 # Time, Speed and Distance: Solved Examples

Example 1: Driving 5/4th of his usual speed, David reached the destination 12 minutes earlier. What is the usual time he takes to travel?
Sol: Let X km/hr be the usual speed and let t hours be the usual time taken.
Speed on this occasion = 5/4 X km/hr. The time taken on this occasion = (t - 12/60) hrs.
Since the distance is the same in both the cases, Xt = 5/4 X x (t - 12/60)
Solving for t, t = 1 hour - the usual time taken.
Example 2: In a cross-country race, a motorist averages a speed of 140 mph during the first 4 hours and then increases his average by 20 mph during the last 3 hours. What was his average speed during the entire race?
Sol: Distance traveled in first 4 hours = 140 x 4 = 560 miles.
Distance traveled in next 3 hours = 160 x 3 = 480 miles.
Therefore, the total distance traveled = 1040 miles.
The total time taken = 7 hours. Therefore, the average speed = 1040/7 = 148 4/7mph.
Example 3: During the onward journey from Bombay to Pune, Deccan Queen travels at an average speed of 80 kmph, while on the return journey, the train is able to average a speed of 100 kmph. What is the average speed of the train on its entire journey?
Sol: Average speed = 2ab/a+b = (2x80x100)/180 = 800/9 = 88 8/9km/hr
Example 4: A train traveling at 100 km/hr crosses a bridge of half a km length completely in 30 seconds. What is the length of the train?
Sol: Speed = 100 km/hr = 100 x 5/18 = 250/9 m/sec. Time taken to cross = 30 seconds.
Therefore, distance traveled = 250/9 x 30 = 2500/3 m. Distance = Length of the train + length of the bridge
2500/3 = Length of the train + 500 Length of the train = 1000/3 m
Example 5: A train crosses a signpost in 6 seconds and a car traveling in the same direction at 50 kmph in 72 seconds. What is the length of train and the speed at which it is traveling?
Sol: Case I: Let X km/hr be the speed of the train. = X x 5/18 m/sec
Time taken to cross a signpost = 6 seconds.
Therefore distance traveled = X x 5/18 x 6 = 5X/3 meter = length of the train
Case II:
The speed of the car = 50 km/hr.
Relative speed of the train w.r.t car = (X – 50) km/hr = (X – 50) x 5/18 m/sec.
Time taken to cross the car = 72 seconds. Therefore, distance traveled = (X – 50) x 5/18 x 72 = 20 (X-50) m = length of train
Equating length of the train in Case I and Case II, we get 5X/3 = 20 (X – 50).
Solving for X, we get X = 600/11 km/hr and the length = 5/3 x 600/11 = 1000/11m
Example 6: An LSS bus and an ordinary bus leave Pune for Chinchwad - a distance of 32 km simultaneously. The ratio between the average speed of the LSS bus and that of the ordinary bus is 3:2. The LSS bus reaches Chinchwad and immediately leaves back for Pune and meets the ordinary bus at Pimpri. What is the distance between Chinchwad and Pimpri?
Sol: Speed of LSS : Speed of ordinary :: 3 : 2.
Since, Distance Speed, Distance traveled by LSS : Distance traveled by ordinary :: 3 : 2
Let distance between Chinchwad and Pimpri be x km.
Then distance traveled by LSS = 32 + x, while the distance traveled by ordinary bus = 32 – x. Therefore, 32 + x : 32 – x :: 3 : 2. Solving for x, we get x = 6.4 km = distance between Pimpri and Chinchwad.
Example 7: Traveling at 6 km/hr, I reach my office 20 minutes late. Traveling at 8 km/hr I reach my office 30 minutes early. What is my usual speed and time taken to reach my office?
Sol: Let my usual speed be S km/hr and my usual time be t hours.
Therefore, Solving for t, we get t = 3 hours.
Since the usual time taken = 3 hours, usual distance traveled = 3S kms,
Equating distance traveled usually, with distance traveled at any of the other two speeds, we get 6 x (3 + 20/60) = 3S. Therefore, S = 6 2/3 km/hr.
Example 8: Raju hikes up a hill at 4 mph and comes down at 6 mph. If the total time taken for the total journey is 3.5 hours, what was the distance between the hilltop and the foothills?
Sol: Average speed = 2ab/a+b = (2x6x4)/10 = 4.8 mph. Time taken = 3.5 hours both ways. So, the two way distance = 4.8 x 3.5 miles = 16.8 miles. Hence, the distance one-way = 8.4 miles.
Example 9: Indrayani leaves Pune for Bombay at 17:30 hours and reaches Bombay at 21:30 hours, while Shatabdi, which leaves Bombay at 17:00 hours, reaches Pune at 20:30 hours. At what time do they pass each other?
Sol: Let the distance between Bombay and Pune = d km.
Indrayani Speed = d/4 km/hr and that of Shatabdi = d/35 km/hr.
Let t be the time in hrs after Shatabdi has left for Pune, when the two trains meet.
Therefore, distance traveled by Shatabdi = d/3.5 x t
And that of Indrayani = d/4 x (t - 30/60) The sum of the distances traveled by the two trains
= distance between Bombay and Pune = d km. Therefore, = d.
Solving for t, we get t = 2.1 hours = 2 hrs 6 min. Hence, the two trains meet at 7:06 pm.