Q1): A train crosses a pole in 12 seconds and a 250 m platform in 32 seconds. What is the speed of the train?
A) 36 km/h
B) 40 km/h
C) 45 km/h
D) 54 km/h
Answer: C) 45 km/h
Explanation:
- Step 1: Let speed = v m/s, train length = L m
- Step 2: Pole crossing: L = 12v
- Step 3: Platform crossing: L + 250 = 32v
- Step 4: Subtract: (L+250) − L = 32v − 12v → 250 = 20v → v = 12.5 m/s
- Step 5: Convert to km/h: 12.5 × (18/5) = 45 km/h
- Final: Speed is 45 km/h, so Option C
Q2): A journey is 450 km. A car covers 3/5 of the distance at 50 km/h and the remaining at 40 km/h. Find the average speed.
A) 44.44 km/h
B) 45.45 km/h
C) 46.15 km/h
D) 47.50 km/h
Answer: B) 45.45 km/h
Explanation:
- Step 1: Total distance = 450 km
- Step 2: Part-1 distance = (3/5)×450 = 270 km, time₁ = 270/50 = 5.4 h
- Step 3: Part-2 distance = 180 km, time₂ = 180/40 = 4.5 h
- Step 4: Total time = 5.4 + 4.5 = 9.9 h
- Step 5: Average speed = 450 ÷ 9.9 = 45.4545… km/h
- Final: Average speed ≈ 45.45 km/h, so Option B
Q3): A boat covers 48 km downstream and 36 km upstream in 7 hours. If its speed in still water is 12 km/h, what is the stream speed?
A) 1 km/h
B) 12/7 km/h
C) 2 km/h
D) 5/2 km/h
Answer: B) 12/7 km/h
Explanation:
- Step 1: Let stream speed = x km/h
- Step 2: Downstream speed = 12 + x, Upstream speed = 12 − x
- Step 3: Total time: 48/(12+x) + 36/(12−x) = 7
- Step 4: Solve gives x(7x − 12) = 0 → x = 12/7 (non-zero)
- Final: Stream speed = 12/7 km/h, so Option B
Q4): A and B start from the same point in opposite directions. A walks at 4 km/h, B walks at 6 km/h. After 1 hour, B turns back and walks at 7 km/h in A’s direction. After how much time from the start will B meet A?
A) 4 hours
B) 4 hours 20 minutes
C) 4 hours 30 minutes
D) 5 hours
Answer: B) 4 hours 20 minutes
Explanation:
- Step 1: In 1 hour: A covers 4 km, B covers 6 km (opposite side)
- Step 2: Separation after 1 hour = 4 + 6 = 10 km
- Step 3: After turning, both move in A’s direction: relative speed = 7 − 4 = 3 km/h
- Step 4: Time to catch after turning = 10/3 h = 3 h 20 min
- Final: Total time = 1 h + 3 h 20 min = 4 h 20 min, so Option B
Q5): A train crosses a 300 m bridge in 45 seconds. If its speed is reduced by 10 km/h, it crosses the same bridge in 60 seconds. What is the length of the train?
A) 150 m
B) 180 m
C) 200 m
D) 240 m
Answer: C) 200 m
Explanation:
- Step 1: Let total distance (train + bridge) = D meters
- Step 2: Speed₁ = D/45, Speed₂ = D/60 (in m/s)
- Step 3: Speed drop = Speed₁ − Speed₂ = D(1/45 − 1/60) = D/180
- Step 4: Given drop = 10 km/h = 10×(5/18) = 25/9 m/s
- Step 5: D/180 = 25/9 → D = 180×25/9 = 500 m
- Step 6: Train length = D − bridge = 500 − 300 = 200 m
- Final: Train length is 200 m, so Option C
Q6): A 180 km trip is covered in three equal parts. Speeds are 30 km/h, 45 km/h, 60 km/h respectively. There is also a 30-minute stop after the second part. Find the average speed.
