ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Men = M; Days = D; Time/Hours = T; Work = W
M₁D₁T₁W₂ = M₂D₂T₂W₁
Note that W₂ is on left side and W₁ is on right side

∴ 20 women x 20 minutes x ? = 400 women x 400 minutes x 20 chapattis
∴ ? = 8000 breads = Are rolled by 400 women in 400 minutes.

Take work done = 1
Let number of sailors who could not board = S.
So sailors who boarded = 2500 - S
∴ 2500 sailors x 40 days x 1 = (2500 - S) sailors x 50 days x 1
∴ S = 500 = Sailors who could not board the ship.

[Men = M; Days = D; Time/Hours = T; Work = W
M₁D₁T₁W₂ = M₂D₂T₂W₁
Note that - W₂ is on left side and W₁ is on right side]

To complete work in 18 days we need either 3 men or 6 boys.
∴ 1 man = 2 boys
Take work done = 1
∴ 3 men x 18 days x 1 = (4 men + 4 boys) x ? days x 1
∴ 6 boys x 18 days x 1 = (8 boys + 4 boys) x ? days x 1 [Convert either all men to boys or all boys to men.]
∴ ? = 9 days = they will need these many days.

Remaining work = 1 - (1/3) = 2/3
Let number of more employees needed be E
Thus (75+E) employees complete 2/3 works in 50 hours.
∴ 75 employes × 40 hours × (2/3) = (75 + E) × 50 hours × (1/3)
∴ E = 45 = these many more employees are needed.

Let each tree take T amount of water every day.
So 72 trees take 72T water in one day.

With 10% reduction each tree will consume 90T/100 amount each day.
So 90 trees take 90 (90T/100) amount in one day
Total water quantity is constant
∴ 72T × 54 = 90 × (90T/100) × D
∴ D = 48 days = Number of days 90 trees can use the water.

We know,
M₁D₁T₁W₂ = M₂D₂T₂W₁ [Men = M; Days = D; Time/Hours = T; Work = W]

Let the former type be E times efficient.
So, 1 former type pump = E x latter type pumps
So 1F = E x L
∴ 12F x 18 hours x 400 = 16L x 24 hours x 300
∴ (12 x E x L) x 18 x 400 = 16 L x 24 x 300 ------------------> Put value of F i.e. 1F
E = 4/3 = these are many times more efficient.

We know,
M₁D₁T₁W₂ = M₂D₂T₂W₁ [Men = M; Days = D; Time/Hours = T; Work = W]

1 child binds half the number of books bound by 1 adult
∴ 1 adult = 2 children
Take work done = 1
∴ (20 adults + 16 children) x 10 days x 1 = (8 adults + 12 children) x ? days x 1
∴ 56 children x 10 days x 1 = 28 children x ? days x 1 [Convert either all adults to children or all children to adults.]
∴ ? = 20 days = they will need these many days.

Let work is done by Rizvi in 1 day = 1/R & Shafi in 1/S day
Both complete work in 24 days. So in 1 day, together they complete = 1/24 = 1/R + 1/S
For 20 days both work together, so work done by them = 20(1/R + 1/S) = 5/6
Remaining work = {1 - (5/6)= 1/6} is done by Rizvi alone in 6 days
Work done by Rizvi in 6 days = 1/6 = 6(1/R)
∴ R = 36 = days needed by Rizvi to complete the work alone

According to the question,
∴ (1/36) + 1/S = 1/24
1/S = (1/24) - (1/36)
1/S = (3 - 2)/72
1/S = 1/72
S = 72 days needed by Shafi to complete the work alone.

We are given that,
3 pumps, working 4 hours a day, can empty a tank in 2 days. Therefore, it means that:
3 pumps take a total of 8 hours to empty the tank.
Hence, 1 pump will take 8 x 3 = 24 hours

As the number of pumps decreases, the time required increases.
So, if 4 pumps work, the time required decreases.
∴ 24/4 = 6hrs. needed to empty the tank in 1 day.

We know,
Indirect proportion: Less cows (↓) More days (↑)
Direct proportion: Less bags (↓) Less days (↓)

20 : x :: { 1 : 20 ------ (Cows); 20 : 1 -------(Bags)}
1 × 20 × x = 20 × 1 × 20
⇒ x = (20 × 1 × 20)/(1 × 20)
⇒ x = 20 Days.

A hostel had provisions for 250 men for 40 days

If 50 men leaves the hostel, remaining men = 250 - 50 = 200
We need to find out how long the food will last for these 200 men.

Let the required number of days = x days.

More men, Less days (Indirect Proportion)
(men) 250 : 200 :: x : 40
⇒ 250 × 40 = 200x
⇒ 5 × 40 = 4x
⇒ x = 5 × 10
= 50

Given that food was sufficient for 2000 people for 54 days
Hence, after 15 days, the remaining food was sufficient for 2000 people for 39 days (∵ 54 - 15 = 39)

Let x number of people come after 15 days.
Then, total number of people after 15 days = (2000 + x)
Then, the remaining food was sufficient for (2000 + x) people for 20 days

More men, Less days (Indirect Proportion)

(men) 2000 : (2000 + x) :: 20 : 39
⇒ 2000 × 39 = (2000 + x) 20
⇒ 100 × 39 = (2000 + x)
⇒ 3900 = 2000 + x
⇒ x = 3900 − 2000
= 1900

Let the required number of rounds be x

More radius, less rounds(Indirect proportion)

Hence we can write as
(radius) 14 : 20 :: x : 70
⇒ 14 × 70 = 20x
⇒ 14 × 7 = 2x
⇒ x = 7 × 7
= 49 days.

