BANK & INSURANCE (SPEED TIME AND DISTANCE) PART 3

Total Questions: 45

1. Two trains namely ‘A’ and ‘B’, whose speeds are in the ratio 3:2, respectively, are travelling from station ‘P’ and ‘Q’ to reach stations ‘Q’ and ‘P’, respectively. Train ‘A’ started 1 hour before train ‘B’ and they meet for the first time after 3 hours of the start of train ‘B’. After reaching their respective destinations they instantly returned towards their source stations. Distance travelled by train ‘A’ is how much percent more than the distance travelled by train ‘B’ when they meet for the second time. It is also known that, while returning back, speed of train ‘A’ decreased by 50% and speed of train ‘B’ increased by 25%.

Correct Answer: (a) (200/17)% 
Solution:

Let the speed of train ‘A’ be ‘3x’ km/hr
So, speed of train ‘B’ = 3x × (2/3) = ‘2x’ km/hr

Speed of train ‘A’ while returning back = 3x × 0.5 = 1.5x km/hr
And, speed of train ‘B’ while returning back = 2x × 1.25 = 2.5x km/hr

Distance travelled by train ‘A’ in first hour = ‘3x’ km

Now, distance travelled by both the trains (‘A’ and ‘B’) in next 3 hours = 3 × (3x + 2x) = ‘15x’ km

Total distance between station ‘P’ and ‘Q’ = 15x + 3x = ‘18x’ km

Time taken by train ‘A’ to travel the distance that train ‘B’ has travelled before meeting train ‘A’ = (6x/3x) = 2 hours

And, time taken by train ‘B’ to travel the distance that train ‘A’ has travelled before meeting train ‘B’ = (12x/2x) = 6 hours

Now, distance covered by train ‘A’ with the reduced speed, in next (6 - 2) hours = (4 × 1.5x) = ‘6x’ km

Now, distance left between both the trains = (18x - 6x) = ‘12x’

Relative speed = (1.5x + 2.5x) = ‘4x’

Time taken in their second meeting = (12x/4x) = 3 hours

Distance travelled by train ‘A’ in these 3 hours = (1.5x × 3) = ‘4.5x’ km

Distance travelled by train ‘B’ in these 3 hours = (2.5x × 3) = ‘7.5x’ km

Total distance travelled by train ‘A’ = (3x + 9x + 6x + 4.5x) = ‘28.5x’ km

And, total distance travelled by train ‘B’ = (6x + 12x + 7.5x) = ‘25.5x’ km

Required percentage = {(28.5x - 25.5x)/25.5 × 100}
= (200/17)%

Hence, option a.

2. Directions (2-3): Answer the questions based on the information given below.

Three friends, Pankaj, Vishal and Rajan were travelling in a car and had to cover a total of 2800 km. They took turns to drive the car such that in beginning Pankaj drove the car with a speed of 180 km/hr for ‘x’ hours. After that Vishal started to drive with a speed which is 25% less than that of Pankaj and drove for 2 hours more than that by Pankaj. After Vishal, Rajan drove the car for ‘y’ hours till the destination with a speed, which is 2.5 km/hr more than the average of the speeds with which Pankaj and Vishal drove the car.
Note: ‘x’ and ‘y’ are integers.

Ques: Find the average of distance for which Pankaj and Rajan drove the car.

Correct Answer: (a) 860 km 
Solution:

Speed with which Vishal drove the car = 0.75 × 180 = 135 km/hr
Speed with which Rajan drove the car = {(135 + 180)/2} + 2.5 = 160 km/hr
According to the question,
180x + 135 × (x + 2) + 160y = 2800
180x + 135x + 160y = 2800 − 270
315x + 160y = 2530
63x + 32y = 506

Since, ‘x’ and ‘y’ are integers
Therefore, only value which ‘x’ and ‘y’ can take is 6 and 4, respectively

Therefore, distance for which Pankaj drove the car = 180x = 1080 km
Distance for which Vishal drove the car = 135x = 1080 km
Distance for which Rajan drove the car = 160y = 640 km

Required average = {(1080 + 640)/2} = 860 km
Hence, option a.

3. If they had to cover the given distance in 4 hours less such that Pankaj had to drive for 10 hours only with 30 km/hr less speed than his original speed and Vishal had to drive for 2 hours only with 20% more speed than his original speed then find the percentage by which Rajan had to increase his speed than his original speed in order to reach the destination in the stipulated time.

