flipped times
• Stoppage per hour = drop/running × 60