Time, Speed and Distance

First, calculate the distance: Speed × Time = 60 km/hr × (40/60) hr = 40 km. To cover the same 40 km in 32 minutes (32/60 = 8/15 hours), the required speed is Distance/Time = 40 ÷ (8/15) = 40 × (15/8) = 75 km/hr. Speed and time are inversely proportional when distance is constant.

Distance = 20 km. Speed 1 = 4 km/hr gives time = 5 hours. Speed 2 = 8 km/hr gives time = 2.5 hours. Difference = 5 - 2.5 = 2.5 hours = 2 hours 30 minutes. Since he was 35 minutes late at speed 1, at speed 2 he would be 35 + (2 hours 30 minutes) = 3 hours 5 minutes early... Wait, that's not an option. Let me recalculate: If 5 hours makes him 35 min late, the fixed time is 4 hours 25 min. At 2.5 hours, he reaches 4:25 - 2:30 = 1 hour 55 min early. Option C is correct.

Let the original speed be v kmph and distance be d km. Original time = d/v hours. When speed increases by 4 kmph: time = d/(v+4) = d/v - 72/60 = d/v - 6/5. When speed decreases by 4 kmph: time = d/(v-4) = d/v + 2. From the equations: d/v - d/(v+4) = 6/5 and d/(v-4) - d/v = 2. Solving these gives v = 20 kmph and d = 96 km. Verification: at 24 kmph, time = 96/24 = 4 hours (72 minutes less than 5 hours); at 16 kmph, time = 96/16 = 6 hours (2 hours more than 4 hours).

Let the distance be D km and stream speed be S km/hr. Upstream speed = 8-S, downstream speed = 8+S. Time = D/(8-S) + D/(8+S) = 20. This gives D(16)/(64-S²) = 20. Without knowing S, we cannot determine D uniquely. Different values of S give different D (e.g., S=1 gives D≈79, S=2 gives D≈60). The data is inadequate.

Let stream speed be s km/h. Downstream speed = 8 + s, upstream speed = 8 - s. Time downstream = 10/(8 + s), time upstream = 10/(8 - s). Given: 10/(8 - s) - 10/(8 + s) = 40/60 = 2/3 hours. Cross-multiplying: 30(8 + s) - 30(8 - s) = 2(64 - s²), giving 60s = 128 - 2s², or s² + 30s - 64 = 0. Solving: (s + 32)(s - 2) = 0, so s = 2 (positive root). Stream speed is 2 km/h.

Let train B's speed be s km/h, so train A's speed is s + 27 km/h. Relative speed (when moving towards each other) = s + (s + 27) = 2s + 27 km/h. Distance covered when they meet = 1872 km in 16 hours, so relative speed = 1872/16 = 117 km/h. Therefore: 2s + 27 = 117, giving 2s = 90 and s = 45 km/h. Train B's speed is 45 km/h.

When rowing upstream, effective speed is (x - 6) kmph, and downstream it's (x + 6) kmph. Taking four times as long upstream means the upstream speed is one-fourth of downstream speed. Solving (x-6) = (x+6)/4 gives x = 10 kmph in still water.

Let v be original speed. Time = distance/speed, so 300/v - 300/(v+5) = 2. Solving: 300(v+5 - v)/v(v+5) = 2, giving 1500 = 2v² + 10v, or v² + 5v - 750 = 0. Factoring: (v+30)(v-25) = 0, so v = 25 km/hr. Check: 300/25 = 12 hr, 300/30 = 10 hr, difference is 2 hr. Option A (30) gives original, not the increased speed.

Downstream speed = speed in still water + speed of current = 15 + 3 = 18 km/hr. Time = 12 minutes = 12/60 hours = 0.2 hours. Distance = Speed × Time = 18 × 0.2 = 3.6 km. This matches option C.

Let boat speed in still water = b and stream speed = s. Speed against stream = b - s, speed with stream = b + s. Since time taken against stream is twice the time with stream, we have d/(b-s) = 2 × d/(b+s), which gives b + s = 2(b - s). Solving this gives 3s = b, so the ratio b : s = 3 : 1.

Speed downstream = 9 + 1.5 = 10.5 kmph, upstream = 9 - 1.5 = 7.5 kmph. Time to travel 105 km downstream = 105/10.5 = 10 hours. Return time = 105/7.5 = 14 hours. Total = 24 hours. Option D matches.

Let the fixed time be T and distance be 20 km. At 5 km/hr: time taken = 20/5 = 4 hours. Since he's 40 minutes late, T = 4 hours - 40 minutes = 3 hours 20 minutes = 200 minutes. At 8 km/hr: time taken = 20/8 = 2.5 hours = 150 minutes. Arrival time = T - 150 = 200 - 150 = 50 minutes early. Therefore he reaches 50 minutes before the fixed time. The key is finding the actual fixed time from the first condition, then comparing with the second travel time.

Let distance = d km. Going time = d/50 hours. Returning time = d/45 hours. Given return takes 1 hour longer: d/45 - d/50 = 1. Finding common denominator (2250): d(50-45)/2250 = 1, so 5d/2250 = 1, giving 5d = 2250, thus d = 450 km. Verification: 450/50 = 9 hours, 450/45 = 10 hours, difference = 1 hour.

Let distance on foot be d km. Then distance on bicycle = (61-d) km. Time on foot = d/4 hours, time on bicycle = (61-d)/9 hours. Total time: d/4 + (61-d)/9 = 9. Solving: 9d + 4(61-d) = 324, so 9d + 244 - 4d = 324, giving 5d = 80, thus d = 16 km.

Distance = Speed × Time. Given average speed is 90 km/hr and time is 4 hours, Distance = 90 × 4 = 360 km.