BANK & INSURANCE (SPEED TIME AND DISTANCE) PART 3

Total Questions: 45

21. Train ‘A’ leaves station ‘X’ for station ‘Y’ at 9:00 a.m. at a speed of 90 km/h. Train ‘B’ leaves station ‘Y’ for station ‘X’ at 11:00 a.m. at a speed of ___ km/h. If the distance between station ‘X’ and ‘Y’ is 828 km and both trains are travelling on parallel tracks, the point where the two trains will meet will be ___ km from station ‘X’.

The values given in which of the following options will fill the blanks in the same order in which is it given to make the statement true:
I. 126, 450  II. 90, 540  III. 72, 504

Correct Answer: (b) Only I
Solution:

Distance covered by train ‘A’ in 2 hours = 90 × 2
= 180 km

Distance between train ‘A’ and train ‘B’ when train ‘B’ leaves the station ‘Y’ = 828 - 180 = 648 km

For statement I:
Speed of train ‘B’ = 126 km/h
So, relative speed of train ‘A’ w.r.t. train ‘B’ = 126 + 90 = 216 km/h
Time taken to meet = 648 ÷ 216 = 3 hours
Time taken by train ‘A’ to reach the meeting point = 2 + 3 = 5 hours
Distance between station ‘X’ and meeting point = 5 × 90 = 450 km
Therefore, statement I is true.

For statement II:
Speed of train ‘B’ = 90 km/h
So, relative speed of train ‘A’ w.r.t. train ‘B’ = 90 + 90 = 180 km/h
Time taken to meet = 648 ÷ 180 = 3.6 hours
Time taken by train ‘A’ to reach the meeting point = 3.6 + 2 = 5.6 hours
Distance between station ‘X’ and meeting point = 5.6 × 90 = 504 km
Therefore, statement II is false.

For statement III:
Speed of train ‘B’ = 72 km/h
So, relative speed of train ‘A’ w.r.t. train ‘B’ = 72 + 90 = 162 km/h
Time taken to meet = 648 ÷ 162 = 4 hours
Time taken by train ‘A’ to reach the meeting point = 2 + 4 = 6 hours
Distance between station ‘X’ and meeting point = 6 × 90 = 540 km
Therefore, statement III is false.

Hence, option b.

22. Time taken by a 450 metre long train ‘A’ to cross a man walking at the speed of 3 m/s in the same direction as that of the train is 20% more than the time taken by it to cross a pole. Train ‘B’ takes 10 seconds less than train ‘A’ to cross a pole and 20 seconds to cross a 110 metre long platform. Find the time taken by the two trains to cross each other while moving on parallel tracks and opposite to each other.

Correct Answer: (d) 19.5 seconds
Solution:

Let the time taken by train ‘A’ to cross a pole = ‘t’ seconds

Then, time taken by train ‘A’ to cross the man = t × 1.2 = 1.2t seconds

Distance covered by train ‘A’ to cross the man = length of the train + distance covered by the man in 1.2t seconds

ATQ:
450 ÷ t = (450 + 3 × 1.2t) ÷ 1.2t

Or, 540t = 450t + 3.6t²
So, t = 90 ÷ 3.6 = 25

So, speed of train ‘A’ = (450/25) = 18 m/s

Let the length of train ‘B’ = ‘k’ metres

Then, distance covered by train ‘B’ in (25 - 10) = 15 seconds = ‘k’ metres

Distance covered by train ‘B’ in 20 seconds = (k + 110) metres

So, distance covered by train ‘B’ in (20 - 15) seconds = k + 110 - k = 110 metres

So, speed of train ‘B’ = (110/5) = 22 m/s

Length of train ‘B’ = 15 × 22 = 330 metres

So, time taken by the trains to cross each other
= (450 + 330) ÷ (22 + 18)
= 780 ÷ 40 = 19.5 seconds

Hence, option d.

23. ‘B’ travels 4 km/h faster than ‘A’. The time taken by ‘B’ to cover 96 km is 72 minutes less than the time taken by ‘A’ to cover the same distance. Which among the following is true based on the given information?

I. If ‘B’ increases his speed by 25%, then he could cover 150 km in 6 hours.
II. Ratio of speeds of ‘A’ and ‘B’ is 4:5, respectively.
III. To cover 60 km, ‘A’ takes 225 minutes.

