Pipes & Cisterns

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

Pipe A can fill in 1 hour(60 minutes) is 1/1 or full of the tank.
Pipe B can empty in 1 hour (1/2) of the tank [120 mins= 2hrs]
Both pipes together can fill the tank in 1 hour = 1/1 - 1/2
= 1/2 of the tank.
Since 1/2 part of the tank is filled in 1 hour, the remaining part left is 1/2 of the tank.
The remaining 1/2 part will be filled in another 1 hour.
So both the pipes take 2 hours(120 minutes) to fill the tank.

Question: A tank can be filled by pipe A in 4 hours and pipe B in 8 hours. At 2 pm pipe A was opened. At what time will the tank be filled if pipe B is opened at 3 pm?

Solution:
পাইপ A এর কাজের হার = 1/4 অংশ/ঘন্টা
পাইপ B এর কাজের হার = 1/8 অংশ/ঘন্টা

বিকাল 2টা থেকে 3টা পর্যন্ত (1 ঘন্টায়), পাইপ A একা ট্যাঙ্কটির 1/4 অংশ পূর্ণ করে।

বাকি কাজ = 1 - 1/4 = 3/4 অংশ

বিকাল 3টার পর পাইপ A ও B একসাথে কাজ করবে।
একসাথে তাদের কাজের হার = (1/4 + 1/8) অংশ/ঘন্টা
= (2 + 1)/8 অংশ/ঘন্টা
= 3/8 অংশ/ঘন্টা

বাকি 3/4 অংশ পূর্ণ করতে সময় লাগবে = (বাকি কাজ)/(একসাথে কাজের হার)
= (3/4)/(3/8) ঘণ্টা
= 3/4 × 8/3 ঘণ্টা
= 2 ঘণ্টা

সুতরাং, বিকাল 3টার পর আরও 2 ঘন্টা সময় লাগবে।
অতএব, ট্যাঙ্কটি পূর্ণ হবে বিকাল 5 টায়।

Pipe A's work in % = 100/10 = 10%
Pipe B's work in % = 100/20 = 5%
Pipe C's work in % = 100/40 = 2.5%
All of them are opened for 2 hours + after 2 hours,
tap C is closed + After the 4th hour, tap B is also closed = 100

According to the question,
(10 + 5 + 2.5) × 2 + (10 + 5) × 2 + work by tap A alone = 100
⇒ 17.5 × 2 + 15 × 2 + work by tap A alone = 100
⇒ 35 + 30 + work by tap A alone = 100
⇒ work by tap A alone = 100 - 65 = 35%.

(m1×h1×t1)/w1 = (m2×h2×t2)/w2
9 taps × 20 min = t taps × 15 min
So, t = 12 taps

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: 2/5 part of the tank is full of water. When 24 liters of water is taken out, the tank becomes empty. The capacity of the tank is -

Solution: 
Let the total capacity of the tank be 5x liters.
Then, water in the tank = 2/5 × 5x = 2x liters.

When 24 liters of water is taken out, the tank becomes empty.
It means the water taken out = water present in the tank.
∴ 2x = 24 litres
⇒ x = 12 litres

∴ Capacity of the tank = 5x
= 5 × 12
= 60 liters.

Let Tap A fill the cistern completely in A hours.
So in 1 hour, it fills 1/A amount of the cistern

Also in 1 hour in Tap B fills in 1/4 amount of the cistern
Together they fill the cistern in 2.4 hours
So, Also in 1-hour together they fill in (1/2.4)amount of the cistern
So, Also in 1-hour cistern filled by both is given by 1/A + 1/4 = 1/2.4
∴ 1/A = 1/2.4 - 1/4 = 1/6

∴ Pipe A can fill 1/6^th tank in 1 hour
∴ Pipe a fills the tank completely in 6 hours.
It has a rate of 100-litre water per hour,
So, in 6 hours it gives out 6 x 100 = 600 litres
In 6 hours cistern is full, so capacity = 600 litres.

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: 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 pipe can fill up an empty tank in 15 minutes. Another pipe flows out 10 liters of water per minute. If the two pipes are opened together and the empty tank is filled up in 30 minutes, how much water does the tank contain?

