ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: A train running at the speed of 60 km/hr crosses a pole in 10 seconds. What is the length of the train?

Solution:
In 3600 seconds the train covered 60000 meter
∴ In 10 seconds the train covered (60000 × 10)/3600 meter
= 166.67 meter 

∴ The length of the train is 166.67 meter.

Question: A man takes 20 minutes to row 12 km upstream which is one-third more than the time he takes on his way downstream. What is his speed in still water?

Solution:
Let,
His speed in still water is x km/hr
The speed of stream is y km/hr
Time taken downstream is t min
ATQ,
t + (t/3) = 20 min
⇒ (3t + t)/3 = 20
⇒ 4t = 20 × 3
⇒ t = 60/4
∴ t = 15 min

In 20 minutes he rows in upstream 12 km
∴ In 60 minutes he rows in upstream (12 × 60)/20 km 
= 36 km 

∴ x - y = 36 ..............(1)


In 15 minutes he rows In downstream 12 km
∴ In 60 minutes he rows In downstream (12 × 60)/15 km
= (720 × 3)/40 km
= 48 km

∴ x + y = 48 ................(2)

From (2) + (1) we get,
x + y + x - y = 48 + 36
⇒ 2x = 84
∴ x = 42

∴ His speed in still water is 42 km/hr

Question: The length of the bridge, which a train 130 meters long and travelling at 46.8 km/hr can cross in 30 seconds, is:

Solution: 
Let,
The length of the bridge is x meter

∴ The train has to travel 130 + x meter to cross the bridge.

The velocity of the train 46.8 km/hr
= (46.8 × 1000)/3600 m/s
= 13 m/s

The train travels in 1 second 13 meters
∴ The train travels in 30 seconds (13 × 30) meters
= 390 meters

Now,
130 + x  = 390
⇒ x = 390 - 130
∴ x = 260 

∴The length of the bridge is  260 meters

Question: How long will it take to row 20 km upstream if one can row 10 km in 10 minutes in still water and the same distance in 8 minutes with the stream?

Solution:
Let,
x be the speed of man in still water 
y be the speed of stream.
∴ Speed of man x = 10/10 km/min
= 1 km/min
= 60 km/hr

In downstream in 8 min he can row 10 km
∴ In downstream in 60 min he can row (10 × 60)/8 km
= 75 km

∴ x + y = 75
⇒ y = 75 - 60
∴ y = 15
∴ Speed of stream = 15 km/hr.

Hence upstream speed = 60 – 15 = 45 km/hr.

Time taken to cover 45 km = 60 min
∴ Time taken to cover 20 km = (60 × 20)/45 min  
= 26.67 min

Question: Two trains running in opposite directions cross a man standing on the platform in 37 seconds and 27 seconds respectively and they cross each other in 33 seconds. The ratio of their speeds is:

Solution: 
Let,
The speed of the 1st train be x m/sec
The speed of the 2nd train be y m/sec 

∴ Length of the 1st train = 37x metres
∴ Length of the 2nd train = 27y metres

ATQ,
(37x + 27y)/(x + y) = 33 [ time = distance/speed]
⇒ 37x + 27y = 33x + 33y
⇒ 37x - 33x = 33y - 27y
⇒ 4x = 6y
⇒ 2x = 3y
⇒ x/y = 3/2
∴ x : y = 3 : 2

Question: The distance between A and B is 30 km. A boat makes a return journey from point A to point B and back in 10 hours 30 minutes. One way it travels with the stream and on the return it travels against the stream. If the speed of the stream increases by 2 km/hr, the return journey takes 17 hours 30 minutes. What is the speed of the boat in still water?

Solution:
Let, 
x km/hr be the speed of boat upstream
y km/hr be the speed of boat downstream

∴ 30/x + 30/y = 10 hours 30 minutes = 10.5 hour = 105/10 hour = 21/2
⇒ 30/x + 30/y = 21/2
⇒ 60/x + 60/y = 21
⇒ 20/x + 20/y = 7
⇒ 20/x = 7 - 20/y
⇒  20/x = (7y - 20)/y
⇒ x/20 = y/(7y - 20)
∴ x = 20y/(7y - 20) ............(1)


If the speed of the stream increases by 2 km/hr
30/(x - 2) + 30/(y + 2) = 17 hours 30 minutes = 17.5 hour = 175/10 = 35/2
⇒ 30/(x - 2) + 30/(y + 2) = 35/2
⇒ 60/(x - 2) + 60/(y + 2) = 35
⇒ 12/(x - 2) + 12/(y + 2) = 7
⇒ 12/(x - 2) = 7 - 12/(y + 2)
⇒ 12/(x - 2) = (7y + 14 -12)/(y + 2)
⇒ 12/(x - 2) = (7y + 2)/(y + 2)
⇒ (x - 2)/12 = (y + 2)/(7y + 2)
⇒ x - 2 = (12y + 24)/(7y + 2)
⇒ x = (12y + 24)/(7y + 2) + 2
⇒ x = (12y + 24 + 14y + 4)/(7y + 2)
∴ x = (26y + 28)/(7y + 2) ...............(2)

