Speed, Time and Distance

The boat covers 24 km downstream in 10 hours, so downstream speed = 24/10 = 2.4 km/hr. It takes 2 hours more to cover the same distance upstream, so upstream time = 12 hours, and upstream speed = 24/12 = 2 km/hr. The speed of the boat in still water is the average of downstream and upstream speeds: (2.4 + 2)/2 = 2.2 km/hr. Since 2.2 km/hr is not an option, the closest given option is 2.8 km/hr (option B). The original correct answer (C) 4 km/hr is incorrect based on these calculations. Therefore, the correct answer should be B (2.8 km/hr) as the closest approximation.

The boatman's speed against the current is 2 km/h, and along the current is 6 km/h. Using these, the speed of the stream is (6 - 2)/2 = 2 km/h, and the boat's speed in still water is (6 + 2)/2 = 4 km/h. To cover 5 km in still water, time = distance/speed = 5/4 = 1.25 hours = 75 minutes. However, the options do not include 75 minutes; rechecking calculations: speed against current = 2 km/h, speed along current = 6 km/h, so speed of boat in still water = (2 + 6)/2 = 4 km/h. Time to cover 5 km = 5/4 = 1.25 hours = 75 minutes (1 hr 15 min). Option C is 1 hr 15 min, which matches the calculation. Therefore, correct answer is C.

Since A beats B by 22.5 m or 6 seconds in a 300 m race, B takes 6 seconds longer than A. A's speed is 300 m divided by (B's time - 6). Using the ratio of distances and times, B's time calculates to 86 seconds.

The speed of the man in still water is the average of the upstream and downstream speeds: (25 + 35) / 2 = 30 kmph. This is because the current's effect cancels out when averaging the two speeds.

The speed of the man in still water is calculated as the average of the downstream and upstream speeds: (18 + 10) / 2 = 14 kmph. The speed of the stream is half the difference between the downstream and upstream speeds: (18 - 10) / 2 = 4 kmph. Therefore, the correct answer is 14 kmph for the man in still water and 4 kmph for the stream, which corresponds to option D.

The speed of the current is half the difference between downstream and upstream speeds: (35 - 25) / 2 = 5 kmph.

The speed of the stream is calculated as half the difference between downstream and upstream speeds: (15.5 - 8.5) / 2 = 7 / 2 = 3.5 km/hr. However, option A is 3.5 km/hr, which matches this calculation, so the correct answer is A, not B. Rechecking options shows option A is 3.5 km/hr, which is correct. Therefore, correct_answer should be A.

The speed of the boat upstream is (9 - 3) = 6 km/h and downstream is (9 + 3) = 12 km/h. Let the distance between A and B be x km. The total time is given as 3 hours, so x/6 + x/12 = 3. Solving this gives x = 12 km.

The speed of the man in still water is the average of his speed against the stream and with the stream. Given: Distance = 0.75 km, Time against stream = 11 ¼ minutes = 11.25/60 = 0.1875 hours, Time with stream = 7 ½ minutes = 7.5/60 = 0.125 hours. Speed against stream = 0.75 / 0.1875 = 4 km/hr. Speed with stream = 0.75 / 0.125 = 6 km/hr. Speed in still water = (4 + 6) / 2 = 5 km/hr. Therefore, the correct answer is D (5 km/hr).

Let the length of the track be x km. Time taken at 150 kmph = x/150 hours; time taken at 180 kmph = x/180 hours. The difference in time is 3 minutes = 3/60 = 1/20 hours. So, x/150 - x/180 = 1/20. Simplifying, (6x - 5x)/900 = 1/20 => x/900 = 1/20 => x = 900/20 = 45 km. However, this calculation shows 45 km, which matches option B, so the initial calculation was correct. Rechecking: x/150 - x/180 = 1/20; LCM of 150 and 180 is 900. So, (6x - 5x)/900 = 1/20 => x/900 = 1/20 => x = 900/20 = 45 km. Therefore, the correct answer is B (45 km). The original correct_answer was B, so it is correct. The explanation was missing and has been added. Difficulty is medium because it involves algebraic manipulation and unit conversion.

Let the speed of the bike be x km/h. Then the speed of the car is 1.5x km/h. The time taken by the bike is 100/x hours, and by the car is 100/(1.5x) hours. The difference in time is 25 minutes = 25/60 hours. So, 100/x - 100/(1.5x) = 25/60. Simplifying gives x = 75 km/h, which is option D.

Let the distance to point A be d km and the time taken at 10 kmph be t hours. Then d = 10t. At 15 kmph, he reaches 2 hours earlier, so time taken is t - 2 hours, giving d = 15(t - 2). Equating the two distances: 10t = 15(t - 2) leads to 10t = 15t - 30, so 5t = 30 and t = 6 hours. Thus, distance d = 10 × 6 = 60 km. To reach at 1 P.M., which is 1 hour earlier than 2 P.M., the time taken should be 5 hours. Speed = distance/time = 60/5 = 12 kmph. Therefore, the correct speed to reach at 1 P.M. is 12 kmph, which corresponds to option C.

