BANK & INSURANCE (BOAT AND STREAM) PART 3

Total Questions: 30

1. Directions [Set of 2 Questions]: Answer the questions based on the information given below.

In stream ‘A’, the downstream speed of a cargo ship is 50% more than its upstream speed. Also, the cargo ship, while travelling in still water takes 32 seconds to cross a pole and 35 seconds to cross a 400 metre long bridge while travelling in downstream in stream ‘A’. Stream ‘B’ flows 2 m/s faster than stream ‘A’. The ratio of downstream speed and upstream speed of a cruise ship travelling in stream ‘B’ is 14:9, respectively and the speed of the cruise ship in still water is ‘y%’ more than that of the cargo ship.

Ques: The cruise ship, while travelling downstream in stream ‘A’, takes 25 seconds to cross a port. What is the sum of the length of the port and the cruise ship?

Correct Answer: (b) 90y metres
Solution:Let the upstream speed of cargo ship in stream ‘A’ = ‘2x’ m/s

Then, downstream speed of cargo ship in stream ‘A’ = 2x × 1.5 = ‘3x’ m/s

So, speed of the cargo ship in still water = (2x + 3x) ÷ 2 = 2.5x m/s

Let the length of the cargo ship = ‘d’ metres

So, distance travelled by the cargo ship in still water in 32 seconds = 2.5x × 32 = 80x = d

Distance travelled by the cargo ship downstream in stream ‘A’ in 35 seconds = 3x × 35 = 105x = (d + 400)

So, 105x - 80x = d + 400 - d

x = 400 ÷ 25 = 16

So, speed of the cargo ship in still water = 16 × 2.5 = 40 m/s

Speed of stream ‘A’ = 16 × 0.5 = 8 m/s

So, speed of stream ‘B’ = 8 + 2 = 10 m/s

Let the speed of the cruise ship in still water = ‘k’ m/s

We have,

(k - 10):(k + 10) = 9:14

14k - 140 = 9k + 90

5k = 230

So, k = 230/5 = 46

So, y = {(46 - 40)/40} × 100 = 15

Downstream speed of the cruise ship in stream ‘A’ = 46 + 8 = 54 m/s

So, distance travelled by cruise ship in 25 seconds

= 54 × 25 = 1350 metres = ‘90y’ metres

Hence, option b.

2. When travelling upstream in stream ‘B’, the cruise ship takes ‘2y’ seconds to cross a light pole. Travelling in still water, if the cruise ship takes 400 seconds to completely overtake the cargo ship, then what was the initial distance between the cruise ship and the cargo ship?

Correct Answer: (a) 40 metres 
Solution:Let the upstream speed of cargo ship in stream ‘A’ = ‘2x’ m/s

Then, downstream speed of cargo ship in stream ‘A’ = 2x × 1.5 = ‘3x’ m/s

So, speed of the cargo ship in still water = (2x + 3x) ÷ 2 = 2.5x m/s

Let the length of the cargo ship = ‘d’ metres

So, distance travelled by the cargo ship in still water in 32 seconds = 2.5x × 32 = 80x = d

Distance travelled by the cargo ship downstream in stream ‘A’ in 35 seconds = 3x × 35 = 105x = (d + 400)

So, 105x - 80x = d + 400 - d

x = 400 ÷ 25 = 16

So, speed of the cargo ship in still water = 16 × 2.5 = 40 m/s

Speed of stream ‘A’ = 16 × 0.5 = 8 m/s

So, speed of stream ‘B’ = 8 + 2 = 10 m/s

Let the speed of the cruise ship in still water = ‘k’ m/s

We have,

(k - 10):(k + 10) = 9:14

14k - 140 = 9k + 90

5k = 230

So, k = 230/5 = 46

So, y = {(46 - 40)/40} × 100 = 15

2y = 2 × 15 = 30

So, distance travelled by the cruise ship in 30 seconds in upstream in stream ‘B’ = 36 × 30 = 1080 metres = length of the cruise ship.

Length of the cargo ship = 40 × 32 = 1280 metres

So, combined length of the cruise ship and the cargo ship = 1280 + 1080 = 2360 metres

In still water, relative speed of the cruise ship with respect to the cargo ship = 46 - 40 = 6 m/s

So, relative distance travelled by the cruise ship in 400 seconds with respect to the cargo ship = 6 × 400 = 2400 metres

So, initial distance between the cruise ship and the cargo ship = 2400 - 2360 = 40 metres

Hence, option a.

