Question 1

A car reached Raipur from sonagarh in 35 min with an average speed of 69 km/h. If the average speed is increased by 36 km/h, how long will it take to cover the same distance?

Accepted Answer

Distance between Sonagarh and Raipur = Average speed x time = 69 x (35/60) km = 161/4 km New speed = (69 + 36) km/h = 105 km/h ∴ Required time = Distance/speed = 161/(4 x 105) h = (161 x 60)/(4 x 105) min = 23 min

Question 2

A student walks from his house at 2 1/ 2 km/h and reaches his school late by 6 min. Next day, he increase his speed by 1 km/h and reaches 6 min before school time. How far is the school form his house?

Accepted Answer

Let the required distance = L According to the question, L/(5/2) - L/(7/2) = 12/60 (difference between two times = 6 + 6 =12 min) ⇒ 2L/5 - 2L/7 = 1/5 ⇒ 14L - 10L = 7 ⇒ 4L = 7 ⇒ L = 7/4 km

Question 3

A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman, chases him. The thief and the policeman run at rate of 10 km/h and 11 km/h, respectively. The distance between them after 6 min will be?

Accepted Answer

Relative speed of policeman with respect to thief = ( 11 - 10 ) = 1 km/h. Now, relative distance covered by policeman in 6 min = Speed x Time = (1 x 6)/60 = 1/10 km = 100 m The distance between the policeman and thief after 6 min = (200 - 100) = 100 m

Question 4

A truck covers a distance of 368 km at a certain speed in 8 h. How much time would a car take at an average speed which is 18 km/h more than that of the speed of the truck to cover a distance which is 16 km more than that travelled by the truck?

Accepted Answer

Speed of truck = Distance/time = 368/8 = 46 km/h Now, speed of car = (speed of truck + 18) km/h = (46 + 18) = 64 km/h and Distance travelled by car = 368 + 16 = 384 km ∴ Time taken by car = Distance/Speed = 384/64 h = 6 h

Question 5

Shantanu drives a motorcycle and covers a distance of 715 km at a constant speed. If the speed of the motorcycle had been 10 km/h more, he would have taken 2 h less to cover the same distance. What is the original speed of the motorcycle?

Accepted Answer

Let the constant speed = V According to the question, 715/V - 715/(V + 10) = 2 ⇒ 1/V - 1/(V + 10) = 2/715 ⇒ [(V + 10) - V] / [V(V + 10)] = 2/715 ⇒ V2 + 10V = 10 x 715/2 ⇒ V2 + 10V - 3575 = 0 ⇒ V2 + 65V - 55V - 3575 = 0 ⇒ V(V + 65) - 55(V + 65) = 0 ⇒ (V - 55) (V + 65) = 0 ∴ V = 55 or V = - 65 Ignoring the negative value, we get V = 55 ∴ Original speed of the motorcycle = 55 km/h

Question 6

The ratio of speeds of a train and a car is 16 : 15 , respectively and a bus covered a distance of 480 km in 8h. The speed of the bus is 3/4th of the speed of train. What will be covered by car in 6 h?

Accepted Answer

Let speed of train = 16k Speed of car = 15k Speed of bus = 480/8 = 60 km/h According to the question, 16k = 60 x 4/3 = 80 ⇒ k = 5 ∴ Speed of car = 15k = 15 x 5 = 75 km/h ∴ Required distance = Speed x Time = 75 x 6 = 450 km

Question 7

Sumit drove at the speed of 45 km/h from home to a resort, returning over the same route, he got stuck in traffic and took an hour longer. Also, he could drive only at the speed of 40 km/h. How many kilometers did he drive each way?

Accepted Answer

Let distance he drove be L. Using Time = Distance/speed From given condition. L/40 - L/45 = 1 ⇒ (45L - 40L)/(40 x 45) = 1 ⇒ 5L = 40 x 45 ⇒ L = 360 km

Question 8

A car runs at the speed of 40 km/h when not serviced and runs at 65 km/h when serviced. After servicing, the car covers a certain distance in 5 h. How much approximate time will the car take to cover the same distance when not serviced?

Accepted Answer

When serviced, the distance covered by the car = 65 x 5 = 325 km. When not serviced, the time taken by the car to cover 325 km at 40 km/h = 325/40 = 8.125 = 8 h (approx)

Question 9

The ratio between the rates of travelling of A and B is 2 : 3 and therefore A takes 20 min more than time taken by B to reach a destination. If A had walked at double the speed, how long would he have taken to cover the distance?

Accepted Answer

We know that speed is inversely proportional to time. Therefore, Time taken by A : Time taken by B = 1/2 : 1/3 Let B takes T min. Then, A takes (T + 20) min. ∴ (T + 20) : T = 1/2 : 1/3 = 3 : 2 ⇒ (T + 20)/T = 3/2 ⇒ 2T + 40 = 3T ∴ T = 40 ∴ A takes (T + 20) = (40 + 20) = 60 min. ∴ At double the speed, A would have covered it in 30 min.

Question 10

A car travels a distance of 75 km at the speed of 25 km/h. It covers the next 25 km of its journey at the speed of 5 km/h and the last 50 km of its journey at the speed of 25 km/h. What is the average speed of the car?

Accepted Answer

Time taken to cover first 75 km = 75/25 = 3 h Time taken to cover next 25 km = 25/5 = 5 h Time taken to cover last 50 km of its journey = 50/25 = 2 h Total distance = 75 + 25 + 50 = 150 km Total time taken = 3 + 5 + 2 = 10 h ∴ Required average speed = Total distance / Total time taken = 150/10 = 15 km/h

Question 11

A car runs at the speed of 50 km/h when not serviced and runs at 60 km/h, when serviced. After servicing, the car covers a certain distance in 6 h. How much time will the car take to cover the same distance when serviced?