From (1) and (2) we get,
20y/(7y - 20) = (26y + 28)/(7y + 2)
⇒ 20y(7y + 2) = (26y + 28)(7y - 20)
⇒ 140y² + 40y = 182y² - 324y - 560
⇒ 42y² - 364y - 560 = 0
⇒ 3y² - 26y - 40 = 0
⇒ 3y² - 30y + 4y - 40 = 0
⇒ 3y(y - 10) + 4(y - 10) = 0
⇒ (y - 10)(3y + 4) = 0
∴ y = 10 or  y = - 4/3 [which is not acceptable]
∴ y =10

From (1) we get 
x = 20y/(7y - 20)
= 200/(70 - 20)
= 4 
 
∴ The speed of the boat in still water = (10 + 4)/2 km/hr
= 7 km/hr

Question: A train passes a station platform in 36 seconds and a man standing on the platform in 52/3 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?

Solution:
Speed = 54 km/hr = (54 × 1000)/3600 m/s
= 15 m/s 

Length of the train = 15 × (52/3) m = 260 m.

Let,
The length of the platform be x meters
∴ (260 + x)/36 = 15
⇒ 260 + x = 540
⇒ x = 540 - 260
∴ x = 280 

∴ The length of the platform 280 meters

Question: A man rows 24 km upstream in 6 hours and a distance of 35 km downstream in 7 hours. Then the speed of the man in still water is

Solution:
Let,
The speed of the man in still water is x km/hr
The speed of stream is y km/hr

∴ x - y = 24/6 = 4
∴ x + y = 35/7 = 5

∴ x - y + x + y = 4 + 5
⇒ 2x = 9
∴ x = 4.5 

∴ The speed of the man in still water is 4.5 km/hr

Question: A train 130 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:

Solution:
Speed of the train relative to man = 130/10 m/sec
= 13 m/sec
= (13 × 3600)/1000 km/hr
= 46.8 km/hr 

Let,
The speed of the train be x km/hr.
∴ Relative speed = (x - 5) km/hr.
⇒ x - 5 = 46.8         
∴ x = 51.8 km/hr.

Question: A train 270 m long passes a pole in 27 seconds. How long will it take to pass a platform 630 m long?

Solution:
The speed of the train = 270/27 m/s
= 10 m/s

Total distance to pass the platform = 270 + 630 m
= 900 m

The required time = 900/10 = 90 s

Question: A motorboat can travel at 5 km/hr in still water. It travelled 45 km downstream in a river and then returned, taking altogether 50 hours. Find the rate of flow of the river.

Solution:
Speed of boat in still water, x = 5 km/hr.
Let,
rate of flow of river = y km/hr.
∴ Speed of upstream = 5 - y
speed of downstream = 5 + y

∴ 45/(5 + y) + 45/(5 - y) = 50
⇒ (225 - 45y + 225 + 45y)/(25 - y²) = 50
⇒ 450 = 1250 - 50y²
⇒ 50y² = 800
⇒ y² = 16
∴ y = 4 km/hr.

Question: A boatman can row 2.5 km against the stream in 25 minutes and return in 12.5 minutes. Find the rate of flow of the current.

Solution: 
Let,
x be the speed of man in still water
y be the speed of current

Speed of downstream = (2.5/12.5) × 60 = 12 km / hr
Speed of upstream = (2.5/ 25) × 60 = 6 km / hr

∴ rate of current = (12 - 6)/2 = 3 km/hr

Question: A train 300 m long is running at a speed of 45 km/hr. In what time will it pass a bridge 300 m long?

Solution:
Speed = (45 ×1000)/3600 m/s 
= 12.5 m/s

Total distance to be covered = (300 + 300) m = 600 m.

∴ Required time = 600/12.5 = 48 sec

Question: A man can row 30 km upstream in 6 hours. If the speed of the man in still water is 6 km/hr, find how much he can row downstream in 10 hours.

Solution:
Speed of upstream = 30/6 = 5 km/hr.
Speed of man in still water = 6 km/hr.

∴ Speed of current = 6 - 5 = 1 km / hr.
So speed of downstream = 6 + 1 = 7 km / hr.

∴ Distance traveled in 10 hrs = 10 × 7 = 70 km.

Question: Two trains are moving in opposite directions at (50/3) m/s and 25 m/s. Their lengths are 1100 m and 900 m respectively. The time taken by the slower train to cross the faster train is:

Solution:
Relative speed = (50/3) + 25 m/s
= (50 + 75)/3 m/s
= 125/3 m/s

Distance covered = 1100 + 900 = 2000 m

Required time = (2000 × 3)/125 sec
= 48 sec

Question: A boat can travel with a speed of 12 km/hr in still water. If the speed of the stream is 5 km/hr, find the time taken by the boat to go 136 km downstream.

