Time, Speed and Distance

Let boy's still water speed = b km/h, current speed = c = 3 km/h. Downstream speed = b + c, upstream speed = b - c. In fixed time t: distance downstream = 2 × distance upstream (given 'swims double'). So (b + c)t = 2(b - c)t, meaning b + c = 2b - 2c. Simplify: c + 2c = 2b - b, so 3c = b. With c = 3: b = 3 × 3 = 9 km/h. Option A is correct.

Avinash travels three equal distances (let each = d) at speeds 10, 20, 60 kmph. Total distance = 3d. Total time = d/10 + d/20 + d/60 = d(1/10 + 1/20 + 1/60) = d(6/60 + 3/60 + 1/60) = d(10/60) = d/6 hours. Average speed = total distance / total time = 3d / (d/6) = 18 kmph. Option A is correct.

Let the total distance be d. The first half (d/2) at 60 km/h takes d/120 hours. The second half (d/2) at 45 km/h takes d/90 hours. Total time is d/120 + d/90 = 7d/360 = 5.25 hours (5 hours 15 minutes). Solving gives d = 270 km. Note that when distances are equal (not time), you cannot use the simple average speed formula.

Let speed = s km/h, then time = 200/s hours. New speed = (s+5) km/h, new time = 200/(s+5) hours. The time difference is 2 hours: 200/s - 200/(s+5) = 2. Solving: 200(s+5 - s) = 2s(s+5), so 1000 = 2s² + 10s, giving s² + 5s - 500 = 0. Therefore (s+25)(s-20) = 0, so s = 20 km/h (speed must be positive).

Let boat speed in still water = b. Current speed = b/4. Upstream speed = b - b/4 = 3b/4. Downstream speed = b + b/4 = 5b/4. Distance = 30 km, upstream time = 6 hours. So (3b/4) × 6 = 30, giving b = 20/3 km/h. Downstream time = 30/(5b/4) = 30 × 4/(5b) = 120/(5 × 20/3) = 120 × 3/100 = 3.6 hours. ✓

Time without stopping = 300/60 = 5 hours. Time with stopping = 300/40 = 7.5 hours. The difference (7.5 - 5 = 2.5 hours) represents total stoppage time. Option C is correct.

By 2 p.m., first man covers 20 km (2 hours running). After 2 p.m., first man runs from 2-3 p.m. (10 km), rests from 3-4 p.m., then cycle repeats. Second man starts at 2 p.m. at 15 km/hr. Relative speed = 5 km/hr. Gap at 2 p.m. = 20 km. Gap closes to 10 km by 3 p.m. From 3-4 p.m., second continues at 15 km/hr while first rests, so gap closes completely. Catch time = 4:40 p.m.

Using d=vt=(v+10)(t-1)=(v+20)(t-1.75), solving yields t=7 hours and v=60 km/h. The distance d=60×7=420 km matches option D. Check: at 60 km/h takes 7 hours, at 70 km/h takes 6 hours, at 80 km/h takes 5.25 hours (difference of 45 minutes).

Speed of boat in still water = 8 km/h, stream speed = 2 km/h. Against the current (upstream), effective speed = 8 - 2 = 6 km/h. In 2 hours, distance covered = speed × time = 6 × 2 = 12 km. The key concept is that upstream speed equals boat speed minus stream speed, while downstream speed equals boat speed plus stream speed. This problem tests understanding of relative motion in streams.

Car speed = 450/15 = 30 kmph. Bus speed = 540/30 = 18 kmph. Sum = 48 kmph. Truck speed = half of sum = 24 kmph. Distance in 5 hours = 24 × 5 = 120 km. Option B is correct.

Downstream: 4 km in same time as 3 km upstream. Let boat speed = b, stream speed = s. Then 4/(b+s) = 3/(b-s). Cross-multiply: 4b - 4s = 3b + 3s, so b = 7s. Total time: 60/(b+s) + 60/(b-s) = 35. Substituting b = 7s: 60/8s + 60/6s = 35. So 7.5/s + 10/s = 35, s = 0.5, b = 3.5.

Downstream speed = 9.5 + 2.5 = 12 kmph. Upstream speed = 9.5 - 2.5 = 7 kmph. Total time = 285 minutes = 4.75 hours = 19/4 hours. Distance = d, so d/12 + d/7 = 19/4. Solving: 7d + 12d = (19 × 21)/4, giving d = 21 km.

Downstream speed = 90/4 = 22.5 km/hr. If boat speed in still water is 20 km/hr, let stream speed be x. Then 20 + x = 22.5, so x = 2.5 km/hr. Upstream speed would be 20 - 2.5 = 17.5 km/hr. To cover 90 km upstream in 4 hours, speed needed = 90/4 = 22.5 km/hr. Increase from 17.5 to 22.5 = 5 km/hr. Percentage increase = (5/20) × 100 = 25%.

Total distance is 440 km and time taken is 4 hours, so relative speed (sum of both speeds) is 440/4 = 110 km/hr, which matches the given information. Train A traveled 200 km in 4 hours, so its speed is 50 km/hr. Train B traveled the remaining 240 km in 4 hours, so its speed is 60 km/hr. The ratio of their speeds is 50:60 = 5:6.

Let boat speed = b and stream speed = s. Upstream: b - s = 75/12 = 6.25 km/h. Downstream: b + s = 75/7.5 = 10 km/h. Solving these equations: 2b = 16.25, so b = 8.125 km/h. Then 2s = 3.75, giving s = 1.875 km/h. Option D is correct.