A) 34 km/h
B) 36 km/h
C) 38 km/h
D) 40 km/h
Answer: B) 36 km/h
Explanation:
- Step 1: Each part distance = 180/3 = 60 km
- Step 2: Time₁ = 60/30 = 2 h
- Step 3: Time₂ = 60/45 = 4/3 h, Time₃ = 60/60 = 1 h
- Step 4: Stop time = 0.5 h
- Step 5: Total time = 2 + 4/3 + 1 + 0.5 = 5 h
- Final: Average speed = 180 ÷ 5 = 36 km/h, so Option B
Q7): In a 600 m race, A runs at 8 m/s. He gives B a start of 60 m and still wins by 10 seconds. What is B’s speed?
A) 6.00 m/s
B) 6.20 m/s
C) 6.35 m/s
D) 6.50 m/s
Answer: C) 6.35 m/s
Explanation:
- Step 1: A’s time = 600/8 = 75 s
- Step 2: A wins by 10 s → B’s time = 75 + 10 = 85 s
- Step 3: B runs only (600 − 60) = 540 m
- Step 4: B’s speed = 540 ÷ 85 = 6.3529… m/s
- Final: B’s speed ≈ 6.35 m/s, so Option C
Q8): Two cyclists start from A and B (72 km apart) towards each other. Their speeds are in ratio 5:7. After meeting, the faster cyclist reaches A in 45 minutes. What is the speed of the slower cyclist?
A) 25 km/h
B) 28.57 km/h
C) 30 km/h
D) 35 km/h
Answer: B) 28.57 km/h
Explanation:
- Step 1: Speed ratio (slower:faster) = 5:7 → distances before meeting also 5:7
- Step 2: Distance from A to meeting point = (5/12)×72 = 30 km
- Step 3: After meeting, faster travels meeting→A = 30 km in 45 min = 0.75 h
- Step 4: Faster speed = 30/0.75 = 40 km/h
- Step 5: If 7k = 40 → k = 40/7 → slower speed = 5k = 200/7 = 28.57 km/h
- Final: Slower speed = 28.57 km/h, so Option B
Q9): A train passes a stationary man in 12 seconds. If the man runs at 9 km/h in the same direction, the train passes him in 18 seconds. What is the length of the train?
A) 80 m
B) 90 m
C) 100 m
D) 108 m
Answer: B) 90 m
Explanation:
- Step 1: Let train speed = v m/s, length = L m
- Step 2: Stationary man: L = 12v
- Step 3: Runner speed = 9 km/h = 9×(5/18) = 2.5 m/s
- Step 4: With running man: L = (v − 2.5)×18
- Step 5: Equate: 12v = 18v − 45 → 6v = 45 → v = 7.5 m/s
- Step 6: L = 12v = 12×7.5 = 90 m
- Final: Train length is 90 m, so Option B
Q10): A person covers a distance at 60 km/h. If he had travelled 10 km/h faster, he would have taken 30 minutes less. What is the distance?
A) 180 km
B) 200 km
C) 210 km
D) 240 km
Answer: C) 210 km
Explanation:
- Step 1: Let distance = D km
- Step 2: Time at 60 km/h = D/60 hours
- Step 3: Time at 70 km/h = D/70 hours
- Step 4: Given difference = 0.5 h → D/60 − D/70 = 0.5
- Step 5: D(1/60 − 1/70) = D(10/4200) = D/420 = 0.5
- Final: D = 210 km, so Option C
Q11): Two trains start from A and B towards each other and meet. After meeting, the train from A takes 2 hours to reach B, and the train from B takes 8 hours to reach A. How long after starting did they meet?
