Question 1

The ratio between the speeds of two cares is 7 : 8. If the 2nd car runs 200 km in 5 h, then find the speed of the 1st car.?

Accepted Answer

Let speed of the 1st car = 7k and speed of the 2nd car = 8k According to the question, 8k = 200/5 = 40 [ using, speed = Distance/ Time ] ∴ k = 40/8 = 5 ∴ Speed of the 1st car = 7k = 7 x 5 = 35 km/h

Question 2

The ratio between the speeds of two buses is 5 : 3. If the 1st bus runs 400 km in 8 h, then find the speed of the 2nd bus.?

Accepted Answer

Let speed of 1st bus = 5k and speed of 2nd bus = 3k According to the question, 5k = 400/8 = 50 ∴ k = 50/5 = 10 ∴ Speed of 2nd bus = 3k = 3 x 10 = 30 km/h

Question 3

The speeds of three cars are in the ratio of 2 : 3 : 5 find the ratio of the time taken by the above cars to travel the same distance.?

Accepted Answer

We know that speed and time are inversely proportional. ∴ Ratio of time taken = 1/2 : 1/3 : 1/5 = 30/2 : 30/3 : 30/5 [ LCM of 2, 3, 5 = 30 ] = 15 : 10 : 6

Question 4

Nilu covers a distance by walking for 6 h. While returning, his speed decreases by 2km/h and he takes 9 h to cover the same distance. What was her speed while returning?

Accepted Answer

Let the speed in return journey = x According to the question, 6( x + 2 ) = 9x ⇒ 6x + 12 = 9x ⇒ 9x - 6x = 12 ⇒ 3x = 12 ∴ x = 12/3 = 4km/h

Question 5

A takes 4 h more than the time taken by B to walk D km. If A doubles his speed, he can make it in 2 h less than that of B. How much time does B require for walking D km?

Accepted Answer

Let B takes x h to walk D km. Then, A takes ( x + 4 ) h to walk D km. With doube of the speed, A will take ( x+ 4)/2 h. According to the question, x - (x + 4)/2 = 2 ⇒ 2x - (x + 4) = 4 ⇒ 2x - x - 4 = 4 ∴ x = 4 + 4 = 8 h

Question 6

If Sohail walks from his home to office at 16 km/h, he is late by 5 min. If he walks at 20 km/h, he reaches 10 min before the office time. Find the distance of his office from his house.

Accepted Answer

Let required distance = L According to the question, L/16 - L/20 = 15/20 ⇒ (5L - 4L)/80 = 1/4 ∴ L = (1/4) x 80 = 20 km

Question 7

Walking with 3/4 of his usual speed, Sachin covers a certain distance in 2 h more than the time he takes to cover the distance at his usual speed. Find the time taken by him to cover this distance with usual speed.

Accepted Answer

Let th distance = D and usual speed = V According to the question, D/(3V/4) - D/V = 2 ∴ D/V = 3 x 2 = 6 Time taken to cover the distance with usual speed = 6 h

Question 8

A car covers a distance of 200 km in 2 h 40 min whereas a jeep covers the same distance in 2 h. What is the ratio of their speeds?

Accepted Answer

Speed of car = Distance/Time = 200/240/60 = 600/8 = 75 km/h Speed of jeep = 200/2 = 100 km/h ∴ Required ratio = 75/100 = 3/4 3 : 4

Question 9

Amit walks at a uniform speed of 4 km/h and 4 h after his start, Brijesh cycles after him at the uniform rate of 20 km/h. How far from the starting point will Brijesh catch Amit?

Accepted Answer

Let Brijesh Catches Amit after x h. Then, distance travelled by Amit in (x + 4) h = Distance travelled by Brijesh in x h. According to the question, 4(x + 4) = 20 x ⇒ 4x + 16 = 20x ⇒ 16x = 16 ∴ x = 16/16 = 1 Distance travelled by Brijesh in1h = 1 x 20 = 20 km

Question 10

A train travel at the rate of 50 km/h without stoppages and it travels at 40 km/h with stoppages. How many minutes does the train stop on an average per hour?

Accepted Answer

Due to stoppages, it covers 10 km less per hour. Time taken to cover 10 km = (10/50) x 60 = 12 min Hence, the train stops on an average 12 min per hour.

Question 11

A bus travels at the rate of 54 km/h without stoppages and it travels at 45 km/h with stoppages. How many minutes does the bus stop on an average per hour?

Accepted Answer

Due to stoppages, bus covers 9 km less per hour. Time taken to cover 9 km =(9/54) x 60 = 10 min Hence, the train stops on an average 10 min per hour.