Solution: 
Let the tank empty in x minute

ATQ,
(1/15) - (1/x) = 1/30
(1/15) - (1/30) = 1/x 
1/30 = 1/x 
x = 30

So the tank emptied by the other pipe in 30 minute

∴ The tank contain = 30 × 10 liter
= 300 liters

Question: A pipe can fill a tank in x hours, and another can empty it in y hours. In how many hours do they together fill it in (y > x)?

Solution:
Pipe fills the tank in x hours
∴ filling rate = 1/x tank/hour
Pipe empties the tank in y hours
∴ emptying rate = 1/y tank/hour
We are told y > x, so the filling pipe is faster than the emptying pipe

When both pipes are open, the net rate = (1/x) - (1/y)

∴ Time to fill the tank = Total work/Net rate
= 1/[(1/x) - (1/y)]
= 1/[(y - x)/xy]
= xy/(y - x) hours

Question: Two pipes, A and B, can fill a cistern together in 3 hours. If each pipe were opened separately, pipe B would take 8 hours more than pipe A to fill the cistern. How long would it take pipe A to fill the cistern on its own?

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
⇒ 3(2x + 8) = x(x+8) [Cross multiply]
⇒ 6x + 24 = x² + 8x
⇒ x² + 2x - 24 = 0
⇒ (x + 6)(x - 4) = 0
So, x = 4 (positive value).

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

Question: Two pipes A and B can fill the tank in 24 and 36 minutes, respectively. Both the pipes are opened together. After how many minutes should the pipe B be turned off, so that the tank be fill in 18 minutes?

Solution:
Given that,
Pipe A fills the tank in 24 minutes.
Pipe B fills the tank in 36 minutes.
Total time to fill the tank = 18 minutes.
Now,
LCM of 24 and 36 = 72 (Total capacity of the tank).
Efficiency of pipe A = 72/24 = 3 units/minute.
Efficiency of pipe B = 72/36 = 2 units/minute.

Let,
pipe B be turned off after x minutes.
Pipe A works for 18 minutes.
Pipe B works for x minutes.
Work done by A in 18 minutes = 3 × 18 = 54 units.
​Work done by B in x minutes = 2x = 2x units.

Total work done = 54 + 2x = 72
⇒ 2x = 72 - 54
⇒ 2x = 18
⇒ x = 18/2
∴ x = 9

∴ Pipe B should be turned off after 9 minutes.

Net part filled in 1 hour
= (1/x − 1/y)
= (y−x)/xy
∴ The tank will be filled in
= xy/(y−x) hours

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: Three pipes A, B, and C can fill a tank from empty to full in 40 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all three pipes are opened. A, B, and C discharge chemical solutions P, Q, and R respectively. What is the proportion of the solution R in the liquid in the tank after 2 minutes?

Solution: 
Part filled by (A + B + C) in 2 minutes
= 2 [(1/40) + (1/20) + (1/10)]
= 2 × (7/40)
= 7/20

Part filled by C (solution R) in 2 minutes = 2/10
= 1/5

∴ Proportion of solution R = (1/5) × (20/7)
= 4/7

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: Two pipes, Pipe X and Pipe Y, can fill a tank in 20 hours and 30 hours, respectively. If both pipes are opened together, after how many hours should Pipe Y be closed so that the tank is completely filled in 15 hours?

Solution:
ধরি, ট্যাঙ্কটির ধারণক্ষমতা হলো LCM (20, 30) = 60 ইউনিট।
পাইপ X-এর কর্মদক্ষতা = 60 / 20 = 3 ইউনিট/ঘণ্টা।
পাইপ Y-এর কর্মদক্ষতা = 60 / 30 = 2 ইউনিট/ঘণ্টা।
পাইপ X এবং Y-এর মিলিত কর্মদক্ষতা = 3 + 2 = 5 ইউনিট/ঘণ্টা।

ধরি, পাইপ X এবং পাইপ Y একত্রে চলে n ঘণ্টা।
∴ পাইপ Y বন্ধ করার পর পাইপ X একা (15 - n) ঘণ্টা চলে।

প্রশ্নানুসারে,
5n + 3(15 - n) = 60
⇒ 5n + 45 - 3n = 60
⇒ 2n + 45 = 60
⇒ 2n = 60 - 45
⇒ 2n = 15
⇒ n = 15/2
⇒ n = 7.5

∴ পাইপ Y-কে 7.5 ঘণ্টা পর বন্ধ করা উচিত।

Question: Two ingoing pipes can fill a tank in 6 hours and 8 hours, respectively. An outgoing pipe is attached to these two pipes and thus the tank was filled in 4 hours. In 48 hours, the outgoing pipe alone can empty how many tanks?

