ব্যাংক নিয়োগ বিষয়ভিত্তিক প্রস্তুতি

Question: A train moving at speed of 72 km/hr crosses a pole in 10 seconds. Find the length of the train.

Solution:
Length of the train is equal to the distance covered by train to cross the pole.
So, we will find the distance travelled by the train in 10 seconds by applying the following formula:
Distance = Speed × Time
Speed is given in Km/hr so we will convert it into m/s
Speed = 72 × (5/18) = 20 m/s
Time = 10 seconds
Distance = 20 × 10 = 200 meters

Question: A cyclist takes 30 minutes to cover 15 km against the wind, which is 50% more than the time taken to cover the same distance with the wind. What is the cyclist's speed in still air?

Solution:
Let the speed in still air = x km/hr.
The cyclist takes 30 min. to cover 15 km against the wind.
∴ Speed against wind = 15/0.5 = 30 km/hr.

Also, the time taken against the wind is 50% more than the time taken with the wind.
∴ Time with wind = 30/1.5 = 20 km/hr.

So, the distance covered with the wind in 20 min. = 15 km.
Speed with wind = 15/(20/60) = 45 km/hr.
So, x + y = 45 ..... (1)
x - y = 30 ...... (2)

Adding the two equations: 
x + y + x - y = 45 + 30
⇒ 2x = 75
∴ x = 37.5
The cyclist's speed in still air is 37.5 km/hr.

Question: A boat can travel with a speed of 20 km/hr in still water. If the speed of the stream is 5 km/hr, find the time taken by the boat to go 75 km downstream.

Solution:
Speed downstream
= Speed of Boat in still water + Speed of the stream
= (20 + 5) km/hr
= 25 km/hr

Time taken to travel 75 km downstream = Distance/Speed
= (75/25) hours
= 3 hours

Question: Two pipes A and B can fill a tank in 10 and 15 hours respectively. If both the pipes are used together, then how long will it take to fill the tank?

Solution:
Part filled by A in 1 hour = 1/10

Part filled by B in 1 hour = 1/15

Part filled by (A + B) in 1 hour
= (1/10) + (1/15)
= (3 + 2)/30
= 1/6

∴ Both pipes can fill the tank in 6 hours

Question: A boat makes a return journey from point A to point B and back in 5 hours 36 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 9 hours 20 minutes. What is the speed of the boat in still water? (The distance between A and B is 16 km.)

Solution:
Let x be speed of u/s
and y be the speed of d/s.

∴ (16/x) + (16/y) = (28/5)
and 16/(y+2) + 16/(x-2) = 28/3

Solving these 2 equations,
we get x = 4km/hr
and y = 10km/hr

∴ speed of boat in still water = (4 + 10)/2 = 7 km/hr.

Question: 40 workers can build 40 engines working 6 hours a day. How many workers need to be appointed extra to boost the production to double if they work 8 hours a days?

Solution:
6 hours to build 40 engines by 50 workers
1 hour to build 1 engine by = (40 × 6)/40 workers
8 hours to build 80 engine by = (6 × 80)/8 workers
= 60 workers

∴ extra workers = 60 - 40 = 20 workers 

Question: A does one-third as much work as B in one-fourth of the time. If together they take 12 days to complete a work, how much time shall B alone take to do it?

Solution:
Let B takes x days to do the work.
A takes 1/4 of x time to do 1/3 of the work.
∴ the work will be done by A in (x/4) × 3 days
= 3x/4 
ATQ,
1/x + 4/3x = 1/12
⇒ 7/3x = 1/12
⇒ x = 28
∴ B alone will take 28 days.

Question: A man takes twice as long to row a distance against the stream as to row the same distance in favour of the stream. The ratio of the speed of the boat (in still water) and the stream is:

Solution:
Let man's rate upstream be x kmph.
Then, his rate downstream = 2x kmph.

∴ (Speed in still water) : (Speed of stream) = {(2x + x)/2} : {(2x - x)/2}
= (3x/2) : (x/2)
= 3 : 1

Question: A train covered first 120 km at a speed of 20 km an hour and then covered the remaining 180 km at a speed of 45 km an hour. Find its average speed.