Question 12

A bus covers a certain distance in 16 h. It covers half the distance at 40 km/h and the rest at 60 km/h. Find the length of the journey.?

Accepted Answer

Let the total distance covered = L = Length of journey According to the question, (L/2 x 1/40) + (L/2 x 1/60) = 16 ⇒ L/80 + L/120 = 16 ⇒ (3L + 2L)/240 = 16 ⇒ 5L = 16 x 240 ∴ L = (16 x 240)/5 = 16 x 48 = 768 km

Question 13

A man started 20 min late and travelling at a speed of 1 1/ 2 times of his usual speed reaches his office in time. The time taken by the man to reach his office at his usual speed is?

Accepted Answer

Let usual speed of a man is V and time is t h. By the formula, V1t1 = V2t2 Vt = 11/2 V(t - 20/60) ⇒ t = 3/2(t - 1/3) ⇒ 3/2 t - t = 1/2 ∴ t = 1 h Hence, the time taken by the man to reach his office at his usual speed is 1 h.

Question 14

X and Y start walking toward each other at 8 : 00 am at the speed 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 T h. then, according to the question, 3T + 4T = 17.5 ⇒ 7T = 17.5 ∴ T = 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 15

If you travel 39 km at a speed of 26 km/h, another 39 km at a speed of 39 km/h and again 39 km at a speed of 52 km/h, what is your average speed for the entire journey?

Accepted Answer

Required average speed = Total distance covered/Total time taken = (3 x 39)/[39/26 + 39/39 +39/52] = 3 x 13/(1/2 + 1/3 + 1/4) = 3 x 13 x 12/13 = 36 km/h

Question 16

Dalbir Singh, a polieman, is 114 m behind a thief. Dalbir Singh runs 21 m and the thief 15 m in a minute. In what time will Dalbir singh catch the thief?

Accepted Answer

Relative speed = 21 - 15 = 6 m per minute. Relative distance = 114 m Time taken to catche = 144/6 min = 19 min

Question 17

A person travels a certain distance at 3 km/h and reaches 15 min late. If he travels at 4 km/h, he reaches 15 min earlier. The distance he has to travel is.?

Accepted Answer

Let the certain distance be d and time t. Now, by given condition, d/3 = ( t + 15 ) min ( t +15 )/60 h ⇒ 20d = t + 15 ⇒ t = 20d - 15 ......(i) And from another condition, d/4 = (t - 15) min = ( t - 15 )/60 h ⇒ 15d = t - 15 ⇒ t = 15d + 15 ...(ii) From Eqs. (i) and (ii), we get 20d - 15 = 15d + 15 ⇒ 5d = 30 ∴ d = 6 km.

Question 18

A man covers a certain distance on scooter. Had he moved 3 km/h faster, he would have taken 40 min less. If he had moved 2 km/ slower, he would have taken 40 min more. The distance (in km) is?

Accepted Answer

Let distance and original speed of the man be d km and s km/h. Then, d/s - d/s + 3 = 2/3 ⇒ [d(s + 3 - s)]/[s(s + 3)] = 2/3 ⇒ 9d = 2s(s + 3) ...(i) And d/(s - 2) - d/s = 2/3 ⇒ [d(s - s + 2)]/[s(s - 2)] = 2/3 ⇒ 3d = s(s - 2) ...(ii) From Eqs.(i) and (ii), we get 3s(s - 2) = 2s(s + 3) ⇒ 3s2 - 6s = 2s2 + 6s ⇒ s2 = 12s ⇒ s = 12 From Eq.(ii), we get 3d = 12(12 - 2) ∴ d = 40 km

Question 19

A man riding a bicycle form his house at 10km/h and reaches his office late by 6 min. He increases his speed by 2km/h and reaches 6 min before. How far is the office form his house?

Accepted Answer

Here, a = 10 km/h b = ( 10 + 2 ) = 12 km/h, t1 = 6 min = 1/10 h and t2= 1/10 h. ∴ Required distance = ab( t1 + t2 )/(b - a) = 10 x 12 x ( 1/10 + 1/10 ) /(12 - 10) = (10 x 12 x 1)/(5 x 2) = 12km

Question 20

A walks at a uniform rate of 2 km/h and 2 h after his start, B cycles after him at the uniform rate of 5 km/h. How far from the starting point will B catch A?

Accepted Answer

Let B catches A after T h. Then, distance travelled by A in (T + 2) h = Distance travelled by B in T h According to the question, 2 (T + 2 ) = 5T ⇒ 2T + 4 = 5T ⇒ 3T = 4 ∴ T = 4/3 h Distance travelled by B in 4/3 h = (4/3) x 5 = 20/3 = 62/3 km

Time and Distance Questions