ব্যাংক ডেইলি কুইজ [লং কোর্সের অংশ]

Question: Two pipes X and Y can fill a tank in 30 minutes and 45 minutes, respectively. Both pipes are opened together. After how many minutes should pipe Y be turned off so that the tank is filled in 20 minutes?

Solution:
Pipe X can fill 1 / 30 part of tank in one minute.
Pipe Y can fill 1 / 45 part of tank in one minute.
Both pipes can fill (1/30 + 1/45) part of tank in one minute.
= (3+2) / 90 = 5/90 = 1/ 18

Let, after time t minutes, we turned off the pipe Y.
according to question, [দুটি পাইপ t সময় পর্যন্ত একসাথে চলতে থাকে, বাকি সময় X পাইপটি চলে এবং সম্পূর্ণ বা 1 অংশ ট্যাংক পূর্ণ করে ]
∴ t/18 + (20-t) / 30 = 1 
⇒ (5t + 60- 3t) / 90 = 1
⇒ 5t + 60 - 3t = 90
⇒ 2t = 30
⇒ t = 15

Question: A tap can fill a tank in 12 hours. After one-third of the tank has been filled, two more similar taps are opened. What is the total time taken to fill the tank completely?

Solution:
A tap can fill a tank in 12 hours. 
∴ A tap can fill 1/12 part in one hour.
 Tap can fill 1 part in 12 hours

⇒ Tap can fill 1/3 part in 12 / 3 = 4 hours
⇒ The rest part = 1 - 1/3 = 2/3. 

After one-third of the tank has been filled, two more identical taps are opened.
3 similar tap can fill 3/12 = 1/4 part in one hour.
3 similar tap can fill 1 part in = 4 hour.
3 similar tap can fill 2/3 part in = 4 × 2/3 = 8/3 hour.

8/3 hours means 2 hours 40 minutes.
∴Total time taken: 6 hours 40 minutes.

Questions: Three taps P, Q, and R can fill a tank in 6 hours. After working together for 2 hours, tap R is closed, and P and Q can fill the rest of the tank in 7 hours. The number of hours taken by R alone to fill the tank is –

Solution:
Three taps P, Q and R can fill a tank in 6 hours.
Three taps can fill in one hour 1/6 part of the tank
Three taps can fill in 2 hours 1/3 part of the tank.

Rest part 1 - 1/3 = 2/3 part
2/3 part can be filled in 7 hours by P and Q

∴ In 1 hour P and Q can fill 2/21 part
∴ In 1 hour P, Q and R can fill 1/6 part

∴ in 1 hour R can fill (1/6 - 2/21) = 1/14 part
Hence, R alone fills the tank in 14 hours.

 Question: Two pipes can fill a tank in 15 and 20 minutes respectively and a waste pipe can empty 2 gallons per minute. All the three pipes working together can fill the tank in 9 minutes. The capacity of the tank is-

Solution:
Let, the waste pipe empty the tank in x minutes.

According to the question,
⇒ (1/15) + (1/20) - (1/x) = (1/9)
⇒ 1/x = (1/15) + (1/20) - (1/9)
⇒ 1/x = (12 + 9 - 20)/180
⇒ 1/x = 1/180
∴ x = 180

A waste pipe can empty 2 gallons per minute In 180 minutes it can empty = 2 × 180 = 360 gallons.
∴ Capacity of the tank = 360 gallons.

 Question: A tank is filled in 9 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank?

Solution:
Suppose,
Pipe A alone takes x hours to fill the tank.
Then, pipes B and C will take x/2 and x/4 hours respectively to fill the tank.

Now,
⇒ (1/x) + (2/x) + (4/x) = 1/9
⇒ 7/x = 1/9
∴ x = 63
∴ Pipe A alone takes 63 hours to fill the tank.

Question: Two pipes A and B together can fill a cistern in 3 hours. If they had been opened separately, B would have taken 8 hours more than A to fill the cistern. How long will A take to fill the cistern separately?

Solution: Let the time taken by A alone be x hours.
Then time taken by B alone = x + 8 hours.
Rate of A = 1/x cistern/hour. Rate of B = 1/(x+8) cistern/hour.
Combined rate = 1/x + 1/(x+8) = 1/3 (since together they fill in 3 hours).

Now,
1/x + 1/(x+8) = 1/3
⇒ (x+8 + x) / [x(x+8)] = 1/3
⇒ (2x + 8) / [x(x+8)] = 1/3

Cross multiply:
3(2x + 8) = x(x+8)
⇒ 6x + 24 = x² + 8x
⇒ x² + 2x - 24 = 0

Factorize:
(x + 6)(x - 4) = 0
So, x = 4 (positive value).

(Other root is negative and discarded.)
Therefore A will take 4 hours alone. 

