Pipes & Cisterns

Question: A tap can fill a cistern in 8 hours. Due to a drain pipe in the bottom, it takes 10 hours to fill the same cistern. If the cistern is full, how much time will the drain pipe take to empty it?

সমাধান:
ট্যাপ দ্বারা 1 ঘন্টায় পূর্ণ হয় 1/8 অংশ।
নল ও ছিদ্র দ্বারা একত্রে 1 ঘন্টায় পূর্ণ হয় 1/10 অংশ।

∴ ছিদ্র দ্বারা 1 ঘন্টায় খালি হয় = (ট্যাপের কাজ - যৌথ কাজ)
= (1/8 - 1/10) অংশ
= (5 - 4)/40 অংশ
= 1/40 অংশ।

∴ ছিদ্রটি সম্পূর্ণ চৌবাচ্চাটি খালি করতে 40 ঘন্টা সময় নেবে।

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: A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time as the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is-

Solution:
Let the time taken by the first pipe = x hours

Then,
The second pipe fills the tank 5 hours faster than the first pipe
∴ Time = x - 5 hours
The third pipe fills the tank 4 hours faster than the second pipe
∴ Time = (x - 5) - 4 = x - 9 hours

ATQ,
(1/x) + {1/(x - 5)} = 1/(x - 9)
⇒ (x - 5 + x)/{x(x - 5)} = 1/(x - 9)
⇒ (2x - 5)(x - 9) = x(x - 5)
⇒ 2x² - 23x + 45 = x² - 5x 
⇒ x² - 18x + 45 = 0
⇒ x² - 15x - 3x + 45 = 0
⇒ (x - 3)(x - 15) = 0
∴ x = 3 or x = 15
Since time cannot be less than 9 hours
Hence, x = 15

Question: One pipe can fill a tank four times as fast as another pipe. If together the two pipes can fill the tank in 24 minutes, then the slower pipe alone will be able to fill the tank in-

Solution:
Let,
the slower pipe alone fill the tank in x minutes.
Then, Faster pipe alone will fill it in x/4 minutes.

ATQ,
(1/x) + (4/x) = 1/24
⇒ 5/x = 1/24
∴ x = 120

∴ The slower pipe alone fill the tank in 120 minutes
= (120/60) hours
= 2 hours

1 / (1/p + 1/q - 1/r)
= 1 / (qr+pr-pq / pqr)
= pqr / (qr + pr - pq)

Question: A tank is 1/3 parts full with water. If 16 liters of water is added, the tank becomes 5/6 parts full. What is the capacity of the tank?
(Janata RC 2022 অনুযায়ী)

Solution:
ধরি,
ট্যাংকের ধারণক্ষমতা = x লিটার

প্রশ্নমতে,
(x/3) + 16 = 5x/6
⇒ (5x/6) - (x/3) = 16
⇒ (5x - 2x)/6 = 16
⇒ 3x = 96
⇒ x = 96/3
⇒ x = 32

অর্থাৎ ট্যাংকের ধারণক্ষমতা = 32 লিটার 

Question: A bucket is 2/7 full. If 18 liters of water are added, it becomes exactly full. What is the capacity of the bucket?

Solution:
Let the capacity of the bucket 'x' liters.
Initially the bucket has (2/7) of x = 2x/7 liters of water
After adding 18 liters then the bucket becomes full. 

So we can form the equation,
(2x/7) + 18 = x
⇒ x - (2x/7) = 18 
⇒ (7x - 2x)/7 = 18
⇒ 5x/7 = 18
⇒ x = (18 × 7)/5
⇒ x = 126/5
∴ x = 25.2 liters

So the capacity of the bucket is 25.2 liters.

Suppose pipe A alone takes x hours to fill the tank in 144 min.
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.

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: 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 tap can completely fill a water tank in 8 hours. The water tank has a hole in it through which the water leaks out. The leakage will cause the full water tank to empty in 10 hours. How much time will it take for the tap to fill the tank completely with the hole?