Accepted Answer

After sevicing speed of car = 60 km/h ∴ Distance covered in 6 h = 60 x 6 km = 360 km. When not serviced, then time taken to cover 360 km = 360/50 = 7.2 h

Question 12

A and B are two station 1000 km apart. A train starts from A and moves towards B at 40 km/h. Another train starts from B at the same time and moves towards A at 60 km/h. How far from A will they cross each other?

Accepted Answer

Let they meet x km from A. Then, according to the question, x/40 = (1000 - x)/60 ⇒ 60 x = 40(1000 - x) ⇒ 3x = 2000 - 2x ⇒ 3x + 2x = 2000 ⇒ 5x = 2000 km ∴ x = 2000/5 = 400 km

Question 13

A train runs for 3 h at the speed of 40 km/h and then for 4 1/ 2 h at the speed of 60 km/h. Thus, it covers 3/5th part of the whole distance. It wants to cover the remaining distance in 4 h, find the average speed of the train for remaining distance?

Accepted Answer

Let total distance = D Now, according to the question, 40 x 3 + 60 x 4.5 = 3D/5 ⇒ 120 + 270 = 3D/5 ∴ D = 390 x 5/3 = 650 km ∴ Remaining distance = [650 - (3/5) x 650] = 650 - 390 = 260 km Required average speed = 260/4 = 65 km/h

Question 14

Two cars A and B start simultaneously from a certain place at the speed of 30 km/h and 45 km/h, respectively.The car B reaches the destination 2 h earlier than A. What is the distance between the starting point and destination?

Accepted Answer

Let the time taken by car B to reach destination is T h. So, the time taken by car A to reach destination is (T + 2) h. Now, S1T1 = S2T2 ⇒ 30(T + 2) = 45 x T ⇒ 30T + 60 = 45T ⇒ 15T = 60 ∴ T = 4 h Now, distance between starting point and destination = S2T2 = 45 x 4 = 180 km

Question 15

Two ladies simultaneousely leave cities A and B connected by a strainght road and travel towards each other. The first lady travels 2km/h faster than the second lady and reaches B one hour before the second lady reaches A. The two cities A and B are…

Accepted Answer

Let the speed of first lady be (v + 2)km/h and speed of second lady will be v km/h. Total distance = 24 km For first lady, Speed = Distance/Time v + 2 = 24/t ...(i) Let time taken by first lady to cover distance of 24 km be t h. Then, time taken by second lady to cover same distance will be(t + 1)h. For second lady, Speed =Distance/time v = 24/(t + 1) ...(ii) Form Eqs.(i) and (ii), we get 24/t - 2 = 24/(t + 1) ⇒ (24 - 2t)/t = 24/(t + 1) ⇒ (24 - 2t)(t + 1) = 24t ⇒ 24t - 2t2 + 24 - 2t = 24t ⇒ 2t2 + 2t - 24 = 0 ⇒ t2 + t - 12 = 0 ⇒ t2 + 4t - 3t - 12 = 0 ⇒ t(t + 4) -3(t + 4) = 0 ⇒ (t - 3)(t + 4) = 0 ⇒ t = 3, -4 ∴ t = 3 (t ≠- 4) ∴ The distance travelled by the first lady in one hour = 24/t = 24/3 = 8 km And distance travelled by the second lady in one hour = 24/t + 1 = 24/3 + 1 = 6 km

Question 16

The speed of a car increases by 2 kms after every one hour. If the distance travelling in the first one hour was 35 kms. what was the total distance travelled in 12 hours?

Accepted Answer

Total distance travelled in 12 hours =(35+37+39+.....upto 12 terms) This is an A.P with first term, a=35, number of terms, n= 12,d=2. Required distance = 12/2[2 x 35+{12-1) x 2] =6(70+23) = 552 kms.

Question 17

A person crosses a 600 m long street in 5 minutes, What is his speed in km per hour?

Accepted Answer

Speed = 600 5 × 60 m / s e c = 2 m/sec = 2 x (18/5) km/hr = 7.2 km/hr

Question 18

A thief is noticed by a policeman from a distance of 200 m. The thief starts running and the policeman chases him. The thief and the policeman run at the rate of 10 km and 11 km per hour respectively. What is the distance between them after 6 minute…

Accepted Answer

Relative speed of the thief and policeman = (11 – 10) km/hr = 1 km/hr Distance covered in 6 minutes = (1/60) x 6 km = 1/10 km = 100 m Therefore, Distance between the thief and policeman = (200 – 100) m = 100 m.

Question 19

Two trains A and B start simultaneously in the opposite direction from two points P and Q and arrive at their destinations 16 and 9 hours respectively after their meeting each other. At what speed does the second train B travel if the first train tr…

Accepted Answer

s 1 s 2 = t 2 t 1 ⇒ 120 s 2 = 9 16 = 3 4 ⇒ s 2 = 160 k m / h

Question 20

Two horses start trotting towards each other, one from A to B and another from B to A. They cross each other after one hour and the first horse reaches B, 5/6 hour before the second horse reaches A. If the distance between A and B is 50 km. what is …

Accepted Answer

If the speed of the faster horse be f s and that of slower horse be s s then f s + s s = 50 1 = 50 and 50 s s - 50 f s = 5 6 Now, you can go through options. The speed of slower horse is 20km/h Since, 20+30=50 and 50 20 - 50 30 = 5 6

Time and Distance Questions