We know,
Speed of Work / Efficiency ∝ 1/Time needed
So more the efficiency, less time is needed to do the task.

Let us denote Akash by 'A' and Biplob by 'B'
Hence,
if A is 5 times faster than B, then B needs 5 times more time than A.
Let A need 'n' days to complete the work.
So B will need 5n days.
Also, 5n – n = 60 ------------------- Given
∴ n = 15 days = Days needed by A to complete work individually
Days needed by B to complete work individually = 5n = 5 x 15 = 75 days.

24 men complete work in 10 days.
In 1 day work done by 24 men = 1/10
In 1 day work done by 30 men = ?
∴ ? × 24 = (1/10) × 30
∴ ? = 1/8 = Amount of work done by 30 men in one day.

Thus in this case,
30 men complete the entire work in 8 days.
[If A completes a work in ''n'' days, in 1 days he completes 1/n amount of work; Conversely, if A completes 1/n amount of work in 1 day, he completes the entire work in 'n' days.]

A can do the work in 3 days. So, in 1 day a does 1/3 amount of work.

B takes double the time as A. So, B can do the work in 6 days. So, in 1 day B does 1/6 amount of work

C takes one more day in addition to the time taken by B. So, C can do the work in 7 days. So, in 1 day C does 1/7 amount of work

When they work together, in 1 day they complete how much work?
In 1 day A, B and C together complete = 1/3 + 1/6 + 1/7 = 9/14 amount of work
A, B and C together complete the entire work in 14/9 days.

April has 30 days. So Protik takes 30 days to build the pavement.
Mahmud is 25% faster than Protik

25% = 25/100 = .25
This means, if Protik is 1, then Mahmud is (1 + 0.25) = 1.25
Protik takes 30 days to do the work.
Mahmud will take = 30/1.24 = 24 days to get the work done.

In a days P does 1/20 work; And in 1 day Q does 1/25 work
As seen above, Q works alone for 10 days.
In 1 days Q completes (1/25) × 10 = 2/5 work
Remaining work = 1 - (2/5) = 3/5 = Done by P and Q together
Total Work done = Total days x Work done by all in 1 day

Let P and Q work together for total K days.

∴ 3/5 = K × (1/20 + 1/25)
K = 20/3 = 6(2/3) days = Days when P and Q worked together
Thus P leaves after 6(2/3) days.

A completes the work in 12 days at 8 hour per day = 12 x 8 = 96 hours
In 1 hour, A completes (1/96) amount of work

B completes the work in 10 days (2 days less than A) working 10 hours every day (i.e. 2 hours more than A per day) = 10 x 10 = 100 hours
In 1 hour, B completes (1/100) amount of work

Working together, in 1 hour A and B complete (1/96) + (1/100) = (49/2400) amount of work
So they complete the entire work in (2400/49) hours

Since they work for 8 hours per day, they need = (2400/49) × (1/8) = 300/49 days.

Pages typed by Anik in 1 hour = 32/6 = 16/3
Pages typed by Shuvo in 1 hour = 40/5 = 8

Pages typed by Anik and Shuvo in 1 hour = 16/3 + 8 = 40/3
Time taken to type 110 pages when Anik and Shuvo work together = 110 × 3 /40 = 33/4
= 8 (1/4) hours = 8 hour 15 minutes.

Let,
The amount of work P does in 1 day = x
Amount of work Q does in 1 day = 2x
Amount of work R does in 1 day = 3x

Amount of work P, Q,R together do in 1 day = x + 2x + 3x = 6x
they can together complete the work in 1 day = (1/6x) days

Given,
1/6x = 2
⇒ 12x = 1
⇒ x = 1/12

Therefore, amount of work Q does in 1 day = 2 × (1/12) = 1/6
That is, Q needs 6 days to complete the work.

Work done by P in 1 day = 1/15
Work done by Q in 1 day = 1/10

Work done by P and Q in 1 day = 1/15 + 1/10 = 1/6
Work done by P and Q in 2 days = 2 × (1/6) = 1/3

Remaining work = 1 - 1/3 = 2/3

Time taken by P to complete the remaining work 2/3 = (2/3) / (1/15) = 10 days

Total time taken = 2 + 10 = 12 days.

1-minute work done by both the punctures = (1/9) + (1/6)
= (4 + 6)/36
= 10/36
= 5/18

So both punctures will make the type flat in = (18/5) min
= 3(3/5) min

Combined work done by Mamun and Tonmoy in 4 + 6 days (4 initial days and last 6 days)= 10/24 + 10/40
= (50 + 30)/120
= 80/120
= 2/3

∴ Remaining work = 1 - 2/3 = 1/3
Robin works for = (1/3) × 30 = 10.

Time taken by A = x days

Time taken by B = x/2 days

Time Taken by C = x/3 days

⇒ {(1/x) + (2/x) + (3/x) = 1/2

⇒ 6/x = 1/2

⇒ x = 12

Time taken by B = x/2 = 12/2 = 6 days.