BANK & INSURANCE (SPEED TIME AND DISTANCE) PART 2

Total Questions: 45

31. A policeman spotted a thief at a distance of 185 metres in from him. If both started running at the same time, then find the time taken by the policeman to catch the thief given that the policeman and the thief are running at the speed of 126 km/hr and 108 km/hr, respectively

Correct Answer: (c) 37 seconds 
Solution:Speed of policeman = 126 × (5/18) = 35 m/s
Speed of Thief = 108 × (5/18) = 30 m/s
Relative speed of policeman with respect to the thief = 35 - 30 = 5 m/s
Distance required to be travelled in order to catch the thief = 185 metres
The time taken by the policeman to catch the thief = (185/5) = 37 seconds
Hence, option c.

32. A man takes 44 seconds to run around a circular field of radius 28 metres. A 384-metre-long train takes 16 seconds to cross a pole. What is the time by the train to overtake the man running in the same direction?

Correct Answer: (d) 19.2 seconds
Solution:Circumference of the circular field = 2 × π × radius
= 2 × (22/7) × 28 = 176 metres
So, speed of the man = 176 ÷ 44 = 4 m/s
Speed of the train = 384 ÷ 16 = 24 m/s
So, relative speed of the train with respect to the man = 24 - 4 = 20 m/s
So, time taken by the train to overtake the man when running in the same direction = 384 ÷ 20 = 19.2 seconds
Hence, option d

33. A’ takes 24 minutes more than ‘B’ to cover a distance of 120 km. If the speed of ‘B’ is 20% more than the speed of ‘A’, then find the time taken by the ‘A’ to cover a distance of 180 km.

Correct Answer: (b) 3 hours and 36 minutes
Solution:Let the speed of ‘A’ = ‘5v’ km/hr
So, the speed of ‘B’ = 5v × (120/100) = ‘6v’ km/hr
ATQ:
(120/5v) - (120/6v) = (24/60)
(24/v) - (20/v) = (24/60)
(4/v) = (24/60)
v = 10
So, the speed of ‘A’ = 5v = 5 × 10 = 50 km/hr
The time taken by the ‘A’ to cover a distance of 180 kilometres = (180/50) hours = 3 hours and 36 minutes.
Hence, option b.

34. If ‘A’ is running 240 metres ahead of ‘B’, then ‘B’ can catch ‘A’ in 16 seconds. ‘A’ can run across a 150 metre long bridge in 6 seconds. If ‘A’ and ‘B’ run a 700 metre long race, then ‘B’ will finish the race how many seconds before ‘A’?

Correct Answer: (b) 10.5 seconds
Solution:

Speed of ‘A’ = 150 ÷ 6 = 25 m/s
Relative speed of ‘B’ with respect to ‘A’ = 240 ÷ 16 = 15 m/s
So, speed of ‘B’ = 25 + 15 = 40 m/s
So, time taken by ‘A’ to finish the 700 metre race = 700 ÷ 25 = 28 seconds
Time taken by ‘B’ to finish the 700 metre long race = 700 ÷ 40 = 17.5 seconds
So, difference in time taken = 28 - 17.5 = 10.5 seconds
Hence, option b.

35. The distance travelled by a boat in downstream in 5 hours is 60 km more than that in upstream in the same time. Find the total time taken by the boat to cover 144 km in upstream and 144 km in downstream, if the downstream speed of the boat is 20% more than the speed of boat in still water.

Correct Answer: (b) 10 hours 
Solution:

Difference between downstream speed and upstream speed = (60/5) = 12 km/hr
Speed of stream = (12/2) = 6 km/hr
Let the speed of boat in still water be ‘5x’ km/hr
So, downstream speed of the boat = 5x + 5x × 0.2 = ‘6x’ km/hr
According to question:
6x = 5x + 6
Or, x = 6
Speed of boat in still water = (5 × 6) = 30 km/hr
Downstream speed of the boat = (6 × 6) = 36 km/hr
And, upstream speed of the boat = (30 - 6) = 24 km/hr
Now, required time taken = (144/24) + (144/36)
= (6 + 4) = 10 hours
Hence, option b.

36. A man completed a journey in 8 hours with an average speed of 50 km/hr. He covered first 20% of the journey in 4 hours. Find at what speed he would have travelled the remaining distance.

Correct Answer: (d) 80 km/hr 
Solution:

Total distance travelled by the man = 50 × 8
= 400 km
Remaining distance = 400 × (80/100) = 320 km
Remaining time = 8 - 4 = 4 hours
Required speed = (320/4) = 80 km/hr
Hence, option d.