Solution: 
Tap alone fills the tank in 8 hours
⇒ Filling rate = 1/8 tank/hour
Leakage alone empties the full tank in 10 hours
⇒ Emptying rate = 1/10 tank/hour

∴ Net rate = Filling rate - Emptying rate
= (1/8) - (1/10)
= (5 - 4)/40
= 1/40

Time to fill the tank with the hole = 1 full tank/Net rate
= 1/(1/40) hours
= 40 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: Pipes A and B take 36 seconds and 45 seconds, respectively, to fill the tank when used separately. Pipe C, which can empty the tank in 30 seconds, starts operating after 7 seconds of A and B being open. How much additional time will be needed to fill the tank completely?
(পাইপ A এবং B আলাদাভাবে ৩৬ সেকেন্ড এবং ৪৫ সেকেন্ডে ট্যাংকটি পূর্ণ করতে পারে। পাইপ C, যা ৩০ সেকেন্ডে ট্যাংকটি খালি করে, A এবং B খোলা থাকা অবস্থায় ৭ সেকেন্ড পরে কাজ শুরু করে। ট্যাংকটি সম্পূর্ণ পূর্ণ করতে কত অতিরিক্ত সময় প্রয়োজন?)

Solution:
এখানে, ট্যাঙ্কের ধারণক্ষমতা (ল.সা.গু) LCM (36, 45, 30) = 180 units
∴ অতএব, পাইপ A এর দক্ষতা = 180/36 = 5 units/second
পাইপ B এর দক্ষতা = 180/45 = 4 units/second
পাইপ C এর দক্ষতা = - 180 / 30 = - 6 units/second

এখন,
প্রথম ৭ সেকেন্ডে, A এবং B খুলে রাখা হয়েছিল।
A এবং B এর যৌথ দক্ষতা = 5 + 4 = 9 units/second 
অতএব, ৭ সেকেন্ডে ট্যাঙ্কের যে অংশটি পূর্ণ হয়েছে তা হচ্ছে = 7 × 9 = 63 units

ট্যাঙ্কের খালি অংশ = 180 - 63 = 117 units

এখন, সব পাইপ খুলে দেওয়া হয়েছে।
সব পাইপের যৌথ দক্ষতা = 5 + 4 - 6 = 3 units/second
অতএব, আরও সময় প্রয়োজন = 117/3 = 39 seconds.

Let, B alone filled the pipe by x hours.

Efficiency ratio of B and C =2 : 1
Time ratio of B and C = 1 : 2

Given,
(1/A + 1/B)- (1/A + 1/C) =7/60 - 1/12
⇒ 1/B - 1/C = 2/60 = 1/30
⇒ 1/x - 1/2x=1/30
⇒ 1/2x=1/30
⇒ 1/x=1/15
⇒ x = 15

B alone filled the pipe by 15 hours.

Question: One tap (A) fills a reservoir four times as fast as another tap (B). If both taps running together can fill the reservoir in 20 minutes, then how long will the slower tap (B) alone take to fill the reservoir?

সমাধান:
ধরি,
ধীরগতির নল B একা চৌবাচ্চাটি পূর্ণ করতে সময় নেয় x মিনিট।
তাহলে, দ্রুতগতির নল A একা চৌবাচ্চাটি পূর্ণ করতে সময় নেবে x/4 মিনিট।

প্রশ্নমতে, তারা একত্রে 20 মিনিটে পূর্ণ করে। অর্থাৎ,
1/x + 1/(x/4) = 1/20
⇒ 1/x + 4/x = 1/20
⇒ (1 + 4)/x = 1/20
⇒ 5/x = 1/20
⇒ x = 5 × 20
⇒ x = 100 মিনিট
∴  x = 1 ঘণ্টা 40 মিনিট [ 60 মিনিট = 1 ঘণ্টা]

∴ ধীরগতির নলটি (B) একা চৌবাচ্চাটি পূর্ণ করতে 1 ঘন্টা 40 মিনিট সময় নেবে।

Question: Two inlet pipes can fill a tank in 10 hours and 20 hours, respectively. An outlet pipe is attached to these two pipes, and thus, the tank was filled in 12 hours. In 90 hours, the outlet pipe alone can empty how many tanks?