Solution:
Let the outgoing pipe take P hours to empty the tank.

So, in 1 hour, total fill-up = (1/6 + 1/8 - 1/P) part
= (3P + 4P - 24)/24P part
= (7P - 24)/24P part

According to the question,
24P/(7P − 24) = 4

⇒ 24P = 28P − 96
⇒ 4P = 96
∴ P = 24

∴ the outgoing pipe take 24 hours to empty the tank.
In 48 hours, it can empty = 48/24 = 2 tanks

Question: Two pipes, A and B, can fill a tank in 37.5 minutes and 45 minutes. If both pipes are open, after how many minutes should pipe B be closed to fill the tank in half an hour?

Solution:
নল A দ্বারা 37.5 মিনিটে পূর্ণ হয় 1 অংশ
∴ 1 মিনিটে পূর্ণ হয় = (1/37.5) অংশ
∴ 30 মিনিটে পূর্ণ হয় = 30/37.5 = 300/375 = 4/5 অংশ 

∴ পূর্ণ হওয়ার বাকি থাকে = 1 - (4/5)
= (5 - 4)/5
= 1/5 অংশ 

B নল দ্বারা ,
1 অংশ পূর্ণ হতে সময় লাগে = 45 মিনিট
∴ 1/5 অংশ পূর্ণ হতে সময় লাগে = 45 × (1/5) = 9 মিনিট 
 
 ∴ B নলটি 9 মিনিট পর বন্ধ করলে  ট্যাংকটি 30 মিনিটে পূর্ণ হবে। 

Shortcut:
(30/37.5) + (x/45) = 1 (whole)
⇒ 0.8 + (x/45) = 1
⇒ x/45 = 1 - 0.8 = 0.2
⇒ x = 0.2 × 45 = 9 

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.
∴ 1/x + 2/x + 4/x = 1/5
⇒ 7/x = 1/5
⇒ x = 35 hours

Question: Two pipes, A and B can fill a tank in 30 and 20 minutes respectively. If both pipes are used together, then how long will it take to fill the tank?

Solution:
Pipe A fill a tank in 30 minutes
So, it fills in one minute (1/30) 

Pipe B fills a tank in 20 minutes
So, it fills in one minute (1/20)

both pipes fill in one minute = (1/30) + (1/20)
= 1/12

so, it will take 1/(1/12) or 12 minutes to fill the tank.

Question: A tank is 1/3 parts full with water. If 9 liters of water is added, the tank becomes 5/6 parts full. What is the capacity of the tank?

Solution:
Let the capacity of the tank = x liters

According to the question,
(x/3) + 9 = 5x/6
⇒ (5x/6) - (x/3) = 9
⇒ (5x - 2x)/6 = 9
⇒ 3x = 54
⇒ x = 54/3
⇒ x = 18

Therefore, the capacity of the tank = 18 liters. 

Question: Water flows through a cylindrical pipe of an internal diameter of 7cm at the rate of 5m/s. The time, in minutes, the pipe would take to fill an empty rectangular tank of 4m × 3m × 2.31m is - 

Solution: 
the total volume of the tank is = 400 × 300 × 231 cc
total water flow per second through the pipe is = πr²h
= (22/7) × (7/2)²× 500

∴ total time = (400 × 300 × 231)/{(22/7) × (7/2)²× 500}
= (400 × 300 × 231 × 4 × 7)/(22 × 49 × 500)
= 1440 s 
= 24 min

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

Let X hours be the time taken to fill a tank by P.
Let Y hours be the time taken to empty the tank by Q.
Then the time taken to fill the tank when P and Q are switched together : XY/(Y - X) hours.
Here, X = 16 minutes And Y = 32 minutes
Therefore,
Required time = (16 × 32)/(32 - 16)
= (32 × 16)/16
= 32 minutes.

