Two trains cross each other (Opposite or Same direction). Do their Lengths always add up?

Answer: Yes, Always. Distance covered = $L_1 + L_2$ in BOTH cases. Only Relative Speed changes ($S_1+S_2$ or $S_1-S_2$).

Two trains cross each other. Train A is stationary. Train B moves. Relative Speed?

Answer: Speed of B. Since A is 0, Relative Speed = $S_B - 0$ (or $+0$) = $S_B$. Logic check.

In a 100m race, A beats B by 10m. B beats C by 10m. Does A beat C by 20m?

Answer: No, 19m. A:B = 100:90 = 10:9. B:C = 10:9. A:B:C = 100 : 90 : 81. Gap A-C = $100-81 = 19$m.

Why is Average Speed 2xyx+y always less than x+y2?

Answer: Harmonic Mean <= Arithmetic Mean. Mathematical property. You spend MORE time at the SLOWER speed, pulling the average down.

A boat moves perpendicular to river flow. Does river speed affect crossing time?

Answer: No (Independent). Crossing time depends ONLY on velocity component perpendicular to bank. River flows parallel, so no effect on perpendicular time. It only causes drift.

A man covers distance at 34 of usual speed and reaches 20 min late. If he had covered it at 45 of usual speed, how late would he be (compared to usual)?

Answer: 15 min. Step 1: Usual Time. Speed $4 \to 3$, Time $3 \to 4$. Gap 1u = 20m. Usual Time (3u) = 60m. Step 2: New Scenario. Speed $5 \to 4$, Time $4 \to 5$. Usual Time (4u) = 60m $\implies$ 1u = 15m. Late gap (5 - 4) = 1u = 15 min.

A man covers distance at certain speed. If he had moved 3 kmph faster, he would have taken 40 min less. If he had moved 2 kmph slower, he would have taken 40 min more. Distance?

Answer: 40 km. Speed Logic: $V = 2 S_1 S_2S_2 - S_1$ for equal time gaps. $V = 2 \times 3 \times 23-2 = 12$ kmph. Case 1: $12 \to 15$ (Faster). Time gap 40m = 23 hr. $D = 12 \times 153 \times 23 = 40$ km.

Excluding stoppages, bus speed 54 kmph. Including stoppages, 45 kmph. Stop time per hour?

Answer: 10 min. Loss in speed = 9 km in 1 hr. Time to cover 9km at original speed (54)? 954 = 16 hr = 10 min.

Two guns fired at interval of 28 mins. Person in train hears 2nd shot 26 mins after 1st. Speed of Sound 325 m/s. Train Speed?

Answer: 90 kmph. Distance traveled by Sound in (28-26)=2 mins equals distance by Train in 26 mins. $S_{Sound} \times 2 = S_{Train} \times 26$. $325 \times 2 = S_T \times 26$. $S_T = 32513 = 25$ m/s. $25 \times 185 = 90$ kmph.

Train crosses pole in 15s, 100m platform in 25s. Length of Train?

Answer: 150 m. Gap 10s is for extra 100m platform. Speed = 10010 = 10 m/s. Length = Speed $\times$ PoleTime = $10 \times 15 = 150$ m.

Theft at 10am. Police chase at 1pm. Thief 42kmph, Police 49kmph. Catch time?

Answer: 7 am. Thief 3h lead. Dist = $42 \times 3 = 126$ km. Rel Speed = $49-42 = 7$ kmph. Time = 1267 = 18 hrs from 1pm. 1pm + 12h = 1am. + 6h = 7am next day.

Two trains same length cross pole in 20s and 30s. Cross each other (Opposite)?

Answer: 24 sec. Formula: 2 T_1 T_2T_1 + T_2. 2(20)(30)50 = 120050 = 24s.

Boat goes 30km upstream and 44km down in 10 hrs. Also 40km up and 55km down in 13 hrs. Speed of Boat?

