Question 1

The length of Lucknow mail is 120m and that of punjab mail is 80 m. These two trains are running in the same direction with velocities of 40 km/h and 50 km/h respectively. The time taken by them to cross each other is?

Accepted Answer

Relative speed = 50 - 40 = 10 km/h = 50/18 m/s ∴ Time taken = Sum of length of the trains / Relative speed = (200/50) x 18 = 72 sec.

Question 2

A train passes an electric pole in 10 seconds and a platform 120 m long in 18 seconds. Its length in meters is?

Accepted Answer

Let the length of train be x m, then x/10 = (120 + x)/18 ⇒ x = 150m

Question 3

The wheel of an engine of 300 cm in circumference makes 10 revolution in 6 seconds. what is the speed of the wheel (in km/h)?

Accepted Answer

Circumferences means one resolutions . Therefore, distance covered in 10 resolutions =300 x 10 = 30 m i.e., 30 meters in 6 seconds. ∴ Speed of wheel = 30/6 m/s = 5 m/s ∴ 5 m/s = 5 x (18/5) = 18 km/h

Question 4

Anjali fires two bullets from the same place at the interval of 6 minutes but Bhagwat sitting in a car approaching the place of firing hears the second fire 5 min 32 sec after the first firing. What is the speed of car, if the speed of sound is 332 …

Accepted Answer

(speed of wind) / (speed of car) = (Time utilised) / (time saved) ⇒ 332/x = 332/28 ∴ x = 28 m/s

Question 5

A car crosses a man walking at 6 km/h. The man can see the things up to 450m only in one direction due to fog. He sees the car which was going in the same direction for 4.5 minutes. What is the speed of the car?

Accepted Answer

Time = Total distance / Relative speed 4.5/60 hr. = (450/1000) / x ⇒ x = 6 km/h Relative speed = Speed of car - Speed of man 6 = x - 6 ⇒ x = 12 km/h

Question 6

X and Y start walking towards each other at 8:00 am at the speeds of 3 km/h and 4 km/h, respectively. They were initially 17.5 km apart. At what time do they meet?

Accepted Answer

Let they meet after x h. Then, according to the question, 3x + 4x = 17.5 ⇒ 7x = 17.5 ∴ x = 17.5 / 7 = 2.5 h So, they meet 2.5 h after 8:00 am. It means they meet at 10:30 am.

Question 7

A train leaves manipur at 6:00 am and reaches Dispur at 10:00 am. Another train leaves Dispur at 8:00 am and reaches Manipur at 11:00 am. At what time do the two trains cross each other?

Accepted Answer

Let distance between Manipur and Dispur = D km Average speed of train from Manipur = D/4 km/h Average speed of train from Dispur = 2D/7 km/h Let they meet T h after 6:00 am. Then, according to the question, (D/4 x T) + 2D/7 x (T - 2) = D ⇒ T/4 + 2(T - 2)/7 = 1 ⇒ 7T + 8(T - 2) = 28 ⇒ 15T = 44 ∴ T = 44/15 h = 2h 56 min Clearly, trains meet 2 h 56 min after 6:00 am. It means the trains meet at 8:56 am.

Question 8

Anurag traveled a distance of 45 km in 8 h 45 min. He traveled partly on foot at 3 km/h and partly by bicycle at 8 km/h .The distance traveled on the bicycle, is?

Accepted Answer

Let the distance traveled on foot be x km Then, distance converted by bicycle = (45 - x) km ∴ x/3 + (45 -x) /8 = 83/4 = 35/4 ⇒ (8x + 135 - 3x)/24 = 35/4 ⇒ 5x + 135 = 210 ⇒ 5x = 75 ⇒ x = 15 ∴ Distance converted by bicycle = (45 - 15) = 300 km

Question 9

Running at a speed of 60 km/h, a train passed through a 1.5 km long tunnel in 2 min. What is the length of the train?

Accepted Answer

Suppose length of the train = L m Speed of the train = 60 km/h = 60 x 1000 = 60000 m/h Length of tunnel = 1.5 km = 1500 m Time taken by train = 2 min = 1/30 h Time = Distance/Speed ∴ 1/30 = L + 1500/60000 ⇒ l = 500 m

Question 10

The average speed of a train in the onward journey is 25% more than that of the return journey. The train halts for one hour on reaching the destination. The total time taken for the complete to and fro journey is 17 h covering a distance of 800 km.…

Accepted Answer

Let the speed of the train during returning journey be x km/h Speed during onward journey = x + 25x/100 = 5x/4 km/h. Distance coverd in onward journey = 800 / 2 = 400 km Total time taken = Covered distance / Speed Time taken by train in onward journey = 400/(5x/4) and time taken in returning journey = 400/x Thus, 400/(5x/4) + 400/x = 16 ⇒ 320/x + 400/x = 16 ⇒ 16x = 720 ∴ x = 45 km/h Speed of the train in the onward journey = 5 x 45/4 = 56.25 km/h

Question 11

The ratio between the speed of Menna and Tenna is 2 : 3. Meena takes 20 min more than Tenna to walk from A to B. If Menna had walked at double the speed, find the time she would take to from A to B.?