A) 2 hours
B) 3 hours
C) 4 hours
D) 6 hours
Answer: C) 4 hours
Explanation:
- Step 1: Let meeting time = t hours, speeds = v₁ (from A), v₂ (from B)
- Step 2: After meeting, remaining for train from A = distance covered by other = v₂t
- Step 3: Given: (v₂t)/v₁ = 2 → v₂t = 2v₁
- Step 4: For train from B: (v₁t)/v₂ = 8 → v₁t = 8v₂
- Step 5: Multiply: (v₂t)(v₁t) = (2v₁)(8v₂) → v₁v₂t² = 16v₁v₂ → t² = 16
- Final: t = 4 hours, so Option C
Q12): A train crosses a 180 m platform in 30 seconds and a 300 m platform in 40 seconds. What is the length of the train?
A) 150 m
B) 160 m
C) 180 m
D) 200 m
Answer: C) 180 m
Explanation:
- Step 1: Let speed = v m/s, train length = L m
- Step 2: First platform: L + 180 = 30v
- Step 3: Second platform: L + 300 = 40v
- Step 4: Subtract: 120 = 10v → v = 12 m/s
- Step 5: Put in L + 180 = 30×12 = 360 → L = 180 m
- Final: Train length is 180 m, so Option C
Q13): A boat travels 24 km downstream and returns 24 km upstream in a total of 5 hours. If stream speed is 2 km/h, what is the speed of the boat in still water?
A) 8 km/h
B) 9 km/h
C) 10 km/h
D) 12 km/h
Answer: C) 10 km/h
Explanation:
- Step 1: Let still water speed = v km/h
- Step 2: Downstream speed = v + 2, Upstream speed = v − 2
- Step 3: Total time: 24/(v+2) + 24/(v−2) = 5
- Step 4: Solve: 24[(v−2)+(v+2)]/(v²−4) = 5 → 48v/(v²−4) = 5
- Step 5: 5v² − 48v − 20 = 0 → v = 10 (valid)
- Final: Still water speed is 10 km/h, so Option C
Q14): Two towns are 180 km apart. A starts from P at 45 km/h and B starts from Q at 30 km/h towards P. After 1 hour, B stops for 15 minutes and then continues. When will they meet (from the start time)?
A) 2 hours 15 minutes
B) 2 hours 30 minutes
C) 2 hours 45 minutes
D) 3 hours
Answer: B) 2 hours 30 minutes
Explanation:
- Step 1: In first 1 hour, distance covered together = 45 + 30 = 75 km
- Step 2: Remaining distance = 180 − 75 = 105 km
- Step 3: B stops for 15 min = 0.25 h, A keeps moving
- Step 4: In 0.25 h, A covers 45×0.25 = 11.25 km → remaining = 105 − 11.25 = 93.75 km
- Step 5: Now both move again, relative speed = 45 + 30 = 75 km/h
- Step 6: Time = 93.75/75 = 1.25 h = 1 h 15 min
- Final: Total time = 1 h + 15 min + 1 h 15 min = 2 h 30 min, so Option B
Q15): A person travels 20 km uphill at 15 km/h, rests for 20 minutes, then travels 30 km downhill at 30 km/h. What is the average speed for the entire trip?
A) 18.00 km/h
B) 18.75 km/h
C) 19.50 km/h
D) 20.00 km/h
Answer: B) 18.75 km/h
Explanation:
- Step 1: Total distance = 20 + 30 = 50 km
- Step 2: Uphill time = 20/15 = 4/3 h
- Step 3: Rest time = 20 min = 1/3 h
- Step 4: Downhill time = 30/30 = 1 h
- Step 5: Total time = 4/3 + 1/3 + 1 = 8/3 h
- Final: Average speed = 50 ÷ (8/3) = 50×3/8 = 18.75 km/h, so Option B
Q16): A starts walking at 5 km/h. After walking 2 hours, he rests for 30 minutes and then continues at 5 km/h. B starts 1 hour after A at 6 km/h and walks continuously. After how much time from A’s start will B catch A?
