Time, Speed and Distance

Let boat speed in still water = b, stream speed = s. Downstream speed = b+s, upstream speed = b-s. Time upstream = 2 × time downstream, so distance/(b-s) = 2 × distance/(b+s). This gives b+s = 2(b-s), so b+s = 2b-2s, meaning 3s = b, or b:s = 3:1. At 3:1 ratio, downstream is 4x, upstream is 2x, giving the required 2:1 time ratio. Option A (2:1) would mean b = 2s, giving speeds 3s and s with time ratio 3:1, not 2:1.

Distance covered in first 7 hours = 7 × 40 = 280 km. Distance covered in next 5 hours = 5 × 60 = 300 km. Total distance in 12 hours = 580 km, which is 4/5 of total distance. So total distance = 580 × 5/4 = 725 km. Remaining distance = 725 - 580 = 145 km. To cover 145 km in 5 hours: average speed = 145/5 = 29 km/hr. Option A (45), B (55), and C (35) are all too high.

Wait, the question says downstream takes 6 hours for 60 km, so downstream speed = 60/6 = 10 km/hr. Then it says 'total 21 hour if it has to go back to the starting point and come back to the destination'. This means: downstream + upstream = 21 hours. So 6 + upstream_time = 21, giving upstream_time = 15 hours for 60 km. Upstream speed = 60/15 = 4 km/hr. Boat speed in still water = (downstream + upstream)/2 = (10 + 4)/2 = 7 km/hr. This matches option C (7).

The train is 16 minutes late initially. With increased speed (v + 10) for 80 km, it reaches on time. The 16 minutes saved means the time difference = 16/60 hours. So: 80/v - 80/(v+10) = 16/60 = 4/15. Simplify: 80(v+10 - v) / v(v+10) = 4/15, so 800 / v(v+10) = 4/15. Cross-multiply: 12000 = 4v(v+10) = 4v² + 40v, so 4v² + 40v - 12000 = 0, giving v² + 10v - 3000 = 0. Solving: v = [-10 ± sqrt(100 + 12000)]/2 = [-10 ± 110]/2, so v = 50 or v = -60. Speed must be positive, so v = 50 km/hr. This matches option A.

Distance = Speed × Time. At 60 km/hr for 6 hours, the bike covers 360 km. At 50 km/hr, the same distance takes 360/50 = 7.2 hours.

Upstream speed = 9/3 = 3 kmph, downstream speed = 18/3 = 6 kmph. Boat speed = (3 + 6)/2 = 4.5 kmph.

Distance covered = (7 × 27) + (9 × 38) = 189 + 342 = 531 km. This is 3/7 of total distance, so total distance = 531 × 7/3 = 1239 km. Remaining distance = 1239 - 531 = 708 km. Speed needed = 708/24 = 29.5 km/hr. Option A is correct.

We know boat speed = 4 km/h and total time = 3 hours. Let distance = d and stream speed = s. Downstream: time = d/(4+s). Upstream: time = d/(4-s). We have d/(4+s) + d/(4-s) = 3. This is one equation with two unknowns (d and s), so we cannot determine d uniquely. Option D is correct.

Let distance = d km and scheduled time = t hours. At 5 km/hr, actual time = d/5 hours (late by 30 min = 0.5 hr). At 6 km/hr, actual time = d/6 hours (late by 5 min = 5/60 = 1/12 hr). Equating: d/5 - 0.5 = d/6 - 1/12. Solving: d/5 - d/6 = 0.5 - 1/12 = 5/12, so d/30 = 5/12, giving d = 12.5 km. Option D is correct.

Relative speed = 18-10 = 8 km/hr = 8×(1000/3600) = 20/9 m/s. Time to catch 500m gap = 500/(20/9) = 225 seconds. Distance thief runs = 10×(1000/3600)×225 = 625 meters. thief runs for same time as police catches up.

Distance = 45 × 8 = 360 km. To cover 360 km in 6 hours, speed needed = 360 ÷ 6 = 60 km/hr. Increase needed = 60 - 45 = 15 km/hr. Distance remains constant, so speed and time are inversely proportional.

Let truck's speed be x km/hr. Then car's speed is (x+10) km/hr. They travel towards each other, so their combined speed is x + (x+10) = 2x + 10. In 4 hours, they cover 360 km together. So 4(2x + 10) = 360, which gives 2x + 10 = 90, so 2x = 80 and x = 40. Therefore, truck's speed is 40 km/hr and car's speed is 50 km/hr. The ratio car:truck = 50:40 = 5:4.

The Hindi version uses 12 km/hr and 15 km/hr, which gives: total time = d/12 + d/15 = 3d/20 = 4 hr 3 min = 81/20 hr. So d=27 km, total distance=2d=54 km. The English version (40 km/hr, 34 km/hr) gives ~149 km, which doesn't match options. Hindi is correct.

Let distance be d and stream speed be c. Downstream: d = (20 + c) × (18/60) = (20 + c) × 0.3. Upstream: d = (20 - c) × (30/60) = (20 - c) × 0.5. Equating: 0.3(20 + c) = 0.5(20 - c). Solving: c = 5 km/h. Then d = (20 + 5) × 0.3 = 7.5 km.

A and B have 3-hour head start. A travels 60 km (20×3), B travels 36 km (12×3). C overtakes B: let C's speed be s. Time to catch B = 36/(s-12). C then overtakes A: time to catch A = 60/(s-20). Given second catch is 120 minutes (2 hours) after first: 60/(s-20) - 36/(s-12) = 2. Solving: 60(s-12) - 36(s-20) = 2(s-20)(s-12). 60s-720-36s+720 = 2(s²-32s+240). 24s = 2s²-64s+480. s²-88s+480=0 doesn't give integer. Alternative approach: let C take t hours to catch B, then t+2 to catch A. Distance same: 12t + s(t+2) = 20t + s(t+4)? No. C catches B at time t: s×t = 12×(3+t). s = 12(3+t)/t. C catches A at time t+2: s×(t+2) = 20×(3+t+2) = 20×(5+t). So 12(3+t)(t+2)/t = 20(5+t). 12(3+t)(t+2) = 20t(5+t). This gives t=4, s=24 km/hr. Answer C is correct.