3. Directions [Set of 3 Questions]: Answer the questions based on the information given below.

A stream flows from point ‘A’ to point ‘B’. ‘C’ is a point in between ‘A’ and ‘B’ such that distance from ‘A’ to ‘C’ and from ‘C’ to ‘B’ is in the ratio 5:3, respectively. The speed of the stream from point ‘A’ to point ‘C’ is 60% of its speed from point ‘C’ to point ‘B’. Boat ‘P’ takes 150 minutes to cover 85 km in still water. Boat ‘P’ takes the same time to travel point ‘A’ to point ‘C’ as it takes to travel from point ‘B’ to point ‘C’. If boat ‘P’ starts from point ‘A’ and travels towards point ‘B’ for 7 hours, then it would have covered a total of 46 km more than the distance it would have covered running for the same time in still water.

The still water speed of boat ‘Q’ is 26 km/h. If the speed of stream when flowing from point ‘A’ to ‘C’ is decreased by ‘? km/h while its speed when flowing from point ‘C’ to ‘B’ remains unchanged, then find the total time taken by boat ‘Q’ to travel from point ‘A’ to ‘B’ and then from point ‘B’ to ‘C’.

Correct Answer: (b) 21 hours 
Solution:Let the total distance from point ‘A’ to point ‘B’ be ‘8d’ km

Then, distance between points ‘A’ and ‘C’ = 8d × (5/8) = ‘5d’ km

Distance between points ‘C’ and ‘B’ = 8d - 5d = ‘3d’ km

Let the speed of the stream from point ‘C’ to point ‘B’ = ‘10x’ km/h

Then, speed of the stream from point ‘A’ to point ‘B’ = 10x × 0.6 = ‘6x’ km/h

Still water speed of boat ‘P’ = 85 ÷ (150/60) = 34 km/h

Speed of boat ‘P’ from point ‘A’ to ‘C’ = (34 + 6x) km/h

Speed of boat ‘P’ from point ‘B’ to ‘C’ = (34 - 10x) km/h

According to the data,

5d ÷ (34 + 6x) = 3d ÷ (34 - 10x)

170d - 50xd = 102d + 18xd

68d = 68xd

So, x = 1

So, speed of the stream from point ‘A’ to ‘C’ = 6 km/h

Speed of the stream from point ‘C’ to ‘B’ = 10 km/h

So, speed of boat ‘P’ when travelling from ‘A’ to ‘C’ = 34 + 6 = 40 km/h

Speed of boat ‘P’ when travelling from ‘C’ to ‘B’ = 34 + 10 = 44 km/h

Distance covered by boat ‘P’ in still water in 7 hours = 34 × 7 = 238

So, distance covered by boat ‘P’ starting from point ‘A’ and travelling towards point ‘B’ for 7 hours = 238 + 46 = 284 km

Let the time for which boat ‘P’ travelled at 40 km/h = ‘k’ hours

Then, time for which boat ‘P’ travelled at 44 km/h = (7 - k) hours

Then, according to the data,

40 × k + 44 × (7 - k) = 284

40k + 308 - 44k = 284

24 - 4k = 0

So, k = 6

So, distance between points ‘A’ and ‘C’ = 40 × 6 = 240 km

And so, distance between points ‘C’ and ‘B’ = 240 × (3/5) = 144 km

So, distance between points ‘A’ and ‘B’ = 240 + 144 = 384 km

Speed of the stream from point ‘A’ to ‘C’ = 6 - 2 = 4 km/h

So, speed of boat ‘Q’ from point ‘A’ to ‘C’ = 26 + 4 = 30 km/h

So, time taken to travel from point ‘A’ to ‘C’ = 240 ÷ 30 = 8 hours

Speed of boat ‘Q’ from point ‘C’ to ‘B’ = 26 + 10 = 36 km/h

So, time taken to travel from point ‘C’ to ‘B’ = 144 ÷ 36 = 4 hours

Speed of boat ‘Q’ from point ‘B’ to point ‘C’ = 26 - 10 = 16 km/h

So, time taken to travel from point ‘B’ to point ‘C’ = 144 ÷ 16 = 9 hours

And so, total time taken for the journey = 8 + 4 + 9 = 21 hours

Hence, option b.

