Question 1

Akbar and Birbal setout at the same time to walk towards each other respectively from Agra and Banaras 144 km apart. Akbar walks at the constant speed of 8 km/h, while Birbal walks 4 km in the first hour, 5 km in the second hour, 6 km in the third h…

Accepted Answer

Distance travelled by them in first hour = 12 km Distance travelled by them in second hour = 13 km Distance travelled by them in third hour = 14 km and so on Thus in 9 hours they will cover exactly 144 km and in 9 h each will cover half the total distance. ( 8 x 9 = 72 and 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 72)

Question 2

Amarnath express left Amritsar for Gorakhpur. Two hours later Gorakhnath express left from Amritsar to Gorakhour. Both trains reached Gorakhpur simultaneously. If Amarnath express had started from Amritsr and Gorakhnath express had started from Gora…

Accepted Answer

Let Amarnath express takes H hours, then Gorakhnath express takes (H - 2) hours. ∴ 1/H + 1/(H - 2) = 60/80 H = 4 hrs.

Question 3

A man goes to the fair in Funcity with his son and faithful dog. Unfortunately man misses his son which he realises 20 minutes later. The son comes back towards his home at the speed of 20 m/min and man follows him at 40 m/min. The dog runs to the s…

Accepted Answer

In 20 minutes the difference between man and his son = 20 x 20 = 400 m Distance travelled by dog when he goes towards son = (400/40) x 60 = 600 m and time required is 10 minutes In 10 minutes the remaining difference between man and son = 400 - (20 x 10) = 200 m Note: Relative speed of dog with child is 40 km/h and the same with man is 100 km/h. Time taken by dog to meet the man = 200/100 = 2 min In 2 minute the remaining distance between child and man = 200 - ( 2 x 20) = 160 m Now, the time taken by dog to meet the child again = 160/40 = 4 min In 4 minutes he covers 4 x 60 = 240 m distance while going towards the son. In 4 minute the remaining distance between man and child = 160 - (4 x 20) = 80 m Time required by dog to meet man once again = 80/100 = 0.8 min In 0.8 min remaining distance between man and child = 80 - (0.8 x 20) = 64 min Now, time taken by dog to meet the child again = (64/40) x (8/5) min ∴ Distance travelled by dog = (8/5) x 60 = 96 m Thus, we can observe that every next time dog just go 2/5th of previous distance to meet the child in the direction of child. So, we can calculate the total distance covered by dog in the direction of child with the help of GP formula. Here, first term (a) = 600 and common ratio (r) = 2/5 Sum of the infinite GP = a/(1 - r) = 600/(1 - 2/5) = (600 x 5)/3 = 1000 m

Question 4

A hunter fired two shots from the branch of a tree at an interval of 76 seconds. A tiger separating too fast hears the two shots at an interval of 83 seconds. If the velocity of the sound is 1195.2 km/h, then find the speed of tiger?

Accepted Answer

In case of increasing gap between two objects. Speed of sound / speed of tiger = Time utilised / Difference in time ⇒ 1195.2 / x = 83 / 7 x = 100.8 km/h

Question 5

A soldier fired two bullets at an interval of 335 seconds moving at a uniform speed v 1. A terrorist who was running ahead of the soldier in the same direction, hears the two shots at an interval of 330 seconds? If the speed of sound is 1188 km/h, t…

Accepted Answer

Speed of wind (sound) / Relative speed of soldier and terrorist = Time utilised / Difference in time 1188/x = 330/5 x = 18 km/h and solder is faster.

Question 6

In reaching the Purnagiri a man took half as long again to climb the second third as he did to climb the first third and a quarter as long again for the last third as for the second third. He took altogether 5 hr 50 minutes. Find the time he spent o…

Accepted Answer

Let the time taken in first third part of the journey be x minutes, then the time required in second third part of the journey is 3x/2 and in the last third part of the journey time required is 15x/8. Therefore, x + (3x/2) + (15x/8) = 350min ⇒ x = 80 min

Question 7

A man travels the first part of his journey at 20 km/h and the next at 70 km/h, covering the entire journey at an average speed of 50 km/h. What is the ratio of the distance that he covered at 20 km/h to that he covered at 70 km/h?

Accepted Answer

x/20 + y/70 = (x + y)/50 ⇒ (70x + 20y)/1400 = (x + y)/50 ⇒ 42x = 8y ⇒ x/y = 4/21

Question 8

A man decides to travel 80 km in 8 h party by foot and party on a bicycle. If his speed on foot is 8 km/h and on bicycle 16 km/h, what distance would he travel on foot?