সমাধান:
ধরি,
ছিদ্র নলটি (Outlet Pipe) একা ট্যাঙ্কটি খালি করতে P ঘন্টা সময় নেয়।

তিনটি নল একত্রে 1 ঘন্টায় পূর্ণ করে = 1/10 + 1/20 - 1/P অংশ।
প্রশ্নমতে, তিনটি নল একত্রে 12 ঘন্টায় পূর্ণ করে।
∴ 1/10 + 1/20 - 1/P = 1/12

১. ছিদ্র নলটির সময় (P) নির্ণয়:
⇒ 1/P = 1/10 + 1/20 - 1/12
হরগুলির (Denominator) ল.সা.গু. (LCM) হলো 60।
⇒ 1/P = (6 + 3 - 5)/60
⇒ 1/P = 4/60
⇒ 1/P = 1/15
⇒ P = 15 ঘন্টা।

90 ঘন্টায় যতগুলি ট্যাঙ্ক খালি করতে পারে = 90 / P
= 90/15
= 6 টি ট্যাঙ্ক।
∴ 90 ঘন্টায় ছিদ্র নলটি একা 6 টি ট্যাঙ্ক খালি করতে পারে।

Question: Pipe A can fill the tank in 8 hours and pipe B can fill it in 12 hours. If pipe A is opened at 8:00 AM and pipe B is opened at 10:00 AM, then at what time will the tank be full ?

Solution: 
A opened 2 hours early to B
In 2 hours A can do 3 × 2 = 6 unit work
Remaining work = 24 - 6 = 18
A + B can do it in
= 18/5 hours
= 3 hours 36 minutes
∴ Tank will be full in 10 A.M. + 3 hours 36 minutes = 1 : 36 P.M.

Question: Two pipes can fill a tank with water in 15 and 12 hours respectively and a third pipe can empty it in 4 hours. If the three pipes be opened, the tank will be emptied in-

Solution:
Part of the tank filled by two pipes in 1 hour = 1/15 + 1/12
= (4 + 5)/60 part
= 9/60 part
= 3/20
Part of the tank emptied by the third pipe in 1 hour = 1/4

∴ Net part of the tank emptied in 1 hour = (1/4 - 3/20) part
= (5 - 3)/20 part
= 2/20 part
= 1/10 part

1/10 Part of tank can be emptied in 1 hour
∴ The whole tank will be emptied in = 10 hours

Question: A cistern is filled by Pipe A and Pipe B together in 2 hours. Pipe A alone can fill the cistern at the rate of 100 litres per hour. Pipe B alone can fill the cistern in 4 hours. What is the capacity of the cistern?

Solution:
Let the capacity of the cistern = x litres.

Pipe A fills at 100 litres per hour
∴ Time taken by A alone = x/100 hours
Pipe B alone fills the cistern in 4 hours
∴ Pipe B's rate = x/4 litres per hour
And Combined rate of A + B = 100 + (x/4) litres per hour

They together fill the cistern in 2 hours, so:
Combined rate = x/2 litres per hour

Therefore, 100 + (x/4) = x/2
⇒ (x/2) - (x/4) = 100
⇒ (2x - x)/4 = 100
⇒ x = 4 × 100
∴ x = 400

So the capacity of the cistern is 400 litres.

Question: Two pipes can fill a tank together in 8 minutes. Both pipes are opened, and after 6 minutes the first pipe is closed. It then takes 6 more minutes for the tank to be completely filled. How long would it take to fill the tank using only the second pipe?