The train travels 600 km at an average speed of 100 km/h, so the running time without stops is 600 ÷ 100 = 6 hours. The train stops for 3 minutes after every 75 km. Since 600 km ÷ 75 km = 8 segments, there are 7 stops (one after each segment except the last). Total stopping time is 7 × 3 = 21 minutes. Therefore, total time = 6 hours + 21 minutes = 6 hours 21 minutes, which corresponds to option A.

Let the length of the journey be x km and the scheduled time be t hours. At 40 kmph, time taken is t = x/40. At 35 kmph, time taken is x/35, which is 15 minutes (0.25 hours) more than t. So, x/35 - x/40 = 0.25. Solving gives x = 70 km.

To find the distance covered, first convert speed from km/h to m/s: 108 km/h = (108 × 1000) / 3600 = 30 m/s. Then, distance = speed × time = 30 m/s × 15 s = 450 metres.

The car covers 42 km in 1 hour 40 minutes 48 seconds at 5/7 of its actual speed. Converting the time to hours: 1 hr + 40/60 hr + 48/3600 hr = 1.68 hours. Speed at 5/7 actual speed = distance/time = 42/1.68 = 25 km/hr. Therefore, actual speed = (7/5) * 25 = 35 km/hr. However, this calculation shows 35 km/hr, matching option B. Rechecking: 1 hr 40 min 48 sec = 1 + 40/60 + 48/3600 = 1 + 0.6667 + 0.0133 = 1.68 hr. Speed = 42/1.68 = 25 km/hr (5/7 actual speed). Actual speed = (7/5)*25 = 35 km/hr. So the correct answer is B, not D. Explanation corrected accordingly.

The man normally walks at 6 kmph and reaches his destination at 9:00 am. Walking slower, he reaches at 10:00 am, which is 1 hour later. The distance to the destination is 12 km. At 6 kmph, the time taken is 12 km ÷ 6 kmph = 2 hours. Walking slower, the time taken is 3 hours (2 + 1). Therefore, the slower speed is distance ÷ time = 12 km ÷ 3 hours = 4 kmph. Hence, the correct answer is 4 kmph, which corresponds to option B.

Let the distance be d km. Asghar's time = d/10 + d/9 hours. Rohit's time = 2d/12 = d/6 hours. The difference in time is 10 minutes = 1/6 hour. So, (d/10 + d/9) - (d/6) = 1/6. Solving gives d = 3.75 km or 3750 meters.

Let the original speed be x kmph and distance be d km. From the problem, (d/(x+3)) = (d/x) - 2/3 hours and (d/(x-2)) = (d/x) + 2/3 hours. Solving these equations yields d = 36 km. Therefore, option A is correct.

Let the distance between the house and school be x km. Time taken to go = x/3 hours, time taken to return = x/2 hours. Total time = x/3 + x/2 = 5 hours. Solving for x: (2x + 3x)/6 = 5 => 5x = 30 => x = 6 km. However, this calculation shows 6 km, but the total time is 5 hours, so let's re-check: x/3 + x/2 = 5 => (2x + 3x)/6 = 5 => 5x = 30 => x = 6 km. So the distance is 6 km, which corresponds to option C. Therefore, the original correct answer C is correct. The explanation has been added for clarity.

Let the total distance be 2x km. The first half (x km) is traveled at 21 km/h, taking x/21 hours, and the second half (x km) at 24 km/h, taking x/24 hours. Total time is x/21 + x/24 = 10 hours. Solving for x gives x = 112 km, so total distance is 224 km.

The total distance covered is calculated by summing the distances covered in each segment of the journey. For the first 2 hours at 120 km/h, the distance is 120 × 2 = 240 km. For the next half an hour (0.5 hours) at 100 km/h, the distance is 100 × 0.5 = 50 km. Adding these gives a total distance of 240 + 50 = 290 km. Therefore, the correct answer is option B (290 km).

The average speed for a round trip where the distances are equal but speeds differ is given by the formula: Average speed = 2 * (speed1 * speed2) / (speed1 + speed2). Here, speed1 = 30 kmph and speed2 = 60 kmph. Plugging in the values, Average speed = 2 * (30 * 60) / (30 + 60) = 3600 / 90 = 40 kmph. This matches option D, so the correct answer is D.

The original speed is 10 km in 12 minutes, which is 50 km/h. Decreasing the speed by 5 km/h gives 45 km/h. Time = Distance / Speed = 10 km / 45 km/h = 4/18 hours = 13 min 20 sec, which matches option D.

The two trains approach each other at a combined speed of 70 + 50 = 120 mph. They start 120 miles apart, so they collide after 120 / 120 = 1 hour. The bird flies continuously at 85 mph for that 1 hour, so it covers 85 miles.