37. Train ‘A’ can cross a pole in 9 seconds and a 150 metres long platform ‘P’ in 16.5 seconds. If a man sitting in train ‘B’ observes that he completely crosses train ‘A’ in ‘x’ seconds, then find the value of ‘x’ given that train ‘B’ can cross a pole and platform ‘P’ in 5 seconds and 11 seconds, respectively and both trains are moving in the same direction on parallel tracks.

Correct Answer: (c) 36
Solution:

Let the length and speed of train ‘A’ be ‘l₁’ metres and ‘s₁’ m/s, respectively.
And, length and speed of train ‘B’ be ‘l₂’ metres and ‘s₂’ m/s, respectively.

According to the question:
l₁ = 9s₁
And, 9s₁ + 150 = 16.5s₁
7.5s₁ = 150
So, s₁ = 20
And, l₁ = 20 × 9 = 180

Now, l₂ = 5s₂
And, 5s₂ + 150 = 11s₂
6s₂ = 150
So, s₂ = 25

Required time taken = x = {180/(25 - 20)}
= (180/5) = 36
Hence, option c.

38. A’ and ‘B’ are running in opposite directions around a circular track whose diameter is 140 metres. If the speeds of ‘A’ and ‘B’ are in ratio 5:6 respectively, then which of the following could be the shortest distance between the starting point of ‘A’ and ‘B’, at the time when ‘A’ reaches the starting point for the fourth time after starting the race. (Take π = 22/7)

Correct Answer: (c) 88 metres 
Solution:Let the speeds of ‘A’ and ‘B’ be ‘5x’ m/s and ‘6x’ m/s, respectively.
Length of the track = (140 ÷ 2) × 2 × (22/7) = 440 metres
Time taken by ‘A’ to reach the starting point for the fourth time = (440/5x) × 4 = (352/x) seconds
Distance covered by ‘B’ in (440/x) seconds = 352/x × 6x = 2112 metres
So, shortest distance between ‘B’ and starting point = 440 × 5 - 2112 = 88 metres
Hence, option c.

39. When Ram walks at a speed equal to 80% of his usual speed, he reaches his school 20 minutes late than he is supposed to. If instead he walks at a speed equal to 125% of his usual speed, how much time Ram would be able to save as compared to the usual time taken by him to reach his school?

Correct Answer: (a) 16 minutes 
Solution:

Let the distance Ram has to travel in order to reach his school be ‘D’ km and usual speed of Ram be ‘S’ km/h.
Let the usual time taken by Ram to reach his school be ‘T’ hours.

ATQ:
(D/S) = T
D = S × T … (I)

And, (D/0.8S) = T + (20/60)
D = 0.8S × {(3T + 1)/3} …… (II)

Using equation (I) and (II), we have;
S × T = 0.8S × {(3T + 1)/3}
3ST = 2.4ST + 0.8S
0.6ST = 0.8S
0.6T = 0.8
T = (4/3) hours = (4/3) × 60 = 80 minutes
So, distance covered to reach the school = D = S × T = (4/3)S km
So, Time taken at 125% of usual speed
= (D/1.25S) = {(4/3)S}/1.25S × 60 = 64 minutes
So, time saved = 80 - 64 = 16 minutes
Hence, option a.

40. Raju, a labourer decides to go to a week - long adult literacy camp set up by the government. At what time will Raju reach the camp on the fourth day if he starts cycling at 10:15 A.M from his home and given the fact that the camp keeps shifting 2 Km farther away from Raju's house each day. Raju's cycling speed is 12Km/hr and the time taken by Raju to reach the camp on the 2nd day was 3hrs.

Correct Answer: (a) 1:35 P.M. 
Solution:

Let the distance between Raju's house and the camp on the 1st day be x Km.
=> The distance between Raju's house and the camp on the 2nd day be (x + 2) Km.
Time taken by Raju on the 2nd day = 3hrs.
(x + 2) / 3 = 12
=> X + 2 = 12 × 3 => X + 2 = 36
=> X = 34Km

Time taken by Raju to reach the camp on the fourth day = [x + (3×2)] / 12
=> [34 + 6] / 12 => 40/12
=> 10/3 hrs
Now 1 hr = 60 minutes
=> (10/3) hrs = (10/3) × 60 minutes
=> 200 minutes
=> 3 hrs and 20 minutes
=> He will reach camp at 3 hrs and 20 minutes from 10:15 A.M
=> 1:35 P.M