Bank Math Master

Question: A, B and C can do a piece of work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day? 

Solution:
A's 2 day's work = (1/20) × 2 = 1/10
(A + B + C)'s 1 day's work = 1/20 + 1/30 + 1/60
= 6/60
= 1/10

Work done in 3 days = 1/10 + 1/10
= 2/10 part
= 1/ 5 part

Now,
1/5 work is done in 3 days.
∴ Whole work will be done in (3 × 5) = 15 days

Question: How long will it take for two pipes to fill a tank together when they can fill it alone in 14 hours and 21 hours respectively?

Solution:
together in one hour they can fill = 1/14 + 1/21 = 5/42

so, the total time to fill the tank is = 42/5 hours =  8.4 hours

Question: A can do a work in 8 days and B in 12 days. If they work on it together for 4 days, then the fraction of the work that is left is-

Solution:
A's 1 day's work = 1/8
B's 1 day's work = 1/12

∴ (A + B)'s 1 day's work = (1/8 + 1/12) part
= (3 + 2)/24 part
= 5/24 part

∴ (A + B)'s 4 day's work = (5/24 × 4) part
= 5/6 part

Therefore, Remaining work = (1 - 5/6) part
= 1/6 part

Question: 15 buckets of water fill a tank when the capacity of each bucket is 16 liters. How many buckets will be needed to fill the same tank, if the capacity of each bucket is 12 liters?

Solution: 
total capacity of the tank is = (15 × 16) = 240 liters.

total buckets of 12 liters = 240/12 = 20 buckets

Question: 10 women can do a work in 8 days. 8 men can complete the same work in 7 days. What is the ratio between the capacity of a man and a woman?

Question: 
(10 × 8) = 80 women can complete the work in 1 day.
∴ 1 woman's 1 day's work = 1/80 part

(8 × 7) = 56 men can complete the work in 1 day.
1 man's 1 day's work = 1/56 part

So, required ratio = 1/56 : 1/80
= (1/56)/(1/80)
= 80/56
= 10/7
= 10 : 7

Question: Two pipes together can fill a tank in 6 hours. One pipe alone can do it in 12 hours. Another pipe alone can fill two tanks in - 

Solution: 
let the second pipe fill the tank in X hours.

in one hour both pipes can fill = 1/12 + 1/X
= (X + 12)/12X

Atq,
12X/(X + 12) = 6
x = 12 hours.

to fill two tanks it will take = 24 hours

Question: If A can do 1/4 of a work in 3 days and B can do 1/9 of the same work in 4 days, how much will A get if both work together and paid Tk 800 in all?

Solution:
Whole work is done by A in (3 × 4) = 12 days
∴ A's 1 day's work = 1/12 part
Whole work is done by B in (4 × 9) = 36 days
∴ B's 1 day's work = 1/36 part

A's 1 day's work : B's 1 day's work
= A's wages : B's wages
= 1/12 : 1/36
= 3 : 1

∴ A's share = (800 × 3/4) Tk
= 600 Tk

Question: An outgoing pipe is attached to a tank that can empty it in 20 hours. An ingoing pipe is connected to the tank that can fill it in 8 hours. How much time will it take to fill the half-full tank?

Solution: 
ingoing pipe in one hour can fill = 1/8
outgoing pipe in one hour can empty = 1/20

in one hour total fill up = 1/8 - 1/20 = 3/40

to fill half the tank it will take = 40/6 = 20/3 hours.

Question: 8 men can complete a piece of work in 20 days. 8 women can complete the same work in 32 days. In how many days will 5 men and 8 women together complete the same work?

Solution:
8 × 20 men = 8 × 32 women
5 men = 8 women
Now, 5 men + 8 women = 8 + 8 = 16 women

ATQ, D₁ × M₁= M₂× D₂
⇒ 8 × 32 women = 16 × D₂
⇒ D₂= (32 × 8)/16
∴ D₂= 16 days

Question: A pipe can fill 1/6th of a tank in 30 minutes. How much time will it take to fill two tanks?

Solution:
full tank is filled in = (6 × 30) = 180 minutes = 3 hours.

two tanks is filled in = 6 hours

Question: To complete a work, X takes 25% more time than Y. If together they take 20 days to complete the work, how much time shall Y take to do it?