Solution:
Together, the two pipes fill the tank in 8 minutes = 1 full tank.
∴ In 1 minute, they fill = 1/8 of the tank.
∴ In 6 minutes, they fill = (1 × 6)/8 = 3/4 of the tank.

∴ Remaining part of the tank = 1 - (3/4) = 1/4.

The second pipe alone fills this 1/4 of the tank in 6 minutes.
∴ Time for the second pipe to fill the whole tank = 6 × 4 = 24 minutes.

So the second pipe alone would take 24 minutes to fill the tank.

Question: A pipe can fill a cistern in 20 hours. Once the cistern is half full, three additional identical pipes are opened. How long will it take to fill the cistern completely?

Solution:

Work done by 1 pipe in 1 hour = 1/20
∴ Time to fill half the cistern with 1 pipe = (1/2) ÷ (1/20) = 10 hours

After cistern is half full,
three additional identical pipes are opened, 
total pipes = 4
Work done by 4 pipes in 1 hour = 4 × (1/20) = 1/5
Time to fill remaining half = (1/2) ÷ (1/5)
= 2.5 hours
= 2 hours 30 minutes

∴ Total time to fill cistern = 10 + 2.5
= 12.5 hours = 12 hours 30 minutes

Three pipes A, B, and C can fill a tank in 8 hours. A, B, and C’s 1 hour work=1/10
A, B and C's 3 hour work= 3/10 Remaining work= 1 – (3/10) = 7/10

The remaining part will be filled by A and B in 14 hours. Then,
⇒ (7/10) × (A + B) = 14
⇒ (A + B)'s whole work= 14 × (10/7)
= 20 hr (A + B)'s 1-hour work
= 1/20

A, B, and C's 1-hour work = 1/10
C's 1 hour work= (A + B + C) – (A + B)
⇒ (1/10) – (1/20)
⇒ 1/20
∴ C can fill the tank in 20 hours.

Let the cistern be filled by pipe A alone in x hours.
Then, pipe B will fill it in (x + 6) hours
∴ 1/ x + 1/ x+6 = 1/ 4
⇒ x+6+x/ x(x+6) = 1/ 4
⇒ x² − 2x−24 = 0
⇒ (x−6)(x+4) = 0
⇒ x = 6 [neglecting the negative value of x].

Question: Two pipes can fill a tank in 36 and 40 minutes respectively, and a waste pipe can empty 3.5 gallons per minutes. All three pipes working together can fill the tank in 30 minutes. The capacity of the tank is-

Solution: 
Work done by the waste pipe in 1 minute = (1/30) - [(1/36) + (1/40)]
= (12 - 10 - 9)/360
= - (7/360) [Negative sign means emptying]

∴Volume of (7/360) part = 3.5 gallons
Volume of whole tank = (360 × 3.5)/7 gallons
= 180 gallons

Rate of flow of water = x cm/minute
∴ The volume of water that flowed in the in 1 minute
= (5 × 4 × x) = 20x cu.cm.

∴ The volume of water that flowed in the tank in 6 hours 18 minutes.
i.e. (6 × 60) + 18 = 378 minutes
= 2x × 378 cu. cm.

According to question,
20x × 378 = 700 × 400 × 450
⇒ x = (700 × 400 × 450)/(20 × 378) cm/minutes
⇒ x = (700 × 400 × 450 × 60)/(20 × 378 × 100000) km/hours
⇒ x = 10 km/hrs.

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: A tank is filled to three-fifths of its capacity with water. When 9 liters of water are added, the tank becomes six-sevenths full. Find the total capacity of the tank.

