ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: If 15 men can reap the crops of a field in 28 days, in how many days will 5 men reap it?

Solution:
Let 5 men can reap a field in x days
So, put the same quantities on the same side.
Men: Days
Now, Men and Days are inversely proportional to each other. If we increase the number of men fewer days will be required to complete the work.
Inversely proportional means = 15 = 1/28, 5 = 1/x
so, 5/15 = 28/x
⇒ 5x = 28 × 15
∴ x = 84

Hence, 5 men can reap a field in 84 days.

Question: Worker A completes a task in 8 days, and worker B completes the same task in 10 days. If both A and B work together, in how many days they will complete the task?

Solution:
Worker A completes the task in 8 days. So, in one day, he will complete 1/8 part of the task.

So, A's one day work = 1/8
Similarly, B's one day work = 1/10

∴ (A+B)'s one day work = 1/8 + 1/10 = (5 + 4)/40 = 9/40

9/40 of the task is completed in one day so both will complete the whole task in 40/9 days

Question: A fort had arrangements for 150 boys for 45 days. After 10 days, 25 boys left the fort. Then after how much time the food will be consumed completely if the consumption of food remains the same for the remaining boys?

Solution:
ATQ,
25 people left the fort after 10 days, but still, the remaining food will be consumed at the same rate.
That means if the 150 boys continue till the end, then the remaining food would last for 150 boys for (45 - 10) = 35 days.
Now
After 10 days the remaining food will be (boys × days) 150 × 35 = 5250 unit
But the 25 boys left the for after 10 days
i.e., 125 boys will consume the 5250 unit food in x days
Now,
x = 5250/125 = 42 days.
That means the remaining food will last for 42 days.

Question: Vikas and Mohan working together can complete a work in 6 days. If Vikas alone completes the same work in 10 days, in how many days Mohan alone can complete the same work?

Solution:
Vikas and Mohan together can complete the task in 6 days.
So, in one day, they will complete 1/6 part of the task.

Therefore, (Vikas + Mohan)'s one day work will be = 1/6
Similarly, Vikas's one day work = 1/10
Therefore, Mohan's one day work = 1/6 - 1/10 = (5 - 3)/30 = 2/30 = 1/15

In one day Mohan completes the 1/15 part of the work so he will complete the entire work in 15 days.

Question: If 30 men can complete a piece of work in 27 days, in what time 18 men can do another piece of work 3 times as greater?

Solution:
30 men can do a piece of work in 27 days, and we know that men × days = total work
So, we can say that the total work = 30 × 27 = 810 units
Now ATQ,
18 men can do work 3 times greater than 810 units.
i.e., 18 men can do a work 3 × 810 = 2430 units in x days
Men × days = total work (2430 units)
⇒ 18 × x = 2430
∴ x = 2430/18 = 135 days

Question: A can do a work in 10 days and B can do the same work in 15 days. If they start working together but stop working after four days, find the fraction of the work that is left.

Solution:
A's one day work = 1/10
B's one day work = 1/15

(A + B)'s one day work = 1/10 + 1/15 = (3 + 2)/30 = 5/30 = 1/6
A and B's four day work = (1/6) × 4 = 2/3

Therefore,the remaining work = 1 - 2/3 = (3 - 2)/3 = 1/3

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: Peter is twice as good as workman as Tom. When they work together they can finish a task in 16 days. If Tom works alone, in how many days he will complete the task?

Solution:
(Peter + Tom)'s one-day work = 1/16
As per the question, Peter can finish twice as much work as finished by Tom in a given duration of time.

Therefore, 2/3 of their one day's work will be completed by Peter and 1/3 of their one day work will be completed by Tom.

So, Tom's one day work will be = (1/16) × (1/3) = 1/48
So, Tom will take 48 days to complete the task.

Question: A contract is to be finished in 46 days and 117 men involved in it, each working 8 hours per day. After 33 days, 4/7 of the work is finished, how many additional men may be employed so that it may be completed in time, each man now working 9 hours a day?