Solution:
Total distance = 120 + 180 = 300 km.

Time taken for the first 120 km = 120/20 = 6 hrs.
Time taken for the next 180 km = 180/45 = 4 hrs.
Total time taken = 6 + 4 = 10 hrs.

Average Speed =Total Distance Travelled/Total Time Taken
= 300/10
= 30 km/hr

Question: There is a leak at the bottom of a cistern. Due to this it takes 8 hours to fill the cistern. Had there not been a leak, it would take one hour less to fill the cistern. How much time does it take for the leak to completely empty the cistern?

Solution:
Normally, the cistern gets filled in one hour less than 8 hours which means 7 hours.
So in 1 hour it fills = 1/7 parts
Due to leak, it takes 8 hours. So in 1 hour it actually fills = 1/8 parts

∴ Water removed by the leak in 1 hour = (1/7) - (1/8)
= (8 - 7)/56
= 1/56
∴ Leak empties the tank in 56 hours.

Question: 45 toymakers can prepare 30 toys per day. Raj wants 360 toys. How many toymakers should he employ to get the job done in 12 days?

Solution:
Let, the required number of toymakers x
45 toymakers make 30 toys per day
So, 1 toymaker makes = 30/45 = 2/3 toys per day
Each toymaker in 12 days makes = (2/3) × 12 = 8 toys
So, x toymakers will make = 8x toys

ATQ,
8x = 360
⇒ x = 360 × (1/8)
∴ x = 45

Question: Two Motorists drove their cars at a speed of 45 km per hour and 50 km per hour respectively. One car took 10 minutes longer than the other to travel a distance. Find the distance travelled.

Solution:
let, the distance is x km

(x/45) - (x/50) = 10/60
⇒ (20x - 18x)/900 = 1/6
⇒ 2x = 150
∴ x = 75 km

Question: Two pipes A and B can fill a cistern in 37.5 minutes and 45 minutes respectively. This cistern will be filled in half an hour if both the pipes are opened together initially and pipe B is then turned off after X minutes. What is X?

Solution:
Let the volume of tank = V liters
Rate of pipe A = (V/37.5) L/min
Rate of pipe B = (V/45) L/min
In X minutes (both pipes open): Part filled = (V/37.5) + (V/45) × X
= V(45 + 37.5)/(37.5 × 45) × X
= V × 82.5/(1687.5) × X
= V × (X/20.45)

In remaining time (30 - X) minutes (only pipe A): Part filled = (V/37.5) × (30 - X)
Total part filled = 1 (complete tank)
X/20.45 + (30-X)/37.5 = 1
⇒ 37.5X + 20.45(30-X) = 37.5 × 20.45
⇒ 37.5X + 613.5 - 20.45X = 766.875
⇒ 17.05X = 153.375
∴ X = 9

Question: A woman and a child received 8400 Tk. as wages for 7 days for the work they did together. The woman's efficiency in the work was triple that of the child. What are the daily wages of the child?

Solution:
Ratio of 1 day's work of woman and child = 3 : 1
Total wages of the child = (8400 × 1/4)
= 2100 Tk.

∴ Daily wages of the child = 2100/7 = 300 Tk.

Question: A fort had provision of food for 200 men for 60 days. After 20 days, 40 men left the fort. The number of days for which the remaining food will last, is-

Solution:
After 20 days : 200 men had food for 40 days.
Suppose 160 men had food for x days.
Now, Less men, More days (Indirect Proportion)
160 : 200 : : 40 : x
⇒ 160/200 = 40/x
⇒ 160 x = 200 × 40
⇒ x = (200 × 40)/160
∴ x = 50 days

Question: A train 300 metres long is running at a speed of 90 km/hr. How many seconds will it take cross a 200 metres long train running in the same direction at a speed of 60 km/hr?