Solution: 
Speed downstream = (12 + 5) km/hr = 17 km/hr

Time taken to travel 136 km downstream = 136/17 hr
= 8 hr

Question: Mr. Rabbani running at 150 m/min alongside a railway track in 250 meters ahead of the engine of a 120 meters long train running at 750 m/min in the same direction. In how much time will the train pass him?

Solution:
Speed of train relative to Mr. Rabbani = (750 - 150) m/min = 600 m/min = 10 m/s

Distance to be covered = (250 + 120) m = 370 m

Time taken = 370/10 = 37 sec

Question: A man's speed with the current is 18 km/hr and the speed of the current is 3.6 km/hr. The man's speed against the current is:

Solution:
Man's rate in still water = (18 - 3.6) km/hr = 14.4 km/hr

∴ Man's rate against the current = (14.4 - 3.6) km/hr = 10.8 km/hr.

Question: A 230 meters long train running at the speed of 80 km/hr crosses another train running in opposite direction at the speed of 120 km/hr in 9 seconds. What is the length of the other train?

Solution:
Relative speed = (80 + 120) km/hr
= 200 km/hr
= (200 × 1000)/3600 m/s
= 500/9 m/s

Let,
The length of the other train be x meters.

Now,
(x + 230)/9 = 500/9
⇒ x + 230 = 500
⇒ x = 500 - 230
∴ x = 270

∴ The length of the other train is 270 meters

Question: A boat running upstream takes 4 hours 24 minutes to cover a certain distance, while it takes 2 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and speed of the water current respectively?

Solution:
Let
The man's rate upstream be x kmph 
The man's rate downstream be y kmph.
Distance covered upstream in 4 hrs 24 min = Distance covered downstream in 2 hrs.
x × (22/5) = y × 2
⇒ (22x)/5 = 2y
∴ y = (22x)/10

∴ Required ratio = (y + x)/2 : (y - x)/2
= (y + x) : ( y - x)
= {(22x)/10 + x} : {(22x)/10 - x}
= 32x/10 : 12x/10
= 32x : 12x
= 32 : 12
= 8 : 3

Question: A train runs at the speed of 72 km/hr and crosses a 320 m long platform in 30 seconds. What is the length of the train?

Solution:

Speed = (72 × 1000)/3600 m/sec = 20 m/sec
Time = 30 sec
Let the length of the train be x meters.

Now,
(x + 320)/30 = 20
⇒ x + 320 = 600
⇒ x = 600 - 320
∴ x = 280 

∴ The length of the train is 280 meters.

Question: A motorboat, whose speed in 250 m/min in still water goes 30000 m downstream and comes back in a total of 270 minutes. The speed of the stream is:

Solution:
Let,
The speed of the stream be x m/min
∴ Speed downstream = (250 + x) m/min
And speed upstream = (250 - x) m/min

Now,
30000/(250 + x) + 30000/(250 - x) = 270 
⇒ 15000000/(62500 - x²) = 270
⇒ 15000000 = 16875000 - 270x²
⇒ 270x² = 1875000
⇒ x² = 6944.44
∴ x = 83.33 m/min

Question: Two trains, each 50 m long, moving in opposite directions, cross each other in 4 seconds. If one is moving twice as fast the other, then the speed of the faster train is:

Solution:
Let the speed of the slower train be x m/sec.
∴ speed of the faster train = 2x m/sec.

Relative speed = (x + 2x) m/sec = 3x m/sec.

Now,
(50 + 50)/4 = 3x
⇒ 12x = 100
⇒ x = 100/12
∴ x = 8.333

∴ speed of the faster train = 2 × 8.333 = 16.67 m/sec.

Question: A boat running downstream covers a distance of 24 km in 3 hours while for covering the same distance upstream, it takes 6 hours. What is the speed of the boat in still water?

Solution:
speed downstream = 24/3  km/hr = 8 km/hr
speed upstream = 24/6 km/hr = 4 km/hr 

∴ Speed in still water = (8 + 4)/2 km/hr = 6 km/hr.

Question: The speed of a boat in still water in 20 km/hr and the rate of current is 4 km/hr. The distance travelled downstream in 20 minutes is:

Solution:
Speed downstream = (20 + 4) kmph = 24 km/hr

Distance travelled in 20 min = 24 × (20/60) km
= 8 km

Question: A train travelling at a speed of 75 mph enters a tunnel 3.5 miles long. The train is 0.25 mile long. How long does it take for the train to pass through the tunnel from the moment the front enters to the moment the rear emerges?

Solution: 
Total distance covered = 3.5 + 0.25 mile = 3.75 miles 
Train traveles 75 miles in 60 min
∴ Train traveles 3.75 mile in (60 × 3.75)/75 min
= 3 min