Correct Answer: (d) 205% 
Solution:

Speed with which Vishal drove the car = 0.75 × 180 = 135 km/hr
Speed with which Rajan drove the car = {(135 + 180)/2} + 2.5 = 160 km/hr

According to the question,
180x + 135 × (x + 2) + 160y = 2800
Or, 180x + 135x + 160y = 2800 − 270
Or, 315x + 160y = 2530
Or, 63x + 32y = 506

Since, ‘x’ and ‘y’ are integers
Therefore, only value which ‘x’ and ‘y’ can take is 6 and 4, respectively

Therefore, distance for which Pankaj drove the car = 180x = 1080 km
Distance for which Vishal drove the car = 135x = 1080 km
Distance for which Rajan drove the car = 160y = 640 km

According to the question,
Distance covered by Pankaj = (180 − 30) × 10 = 1500 km
Distance covered by Vishal = 1.2 × 135 × 2 = 324 km

Distance to be covered by Rajan = 2800 − (1500 + 324) = 976
Time to be taken by Rajan = 14 − 10 − 2 = 2 hours

Therefore, speed of Rajan = 976/2 = 488 km/hr
Required percentage change = {(488 − 160)/160} × 100 = 205%

Hence, option d.

4. Directions (4-5): Answer the questions based on the information given below

While travelling from city ‘A’ to city ‘B’, train ‘P’ can cross a pole in 32 seconds and while travelling from city ‘B’ to city ‘C’, it takes 48 seconds to cross a 100 metre long bridge given that total distance between cities ‘A’ and ‘C’ is 594 km and train ‘P’ takes 90% as much time to travel from city ‘A’ to ‘B’ as it takes to travel from city ‘B’ to ‘C’. If the distance between cities ‘A’ and ‘B’ had been 4 km less while that between cities ‘B’ and ‘C’ had been 10 km more, then the ratio of the distance between ‘A’ and ‘B’ and that between ‘B’ and ‘C’ would have been 8:7, respectively. Train ‘Q’ is 272 metres longer than train ‘P’ and the cities ‘A’, ‘B’ and ‘C’ are located in a straight line in the same order.

Ques: While travelling in opposite directions, train ‘P’ and ‘Q’ can cross each other completely in 36 seconds. Find the speed of train ‘Q’ given that speed of train ‘P’ is equal to its speed while travelling from city ‘A’ to city ‘B’.

Correct Answer: (a) 27 m/s 
Solution:

Let the length of train ‘P’ = ‘d’ metres
Then, speed of train ‘P’ between cities ‘A’ and ‘B’ = (d/32) m/s
Speed of train ‘P’ between cities ‘B’ and ‘C’ = {(d + 100)/48} m/s

Let the distance between city ‘A’ and ‘B’ = ‘x’ metres
Let the distance between cities ‘B’ and ‘C’ = ‘y’ metres

According to the question,
(x − 4):(y + 10) = 8:7
7x − 28 = 8y + 80
8y = 7x − 108

Also, (x + y) = 594
y = (594 − x)

So, 8y = 4752 − 8x = 7x − 108
4860 = 15x
So, x = 4860 ÷ 15 = 324

So, distance between cities ‘A’ and ‘B’ and between cities ‘B’ and ‘C’ are 324 km and 270 km, respectively.

So, ratio of distance between cities ‘A’ and ‘B’ and that between cities ‘B’ and ‘C’ = 324:270 = 6:5

Ratio of time taken by train ‘P’ to travel between cities ‘A’ and ‘B’ and that between cities ‘B’ and ‘C’ = 9:10

So, ratio of speed of train ‘P’ when travelling between cities ‘A’ and ‘B’ and that between cities ‘B’ and ‘C’ = (6/9):(5/10) = 4:3

So, (d/32):{(d + 100)/48} = 4:3
(d/128) = {(d + 100)/144}
144d = 128d + 12800
So, d = 12800 ÷ 16 = 800

So, length of train ‘P’ = d = 800 metres

So, speed of train ‘P’ while travelling from city ‘A’ to ‘B’ = (800/32) = 25 m/s = (25 × 18/5) = 90 km/h

Time taken to travel from ‘A’ to ‘B’ (consider length of the train as negligible) = 324 ÷ 90 = 3.6 hours

Speed of train ‘P’ while travelling from city ‘B’ to city ‘C’ = 90 × (3/4) = 67.5 km/h

Time taken to travel from city ‘B’ to ‘C’ (consider length of the train as negligible) = 3.6 ÷ 0.9 = 4 hours

Length of train ‘Q’ = 800 + 272 = 1072 metres

According to the question,
Relative speed of train ‘Q’ with respect to train ‘P’ = (800 + 1072) ÷ 36 = 52 m/s

So, speed of train ‘Q’ = 52 − 25 = 27 m/s

Hence, option a.

5. 5. Find the total time taken by train ‘P’ to travel from city ‘A’ to city ‘C’.