Question: A petrol tank is initially one-third full. After removing 5 gallons of petrol, the tank becomes one-fifth 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/3) - 5 = x/5
⇒ (x - 15)/3 = x/5
⇒ 5(x - 15) = 3x
⇒ 5x - 75 = 3x
⇒ 5x - 3x = 75
⇒  2x = 75
∴ x = 37.5 gallons

Let Tap A take T minutes to fill the tank alone.
Since Tap A is 5 times faster than Tap B, Tap B takes 5 times more time.
So time taken by Tap B = 5T minutes
Also, 5T-T = 32 ----------- Given
∴ T = 8 minutes = Time taken by A
Time taken by B = 5 x 8 = 40 minutes.

In 1 min, A + B fills = 1/8 + 1/40 = 3/20 parts
So entire tank is filled in = 20/3 hours.

Question: Pipe A fills a tank in 40 seconds, pipe B in 60 seconds, and pipe C empties it in 30 seconds. Initially, A and B are opened, and after 8 seconds, C is also opened. In how much more time will the tank be completely filled?

Solution:
Let the capacity of the tank be LCM of 40, 60, 30 = 120 units.

Efficiency of A = 120/40 = 3 units/second
Efficiency of B = 120/60 = 2 units/second
Efficiency of C = –120/30 = –4 units/second

(A + B open) for first 8 seconds,
Combined efficiency = 3 + 2 = 5 units/second
Work done in 8 seconds = 8 × 5 = 40 units
Remaining work = 120 – 40 = 80 units

Now when all three pipes open,
Combined efficiency = 3 + 2 – 4 = 1 unit/second
More time required = 80 ÷ 1 = 80 seconds

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

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
∴ 1/x + 2/x + 4/x = 1/5
7/x = 1/5
x = 35 hours.

Question: A water tank has two taps (Tap-1 and Tap-2). Tap-1 can fill a tank in 8 hours and Tap-2 can empty the tank in 16 hours. How long will they take to fill the tank if both taps are opened simultaneously but Tap-2 is closed after 8 hours?

 
Solution:
Tap-1, in 1 hour it fills = 1/8 part
Tap-2, in 1 hour it empties = 1/16 part

When both taps are open, in 1 hour it fills
= (1/8 - 1/16) part
= (2 - 1)/16 part
= 1/16 part

When both taps are open, in 8 hours it fills
= (1/16 × 8) part
= 1/2 part

∴ Remaining part = (1 - 1/2)
= 1/2 part

As Tap-2 is closed after 8 hours,

∴ Tap-1 alone can fill 1 part in = 8 hours
So, remaining 1/2 part will be filled in
= 8 × 1/2
= 4 hours

∴ Total time required = 8 + 4 = 12 hours

Assume that the reservoir is filled by the first pipe in 'x' hours.
So, the reservoir is filled by a second pipe in 'x + 20' hours.

Now, from these above conditions,
we can form the equations as,
1/x + 1/(x + 20) = 1/24
[x + 20 + x]/[x(x + 20)] = 1/24
x²– 28x – 480 = 0

By solving this quadratic equation , we get the factors (x – 40) (x+12) = 0
Hence, we get two values :
(x – 40) = 0 and (x+12) = 0
⇒ x = 40 and x = -6

Since the filling of the reservoir is positive work, we can neglect the negative value of 'x'.
Thus, x = 40

This means that the second pipe will take (x+ 20) hrs = 40 + 20 = 60 hrs to fill the reservoir.

Question: Two pipes can fill a tank in 12 minutes working together. After working together for 8 minutes, the first pipe is closed. It then takes 10 more minutes for the second pipe to fill the tank completely. How long would the second pipe take alone to fill the tank?

Solution: 
Let the first pipe fill the tank at rate A tanks per minute.
Let the second pipe fill the tank at rate B tanks per minute.
Both pipes together fill the tank in 12 minutes. so,
A + B = 1/12  ....... (1)

They work together for 8 minutes.
Work done in 8 minutes = 8 × (1/12) = 8/12 = 2/3 of the tank

∴ Remaining work = 1 - (2/3) = 1/3 of the tank
This remaining 1/3 is filled by the second pipe alone in 10 minutes.
⇒ B × 10 = 1/3
∴ B = (1/3)/10 = 1/30 tank per minute

Therefore, time taken by the second pipe alone to fill the full tank = 1/1/30 = 30 minutes.