Accepted Answer

Since, ratio of speeds of Meena and teena is 2 : 3. ∴ Ratio of time taken = 3 : 2 If, Teena takes x min to walk from A to B, then Meena takes (x + 20) min. ∴ (x + 20)/x = 3/2 ⇒ 2x + 40 = 3x ∴ x = 40 min Hence, Meena takes 60 min walking at her usual speed and at double the speed, she would take 30 min.

Question 12

One local and another express train were proceeding in the same direction on parallel tracks at 29 km/h and 65 km/h, respectively. The driver of the former noticed that it took exactly 16 s for the faster train to pass by him. What is the length of …

Accepted Answer

Relative speed of express train to local train = 65 - 29 = 36 km/h = 36 x (5/18) m/s = 10 m/s ∴ Length of faster train = 10 x 16 = 160 m

Question 13

A man performs 2/25 of his total journey by bus, 21/50 by car and the remaining 2 km on foot. Find the total distance covered in the whole journey.?

Accepted Answer

Let total distance covered in the whole journey = L km ∴ 2L/25 + 21L/50 + 2 = L ⇒ L = 4 ∴ Total distance covered = 4 km

Question 14

A man in car notices that he can count 25 telephone posts in 1 min. If they are known to be 40 m apart, then at what speed is the car travelling?

Accepted Answer

Number of gaps between 25 telephone posts = 24 Distance travelled in 1 min = 40 x 24 = 960 m Speed = (960 x 60)/1000 = 57.6 km/h

Question 15

A man walks 7.5 km at a speed of 3 km/h. At what speed would the man need two walk during the next 2 h to have an average of 4 km/h for the entire session?

Accepted Answer

Let speed of walking be V km/h. Total time taken = (7.5/3) + 2 = 4.5 h Total distance covered = (7.5 + 2V) km ∴ (7.5 + 2V)/4.5 = 4 ⇒ 7.5 + 2V = 18 ⇒ 2V = 10.5 ⇒ V = 5.25 ∴ Speed of walking = 5.25 km/h

Question 16

Ram and Sita are walking towards each other with the speed of 8 km/h and 2 km/h, respectively over a road 160 km long. How soon will they meet?

Accepted Answer

Distance = 160 km Relative Speed = 8 + 2 = 10 Time = Distance/Relative speed = 160/10 = 16 h

Question 17

Two train leaves New Delhi at the same time. One travels towards North at 60 km/h and the other travels towards South at 40 km/h . After how many hours will the trains be 150 km apart?

Accepted Answer

Relative speed = 60 + 40 = 100 km/h Time = Distance/speed = 150/ 00 = 3/2 h

Question 18

Walking 5/7 of his usual rate, a boy reaches his school 6 min late. Find his usual time to reach the school?

Accepted Answer

Since, the boys now walks at 5/7 of usual speed, so he will take 7/5 of his usual time. ⇒ Extra time = (7/5 - 1) x Usual time = 6 min (known) ⇒ 2/5 x Usual time = 6 ∴ Usual time = 15 min

Question 19

A car during its journey travels 30 min at a speed of 40 km/h, another 45 min at a speed of 60 km/h and 2 h at a speed of 70 km/h. Find its average speed?

Accepted Answer

Average speed = [(30/60) x 40 + (45/60) x 60 + (2 x 70)] / [(30/60) + (45/60) + 2] = (20 + 45 + 140)/[(30 + 45 + 120)/60] = (205/195) x 60 km/h = 63.07 km/h = 63 km/h

Question 20

A train running at 8/11 of its own speed reached a place in 5 1/ 2 h. How much time could be saved, if the train would have run as its own speed?

Accepted Answer

New speed = 8/11 of usual speed New time = 11/8 of usual time ∴ 11/8 of usual time = 11/2 h ⇒ Usual time = (11 x 8)/(2 x 11) = 4 h ∴ Time saved = 51/2- 4 = 11/2

Time and Distance Questions