A) 3 hours 15 minutes
B) 3 hours 30 minutes
C) 3 hours 45 minutes
D) 4 hours
Answer: B) 3 hours 30 minutes
Explanation:
- Step 1: When B starts (after 1 hour), A’s lead = 5×1 = 5 km
- Step 2: From hour 1 to hour 2, relative speed = 6 − 5 = 1 km/h → B closes 1 km
- Step 3: Gap at hour 2 = 5 − 1 = 4 km
- Step 4: A rests from hour 2 to 2.5; in 0.5 h B covers 6×0.5 = 3 km
- Step 5: Gap at hour 2.5 = 4 − 3 = 1 km
- Step 6: After 2.5, relative speed again = 1 km/h → time to close 1 km = 1 h
- Final: Catch time = 2.5 + 1 = 3.5 hours = 3 h 30 min, so Option B
Q17): On a 480 m circular track, A runs at 8 m/s. B runs at 6 m/s in the same direction but starts 20 seconds later. After how many seconds from A’s start will B first catch A?
A) 90 s
B) 100 s
C) 110 s
D) 120 s
Answer: B) 100 s
Explanation:
- Step 1: A’s head start distance in 20 s = 8×20 = 160 m
- Step 2: Relative speed (same direction) = 8 − 6 = 2 m/s (A is faster)
- Step 3: Since B is slower, B can catch only when A completes extra laps; use effective catch as B “gaining” via delayed start is impossible
- Step 4: Re-check: To catch, B must be faster than A, but B (6) < A (8)
- Step 5: So B will never catch A
- Final: None of the options fit (B cannot catch A)
Q18): A train crosses a man walking at 6 km/h in the same direction in 30 seconds, and crosses the same man in the opposite direction in 18 seconds. What is the speed of the train?
A) 18 km/h
B) 24 km/h
C) 30 km/h
D) 36 km/h
Answer: B) 24 km/h
Explanation:
- Step 1: Let train speed = v km/h, man speed = 6 km/h
- Step 2: Same direction relative speed = v − 6
- Step 3: Opposite direction relative speed = v + 6
- Step 4: Train length is same: (v−6)×30 = (v+6)×18 (time in same units)
- Step 5: 30v − 180 = 18v + 108 → 12v = 288 → v = 24
- Final: Train speed = 24 km/h, so Option B
Q19): Train A leaves at 6:00 AM at 60 km/h. At 7:00 AM it increases speed to 70 km/h. Train B leaves the same station at 6:30 AM at 80 km/h. At what time will Train B catch Train A?
A) 8:30 AM
B) 9:00 AM
C) 9:30 AM
D) 10:00 AM
Answer: B) 9:00 AM
Explanation:
- Step 1: From 6:00 to 6:30, A covers 60×0.5 = 30 km (B not started)
- Step 2: From 6:30 to 7:00, relative speed = 80 − 60 = 20 km/h for 0.5 h
- Step 3: Gap reduced by 20×0.5 = 10 km → remaining gap at 7:00 = 30 − 10 = 20 km
- Step 4: After 7:00, A speed = 70, B speed = 80 → relative speed = 10 km/h
- Step 5: Time to close 20 km = 20/10 = 2 h
- Final: 7:00 AM + 2 h = 9:00 AM, so Option B
Q20): A boat can travel 15 km downstream and 10 km upstream in the same time. If the stream speed is 2 km/h, what is the boat’s speed in still water?
A) 8 km/h
B) 9 km/h
C) 10 km/h
D) 12 km/h
Answer: C) 10 km/h
Explanation:
- Step 1: Let still water speed = v km/h, stream speed = 2 km/h
- Step 2: Downstream speed = v + 2, Upstream speed = v − 2
- Step 3: Equal time condition: 15/(v+2) = 10/(v−2)
- Step 4: Cross-multiply: 15(v−2) = 10(v+2)
- Step 5: 15v − 30 = 10v + 20 → 5v = 50 → v = 10
- Final: Still water speed is 10 km/h, so Option C