Time, Speed and Distance

Speed in still water = 5 km/hr, stream speed = 4 km/hr. Upstream speed = 5 - 4 = 1 km/hr. Downstream speed = 5 + 4 = 9 km/hr. Time to go 18 km upstream = 18/1 = 18 hours. Time to return 18 km downstream = 18/9 = 2 hours. Total time = 18 + 2 = 20 hours. This is a standard upstream-downstream time calculation.

Distance = speed × time = 15 × 12 = 180 km. Return trip time = distance/speed = 180/20 = 9 hours. The return journey takes 9 hours.

A is ahead by 15 km. Since A is slower (3 km/h) than B (8 km/h), B will catch A, not the other way around. Relative speed = 8-3 = 5 km/h (B catching A). Time = 15/5 = 3 hours. The question asks when 'A crosses B' which incorrectly implies A is faster, but the calculation gives 3 hours. Option B is numerically correct for the catch-up time.

Perimeter of rectangular field = 2 × (length + breadth) = 2 × (160 + 140) = 600 m. Distance for 4 revolutions = 4 × 600 = 2400 m = 2.4 km. Speed = 8 km/h, so time = distance/speed = 2.4/8 = 0.3 hours = 18 minutes. Since 18 minutes is not among options A-D, the correct answer is 'None of these'. The options 10, 15, and 20 minutes are distractors.

Speed = 4 km/hr, Time = 2 hr 45 min = 11/4 hr. Distance = 4 × 11/4 = 11 km. New speed = 12 km/hr. Time = 11/12 hr = 55 minutes. Option B is correct.

Let distance be d and stream speed be v. Upstream: d/(4-v) = 9, downstream: d/(4+v) = 3. Equating distances: 9(4-v) = 3(4+v), so 36-9v = 12+3v, giving 12v = 24, v = 2 km/hr. Option B (3 km/hr) would make upstream speed negative. Option D (2.5 km/hr) and C (1.5 km/hr) don't satisfy the time ratio of 3:1.

Bus speed = 312/8 = 39 km/hr. Car speed = 468/6 = 78 km/hr. Difference = 78 - 39 = 39 km/hr. Options A/B/C are computational errors, E is wrong.

We know the train covers a distance in 70 minutes, and speed decreases by 10 km/hr. However, we don't know the original speed or the distance. Without either of these values, we cannot determine the new time. The time increase depends on how much 10 km/hr affects the original speed, which varies based on the initial speed.

When trains meet, they travel for the same time. Let time = t hours. Distance by first train = 50t. Distance by second = 60t. Given: 60t = 50t + 120, so 10t = 120, giving t = 12 hours. Total distance = 50t + 60t = 110t = 110 × 12 = 1320 km. The speed difference of 10 km/hr covers the 120 km excess.

A starts at 12 noon at 8 km/h. After 1 hour, A has traveled 8 km. At 1 PM, B starts at 7 km/h towards A. Remaining distance = 68 - 8 = 60 km. Relative speed = 8 + 7 = 15 km/h. Time to meet = 60/15 = 4 hours. They meet at 1 PM + 4 hours = 5 PM. Option A matches.

By 10 AM, Train A has traveled 20 × 2 = 40 km. Remaining distance = 220 - 40 = 180 km. Relative speed when both trains move toward each other = 20 + 25 = 45 km/h. Time to meet after 10 AM = 180/45 = 4 hours. They meet at 10 AM + 4 hours = 2 PM. Answer A is correct.

Let the distance be d km. Time going to office = d/30 hours, time returning = d/40 hours. Total time = d/30 + d/40 = 35/60 hours. Solving: d(1/30 + 1/40) = 7/12, so d(7/120) = 7/12, giving d = 10 km. Option A is correct.

Let usual speed = S, usual time = T. Distance = S × T. New speed = 1.25S (25% more). New time = Distance/(1.25S) = T/1.25 = 0.8T. Time saved = T - 0.8T = 0.2T = 15 minutes. So T = 15/0.2 = 75 minutes = 1 hour 15 minutes. Option C is correct.

Distance = 4 × (2 + 45/60) = 4 × 2.75 = 11 km. Time = 11/16.5 = 0.666... hours = 40 min. Check: 16.5 kmph means 16.5/60 = 0.275 km/min, so 11/0.275 = 40 min exactly.

Total distance = 50 + 70 = 120 km. Time for first part = 50/25 = 2 hours. Time for second part = 70/35 = 2 hours. Total time = 4 hours. Average speed = Total distance/Total time = 120/4 = 30 km/h.