4. Two boats ‘M’ and ‘N’ with still water speeds in the ratio of 6:7, respectively, start from point ‘A’ and point ‘B’ respectively, at the same time and travel towards each other such that they meet 8 hours after starting. Find the still water speed of boat ‘M’.

Correct Answer: (c) 24 km/h
Solution:Let the total distance from point ‘A’ to point ‘B’ be ‘8d’ km

Then, distance between points ‘A’ and ‘C’ = 8d × (5/8)

= ‘5d’ km

Distance between points ‘C’ and ‘B’ = 8d - 5d

= ‘3d’ km

Let the speed of the stream from point ‘C’ to point ‘B’

= ‘10x’ km/h

Then, speed of the stream from point ‘A’ to point ‘B’

= 10x × 0.6 = ‘6x’ km/h

Still water speed of boat ‘P’ = 85 ÷ (150/60)

= 34 km/h

Speed of boat ‘P’ from point ‘A’ to ‘C’ = (34 + 6x) km/h

Speed of boat ‘P’ from point ‘B’ to ‘C’

= (34 - 10x) km/h

According to the data,

5d ÷ (34 + 6x) = 3d ÷ (34 - 10x)

170d - 50xd = 102d + 18xd

68d = 68xd

So, x = 1

So, speed of the stream from point ‘A’ to ‘C’ = 6 km/h

Speed of the stream from point ‘C’ to ‘B’ = 10 km/h

So, speed of boat ‘P’ when travelling from ‘A’ to ‘C’

= 34 + 6 = 40 km/h

Speed of boat ‘P’ when travelling from ‘C’ to ‘B’

= 34 + 10 = 44 km/h

Distance covered by boat ‘P’ in still water in 7 hours

= 34 × 7 = 238

So, distance covered by boat ‘P’ starting from point ‘A’

and travelling towards point ‘B’ for 7 hours

= 238 + 46

= 284 km

Let the time for which boat ‘P’ travelled at 40 km/h

= ‘k’ hours

Then, time for which boat ‘P’ travelled at 44 km/h

= (7 - k) hours

Then, according to the data,

40 × k + 44 × (7 - k) = 284

40k + 308 - 44k = 284

24 - 4k = 0

So, k = 6

So, distance between points ‘A’ and ‘C’ = 40 × 6 = 240 km

And so, distance between points ‘C’ and ‘B’ = 240 × (3/5) = 144 km

So, distance between points ‘A’ and ‘B’ = 240 + 144 = 384 km

Let the still water speed of boat ‘M’ = ‘6n’ km/h

Then still water speed of boat ‘N’ = ‘7n’ km/h

Since they meet after 6 hours, we have,

(6n + 6) × 8 + (7n - 10) × 8 = 384

48n + 48 + 56n - 80 = 384

104n = 416

So, n = 416 ÷ 104 = 4

And so, still water speed of boat ‘M’ = 6 × 4 = 24 km/h

Hence, option c.

5. A person fell into the stream at point ‘C’. A rescue boat with still water speed of 90 km/h started from point ‘A’ 93 minutes after the person fell into the stream. After travelling how much distance would the boat rescue the person?

Correct Answer: (a) 285 km 
Solution:Let the total distance from point ‘A’ to point ‘B’ be ‘8d’ km

Then, distance between points ‘A’ and ‘C’ = 8d × (5/8)

= ‘5d’ km

Distance between points ‘C’ and ‘B’ = 8d - 5d

= ‘3d’ km

Let the speed of the stream from point ‘C’ to point ‘B’

= ‘10x’ km/h

Then, speed of the stream from point ‘A’ to point ‘B’

= 10x × 0.6 = ‘6x’ km/h

Still water speed of boat ‘P’ = 85 ÷ (150/60) = 34 km/h

Speed of boat ‘P’ from point ‘A’ to ‘C’ = (34 + 6x) km/h

Speed of boat ‘P’ from point ‘B’ to ‘C’ = (34 - 10x) km/h

According to the data,

5d ÷ (34 + 6x) = 3d ÷ (34 - 10x)

170d - 50xd = 102d + 18xd

68d = 68xd

So, x = 1

So, speed of the stream from point ‘A’ to ‘C’ = 6 km/h

Speed of the stream from point ‘C’ to ‘B’ = 10 km/h

So, speed of boat ‘P’ when travelling from ‘A’ to ‘C’

