ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Distance covered downstream = 64 km
Time is taken in downstream = 12 hours.
Rate of downstream = distance/time = a = 64 km/12 hours
= 16/3 km/hr.
Distance covered in upstream = 32km
Time is taken upstream = 8 hours.
Rate of upstream = distance/time = b = 32km/8 hours
= 4 km/hr.
Speed in still water = (a + b)/2 = (1/2){(16/3) + 4} km/hr
= (1/2)(28/3)
= 14/3 km/hr.

Given that, time is taken to travel upstream = 2 × times taken to travel downstream
When the distance is constant, speed is inversely proportional to the time
Hence, 2 × speed upstream = speed downstream
Let speed upstream = x
Then speed downstream 2x
we have,
1/2(x + 2x) = speed in still water
⇒ 1/2(3x)=7.5
⇒ 3x = 15
⇒ x = 5
i.e., speed upstream = 5 km/hr
Rate of stream = 1/2(2x - x)
= x/2
= 5/2
= 2.5 km/hr.

Speed downstream = 1/(5/60)
= 60/5
= 12 km/hr
Speed upstream = 4/1
= 4 km/hr
velocity of the current = 1/2(12 - 4)
= (1/2) × 8
= 4 km/hr.

Let the speed of the boat in still water=x km/hr
Speed of the current = 2 km/hr
Then, speed downstream = (x + 2) km/hr
speed upstream = (x - 2) km/hr
Total time taken to travel 10 km upstream and back = 55 minutes
= (55/60) hr
= 11/12 hr.
According to question,
10/(x - 2) + 10/(x + 2) = 11/12
⇒ 120(x+2)+120(x−2) = 11(x² - 4)
⇒ 240x = 11x² - 44
⇒ 11x² - 240x - 44 = 0
⇒ 11x(x -22) + 2(x - 22) = 0
⇒ (x -22) (11x + 2) = 0
Since x cannot be negative.
So, x = 22 km/hr.
Hence, the Speed of the motorboat is 22 km/hr.

Let the length of the tunnel = x meter
Then, distance = (800 + x) meter.
Time = 1 minute = 60 seconds.
Speed = 78 km/hr
= 78 × (5/18)
= (65/3) m/s
According to the question,
800 + x = 60 × (65/3)
⇒ 800 + x = 1300
⇒ x = 500 meter.

Relative speed = (45 + 30) km/hr.
= 75 km/hr.
= 75 × (5/18) km/hr.
= (125/6) m/s.
We are calculating the time taken by the slower train to pass the driver of the faster one.
Hence, distance = length of the slower train = 500 meter.
Time = 500/(125/6)
= 500 × (6/125)
= 24 seconds.

Distance covered = (120 + 120)
= 240 meter.
Time = 12 seconds.
Relative speed = 240/12
= 20 m/s.
= 20 × (18/5) km/hr.
= 72 km/hr.
The relative speed, in this case, is the sum of the speeds of the trains and each train has the same speed,
speed as each train = 72/2 km/hr.
= 36 km/hr.

Distance covered = (108 + 112)
= 220 meter.
Time = 6 seconds.
Relative speed = 220/6 = 110/3 m/s.
= (110/3) × (18/5) km/hr
= 132 km/hr.
50 + Speed of second train = 132 km/hr.
Speed of second train = (132 - 50)
= 82 km/hr.

Distance = 110 meter.
Since the train and man move in opposite directions, the relative speed
= (60 + 6) km/hr.
= 66 km/hr.
= 66 × (5/18) m/s.
= 110/6 m/s.
Time = 110/(110/6)
= 6 seconds.

Speed downstream = 15 km/hr
Rate of the current = 1(1/2) km/hr.
Speed in still water = {15 - 1(1/2)}
= 13(1/2) km/hr.
Rate against the current = {13(1/2) - 1(1/2)} km/hr
= 12 km/hr.

Speed upstream = 8/(24/60)
= 20 km/hr.
Speed of the stream = 4 km/hr.
Speed of boat in still water = (20 + 4)
= 24 km/hr.

Distance covered in 26 seconds.
= 26 × 72 × (5/18)
= 520 meter.
Length of the train = (520 - 250) meter
= 270 meter.

Let the speed of the train be v km/hr.
(v - 2) : (v - 4) = 10 : 9 [speed and time are inversely proportional]
⇒ 9v - 18 = 10v - 40
⇒ v = 22 km/hr.
Length of the train
= (22 - 2) × (5/18) × 9
= 50 meter.

Speed of the first train = 48 km/hr.
Let the length of the first train = 2x meter.
Speed of the second train = 42 km/hr.
Let the length of the second train = x meter.
Distance = (2x + x) = 3x meter.
Time = 12 seconds
Relative speed = 48 + 42 = 90 km/hr.
= 90 × (5/18) = 25 m/s.
3x = 25 × 12
⇒ x = 100 meter.
Length of the first train = 200
Time is taken to cross the platform = 45 seconds.
Speed of first train = 48 km/hr
= 48 × (5/18)
= 40/3 m/s.
Let the length of the platform = y meter.
Distance = 200 + y meter.
⇒ 200 + y = 45 × (40/3)
⇒ 200 + y = 600
⇒ y = 400 meter.

Relative speed = 120 + 80
= 200 km/hr.
= 200 × (5/18)
= 500/9
Time = 9 seconds
Distance covered = (500/9) × 9
= 500 meter.
Length of the train = (500 - 270) meter.
= 230 meter.

Speed of the man in still water = 12 km/hr.
Speed of the stream = 3 km/hr.
Speed downstream = 12 + 3
= 15 km/hr.
Speed upstream = (12 - 3)
= 9 km/hr.
Average speed = (Speed downstream × Speed upstream)/Speed in still water
= (15 × 9)/12
= (15 × 3)/4
= 45/4
= 11(1/4) km/hr.

Speed upstream = 3/(20/60)
= 9 km/hr.
Speed downstream = 3/(18/60)
= 10 km/hr.
Rate of current = (10 - 9)/2
= 1/2 km/hr.

Velocity of the stream = 4 km/hr.
The speed of the boat in still water is 14 km/hr.
Speed downstream = 14 + 4 = 18 km/hr.
Speed upstream = 10 km/hr.
Let the distance between A and B be x km.
Time taken to travel downstream from A to B + Time taken to travel upstream from B to C(mid of A and B)
= 38 hours.
⇒ x/18 + (x/2)/10 = 38
⇒ x/18 + x/20 = 38
⇒ 19x/180 = 38
⇒ x/180 = 2
⇒ x = 360 km.