ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: A train running at the speed of 120 km/h crosses a pole in 12 seconds. What is the length of the train?

Solution:
Speed = 120 km/h
= (120 × 5/18) m/sec
= 100/3 m/sec

∴ Length of the train = (100/3 × 12) m
= 400 m

Question: Two trains are moving in opposite directions  80 km/h and 100 km/h. Their lengths are 1100 m and 900 m respectively. The time taken by the slower train to cross the faster train in seconds is:

Solution:
Relative speed= (80 + 100) km/h
= 180 × (5/18) m/sec
= 50 m/sec

Distance covered= (1100 + 900) m
=2000 m

Required time= (2000 ÷ 50) sec
= 40 sec

Question: A man takes 8 hours in walking a distance and riding back to starting place. He could walk both ways in 10 hours. The time taken by him to ride back both ways is:

Solution:
Time taken in walking both the ways = 10 hours.................(i)
Time taken in walking one way and riding back = 8 hours...................(ii)

By the equation (ii) × 2 - (i), we have,
Time taken by the man in riding both ways,
= 16 hours - 10 hours
= 6 hours

Question: A train moves past a telegraph post and a bridge 240 m long in 8 seconds and 20 seconds respectively. What is the speed (km/h) of the train?

Solution:
Let,
the length of the train = x metres
and its speed = y m/sec

ATQ,
x/y = 8
⇒ x = 8y

and,
(x + 240)/20 = y
⇒ x + 240 = 20y
⇒ 8y + 240 = 20y
⇒ 240 = 20y - 8y
⇒ 12y = 240
∴ y = 20

∴ The speed of the train = 20 m/sec
= 20 × (18/5) km/h
= 72 km/h

Question: How many seconds will a 800 meter long train take to cross a man walking with a speed of 6 km/h in the direction of the moving train if the speed of the train is 78 km/h?

Solution:
Speed of the train relative to man = (78 - 6) km/h
= 72 km/h
= (72 × 5/18) m/sec
= 20 m/sec

Time taken to pass the man = 800/20 sec
= 40 sec

Question: A runs twice as fast as B and B runs thrice as fast as C. The distance covered by C in 144 minutes, will be covered by A in:

Solution:
Let,
The speed of C = x
So the speed of B = 3x
and the speed of A = 3x × 2
= 6x

Ratio of speed A : B : C = 6x : 3x : x
= 6 : 3 : 1

Then ratio of time taken = 1/6 : 1/3 : 1
= 1 : 2 : 6

Hence, time taken by A = 144/6 minutes
= 24 minutes

Question: A 220 meters long train running at the speed of 108 kmph crosses another train running in opposite direction at the speed of 90 kmph in 9 seconds. What is the length of the other train?

Solution:
Relative speed =(108 + 90) km/h
= (198 × 5/18) m/sec
= 55 m/sec

Let,
the length of the other train be x metres.
Then,
(x + 220)/9 = 55
⇒ x + 220 = 55 × 9
⇒ x = 495 - 220
∴ x = 275

Question: A boat takes 30 minutes less to travel 12 miles downstream than to travel the same distance upstream. If the speed of the boat in still water is 10 mph, the speed of the stream is-

Solution:
Let,
The speed of the stream = x mph

Then,
Speed downstream = (10 + x) mph
Speed upstream = (10 - x) mph

ATQ,
{12/(10 - x) - 12/(10 + x)} = 30/60
⇒ (24x × 60) = 30(100 - x²)
⇒ x² + 48x - 100 = 0
⇒ (x + 50)(x - 2) = 0
∴ x = 2 mph

Question: Two trains of equal length are running on parallel lines in the same direction at 52 km/h and 40 km/h. The faster train passes the slower train in 36 seconds. The length of each train is-

Solution:
Let,
The length of each train = x metres.
Then, distance covered = 2x metres.
Relative speed = (52 - 40) km/hr
= {12 × (5/18)} m/sec
= 10/3 m/sec

ATQ,
2x/36 = 10/3
⇒ 2x = 120
∴ x = 60 m

Question: The speed of a boat in still water in 25 km/h and the rate of current is 3 km/h. The distance travelled downstream in 15 minutes is-

Solution:
Given,
The speed of a boat in still water = 25 km/h
The rate of current = 3 km/h

∴ Speed downstream = (25 + 3) kmph
= 28 kmph.