Solution:
Length of 1st train 300 metres
Length of 2nd train 200 metres

∴ Total distance to cross each other = 300 + 200 metres
= 500 metres

Relative speed for travelling same direction = 90 - 60 km/hr
= 30 km/hr
= (30 × 1000)/3600 m/s
= 300/36 m/s

Required time to cross = 500/(300/36) s
= (500 × 36)/300 s
= 60 s

Question: Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is:

Solution:
Let the length of each train be x metres.
Then, distance covered = 2x metres.
Relative speed = (46 - 36) km/hr
= 10 × (5/18) m/sec
= 25/9 m/sec

ATQ,
2x/36 = 25/9
⇒ 18x = 25 × 36
⇒ x = (25 × 36)/18
= 50

Question: A motorboat can travel at 5 km/hr in still water. It travelled 90 km downstream in a river and then returned, taking altogether 100 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 u/s = 5 - y
and speed of d/s = 5 + y

∴ 90/(5 + y) + 90/(5 - y) = 100
⇒ 450 - 9y + 450 + 9y = 100(25 - y²)
⇒ 9 = 25 - y²
⇒ y² = 16
⇒ y = 4 km/hr.

Question: One pipe can fill a tank four times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in:

Solution:
Let the faster pipe can fill it in X minutes
in one minute it can fill up = 1/X of the tank

so the slower pipe can do it in 4X minutes
in one minute it can fill up = 1/4X of the tank

so in one minute both can fill = (1/X) + (1/4X)
= 5/4X
the full tank will be filled in = 4X/5 minutes
ATQ,
4X/5 = 36
X = 45
so the slower pipe can do it in = 4 × 45 = 180 minutes
= 3 hours

Question: To complete a work, A takes 50% more time than B. If together they take 18 days to complete the work, how much time shall B take to do it?

Solution:
Let B takes x days to complete the work
Then A will take 50% more i.e 150% of x days i.e 3/2x days.

So the one day work of A and B together will be
(1/x) +{1/(3/2x)} = 1/18
⇒ (1/x) + (2/3x) = 1/18
⇒ 5/3x = 1/18
⇒ x = 30
∴ B takes 30 days to complete the work.

Question: If 9 engines consume 24 metric tons of coal, when each is working 8 hours per day, how much coal should be available for 8 engines, each running 13 hours per day, it is given that 3 engines of the former type consume as much as 4 engines of later type.

Solution:
We have:
The lesser engines, less coal consumed
More working hours, more coal consumed
Both the cases are directly proportional.
If three engines of former type consume 1 unit, 1 engine will consume 1/3 unit.
If four engines of latter type consume 1 unit, 1 engine will consume ¼ units.
And, less rate of consumption, less coal consumed.
Now, Number of engines = 9: 8
Working hours = 8:13
Therefore, rate of consumption = 1/3 : 1/4

Let the coal consumed by 8 engines is x metric tones
9 × 8 × 1/3 : 8 × 13 × 1/4 = 24 : x
⇒ 9 × 8 × (1/3) × x = 8 × 13 × (1/4) × 24
⇒ 24x = 624
∴ x = 26

Question: The distance between two places A and B is 370 km. the 1st car departs from place A to B, at a speed of 80 kmph at 10 am and 2nd car departs from place B to A at a speed of 50 kmph at 1 pm. At what time both cars meet each other?

Solution:
Total distance = 370 km
Now, the distance covered by first car in (10 am to 1 pm =) 3 hours
= 80 × 3
= 240 km

Remaining distance = 370 - 240 = 130 km
Relative speed = 80 + 50 = 130 kmph
Now, they cover 130 km in (130/130) = 1/1 hrs  = 60 mins

So, they meet 60 mins after 1 pm.
So, requred answer = 2:00 pm.

Question: An outgoing pipe pours water at half the amount of an ingoing pipe. After 6 hours of running both pipes, a tank was filled. If the outgoing pipe was closed, how much time would it take to fill the tank with the ingoing pipe?

Solution:
Let,
ingoing pipe needs X hours,
The outgoing pipe needs 2X hours.

together in one hour, these pipes can fill = (1/X) - (1/2X) = 1/2X

ATQ,
2X = 6
X = 3
∴ Ingoing pipe will take 3 hours to fill the tank.