= 34 + 6 = 40 km/h

Speed of boat ‘P’ when travelling from ‘C’ to ‘B’

= 34 + 10 = 44 km/h

Distance covered by boat ‘P’ in still water in 7 hours = 34 × 7 = 238

So, distance covered by boat ‘P’ starting from point ‘A’

and travelling towards point ‘B’ for 7 hours

= 238 + 46

= 284 km

Let the time for which boat ‘P’ travelled at 40 km/h

= ‘k’ hours

Then, time for which boat ‘P’ travelled at 44 km/h

= (7 - k) hours

Then, according to the data,

40 × k + 44 × (7 - k) = 284

40k + 308 - 44k = 284

24 - 4k = 0

So, k = 6

So, distance between points ‘A’ and ‘C’ = 40 × 6 = 240 km

And so, distance between points ‘C’ and ‘B’ = 240 × (3/5) = 144 km

So, distance between points ‘A’ and ‘B’ = 240 + 144 = 384 km

Distance travelled by the person in 93 minutes (from point ‘C’ towards point ‘B’ because of the stream)

= (93/60) × 10 = 15.5 km

Time taken by the boat to reach point ‘C’ = 240 ÷ (90 + 6) = 2.5 hours

Distance travelled by the person in those 2.5 hours

= 2.5 × 10 = 25 km

Now, distance between the rescue boat and the person when the boat reaches point ‘C’ = 15.5 + 25

= 40.5 km

Relative speed of the boat with respect to the person = 90 km/h

So, time taken to rescue the person once the boat reaches point ‘C’ = 40.5 ÷ 90 = 0.45 hours

Distance travelled by the person in 0.45 hours = 0.45 × 10 = 4.5 km

So, total distance travelled by the person from point ‘C’ = 15.5 + 25 + 4.5 = 45 km

And so, total distance travelled by the boat = 240 + 45 = 285 km

Hence, option a.

6. River ‘M’ flows 25% faster than river ‘N’. Two boats ‘A’ and ‘B’ are travelling in river ‘M’ and ‘N’, respectively. Speed of boat ‘A’ in still water is __ km/hr more than that of boat ‘B’. Time taken by boat ‘A’ to cover certain distance in upstream is equal to the time taken by boat ‘B’ to cover the same distance in downstream. Time taken by boat ‘A’ to travel 150 km in downstream is __ hours less than time taken by the boat ‘B’ to cover __ km in upstream. (Speeds of both the boats in still water have integral values).

The values given in which of the following options will not fill the blanks in the same order in which it is given to make the statement true:

I. 9, 144, 1.5  II. 15, 150, 0  III. 18, 200, 0

Correct Answer: (d) Only I 
Solution:Let the speed of stream in river ‘M’ be ‘5y’ km/h.

So, speed of stream in river ‘N’ = 5y × 0.8 = ‘4y’ km/h

For I:

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

So, speed of boat ‘A’ in still water = (x + 9) km/h

Upstream speed of boat ‘A’ = (x + 9 - 5y) km/h

Downstream speed of boat ‘B’ = (x + 4y) km/h

ATQ:

9 + x - 5y = x + 4y

9 = 9y

So, y = 1

So, speed of stream in river ‘M’ = 5 km/h

And speed of stream in river ‘N’ = 4 km/h

ATQ:

150/(x + 14) = 144/(x - 4) × 1.5

150/(x + 14) = {144 - (1.5x-6)}/(x - 4)

150(x - 4) = (x + 14)(150 - 1.5x)

150x - 600 = 150x + 2100 - 1.5x² - 21x

1.5x² + 21x - 2700 = 0

1.5x² + 75x - 54x - 2700 = 0

1.5(x + 50) - 54(x + 40) = 0

(1.5x - 54)(x + 50) = 0

So, x = 36 or x = -50

So, speed of boat ‘B’ in still water = 36 km/h

So, ‘I’ is true.