Solution:
Remaining work after 33 days = 1 - 4/7 = 3/7
Remaining period = 46 - 33 = 13 days
Now, we have
Less work, less man (directly proportion)
Less days, more men (inverse proportion)
More hours/days, less man (inverse proportion)
Now,
we can say that work = 4/7 : 3/7
Therefore, days = 13 : 33
And, hours/day = 9 : 8

(4/7) × 13 × 9 : (3/7) × 33 × 8 = 117 : x   [Where, x is total number of men after 33 days.]
⇒ (4/7) × 13 × 9 × x = (3/7) × 33 × 8 × 117
⇒ x = (3 × 33 × 8 × 117)/(4 × 13 × 9)
∴ x = 198

Therefore, extra men to be employed = 198 - 117 = 81

Question: A can do a job in 12 days and B can do the same job in 10 days. With the help of C they can do the same job in 4 days. In how many days C alone can do this job?

Solution:
A's one day work = 1/12

B's one day work = 1/10

(A+B+C)'s one day work = 1/4

Therefore, C's one day work = (A+B+C)'s one day work - (A+B)'s one day work
So, C's one day work = 1/4 - (1/12 + 1/10) = 1/4 - {(5 + 6)/60} = 1/4 - 11/60 = (15 - 11)/60 = 4/60 = 1/15

So, C will complete the work in 15 days.

Question: 24 women or 15 men or 36 boys can finish a piece of work in 12 days, working 8 hours per day, how many men must be associated with 12 women and 6 boys to finish another piece of work 9/4 times as greater in 30 days working 6 hours a day?

Solution:
We have 15 men = 24 women
Or, 12 women = 7.5 men
Also, 36 boys = 15 men
6 boys = 15/6 = 5/2 = 2.5 men
Therefore, 12 women + 6 boys = 7.5 + 2.5 men= 10 men
Now,
The number of day's ratio is 30 : 12
The hour's ratio is 6 hours : 8 hours
The work ratio is 1 work : 9/4 work
Now,
we can say that
(30 × 6 × 1) : (12 × 8 × 9/4) = 15 : x [Where x is the total number of men]
⇒ (30 × 6 × 1) × x = (12 × 8 × 9/4) × 15
⇒ x = 18

Therefore, the total number of men = 18
So, 18 -10 = 8 men must be associated.

Question: A, B, C can do a job in 10, 20 and 40 days respectively. In how many days A can complete the job if he is assisted by B and C on every third day?

Solution:

Question: The price of 357 mangoes is Tk. 1517.25. Find the approximate price of 49 dozens of such mangoes?

Solution:
We know that 1 dozen = 12 piece
49 dozens = 49 × 12 = 588 mangoes
Let,
x is the price for 588 mangoes.
Now,
Put the same unit on the same side
Price and mangoes are directly proportional to each other, so
⇒ 357 mangoes : 588 mangoes = 1517.25 : x
∴ x = (1517.25 × 588)/ 357 = 2499
Hence, The approximate value, x = Tk. 2500

Question: If 5 men can colour 50-meter long cloth in 5 days, in many days 4 men can color a 40-meter long cloth?

Solution:
M₁D₁W₂ = M₂D₂W₁
⇒ 5 × 5 × 40 = 4 × D₂ × 50
⇒ D₂ = (5 × 5 × 40)/(4 × 50)
∴ D₂ = 5

Question: If 2 kg of almonds cost as much as 8 kg of walnuts, and the cost of 5 kg of almonds and 16 kg of walnuts is Tk. 1080, what is the cost of almonds per kg?

Solution:
Let the cost of almond per kg be Tk. x, and the cost of walnuts per kg be Tk. y.
Now, ATQ,
2x = 8y
or, x = 4y

Now,
5x + 16y = 1080
Or, 5x + 4x = 1080
Or, x = 120

Therefore, the cost price of almond per kg = Tk. 120

Question: If 16 men working 7 hours a day can plow a field in 48 days, in how many days will 14 men working 12 hours a day plow the same field?

Solution:
Let the one-day work = number of men ×  total working hours per day
Now, the ratio of total work = (14 × 12) : (16 × 7)
Now, the ratio of days = 48 : x
Where x is the required number of days
Now,
one day work is inversely proportional to the number of days:
So, (14 × 12) : (16 × 7) = 48: x
Or, x = (48 ×16 × 7)/ (14 × 12) = 32

Therefore, 32 days are required to plow the same field.