Answer: 8 kmph. Common factors. Down: 44, 55 $\to$ likely factor 11. Up: 30, 40 $\to$ likely factor 5 or 10. Check: If D=11, U=5. Case 1: 305 + 4411 = 6 + 4 = 10. Matches! $B = 11+52 = 8$.

A and B run 5km race on 400m track. Speeds 5:4. How many times A passes B?

Answer: 2 times. Passes = Floor(Lead / Track Length). When A runs 5000m, B runs 4000m. Lead = 1000m. Number of times = 1000400 = 2.5 $\to$ 2 complete passes.

A man covers 13rd of distance at 20 kmph, 14th at 30 kmph and the remaining at 50 kmph. Avg Speed for whole journey?

Answer: 30 kmph. LCM of (3, 4) and Speeds (20, 30, 50) $\to$ Assume Total Dist = 600 km. 1. 200 km @ 20 kmph = 10 hrs. 2. 150 km @ 30 kmph = 5 hrs. 3. 250 km @ 50 kmph = 5 hrs. Total Time = 20 hrs. Avg Speed = 60020 = 30 kmph.

Fog Logic: Man (4kmph) sees carriage for 3 min. Visible dist 100m. Carriage speed?

Answer: 8 kmph. Relative Displacement = $100m + 100m = 200m$ (from exactly behind visible to strictly ahead visible) OR just 200m total stretch. Rel Speed = $x-4$. Time = 360 h. $(x-4) \times 120 = 0.2 x-4 = 4 \implies x=8$.

Accident Method: Train meets accident, speed becomes 45, 45 m late. If accident 50km further, 30 m late.

Answer: 50 kmph. Gap 15m. Stretch 50km. Speed 5:4. Time 4:5. Gap 1u = 15m. Normal Time (4u) = 60m = 1 hr. Speed = 50km1h = 50 kmph.

Monkey climbs 6m, slips 3m. 60m pole.

Answer: 37 min. Positive cycle ends work. Target = 60 - 6 = 54m. Net jump = 3m/min. 543 = 18 cycles. (36 min). Next jump is +6 (1 min). Total 37 min.

In 1000m race, A beats B by 100m. In 400m race, B beats C by 40m. In 500m race, A beats C by?

Answer: 95 m. A:B = 1000:900 = 10:9. B:C = 400:360 = 10:9. Combine A:B:C = 100:90:81. For 500m race (A runs 500): Scale by 5. A=500, C=$81 \times 5 = 405$. Margin = $500-405 = 95$ m.

Two trains 100m, 120m len. Same direction cross in 72s. Opposite in 12s. Speed of faster train?

Answer: 10.7 m/s (Approx). Total length 220m. Same ($S_1-S_2$) = 22072. Opposite ($S_1+S_2$) = 22012. $S_1 = 110(772) \approx 10.7$ m/s.

A goes A to B at 40 kmph, returns B to A. Avg Speed 48. Return Speed?

Answer: 60 kmph. Formula: $48 = 2(40)y40+y$. $48(40+y) = 80y$. $3(40+y) = 5y$. $120 + 3y = 5y \implies 2y=120 \implies y=60$.

Train crosses man running 6kmph (Same Dir) in 10s. Runs 6kmph (Opposite) in 5s. Length?

Answer: 33.3 m. Same: $10(S-6)$. Opp: $5(S+6)$. $2(S-6) = S+6 \implies S=18$. Length = $5(24) \times 518 = 33.3$m.

Walk one way, Drive other = 6 hrs. Drive both = 4 hrs. Walk both?

Answer: 8 hrs. $W+D=6$. $2D=4 \to D=2$. $2W = 2(Total - DriveOne) = 2(6 - 2) = 8$.

Car covers four 3km stretches at 10, 20, 30, 60 kmph. Avg Speed?