Accepted Answer

Here, S1 = 8 km/h and S2 = 16 km/h ∴ d1 =s1 x t1 = 8t1 ...(i) and d2 = s2 x t2 = 16t2 ...(ii) We know that, t1 + t2 = 8 ...(iii) and d1 + d2 = 80 (given) ...(iv) From Eqs.(i) and (ii) put the value of d1 and d2 in Eq. (iv), we get d1 + d2 = 80 8t1 + 16t2 = 80 ⇒ 8t1 + 8t2 + 8t2 = 80 8(t1 + t2) + 8t2 = 80 From Eq (iii), 8 x 8 + 8t2 = 80 8t2 = 80 - 64 = 16 ⇒ t2 = 16/8 = 2h ∴ t1 = 8 - 2 = 6 h ∴ Distance travelled by foot = d1 = 8 x 6 = 48 km.

Question 9

A car driver covers a distance between two cities at a speed of 60 km/h and on the return his speed is 40 km/h. He goes again from the 1st to the 2nd city at twice the original speed and returns at half the original return speed. Find his average sp…

Accepted Answer

Required average speed = 4/(1/60 + 1/40 + 1/120 +1/20) = (4 x 120)/12 = 40 km/h

Question 10

The distance between two station 'X' and 'Y' is 450 km. A train L starts at 6:00 pm from X and moves towards Y at an average speed of 60 km/h. Another train M starts from Y at 5:20 pm and moves towards X at an average speed of 80 km/h How far from X…

Accepted Answer

Let two trains meet at a km from 'X' [Time taken by M to cover (450 - a)km] - [Time taken by the L to cover a km ] = 40/60 ⇒ (450 - a)/80 - a/60 = 40/60 ⇒ (450 - a)/80 = 40/60 + a/60 ⇒ (450 - a)/8 - (a + 40) / 6 = 0 ⇒ 3(450 - a) - 4(a + 40) = 0 ⇒ 7a = 1190 ∴ a = 1190 / 7 = 170 Time taken by L to cover 170 km = 170/60 h = 2 h 50 min So, the two trains will meet 2 h 50 min after 6:00 pm. it means that the two trains will meet at 8 : 50 pm .

Question 11

Two cars A and B are running towards each other from two different place 88 km apart. If the ratio of the speeds of the cars A and B is 5:6 and the speeds of the cars B is 90 km/h, after what time will they meet each other?

Accepted Answer

Let speed of A = 5x Then, spend of B = 6x Given that,speed of B = 6x = 90 ∴ x = 90/6 = 15 ∴ Speed of A = 5x = 5x = 5 x 15 = 75 km/h Let A and B meet after T h. Then, 75 x T + 90 x T =88 ∴ T = 88/165 h = (88 x 60)/165 = 32 min

Question 12

A car starts running with the initial speed of 40 km/h with its speed increasing every hour by 5 km/h. How many hours will it taken to cover a distance of 385 km?

Accepted Answer

Required number of hours is the number of terms of the series 40 + 45 + 50 +... as speed increases every hour. Given, sum of the series is 385 ∴ a = 40, d=5, s = 385 and n= ? Using S = n/2[2a + (n - 1)d] ⇒ 385 = n/2[80 + 5n - 5] ⇒ 770 = 5n2 + 75n ⇒ n2 + 15n - 154 = 0 ⇒ n2 + 22n - 7n - 154 = 0 ⇒ n(n + 22) - 7(n + 22) = 0 ⇒ (n + 22) (n - 7) = 0 ∴ n = 7 Remaining part = 1/12 - 1/20 = (5 - 3)/60 = 2/60 = 1/30 i.e., 1/30 th part is filled by B in 1 min Hence, required time to fill the whole tank = (165 + 1 + 1) min = 167 min

Question 13

An express train travelled at an average speed of 100 km/h, stopping for 3 min after every 75 km. A local train travelled at a speed of 50 km/h, stopping for 1 min after every 25 km. If the trains began travelling at the same time, how many kilometr…

Accepted Answer

Time taken by express train to cover 75 km including stoppage = (60 / 100 x 75) min + = 48 min Time taken by express train to cover 600 km. = Time taken by it to cover 75 km = (48 / 75 x 525 ) min + (60 / 100 x 75 ) min = (336 + 45) min = 381 Time taken by local train to9 cover 25 km including stoppage = (60 / 50 x 25) min + 1 min = 31 min In 31 min, distance covered = 25 km In(31 x 12) min. distance covered = (25 / 31 x 31 x 12 ) = 300 km In last 9 min, distance covered = (25 / 31 x 9 ) = 7.25 km ∴ Required total distance = 300 + 7.25 = 307.25 km