Question: A can do a piece of work in 10 days. B is 50% more efficient than A. In how many days B alone can do the same job?

Solution:
B is 50% more efficient than A so he will take less time to do a piece of work.

Therefore, the ratio of the time taken by A and B = 150/100 = 3 : 2

Let B takes X days to do the job.
3 : 2 = 10 : X
⇒ 3X =20
∴ X = 6.6 days

Question: A tower 17.5 m high casts a shadow of 40.25 m. What is the height of the building which casts a shadow 28.75 m long under similar conditions?

Solution:
Let the height of the building is x
Now, the shadow ratio = building ratio
Height is directly proportional to shadow, so:
40.25 : 28.75 = 17.5 : x
Now, x = (28.75 × 17.5)/ 40.25 = 12.5 m

Question: A can do a job in 30 days. B alone can do the same job in 20 days. If A starts the work and joined by B after 10 days, in how many days the job will be done?

Solution:
A's one day work = 1/30
A's ten-day work = (1/30) × 10 = 1/3

So the remaining work would be = 1 - 1/3 = 2/3

B's one day work = 1/20
A and B's one day work = 1/30 + 1/20 = (2 + 3)/60 = 5/60 = 1/12

1/12 of the job will be done by them in one day.
So, the remaining job 2/3 will be done in = (2/3) × 12 = 8 days

Therefore, the total number of days required to do the job would be = 10 + 8 = 18 days

Question: One army camp had ration for 560 soldiers for 20 days, 560 soldiers reported for the camp, and after 12 days, 112 soldiers were sent to another camp. For how many days, the remaining soldiers can stay in the camp without getting any new ration?

Solution:
After 12 days, there was ration for 560 soldiers for 8 days.
Remaining persons = (560-112) = 448
Less soldiers, more days (inverse proportion)
Let the x is the required number of days
Then, 448 : 560 = 8 : x
Or, x = (560 × 8)/ 448 = 10
Hence, the required number of days is 10.

Question: If I would have been twice as efficient as today, I would have finished work in 12 days. If my efficiency is reduced to one-third of what it is at present, in how many days, I would be able to finish the work?

Solution:
Let I finish work in x days.
With double efficiency, the time taken = x/2 days
That means when the efficiency was double (2x),
then the time taken to finish the work is 12 days
Now,
with the present efficiency time taken to complete the work = 12 × 2 =24 days
With one-third efficiency, the days required to finish the work = 3 × 24 = 72 days,
as efficiency is inversely proportional to days.
Hence, when the efficiency gets one-third, then the work will be finished in 72 days.

Question: 5 men undertook a piece of work and finished half the work in 18 days if two men drop out, in how many days the remaining work will be completed?

Solution:
5 men completed half of the work in 18 days so the entire work will be completed in 36 days.
5 men' one day work will be = 1/36
One man's one day work = 1/(36 × 5) = 1/180

Two men drop out, so the three men have to complete the remaining work.
Three men's one day work will be = (1/180) × 3 = 1/60

1/60 part of the work is completed by three men in one day
1 or full part of the work is completed by three men in 60 day
1/2 part of the work is completed by three men in 60/2 = 30 day

Question: Pervez and Sunny can complete a piece of work in 20 and 15 days respectively. They worked together for 6 days, after which Sunny was replaced by Ashu. If the work would be finished in next 4 days, find the number of days in which Ashu alone could complete the work.

Solution:
Pervez can complete the work in 20 days
Sunny can complete the same job in 15 days
Let the total work = LCM of both
Therefore, the LCM of 20 and 15 is 60, that means total work = 60 units
Now, Pervez's one-day work efficiency = 60/20 = 3 units
Sunny's one-day work efficiency = 60/15 = 4 units
ATQ,
(Pervez + Sunny) together works for 6 days
That means (Pervez+Sunny) have done 7 × 6 = 42 units work in 6 days
Now, the remaining work will be 60 - 42 = 18 units, and it will finish in 4 days.
Now, Sunny is replaced with Ashu:
So, per days work required to be finished = 18 units / 4 days = 4.5 units
But, we know that Pervez's one-day work = 3 units, so Ashu's one-day work = 4.5 - 3 = 1.5 units
Therefore, Ashu alone can finish the total work in 60 units /1.5 days = 40 days