Correct Answer: (c) 7.6 hours
Solution:

Let the length of train ‘P’ = ‘d’ metres
Then, speed of train ‘P’ between cities ‘A’ and ‘B’ = (d/32) m/s
Speed of train ‘P’ between cities ‘B’ and ‘C’ = {(d + 100)/48} m/s

Let the distance between city ‘A’ and ‘B’ = ‘x’ metres
Let the distance between cities ‘B’ and ‘C’ = ‘y’ metres

According to the question,
(x − 4):(y + 10) = 8:7
7x − 28 = 8y + 80
8y = 7x − 108

Also, (x + y) = 594
y = (594 − x)

So, 8y = 4752 − 8x = 7x − 108
4860 = 15x
So, x = 4860 ÷ 15 = 324

So, distance between cities ‘A’ and ‘B’ and between cities ‘B’ and ‘C’ are 324 km and 270 km, respectively.

So, ratio of distance between cities 'A' and 'B' and that between cities 'B' and 'C' = 324:270 = 6:5
Ratio of time taken by train 'P' to travel between cities 'A' and 'B' and that between cities 'B' and 'C' = 9:10
So, ratio of speed of train 'P' when travelling between cities 'A' and 'B' and that between cities 'B' and 'C' = (6/9):(5/10) = 4:3
So, (d/32):{(d + 100)/48} = 4:3
(d/128) = {(d + 100)/144}
144d = 128d + 12800
So, d = 12800 ÷ 16 = 800
So, length of train 'P' = d = 800 metres

So, speed of train 'P' while travelling from city 'A' to 'B' = (800/32) = 25 m/s = 25 × (18/5) = 90 km/h
Time taken to travel from 'A' to 'B' (consider length of the train as negligible) = 324 ÷ 90 = 3.6 hours

Speed of train 'P' while travelling from city 'B' to city 'C' = 90 × (3/4) = 67.5 km/h
Time taken to travel from city 'B' to 'C' (consider length of the train as negligible) = 3.6 ÷ 0.9 = 4 hours

Length of train 'Q' = 800 + 272 = 1072 metres
Total time taken = 3.6 + 4 = 7.6 hours
Hence, option c.

6. A man walks at the speed of 3 m/s and train ‘A’ takes 21.25 seconds to cross him when he is walking in the same direction as that of train and 17 seconds to cross him when he is walking in the opposite direction as of the train. If train ‘A’ takes 20 seconds to cross train ‘B’ which is coming from opposite direction and 160 seconds to completely overtake train ‘B’, then find the length of train ‘B’.

Correct Answer: (e) 450 metres
Solution:

Let the length of train 'A' = 'd' metres
Let the speed of train 'A' = 'y' m/s

According to the question,
{d ÷ (y - 3)} = 21.25
21.25y - 63.75 = d

Also, {d ÷ (y + 3)} = 17
17y + 51 = d

So, 17y + 51 = 21.25y - 63.75
114.75 = 4.25y
So, y = 114.75 ÷ 4.25 = 27

And, d = (27 + 3) × 17 = 510

Let the length of train 'B' = 'k' metres
Let the speed of train 'B' = 'z' m/s

According to the question,
(510 + k) ÷ (27 + z) = 20
540 + 20z = (510 + k)
30 + 20z = k

Also, (510 + k) ÷ (27 - z) = 160
4320 - 160z = 510 + k
3810 - 160z = k

So, 30 + 20z = 3810 - 160z
180z = 3780
So, z = 3780 ÷ 180 = 21

And, k = 30 + 20 × 21 = 450

So, length of train 'B' = 450 metres
Hence, option e.

7. A train running at a speed of 72 km/h can cross a pole and a bridge ___ meters long in ___ seconds and 24 seconds respectively

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 above statement true:
A. 320, 8  B. 280, 10  C. 250, 11

Correct Answer: (c) Only A and B 
Solution:

Speed of the train = 72 × 5/18 = 20 m/s

For option A:
Length of the train = 20 × 8 = 160 m
So, the length of the bridge = 20 × 24 - 160 = 320
480 - 160 = 320
320 = 320
So, option A can be the answer.

For option B:
Length of the train = 20 × 10 = 200 m
So, the length of the bridge = 20 × 24 - 200 = 280
480 - 200 = 280
280 = 280
So, option B can be the answer.

For option C:
Length of the train = 20 × 11 = 220 m
So, the length of the bridge = 20 × 24 - 220 = 250
480 - 220 = 250
260 ≠ 250
So, option C can't be the answer.

Hence, option c.

