Time, Speed and Distance

Downstream speed = 15+5 = 20 km/hr. Upstream speed = 15-5 = 10 km/hr. Time equation: (x+60)/20 + (x-35)/10 = 22. Solving: (x+60) + 2(x-35) = 440 → 3x - 10 = 440 → x = 150km. Pond distance = 150+30 = 180km. At 15km/hr in still water, time = 180/15 = 12 hours. However, this conflicts with checking: (150+60)/20 + (150-35)/10 = 10.5 + 11.5 = 22 hours ✓. So pond time should be 12 hours, not the claimed 12 hours which matches Option B.

Cars moving toward each other have combined speed 30+50=80 km/hr. Time to meet = 800/80 = 10 hours. Distance from first town = Car A's speed × time = 30 × 10 = 300 km. This is the point where both cars meet after 10 hours.

This is a classic trick question. We know time = distance/speed. If the original time is 1 hour, distance = speed (let's call it s). New speed = s+10, so new time = s/(s+10) hours. Without knowing the original speed (or distance), we cannot determine the new time. For example: if s=50 km/h, time = 50/60 = 50 minutes; if s=100 km/h, time = 100/110 ≈ 54.5 minutes. The answer varies, so 'Cannot be determined' is correct.

Relative speed = 18 - 15 = 3 km/hr = (3 × 1000/3600) m/s = 5/6 m/s. Distance closed in 18 seconds = (5/6) × 18 = 15 m. Initial distance was 240 m, so remaining distance = 240 - 15 = 225 m. Option B is correct.

The cyclist and runner move in the same direction. The relative speed covers the visibility distance of 100 m in 5 minutes. Relative speed = 100m/5min = 20 m/min = 20 × 60/1000 km/h = 1.2 km/h. Since runner's speed is 6 km/h, cyclist's speed = 6 + 1.2 = 7.2 km/h. Option A is correct.

Let the faster train's speed be x km/hr. The slower train's speed is (x - 12.5) km/hr. They meet after 8 hours, so combined distance = 8x + 8(x - 12.5) = 276. Solving: 16x - 100 = 276, so 16x = 376, x = 23.5 km/hr. Therefore slower train speed = 23.5 - 12.5 = 11 km/hr. None of the options match this answer.

Let original speed = v km/h. After 555 km, speed becomes 0.85v. If accident was 111 km later (at 666 km), delay would be 1 hour less (difference between 6 hours and 5 hours late). The extra 111 km at original speed vs reduced speed accounts for this 1-hour difference. Time difference = 111/v - 111/0.85v = 111/v × (1 - 1/0.85) = 111/v × 0.15/0.85 = 1 hour. Solving gives v = 19.58 km/h. Total distance = 555 + 666 = 1221 km.

Let original speed be x km/hr. Original time = 200/x hours. New speed = x + 5 km/hr. New time = 200/(x+5) hours. Difference is 2 hours: 200/x - 200/(x+5) = 2. Solving: 200(x+5) - 200x = 2x(x+5). 1000 = 2x² + 10x. x² + 5x - 500 = 0. (x + 25)(x - 20) = 0. x = 20 (positive root). Original speed is 20 km/hr.

Downstream speed = 12 kmph, upstream = 8 kmph. Distance = 12 × 91.5 = 1098 km. In still water, speed = (12+8)/2 = 10 kmph. Time in still water = 1098/10 = 109.8 hours. Still water speed is the average of downstream and upstream speeds.

Let distance = d km. A's time = d/40 hours, B's time = d/60 hours. A takes 15 minutes (0.25 hours) longer: d/40 = d/60 + 0.25. Multiplying by 120: 3d = 2d + 30, so d = 30 km.

Let distance between A and B = d km. Time to go = d/30 hours. Time to return = d/12 hours. Total time = 9 hours 20 min = 28/3 hours. Equation: d/30 + d/12 = 28/3. Multiplying by 60: 2d + 5d = 560, so 7d = 560, giving d = 80 km.

Speed in still water = 7 km/h, current = 3 km/h. Downstream speed = 7+3 = 10 km/h. Upstream speed = 7-3 = 4 km/h. Time difference: d/4 - d/10 = 9 hours. Solving: (5d - 2d)/20 = 9, so 3d = 180, d = 60 km.

Let each quarter = d km. Time for each quarter = d/10 + d/20 + d/30 + d/40 = d(12/120 + 6/120 + 4/120 + 3/120) = 25d/120. Total distance = 4d. Average speed = 4d / (25d/120) = 480d/25d = 19.2 km/hr.

This is a boat and stream problem requiring still water speed. Statements I and II give us two equations: upstream and downstream distances with total times. Let boat speed in still water be b and stream speed be c. From I: 25/(b-c) + 45/(b+c) = 10. From II: 40/(b-c) + 55/(b+c) = 12. Two equations, two unknowns (b and c) - solvable. Statement III says upstream=downstream distance, which is not needed and doesn't help solve for b. So only I and II are required.

The boat aims from A to B (24 km apart) but drifts to C, 10 km from B. This forms a right triangle with boat displacement and river drift. Using boat speed 24 kmph and the drift, river speed = (distance drifted)/(time taken). If boat crosses directly at 24 kmph for distance 24 km, time = 1 hour. In that time, river pushes boat 10 km sideways, so river speed = 10 kmph. Option A is correct. The geometry creates a right angle where river flow is perpendicular to crossing direction.