For II:

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

So, speed of boat ‘A’ in still water = x + 15 km/h

Upstream speed of boat ‘A’ = (x + 15 - 5y) km/h

Downstream speed of boat ‘B’ = (x + 4y) km/h

ATQ:

15 + x - 5y = x + 4y

15 = 9y

So, y = (5/3)

So, speed of stream in river ‘M’ = (5/3) km/h

And speed of stream in river ‘N’ = (5/3) km/h

So, downstream speed of boat ‘A’ = (x + 15) + (25/3)

= {(3x + 70)/3} km/h

And, upstream speed of boat ‘B’ = {x - (20/3)}

= {(3x - 20)/3} km/h

ATQ:

[150 × {3/(3x + 70)}] = [150 × {3/(3x - 20)}]

3x - 20 = 3x + 70

So, -20 = 70

Since, we can’t find the value of ‘x’.

So, ‘II’ is false.

For III:

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

So, speed of boat ‘A’ in still water = x + 18 km/h

Upstream speed of boat ‘A’ = (x + 18 - 5y) km/h

Downstream speed of boat ‘B’ = (x + 4y) km/h

ATQ:

18 + x - 5y = x + 4y

18 = 9y

So, y = 2

So, speed of stream in river ‘M’ = 10 km/h

And speed of stream in river ‘N’ = 8 km/h

So, downstream speed of boat ‘A’ = (x + 28) km/h

And, upstream speed of boat ‘B’ = (x - 8) km/h

ATQ:

150/(x + 28) = 200/(x - 8)

3(x - 8) = 4(x + 28)

3x - 24 = 4x + 112

x = -136

Since, value of ‘x’ cannot be negative, this case is also invalid.

So, ‘III’ is false.

Hence, option d.

7. The speed of a boat in still water is ‘x’ km/hr, and speed of the stream is 8 km/hr. If the speed of the boat in still water had been 4 km/hr more, then the total time taken by the boat to cover 120 km downstream and same distance in upstream would have been 20 hours. The time taken by the boat to cover 120 km in downstream with the original speed can be:

I. (x − 2.5) hours  II. (0.5x + 2) hours
III. (1.5x − 12) hours

Correct Answer: (a) Only III 
Solution:According to the question,

{120/(x + 4 + 8)} + {120/(x + 4 - 8)} = 20

6(x - 4) + 6(x + 12) = (x + 12)(x - 4)

6x - 24 + 6x + 72 = x² + 12x - 4x - 48

x² - 4x - 96 = 0

x² - 12x + 8x - 96 = 0

x(x - 12) + 8(x - 12) = 0

(x - 12)(x + 8) = 0

x = 12, -8

Since, the speed of boat cannot be negative

Therefore, x = 12

Time taken by boat to cover 120 km in downstream with original speed

= 120/(x + 8) = 120/20 = 6 hours

For I:

Time taken by boat to cover 120 km in downstream with original speed = (x - 2.5) = (12 - 2.5) = 9.5 hours

Therefore, I cannot be the answer.

For II:

Time taken by boat to cover 120 km in downstream with original speed = (0.5x + 2) = (6 + 2) = 8 hours

Therefore, II cannot be the answer.

For III:

Time taken by boat to cover 120 km in downstream with original speed = (1.5x - 12) = (18 - 12)

= 6 hours

Therefore, III can be the answer.

Hence, option a.

8. Atul can swim a certain course against the river flow in 150 minutes; he can swim the same course with the river flow in 20 minutes less than he can swim in still water. How much time would he take to swim the course with the river flow?

Correct Answer: (a) 100 minutes 
Solution:Let speed of Atul be ‘a’ km/h and speed of river be ‘s’ km/h

So, speed of Atul in upstream, in still water and in downstream will be ‘a - s’ km/h, ‘a’ km/h and ‘a + s’ km/h, respectively.

Since, speed of Atul in upstream, in still water and in downstream are in arithmetic progression, so, the time taken by Atul to go the desired distance in upstream, in still water and in downstream will be in harmonic progression.

Let time taken by the boat to cover the desired distance in downstream be ‘t’ minutes.

So, time taken by Atul to cover the desired distance in still water = (t + 20) minutes

Therefore, 150, t + 20 and t are in harmonic progression.

If three terms given as ‘a’, ‘b’ and ‘c’ are in harmonic progression, then b = 2ac/(a + c)

So, t + 20 = (2 × 150 × t)/(t + 150)

(t + 20)(t + 150) = 300t

t² + 170t + 3000 = 300t

t² - 130t + 3000 = 0

t² - 100t - 30t + 3000 = 0

t(t - 100) - 30(t - 100) = 0

(t - 100)(t - 30) = 0

t = 30 or t = 100

Hence, option d.