8. The distance between points ‘A’ and ‘C’ is 60 metres more than the distance between points ‘A’ and ‘B’. A bike and a jeep are 2 m/s slower and 1 m/s faster than a car, respectively. The car takes 2 seconds less than the bike to travel from point ‘A’ to ‘B’, and takes 1 second more than the jeep to travel from point ‘A’ to ‘C’. If a train’s length is equal to the distance between point ‘A’ and ‘B’, then what must be its speed so that it can cross a 228 metre long platform in 21 seconds?

Correct Answer: (d) 28 m/s 
Solution:

Let the distance between points 'A' and 'B' = 'd' metres
Then, distance between points 'A' and 'C' = (d + 60) metres

Let the speed of the car = 'x' m/s
Then, speed of the bike = (x - 2) m/s
Speed of the jeep = (x + 1) m/s

According to the question,
{d/(x - 2)} ÷ (d/x) = 2
(dx - dx + 2d) ÷ (x² - 2x) = 2
d = (x² - 2x) …… (I)

Also, we have,
{(d + 60)/x} ÷ {(d + 60)/(x + 1)} = 1
(dx + 60x + d + 60 - dx - 60x) ÷ (x² + x) = 1
(d + 60) = x² + x
d = (x² + x - 60) …… (II)

So, x² + x - 60 = x² - 2x
3x - 60 = 0
So, x = 20

So, {d/(x - 2)} - (d/x) = (d/18) - (d/20) = 2
(10d - 9d) ÷ 180 = 2
So, d = 180 × 2 = 360

So, length of the train = 360 metres
So, speed of the train = (360 + 228) ÷ 21 = 28 m/s

Hence, option d.

9. In a race of 2700 metres, Aman beats Barkha by 200 metres, while in another race of 3300 metres, Chetna beats Dhani by 300 metres. Speed of Barkha and Dhani is same. (Assume that Aman, Barkha, Chetna and Dhani run with uniform speed in all the events). If Aman and Chetna participate in a race of 4400 metres, then which one of the following is correct?

Correct Answer: (b) Chetna beats Aman by 80 metres
Solution:

Let speed of Barkha = Speed of Dhani = 'x' m/s

In a race of 2700 m, Aman beats Barkha by 200 metres
Hence distance covered by Barkha = 2700 - 200 = 2500 metres

Since, time taken by Aman and Barkha will be the same
So, time taken by Aman = time taken by Barkha
(2700/speed of Aman) = (2500/speed of Barkha)
(2700/speed of Aman) = (2500/x)

Speed of Aman = (27x/25) m/s

Similarly, in the race of 3300 m, Chetna beats Dhani by 300 metres
Hence distance covered by Dhani = 3300 - 300 = 3000 metres

Since, time taken by Chetna and Dhani will be the same
So, time taken by Chetna = time taken by Dhani
(3300/speed of Chetna) = (3000/speed of Dhani)
(3300/speed of Chetna) = 3000/x

Speed of Chetna = (11x/10) m/s
Time taken by Chetna to complete the race
= 4400/(11x/10) = 4000/x sec

Distance covered by Aman in the same time
= (27x/25) × (4000/x) = 4320 metres

So, Chetna beats Aman by (4400 − 4320)
= 80 metres

Hence, option b.

10. A bus travelling with a speed ‘x – 2’ km/h while speed of a truck is ‘x + 10’ km/h. Time taken by the bus to cover a distance of ‘2x + 44’ km is 48 minutes more than time taken by the truck to cover a distance of ‘5x – 118’ km. Which of the following statements can be true?

I. 75% of (8x + 32) = 182
II. A 200 metres long train moving with a speed of ‘x + 22’ km/h can cross a pole in 10 seconds.
III. Distance travelled by a man in 48 minutes is 40 km if he is walking with a speed of ‘x’ km/h.

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

{(2x + 44)/(x − 2)} − {(5x − 118)/(x + 10)} = 48/60
= 4/5

19x² − 928x − 1100 = 0
19x² − 950x + 22x − 1100 = 0
19x(x − 50) + 22(x − 50) = 0
(19x + 22)(x − 50) = 0

x = 50 (Speed and distance can’t be negative)

For I:
75% of (8x + 32) = 0.75 × (8 × 50 + 32)
= 0.75 × 432 = 324 = 182
So, ‘I’ is true.

For II:
Speed of train = x + 22 = 72 km/h = 72 × (5/18)
= 20 m/s

So, length of train = 20 × 10 = 200 metres
So, ‘II’ is true.

For III:
Speed of man = x = 50 km/h = (50/60)
= 5/6 km/minute

Distance travelled by man in 48 minutes = (5/6) × 48 = 40 km

So, ‘III’ is true.
Hence, option e.