Question 14

A student goes to his school from his house walking at 4 km/h and reaches his school 10 min late. Next day, starting at the same time he walks as 6 km/h and reaches his school 5 min earlier than the scheduled time. Find the distance between school a…

Accepted Answer

Let the distance be x km. ∴ Time = Distance/speed 1st day time taken = x/4 h 2nd day time taken = x/6 h Difference in time = (x/4 - x/6) h Actual difference between these two times = 15 min = 1/4 h ∴ x/4 - x/6 = 1/4 ⇒ (3x - 2x)/12 = 1/4 ∴ x = 3 km

Question 15

Two train of length 250 m and 140 m are running on parallel lines in the same direction at 58 km/h and 32 km/h, respectively. Find the time taken by the slower train to pass the driver of the faster one?

Accepted Answer

Relative speed = (58 - 32) = 26 km/h = (26 x 5/18) m/s = 65/9 m/s Time taken to cross each other = (250 + 140)/(65/9) = (390 x 9)/65 = 54 s

Question 16

A person travelled a distance of 50 km in 8 h. He covered a part of the distance on foot at the rate of 4 km/h and a part on a bicycle at the rate of 10 km/h. How much distance did he travel on foot?

Accepted Answer

Let T h, he travels on foot. ∴ 4T + 10(8 - T) = 50 ⇒ 80 - 6T = 50 ⇒ 6T = 30, ∴ T = 5 h ∴ The distance travelled on foot = 4 x 5 = 20 km

Question 17

Total time taken by a person in going to a place by walking and returning on cycle is 5 h 45 min. He would have gained 2 h by cycling both ways. The time taken by him to walk both ways, is?

Accepted Answer

Walking time + Cycling time = 5 h 45 min = 345 min ....(i) If he had cycled both way he would have gained 2 h (120 min). ∴ 2 x Cycling time = 345 - 120 = 225 min ...(ii) Walking time = 2 x 345 - 225 = 690 - 225 = 465 min Time taken by him to walk both ways = 7 h 45 min

Question 18

Amit start from a point A and walks to another point B and then return from B to A by his car and thus takes a total time of 6 h 45 min. If he had driven both ways in his car, he would have taken 2 h less. How long would it take for him to walk both…

Accepted Answer

Let w be the time taken in one way by walking and c be the time taken in one way by car. Now, according to the question, In first case, w + c = 6 h 45 min or 2w + 2c = 13 h 30 min ..(i) In second case, 2c = 4 h 45 min ....(ii) From Eqs. (i) and (ii), we get 2w + 2c = 13 h 30 min ⇒ 2w + 4h 45 min = 13 h 30 min ⇒ 2w = 13 h 30 min - 4 h 45 min ⇒ 2w = 8 h 45 min ∴ If he walks both ways, then time taken = 8 h 45 min

Question 19

Two cars X and Y starts from two places A and B respectively which are 700 km apart at 9 am. Both the cars run at an average speed of 60 km/h, Car X stops at 10 am and again starts at 11 am, while the other car Y continues to run without stopping. W…

Accepted Answer

Since, the speed of X and Y are 60 km/h each. The distance between A and B is 700 km. Distance traveled by X upto 11 am is 60 km. Since, X stops at 10 am and distance traveled by up to 11 am is 120 km. Now, the distance between them = 700 - 180 = 520 km Now, let they will meet at a distance of x km from X's position. ∴ Distance traveled by X = x km and distance traveled by Y = (520 - x) km ∴ x/60 = (520 - x)/60 ⇒ 2x = 520 ⇒ x = 260 ∴ T = 260/60 = 41/3 h Thus, they will cross each other after, 41/3 h after 11 am i.e., at 3 : 20 pm.

Question 20

A car covers a distance from town A to town B at the speed of 58 km/h and covers the distance from town B to town A at the speed of 52 km/ h. What is the Approximate average speed of the car?

Accepted Answer

We know that, if two equal distances are covered at two different speeds A and B, then Average speed = 2AB/(A + B) = (2 x 58 x 52)/(58 + 52) = 54.8 km/h = 55 km/h (Approx)

Time and Distance Questions