Answer: 20 kmph. Total Dist 12. Terms constant. Let Dist per stretch = 60. Time: 6, 3, 2, 1. Total 12. Avg = 4 \times 6012 = 20 kmph.

In 100m race, A beats B by 20m and B beats C by 5m. In the same race, by what distance does A beat C?

Answer: 24 m. A:B = 100:80. B:C = 100:95. Combine: A:B:C = $100 : 80 : 80 \times 95100 = 100 : 80 : 76$. Gap A-C = 100 - 76 = 24m.

Train A leaves 5am, reaches 9am. Train B leaves 7am, reaches 10:30am. Meet when?

Answer: 7:56 am. T1 (4h), T2 (3.5h). Dist 28km. S1=7, S2=8. At 7am: T1 done 14km. Rem 14km. Rel Speed 15. Time 1415 hrs = 1415 \times 60 = 56 min. 7:00 + 56m = 7:56 am.

Train travels 50km, accident, speed 34, late 35 min. If accident 24km further, late 25 min. Speed?

Answer: 48 kmph. Gap 10 min for 24 km. Ratio 4:3. Time 3:4. Gap 1u = 10m. Normal Time 30m = 0.5h. Speed = 240.5 = 48 kmph.

Man walks up moving escalator in 30s. Walks down same in 90s. Time if escalator stationary?

Answer: 45 s. Use Harmonic Mean Logic. $M+E = D/30$. $M-E = D/90$. $2M = D(130 + 190) = D(490)$. $M = D45$. Time = 45s.

A, B circular track (12km). Speeds 3, 7, 13 kmph. Meet at start after?

Answer: 12 hrs. Times: 4h, 127 h, 1213 h. LCM Numerators(4,12,12)=12. HCF Denom(1,7,13)=1. LCM = 121 = 12 hrs.

Car A chase Car B. Speeds 40, 50. Car B starts 1h early. Dog runs back and forth at 60. Dog Dist?

Answer: 240 km. Gap 40km. Rel Speed 10. Catch Time = 4010 = 4 hrs. Dog runs for 4 hrs at 60. Dist = $60 \times 4 = 240$ km.

Why is Average Speed 2xyx+y always less than x+y2?

Harmonic Mean <= Arithmetic Mean

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Time, Speed & Distance

Q: A boat moves perpendicular to river flow. Does river speed affect crossing time?

Answer: No (Independent). Crossing time depends ONLY on velocity component perpendicular to bank. River flows parallel, so no effect on perpendicular time. It only causes drift.

Q: A man covers distance at 34 of usual speed and reaches 20 min late. If he had covered it at 45 of usual speed, how late would he be (compared to usual)?

Answer: 15 min. Step 1: Usual Time. Speed $4 \to 3$, Time $3 \to 4$. Gap 1u = 20m. Usual Time (3u) = 60m. Step 2: New Scenario. Speed $5 \to 4$, Time $4 \to 5$. Usual Time (4u) = 60m $\implies$ 1u = 15m. Late gap (5 - 4) = 1u = 15 min.

Q: A man covers distance at certain speed. If he had moved 3 kmph faster, he would have taken 40 min less. If he had moved 2 kmph slower, he would have taken 40 min more. Distance?

Answer: 40 km. Speed Logic: $V = 2 S_1 S_2S_2 - S_1$ for equal time gaps. $V = 2 \times 3 \times 23-2 = 12$ kmph. Case 1: $12 \to 15$ (Faster). Time gap 40m = 23 hr. $D = 12 \times 153 \times 23 = 40$ km.

Q: Excluding stoppages, bus speed 54 kmph. Including stoppages, 45 kmph. Stop time per hour?

Answer: 10 min. Loss in speed = 9 km in 1 hr. Time to cover 9km at original speed (54)? 954 = 16 hr = 10 min.

Q: Two guns fired at interval of 28 mins. Person in train hears 2nd shot 26 mins after 1st. Speed of Sound 325 m/s. Train Speed?