Question: A booster pump can be used to fill as well as to empty the tank. The capacity of the tank is 1200 m³. The emptying capacity of the tank is 10 m³ per minute higher than its filling capacity and the pump requires 4 minutes less to vacant the tank than it requires to fill it. Calculate the filling capacity of the pump is-

Solution:
Let, the filling capacity of the pump = x m³/min

Given that,
Capacity of the tank = 1200 m³
Emptying rate is 10 m³/min more than filling rate
Emptying time is 4 minutes less than filling time

Then,
Filling time = 1200/x​ minutes
Emptying capacity = x+10 m³/min
Emptying time = 1200/(x + 10)

According to question,
(1200/x​) - {1200/(x + 10)} = 4
(1/x) - {1/(x + 10)} = 1/300
(x + 10 - x)​/x(x + 10) = 1/300
x2 + 10x - 3000 = 0
(x + 60) (x - 50) = 0

So, possible values is,
x = 50 And x = - 60 [not valid]

So, the filling capacity of the pump is 50 m³/min

Question: It takes two pipes X and Y, running together, to fill a tank in 6 minutes. It takes X, 5 minutes less than Y to fill the tank, then what will be the time taken by Y alone to fill the tank?

Solution:
Let the time taken by pipe X to fill the tank be a minutes
Time is taken by pipe Y to fill the tank = a + 5 minutes

So,
⇒ (1/a) + {1/(a + 5)} = 1/6
⇒ (2a + 5)/a(a + 5) = 1/6
⇒ a2 + 5a - 12a - 30 = 0
⇒ a2 - 7a - 30 = 0
⇒ (a - 10)(a + 3) = 0
⇒ a = 10, - 3
∴ a = 10  ; [neglecting the negative value of a]

Thus, time taken by Y alone to fill the tank is 10 + 5 = 15 minutes

Question: A cistern can be filled by two taps X and Y in 15 hours and 20 hours respectively. The full cistern can be emptied by a third tap Z in 10 hours. If all the taps are turned on at the same time, in how much time will the empty cistern be filled completely?

Solution :

X’s 1 hour work = 1/15
Y’s 1 hour work = 1/20
Z’s 1 hour work = 1/10 (emptying → negative)

Net 1 hour work = 1/15 + 1/20 – 1/10

Find LCM of denominators (15, 20, 10) → 60

Convert:
1/15 = 4/60, 1/20 = 3/60, 1/10 = 6/60

Net work = 4/60 + 3/60 – 6/60 = 1/60

Time taken = 1 ÷ (1/60) = 60 hours.

Question: A water tank is one-third full. Pipe A can fill the tank in 6 minutes, and pipe B can empty it in 12 minutes. If both pipes are open together, how long will it take to fill the tank completely?

Solution:
Let total tank = 1 unit.
Current water = 1/3

A’s 1 minute work = 1/6 (filling)
B’s 1 minute work = 1/12 (emptying → negative)

Net work per minute = 1/6 – 1/12 = (2 – 1)/12 = 1/12

Remaining to fill = 1 – 1/3 = 2/3

Time to fill = (2/3) ÷ (1/12) = (2/3) × 12 = 8 minutes

Question: A tank was 20% full of oil. The oil was poured into an empty 100-liter bucket, filling it halfway. What is half of the tank’s total capacity (in liters)?

Solution:
Let the total capacity of the tank = T liters.
The tank was 20% full 
now, oil volume = 20% of T = 0.2 × T

The oil fills 50% of the bucket
50% of 100 liters = 50 liters

So, 0.2 × T = 50 → T = 50 / 0.2 = 250 liters

Half of the tank = 250 ÷ 2 = 125 liters.

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

Solution:

Let the tank capacity = 1 unit.
Filling pipe rate = 1/p (tank per hour)
Emptying pipe rate = 1/q (tank per hour → negative)

Net rate when both pipes are open = 1/p – 1/q = (q – p) / (pq)
Time to fill the tank = Total tank ÷ Net rate = 1 ÷ [(q – p)/ (pq)] = pq / (q – p) hours

Question: A petrol tank is half full. If 10 gallons of petrol are removed, the tank becomes one-tenth full. What is the total capacity of the tank in gallons?

Solution:
Let,
The capacity of the tank in gallons is x gallons.

According to question,
⇒ (x/2) - 10 = x/10
⇒ (x - 20)/2 = x/10
⇒ 10(x - 20) = 2x
⇒ 10x - 200 = 2x
⇒ 10x - 2x = 200
⇒  8x = 200
∴ x = 200/8 = 25 gallons

Question: 9 pumps working 8 hours a day can empty a reservoir in 20 days. How many such pumps are needed to empty the same reservoir working 6 hours a day in 16 days? 

Solution:
The total work required to empty the reservoir is equal to the number of pumps multiplied by the hours they work per day and the number of days.
So, total work = 9 pumps × 8 hours/day × 20 days = 1440 pump-hours.

Let x be the number of pumps needed. These pumps will work 6 hours per day for 16 days. So, total work done by these pumps = x pumps × 6 hours/day × 16 days = 96 x pump-hours.

Since the total work is the same, 1440 pump-hours = 96 × x pump-hours.

Divide 1440 by 96
We get, x = 15.