∴ Distance travelled = (28 × 15/60) km
= 7 km

Question: Two trains start at the same time from Dhaka and Cumilla and proceed towards each other at 70 kmph and 90 kmph respectively. When they meet, it is found that one train has travelled 60 km more than the other. Find the distance between Dhaka and Cumilla.

Solution:
Let,
They meet after t time.

ATQ,
(90t) = (70t) + 60
⇒ (90t) - (70t) = 60
⇒ 20t = 60
∴ t = 3 hours

So the distance between Dhaka and Cumilla = (90 × 3) + (70 × 3)
= 270 + 210
= 480 km

Question: Subarna express normally reaches its destination at 60 km/h in 8 hours. Find the speed at which it travels to reduce the time by 3 hours?

Solution:
Distance to be covered = Speed × Time
= 60 × 8
= 480 km

Time = (8 - 3) hours
= 5 hours

∴ Required Speed = 480/5
= 96 km/h

Question: Two trains starting at the same time from two-stations 250 km apart and going in opposite directions, cross each other at a distance of 140 km from one of them. The ratio of their speeds is -

Solution:
In the same time, they cover 140 km and 110 km respectively.
Therefore, the ratio of their speeds = 140 : 110
= 14 : 11

Question: A wheel of an engine of 450 cm in circumference makes 20 revolutions in 6 seconds. What is the speed of the wheel in km/h?

Solution:
Total distance = (450 × 20) cm
= 9000 cm
= 9000/100 m
= 90 m

We know,
Speed = (Total distance ÷ Time)
= (90 ÷ 6) m/sec
= 15 m/sec
= (15 × 18/5) km/h
= 54 km/h

Question: A train 340 m long passes a pole in 17 seconds. How long will it take to pass a platform 660 m long?

Solution:
Speed of the train = (Distance ÷ Time)
= (340 ÷ 17) m/sec
= 20 m/sec

∴ Required time = (340 + 660)/20 seconds
= 50 seconds

Question: A man can row 1.8 kilometer against the stream in 15 minutes and down the stream in 3 minutes. The speed (in m/sec) of the man in still water is-

Solution:
1.8 kilometer = 1800 meters
15 minutes = 15 × 60
= 900 sec
3 minutes = 3 × 60
= 180 sec
Rate upstream = 1800/900 m/sec
= 2 m/sec
Rate downstream = 1800/180 m/sec
= 10 m/sec
∴ Rate in still water = (1/2)(2 + 10) m/sec
= 6 m/sec

Question: A 100 m long train is going at a speed of 108 km/h. It will cross a 140 m long tunnel in-

Solution:
Given,
Speed of the train = 108 km/h
= (108 × 5/18) m/sec
= 30 m/sec

∴ Total distance covered = (100 + 140) m
= 240 m

∴ Required time = (240 ÷ 30) sec
= 8 sec

Question: An Aeroplan covers a certain distance at a speed of 300 kmph in 5 hours. To cover the same distance in 5/3 hours, it must travel at a speed of-

Solution:
Given,
Distance = (300 × 5)
= 1500 km

We know,
Speed = (Distance ÷ Time)

∴ Required speed = (1500 ÷ 5/3) km/h
= (1500 × 3/5) km/h
= 900 km/h

Question: A man's speed with the current is 18 km/h and the speed of the current is 3 km/h. The man's speed against the current is-

Solution:
Given,
man's speed with the current is 18 km/h
the speed of the current is 3 km/h

∴ Man's rate in still water = (Man's speed with the current - The speed of the current)
= (18 - 3) km/h
= 15 km/h

∴ Man's rate against the current = (Man's rate in still water - The speed of the current)
= (15 - 3) km/h
= 12 km/h