Solution:
Let Y takes x days to complete the work
Then X will take 25% more i.e 125% of x days i.e 5/4x days.
So the one day work of X and Y together will be
(1/x) +{1/(5/4x)} = 1/20
⇒ (1/x) + (4/5x) = 1/20
⇒ 9/5x = 1/20
⇒ x = 36
∴ Y takes 36 days to complete the work.

Question: Time required by two pipes A and B working separately to fill a tank is 36 seconds and 45 seconds respectively. Another pipe C can empty the tank in 30 seconds. Initially, A and B are opened and after 7 seconds, C is also opened. In how much more time the tank would be completely filled?

Solution:
Let the capacity of the tank be LCM (36, 45, 30) = 180 units
∴ Efficiency of pipe A = 180/36 = 5 units/second
Efficiency of pipe B = 180/45 = 4 units/second
Efficiency of pipe C = - 180 / 30 = - 6 units/second

Now,
for the first 7 seconds, A and B were open. 
Combined efficiency of A and B = 5 + 4 = 9 units/second 
∴ Part of the tank filled in 7 seconds = 7 × 9 = 63 units

Part of tank empty = 180 - 63 = 117 units

Now, all pipes are opened.
Combined efficiency of all pipes = 5 + 4 - 6 = 3 units/second
Therefore, more time required = 117/3 = 39 seconds.

Question: If the workforce is doubled, how much longer will it take to finish the task?
​
Solution:
ধরি,
শ্রমিক সংখ্যা = x, এর দিগুণ = 2x,
সময় = n
x জন কাজটি করে n সময়ে

১ জন কাজটি করে = xn সময়ে
২x জন কাজটি করে = xn/২x
= n/২ সময়ে বা ১/২ সময়ে।

Question: 6 pipes, working 10 hours a day, can empty a cistern in 3 days. How many hours a day must 9 pipes work to empty the cistern in one day?

Solution:
By applying the MDH method,
it can be written as,

6 × 10 × 3 = 9 × x × 1
⇒ x = 20 hours

Question: 3/4 part of the tank is full of water. When 30 litres of water is taken out, the tank becomes empty. The capacity of the tank is -

Solution:
Let us consider,
The tank has 4x litres of total capacity and holds 3x litres of water.
And if 30 litres of water is taken out, then the tank becomes empty.

It means 3x litres of water is taken out.
∴ 3x = 30 litres
⇒ x = 10 litres

Capacity of tank
= 4x
= 4 × 10
= 40 liters.

Question: P can build a wall in 24 days and Q can do it in 20 days. With help of R, they completed the work in 6 days. Find in how many days R alone can do the work.

Solution:
P can do 1/24 part per day
Q can do 1/20 part per day
R can do per day = (1/6) - {(1/24) + (1/20)}
= (1/6) - (11/120)
= (20/120) - (11/120)
= 3/40
Time taken by R alone = 40/3 days

Question: A man completes 5/8 of a job in 10 days. At this rate, how many more days will it takes him to finish the job?

Solution:
Work done = 5/8
Balance work = 1 - (5/8) = 3/8

Let the required number of days be x
Then,
(5/8) : (3/8) :: 10 : x
⇒ (5/8) × x = (3/8) × 10
⇒ x = (3/8) × 10 × (8/5)
∴ x = 6

Question: A pipe can fill a tank in x hours and another pipe can empty it in y (y > x) hours. If both pipes are open, in how many hours will the tank is filled?

Solution:
Net part filled in 1 hour
= (1/x) - (1/y)
= (y - x)/xy hours

∴ The tank will be filled in = xy/(y - x) hours.

Question: A takes twice as much time as B or thrice as much time as C to finish a piece of work. Working together, they can finish the work in 4 days. B can do the work alone in-

Solution:
Suppose A, B and C take x, x/2 and x/3 days respectively to finish the work.
Then,
1/x + 2/x + 3/x = 1/4
⇒ 6/x = 1/4
⇒ x/6 = 2
∴ x = 24

So, B takes (24/2) = 12 days to finish the work.

Question: X can complete a task in 12 days and Y in 18 days. If they work together for 3 days, what fraction of work is left?

Solution:
The work rate of X is 1/12 (since X can complete the work in 12 days).
The work rate of Y is 1/18 (since Y can complete the work in 18 days).
In 3 days they complete = 3{(1/12) + (1/18)}
= 3 × (5/36)
= 5/12

Then the fraction of work that is left is = 1 - (5/12)
= (12 - 5)/12
= 7/12