Time, Speed and Distance

Let total distance be D. Bus covers 40% (0.4D) at 60 km/hr in 0.4D/60 hr. Remaining is 0.6D. Auto covers half of remaining (0.3D) at 40 km/hr in 0.3D/40 hr. New remaining is 0.3D. E-rikshaw covers 80% of this (0.24D) at 30 km/hr in 0.24D/30 hr. Last 3 km on foot at 5 km/hr (0.5D = 3 km, so D = 6 km and walking speed = 3/0.08 = 37.5? No, working backwards: remaining after e-rikshaw = 0.06D = 3 km, so D = 50 km). Times: 12/60 = 0.2 hr (12 min), 15/40 = 0.375 hr (22.5 min), 12/30 = 0.4 hr (24 min), 3/6 = 0.5 hr (30 min). Total = 88.5 min = 1 hr 28.5 min. Wait, let me recalculate: 0.4D = 20 km, 0.3D = 15 km, 0.24D = 12 km, remaining = 3 km. Walking speed from remaining 0.06D in 6 min? 3 km in 0.08 hr means 37.5 km/hr. Times: bus = 20/60 = 1/3 hr = 20 min, auto = 15/40 = 0.375 hr = 22.5 min, e-rikshaw = 12/30 = 0.4 hr = 24 min, walk = 3/6 = 0.5 hr = 30 min. Total = 96.5 min? Actually: 20 + 22.5 + 24 + 30 = 96.5 min = 1 hr 36.5 min. Hmm. Let me re-read: last 3 km on foot, and remaining after e-rikshaw is 3 km = 0.06D, so D = 50 km. Then: bus = 20 km at 60 = 20 min, auto = 15 km at 40 = 22.5 min, e-rikshaw = 12 km at 30 = 24 min, walk = 3 km at (3 km / time). But we need walking speed. From 20% of 50 = 10 km covered on foot taking 6 min? So 10 km in 6 min = 100 km/hr? That's wrong. Let me recalculate more carefully. Total distance = D. 40% = 0.4D by bus. Remaining = 0.6D. Half of remaining = 0.3D by auto. Remaining = 0.3D. 80% of remaining = 0.24D by e-rikshaw. Remaining = 0.06D. This 0.06D = 3 km. So D = 3/0.06 = 50 km. Now we need walking speed. 3 km covered in time T at speed S. But we don't have T or S directly. Oh wait, re-reading: 3 km on foot, no time given. The 6 min from calculation: 0.06D in 6 minutes means 0.06 × 50 = 3 km in 6 min = 30 km/hr. Times: bus = 20/60 = 20 min, auto = 15/40 = 22.5 min, e-rikshaw = 12/30 = 24 min, foot = 3/30 = 6 min. Total = 20 + 22.5 + 24 + 6 = 72.5 min = 1 hr 12.5 min? That's not matching. Actually 3/30 = 0.1 hr = 6 min. Total = 20 + 22.5 + 24 + 6 = 72.5 min. But answer C is 1 hr 42 min 30 sec = 102.5 min. Let me reconsider: maybe the 6 minutes I calculated should be different. If foot speed is 5 km/hr (reasonable), then 3/5 = 0.6 hr = 36 min. Total = 20 + 22.5 + 24 + 36 = 102.5 min = 1 hr 42 min 30 sec. This matches option C! So walking speed must be 5 km/hr, not calculated from given info but as standard walking speed.

Volvo: 80 km/hr with 10 min stop after every hour. For 400 km: 400/80 = 5 hours driving time. Stops: after hours 1, 2, 3, 4 = 4 stops × 10 min = 40 min. Total time = 5 hr 40 min = 340 min. Normal bus: 50 km/hr with 6 min stop after every 50 km. In 340 min, it travels: 50 km in 60 + 6 = 66 min, another 50 km in 66 min (total 132 min, 100 km). Remaining time = 340 - 132 = 208 min. 50 km in 66 min uses 66 min, leaving 142 min for 50 km? Actually in remaining 208 min: next 50 km in 66 min (total 198 min, 150 km). Remaining 142 min. Next segment: can travel 50 km in 66 min (total 264 min, 200 km) but we only have 142 min. In 142 min: first 60 min covers 50 km (total 258 min, 250 km) - no, that exceeds. So in remaining 142 min: 60 min driving covers 50 km, but we have 142 - 60 = 82 min for driving. Distance in 82 min = 82/60 × 50 = 68.33 km. Total distance = 200 + 68.33 = 268.33 km. That's not matching option B (258.33). Let me recalculate: 340 min available. Normal bus covers 50 km in 66 min. 340/66 = 5.15 segments. 5 segments = 5 × 50 = 250 km in 5 × 66 = 330 min. Remaining 10 min can cover 10/60 × 50 = 8.33 km. Total = 258.33 km. Yes! This matches option B.

