**Example 1:** Driving 5/4^{th} 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

Must Read Time Speed and Distance Articles

**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.