Answer: 90 kmph. Distance traveled by Sound in (28-26)=2 mins equals distance by Train in 26 mins. $S_{Sound} \times 2 = S_{Train} \times 26$. $325 \times 2 = S_T \times 26$. $S_T = 32513 = 25$ m/s. $25 \times 185 = 90$ kmph.

Q: Train crosses pole in 15s, 100m platform in 25s. Length of Train?

Answer: 150 m. Gap 10s is for extra 100m platform. Speed = 10010 = 10 m/s. Length = Speed $\times$ PoleTime = $10 \times 15 = 150$ m.

The heaviest topic in arithmetic. Mastering Relative Speed and Ratio Methods is key to solving Trains and Boats problems without long equations.

Expert Answer & Key Takeaways

The heaviest topic in arithmetic. Mastering Relative Speed and Ratio Methods is key to solving Trains and Boats problems without long equations.

1. The Ratio Method (Constant Distance)

When Distance is same.

Rule: $S_1 : S_2 = a : b \implies T_1 : T_2 = b : a$ .
Use for 'Late/Early' problems to find exact time difference units.

Example:

Q: Speed 34th, reaches 20 min late.

Solution: Speed 4:3. Time 3:4. Gap 1u = 20m.
Actual Time (3u) = 60 min.

2. Average Speed

Not

(S_1+S_2)/2

Case A (Equal Distances): Harmonic Mean 2xyx+y.
Case B (Equal Time): Arithmetic Mean x+y2.
General: Total Distance / Total Time.

Example:

Q: Goes 60 kmph, returns 40 kmph.

Solution: 2 \times 60 \times 40100 = 48 kmph.

3. Relative Speed

Moving bodies.

Opposite Direction: Add Speeds ( $S_1 + S_2$ ).
Same Direction: Subtract Speeds ( $S_1 - S_2$ ).
Used for Meeting Times and Overtaking.

Example:

Q: A (40) and B (50) move towards each other. Gap 180km.

Solution: Rel Speed = 90. Time = 18090 = 2 hrs.

4. Trains (Length Logic)

Distance is never zero.

Crossing Pole: Dist = Train Length ( $L_T$ ).
Crossing Platform/Train: Dist = $L_T + L_P$ .
Speed is Relative Speed if object moves.

Example:

Q: Train (100m) crosses Bridge (200m) in 20s.

Solution: Total Dist 300m. Speed = 30020 = 15 m/s = 54 kmph.

5. Boats & Streams

River flow adds or subtracts.

Downstream (Along): $B + S$ .
Upstream (Against): $B - S$ .
Formulas: $B = (D+U)/2$ , $S = (D-U)/2$ .

Example:

Q: Down 20 kmph, Up 10 kmph.

Solution: Boat Speed = 20+102 = 15. Stream = 5.

6. Linear Races & Head Starts

Giving a 'start'.

Start of distance: A beats B by $x$ meters.
Start of time: A beats B by $t$ seconds.
Dead Heat: Both reach at same time.

Example:

Q: A beats B by 10m in 100m race.

Solution: Ratio Dist A:B = 100:90 = 10:9. Speed Ratio is same.

7. Circular Motion

Meeting on track.

First Meeting: $L / S_{rel}$ .
Meeting at Start: LCM( $T_1, T_2$ ).
Distinct Points: $S_1/S_2$ reduced ratio $(a/b)$ . Points = $a+b$ (Opp) or $|a-b|$ (Same).

Example:

Q: Speeds 3:2, Circular Track. How many distinct meeting points?

Solution: Opposite:

3+2=5

. Same Dir:

3-2=1

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Time, Speed & Distance

Expert Answer & Key Takeaways

1. The Ratio Method (Constant Distance)

2. Average Speed

3. Relative Speed

4. Trains (Length Logic)

5. Boats & Streams

6. Linear Races & Head Starts

7. Circular Motion

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