9. Directions (9-10): Answer the questions based on the information given below.

The ratio of the speeds of two boats ‘A’ and ‘B’ rowing in the same river, in still water is 5:4, respectively and the speed of the current is 6 km/h. Boat ‘A’ starts from point ‘X’ in downstream and after 1 hour boat ‘B’ also starts in same direction from the same point. After 6 hours of the starting of boat ‘A’ it is found that boat ‘A’ is 66 km ahead of boat ‘B’. In another case, if boat ‘A’ starts from point ‘Y’ in upstream and boat ‘B’ starts from point ‘X’ in downstream at the same time and distance between ‘X’ and ‘Y’ initially is ‘a’ km, then boat ‘A’ and boat ‘B’ will meet each other in 4.5 hours.

Ques. Which of the following statement is true?
I: The value of ‘a + 27’ is 280.
II: The speed of the boat ‘B’ in downstream is 30 km/h
III: The sum of the speeds of the boat ‘A’ and ‘B’ in upstream is 42 km/h
IV: Time taken by boat ‘A’ to cover 90 km in still water is 1.5 hours.

Correct Answer: (b) Only II and III 
Solution:Let the speed of the boat ‘A’ = ‘5x’ km/h

The speed of the boat ‘B’ = ‘4x’ km/h

Speed of the stream = 6 km/h

Speed of the boat ‘A’ in downstream = (5x + 6) km/h

Speed of the boat ‘B’ in downstream = (4x + 6) km/h

According to the question,

6(5x + 6) - 5(4x + 6) = 66

30x + 36 - 20x - 30 = 66

10x = 60

x = 6

The speed of the boat ‘A’ in still water = 30 km/h

The speed of the boat ‘B’ in still water = 24 km/h

Speed of the boat ‘A’ in upstream = 30 - 6 = 24 km/h

Speed of the boat ‘B’ in downstream = 24 + 6 = 30 km/h

Relative speed of boat ‘A’ and ‘B’ = 24 + 30 = 54 km/h

According to the question,

a = 54 × 4.5

a = 243 km

For ‘I’:

Since, a + 27 = 243 + 27 = 270 ≠ 280

So, this statement is not true.

For ‘II’:

Since, the speed of the boat ‘B’ in downstream = 30 km/h

So, this statement is true.

For ‘III’:

Since, the sum of the speed of the boat ‘A’ and ‘B’ in upstream = 24 + 18 = 42 km/h

So, this statement is true.

For ‘IV’:

Required time = 90/30 = 3 hours ≠ 1.5 hours

So, this statement is not true.

Hence, option b.

10. If the speed of boat ‘A’ in still water is decreased by 10% and the speed of the boat ‘B’ in still water is increased by 50% and the speed of stream remains same, then find the difference between the time taken by boat ‘A’ and ‘B’ to cover 462 km in downstream separately

Correct Answer: (c) 3 hours
Solution:Let the speed of the boat ‘A’ = ‘5x’ km/h

The speed of the boat ‘B’ = ‘4x’ km/h

Speed of the stream = 6 km/h

Speed of the boat ‘A’ in downstream = (5x + 6) km/h

Speed of the boat ‘B’ in downstream = (4x + 6) km/h

According to the question,

6(5x + 6) - 5(4x + 6) = 66

30x + 36 - 20x - 30 = 66

10x = 60

x = 6

The speed of the boat ‘A’ in still water = 30 km/h

The speed of the boat ‘B’ in still water = 24 km/h

Speed of the boat ‘A’ in upstream = 30 - 6 = 24 km/h

Speed of the boat ‘B’ in downstream = 24 + 6 = 30 km/h

Relative speed of boat ‘A’ and ‘B’ = 24 + 30 = 54 km/h

According to the question,

a = 54 × 4.5

= 243 km

New speed of the boat ‘A’ in still water = 30 × 90%

= 27 km/h

New speed of the boat ‘B’ in still water = 24 × 150%

= 36 km/h

Speed of the boat ‘A’ in downstream = 27 + 6

= 33 km/h

Speed of the boat ‘B’ in downstream = 36 + 6

= 42 km/h

Time taken by boat ‘A’ to cover 462 km in downstream

= 462/33 = 14 hours

Time taken by boat ‘B’ to cover 462 km in downstream

= 462/42 = 11 hours

Required difference between the time = 14 - 11

= 3 hours

Hence, option c.