Correct Answer: (d) All of I, II and III
Solution:

Let the speed of ‘A’ = ‘y’ km/h
Then, speed of ‘B’ = (y + 4) km/h

According to the question,
(96/y) - (96 ÷ (y + 4)) = (72/60)

(96y + 384 - 96y) = (6/5) × (y² + 4y)

y² + 4y - 320 = 0

y² - 16y + 20y - 320 = 0

y(y - 16) + 20(y - 16) = 0

(y + 20)(y - 16) = 0

y = -20 or y = 16

Since, speed cannot be negative. So, y = 16

So, speed of ‘A’ = 16 km/h
And, speed of ‘B’ = 16 + 4 = 20 km/h

For statement I:
Increased speed of ‘B’ = 20 × 1.25 = 25 km/h
Required time = 150 ÷ 25 = 6
So, statement I is true.

For statement II:
Ratio of speeds of ‘A’ and ‘B’ = 16 : 20 = 4 : 5
So, statement II is true.

For statement III:
Time taken by ‘A’ to cover 60 km = 60 ÷ 16 = 3.75 hours
= 3.75 × 60 = 225 minutes
So, statement III is also true.

Hence, option d.

24. Distance between point ‘A’ and point ‘B’ is 220 km. A sailor is going from point ‘A’ to ‘B’ in his boat at 44 km/h. After travelling for 3 hours, the sailor realises that he had dropped his hat in the water exactly at the mid-way (mid-point of ‘AB’). So, he turns back to get his hat which was flowing with the stream towards point ‘B’. If the speed of the stream is 10% the speed of the boat in still water, then find the total time taken by the sailor to reach point ‘B’. [After taking the hat the sailor instantly started to move towards point ‘B’]

Correct Answer: (e) None of these
Solution:

Let speed of boat in still water is ‘x’ km/h

So, speed of stream = 0.10 × x = 0.1x km/h

Speed of boat in still water = x + 0.1x = 1.1x km/h

So, 1.1x = 44
x = 40

So, speed of boat in still water = 40 km/h
And, speed of stream = 4 km/h

Speed of boat in upstream = 40 - 4 = 36 km/h
Speed of boat in downstream = 40 + 4 = 44 km/h

Distance travelled by the boat in 3 hours = 44 × 3 = 132 km

Time taken by the boat to reach midpoint of ‘A’ and ‘B’
= 220 ÷ 44 = 2.5 hours

So, the hat travelled in water for 0.5 hours (3 - 2.5 = 0.5) before the sailor realised, that he had dropped the hat.

Distance covered by the hat with the stream in 0.5 hours = 4 × 0.5 = 2 km

Distance between boat and hat when the sailor turned back = 132 - 110 - 2 = 20 km

Time taken by the sailor to get the hat = 20 ÷ (36 + 4) = 0.5 hours

Distance covered by the sailor in 0.5 hours = 36 × 0.5 = 18 km

Total distance remaining = 220 - 132 + 18
= 106 km

Time taken by the sailor to cover 106 km
= (106/44) = (53/22) hours

Total time taken by the sailor = 3 + (1/2) + (53/22)
= (65/11) hours

Hence, option e.

25. Towns ‘A’ and ‘B’ are 180 km apart. A bus leaves town ‘A’ for town ‘B’ at 8:00 a.m. at 30 km/h. A taxi leaves town ‘B’ for town ‘A’ after every 30 minutes at 45 km/h. If the first taxi leaves town ‘B’ 60 minutes after the bus leaves town ‘A’, then find number of taxies that the bus will meet before first taxi reaches town ‘A’.

Correct Answer: (a) 5 
Solution:

The first taxi will reach town ‘A’ after (180/45)
= 4 hours

Time taken by the bus to reach town ‘B’ = 180/30
= 6 hours i.e., 2:00 p.m.

The bus and the first taxi will meet after {(180 - 30)/(45 + 30)} = 2 hours

So, the first bus and first taxi will meet at 11:00 a.m.,
i.e., 2 hours after the first taxi leaves town ‘B’

Now the bus will keep meeting a taxi every 30 minutes for the remaining time.

But since the first taxi will reach town ‘A’ at 1:00 p.m.
So, we will count only till 1:00 p.m. i.e. for 2 hours.

So, the bus will meet (2/30) × 60 = 4 taxis

So, total number of taxis that the bus will meet
= 4 + 1 = 5

Hence, option a.

26. The road by which a car travels from city ‘A’ to city ‘B’ consists of uphill journey, downhill journey and journey on the flat surface. Time taken by the car while going from city ‘A’ to ‘B’ is 3 hours while it took 3 hours and 40 minutes to travel from city ‘B’ to ‘A’ via the same path. If average speed of the car in uphill, downhill and on flat surface was 40 km/h, 60 km/h and 48 km/h, respectively, then find the distance between city ‘A’ and ‘B’.