After 2 hours, A travels 14 km and B travels 10 km. The minimum distance occurs when A starts behind B. In this case, A's position is 14 km from his start and B's position is 11 km from A's start (10 km from B's start + 1 km initial gap). The distance between them is 14 - 11 = 3 km.

The question states the ratio of time taken is 6:2:3. Since speed = distance/time and distance is constant, speed is inversely proportional to time. Therefore, the ratio of speeds is the reciprocal of the time ratio: 1/6 : 1/2 : 1/3 = 1 : 3 : 2 when normalized.

Given: Distance XY = 360 km, Speed of B = 1.5 × Speed of A. Let Speed of A = v, then Speed of B = 1.5v. Relative speed when moving toward each other = v + 1.5v = 2.5v. The difference from 72 kmph is |1.5v - v| = 0.5v. The percentage difference to 72 depends on v's value. If 72 represents a reference speed, we compare 0.5v to 72 and express as percentage.

For downstream: speed = boat speed + current speed. Distance = 48 km, time = 4 hrs, so downstream speed = 48/4 = 12 km/h. For upstream: speed = boat speed - current speed. Distance = 20 km, time = 4 hrs, so upstream speed = 20/4 = 5 km/h. Adding: (boat + current) + (boat - current) = 12 + 5 = 17, giving 2 × boat speed = 17, so boat speed = 8.5 km/h. Subtracting: (boat + current) - (boat - current) = 12 - 5 = 7, giving 2 × current speed = 7, so current speed = 3.5 km/h.

Suresh travels same distance both ways: 12 × 5 = 15 × 4 = 60 km. Average speed = total distance/total time = 120/9 = 13.33 km/h. Anil's speed = 4/5 × 13.33 = 10.67 km/h. Time for 80 km = 80/10.67 = 7.5 hours. Alternatively, if using weighted average: (60+60)/(5+4) = 120/9 = 13.33 km/h, same result.

Let total distance be D. Time for first third: (D/3) ÷ 12 = D/36 hours. Time for remainder (2D/3): (2D/3) ÷ 14 = 2D/42 = D/21 hours. Total: D/36 + D/21 = 7D/252 + 12D/252 = 19D/252 = 19 hours. Solving: D = 252 km.

Average speed = total distance ÷ total time. Total distance = 18 + 16 + 30 = 64 km. Total time = (18/6) + (16/8) + (30/6) = 3 + 2 + 5 = 10 hours. Average speed = 64 ÷ 10 = 6.4 km/hr. The key is to use total distance and total time, not the individual speeds.

Let stream speed = s kmph. Downstream speed = (12 + s), upstream speed = (12 - s). Distance is same. Time downstream = 12 hours, upstream = 18 hours. Distance = (12 + s) × 12 = (12 - s) × 18. 144 + 12s = 216 - 18s. 30s = 72. s = 72/30 = 2.4 kmph.

Quantity I: Speed = 12 km/hr, distance = 28 km. Time = 28/12 = 7/3 ≈ 2.33 hours. Quantity II: Downstream speed = 12 + 4 = 16 km/hr, distance = 32 km. Time = 32/16 = 2 hours. 2.33 hours > 2 hours, so Quantity I > Quantity II.

Speed downstream = 8 + 2 = 10 km/h, upstream = 8 - 2 = 6 km/h. Let distance be d km. Time downstream = d/10, upstream = d/6. Given: d/6 - d/10 = 2 hours. Solving: d(1/6 - 1/10) = 2, so d(5/30 - 3/30) = 2, d(2/30) = 2, d = 30 km. Verification: Downstream 30/10 = 3h, upstream 30/6 = 5h, difference = 2h.

Train A travels alone from 8 AM to 9:30 AM = 1.5 hours at 80 km/hr = 120 km. Remaining distance = 545 - 120 = 425 km. Relative speed when both travel = 80 + 90 = 170 km/hr. Time to meet after 9:30 AM = 425/170 = 2.5 hours. Meeting time = 9:30 AM + 2.5 hours = 12:00 PM. Option C is correct. Options A (11:30) misses 0.5 hours, B (12:30) is 0.5 hours late, D (1:00) is 1 hour late, E (11:00) is 1 hour early.

For boat and stream problems, if upstream time is twice downstream time, then speed upstream = 1/2 × speed downstream. Let boat speed = 60 kmph, current speed = c. Downstream: 60 + c. Upstream: 60 - c. Since time upstream = 2 × time downstream, (60 - c) = (1/2)(60 + c). Solving: 60 - c = 30 + c/2, giving 30 = 3c/2, so c = 20 kmph. Option B is correct.

Upstream speed = 9 - 3 = 6 km/hr. Downstream speed = 9 + 3 = 12 km/hr. Total time = 3 hours for both trips. If d is distance: d/6 + d/12 = 3. Solving: 2d/12 + d/12 = 3, so 3d/12 = 3, giving d = 12 km. The faster downstream speed covers more distance in the same time.