Solution:
ধরি,
ট্যাংকটির ধারণ ক্ষমতা = x লিটার

প্রশ্নমতে,
(3x/5) + 9 = 6x/7
⇒ (6x/7) - (3x/5) = 9
⇒ (30x - 21x)/35 = 9
⇒ 9x/35 = 9
⇒ 9x = 9 × 35
⇒ x = (9 × 35)/9
⇒ x = 35

∴ ট্যাংকটির ধারণ ক্ষমতা = 35  লিটার

Pipe 1 can fill 1/2 of the cistern in 1 hour
Pipe 2 can fill 1/6 of the cistern in 1 hour
Pipe 3 can empty 1/9 of the cistern in 1 hour
According to the question,
Time taken to full the cistern = 1/2 + 1/6 - 1/9
= 5/9
5/9 of the cistern will be filled in 1 hour.
Full cistern will be filled in = (5/9)/1
= 9/5 x 1
= 1.8 hours

Given that,
A takes 20 minutes to fill and B takes 12 minutes to empty
Clearly,
tap B is faster than tap A.
And so, the tank will be emptied.
Half of the tank or 1/2 part of the tank is already filled.
Therefore,
we have to find the time taken to empty that 1/2 part.
Part filled by A in 1 minute = 1/20
Part emptied by B in 1 minute = 1/12.
Part emptied by (A + B) in 1 minute
= (1/12) – (1/20)
= (5 - 3)/60
= 1/30.
Therefore,
The time taken by (A + B) to empty the full tank is 30 minutes.
Time taken to empty 1/2 part of the tank is
= 30/2
= 15 minutes.

Question: Two pipes P and Q can fill a reservoir in 15 and 20 hours respectively. Both pipes are opened together. After how many hours should pipe P be turned off so that the reservoir is filled in 12 hours?

সমাধান:
ধরি, মোট সময় 12 ঘন্টা পর চৌবাচ্চাটি পূর্ণ হয়। এই সম্পূর্ণ সময়ে কেবল নল Q খোলা ছিল।

নল Q, 20 ঘন্টায় চৌবাচ্চাটি পূর্ণ করতে পারে।
1 ঘন্টায় Q পূর্ণ করে 1/20 অংশ।
12 ঘন্টায় Q পূর্ণ করে = 12/20 অংশ
= 3/5 অংশ।

অবশিষ্ট অংশ যা P পূর্ণ করেছিল = 1 - 3/5 অংশ
= 2/5 অংশ।

নল P, 15 ঘন্টায় পূর্ণ করে 1 অংশ।
1 অংশ পূর্ণ করে 15 ঘন্টায়।
∴ 2/5 অংশ পূর্ণ করে = (15 × 2/5) ঘন্টা
= 6 ঘন্টা।

অর্থাৎ, নল P, 6 ঘন্টা কাজ করার পর বন্ধ করা হয়েছিল।
∴ নল P কে 6 ঘন্টা পর বন্ধ করতে হবে।

A opened 2 hours early to B
In 2 hours A can do 3 × 2 = 6 unit work
Remaining work = 24 - 6 = 18
A + B can do it in
= 18/5 hours
= 3 hours 36 minutes
∴ Tank will be full in 9 A.M. + 3 hours 36 minutes = 12.36 P.M.

Question: A reservoir has two pipes, A and B. A can fill the reservoir 5 hours faster than B. If both together fill the reservoir in 6 hours, the reservoir will be filled by A alone in-

Solution: 
Let, A alone can fill the reservoir in x hours 
B can fill in x + 5 hours 

Both complete in 1 hour = (1/x) + (1/ x + 5)
= (2x + 5)/(x2 + 5x)

Now
1/{(2x + 5)/(x2 + 5x)} = 1/6
 (x2 + 5x)/(2x + 5) = 6 
⇒ x2 + 5x = 12x + 30 
⇒ x2 - 7x - 30 = 0
⇒ x2 - 10x + 3x - 30 = 0 
⇒ x (x - 10) + 3 (x - 10) = 0
⇒ (x - 10) (x + 3) = 0 
∴ x = 10 or, x = -3 , negative value not possible 

A alone can fill the reservoir in 10 hours