Correct Answer: (a) 160 km 
Solution:

Let total distance between city ‘A’ and ‘B’ = ‘d’ km

And, distance travelled in uphill and downhill is ‘x’ km and ‘y’ km, respectively

So,
x/40 + (d - (x + y))/48 + y/60 = 3 …(I)

And,
y/40 + (d - (x + y))/48 + x/60 = 3 …(II)

By adding equation (I) and equation (II), we get
(x/40 + y/60) + (y/40 + x/60) + (d - (x + y))/24 = 6

(5x/120) + (5y/120) + (d - (x + y))/24 = 6

(x + y)/24 + (d - (x + y))/24 = 6

d/24 = 6

d = 160

Therefore, distance between city ‘A’ and city ‘B’
= 160 km

Hence, option a.

27. Travelling at its usual speed, a train takes ‘x’ seconds to cross a pole. Travelling at a speed which is 25% more than its usual speed, the train takes (x + 2) seconds to cross a 168 metre long platform. Travelling at a speed which is 50% more than its usual speed, the train takes ‘1.5x’ seconds to cross a 540 metre long bridge. Find the usual speed of the train.

Correct Answer: (d) 72 m/s 
Solution:

Let the usual speed of the train = 4y m/s
Let the length of the train = ‘d’ metres

We have, (d/4y) = x
4xy = d

Also, (d + 540) ÷ 4y × 1.5 = 1.5x
(d + 540)/6y = 1.5x

9xy = d + 540

So, 9xy - 4xy = 5xy = d + 540 - d = 540

xy = (540/5) = 108

So, d = 4xy = 432

Also, we have (d + 168) ÷ 4y × 1.25 = (d + 168)/5y = (x + 2)

5xy + 10x = d + 168

540 + 10x = 432 + 168 = 600

x = (600 - 540)/10 = 6

Therefore, usual speed of the train = 432/6 = 72 m/s

Hence, option d.

28. Car ‘A’ has a speed of 151.2 km/h. The relative speed of car ‘A’ and car ‘B’ when both cars are travelling in the same direction is 40% of the relative speed of car ‘A’ and car ‘B’ when they are travelling towards each other. If cars ‘A’ and ‘B’ start a race at the same time, then 5 hours after the start of race, car ‘A’ would be how many kilometres ahead of ‘B’?

Correct Answer: (c) 432 km
Solution:

Speed of car ‘A’ = 151.2 ÷ 3.6 = 42 m/s
Let the speed of car ‘B’ = ‘x’ m/s

We have, (42 + x) × 0.4 = (42 - x)

16.8 + 0.4x = 42 - x

25.2 = 1.4x

So, x = 25.2 ÷ 1.4 = 18 m/s

So, relative speed of car ‘A’ with respect to car ‘B’ in the race (when travelling in the same direction)
= 42 - 18 = 24 m/s = 24 × 3.6 = 86.4 km/h

So, distance between cars ‘A’ and ‘B’ after 5 hours
= 86.4 × 5 = 432 km

Hence, option c.

29. A driver of a bus covers a distance of 39 km at a certain speed in certain time. If the driver increases the speed of the bus by 8 km/h then he takes 1 hour 18 minutes less to cover the same distance. Find the distance travelled by the bus driver in 6.5 hours with the original speed.

Correct Answer: (c) 78 km
Solution:

Let the original speed of the bus be ‘x’ km/h

1 hour 18 minutes = 1.3 hours

According to question,
39/x - 39/(x + 8) = 1.3

39[(x + 8 - x)/(x(x + 8))] = 1.3

8/(x(x + 8)) = 1/30

x² + 8x - 240 = 0

x = 12 km/h

Desired Distance = 12 × 6.5 = 78 km

Hence, option c.

30. Travelling with a speed of ‘x + 4’ km/h, Sourav reaches his office which is 60 km away, 30 minutes early while travelling with a speed of ‘x – 8’ km/h, Sourav reaches his office 120 minutes late. Find the time taken by Sourav to travel 180 km while he is travelling with a speed of ‘x + 20’ km/h.

Correct Answer: (b) 4.5 hours
Solution:

Let, time taken to reach office (on schedule time) be ‘t’ hours

So, 60/(x + 4) = t - 1/2 …(1)

And, 60/(x - 8) = t + 2 …(2)

Equation (2) - Equation (1)

So, 60/(x - 8) - 60/(x + 4) = 2 + 1/2 = 5/2

x² - 4x - 320 = 0

x² - 20x + 16x - 320 = 0

x(x - 20) + 16(x - 20) = 0

(x - 20)(x + 16) = 0

x = 20

Desired time = 180/(20 + 20) = 4.5 hours

Hence, option b.