Pipes & Cisterns

Question: An inlet pipe can fill a tank completely in 18 hours. In what time will the pipe fill 2/3 part of the tank?

সমাধান:
দেওয়া আছে,
ইনলেট পাইপটি সম্পূর্ণ ট্যাঙ্কটি পূর্ণ করতে সময় নেয় 18 ঘন্টা।

∴ ট্যাঙ্কটির 2/3 অংশ পূর্ণ করতে সময় লাগবে = (টোটাল সময় × অংশের পরিমাণ)
= (18 × 2/3) ঘন্টা
= (6 × 2) ঘন্টা
= 12 ঘন্টা।
∴ ট্যাঙ্কটির 2/3 অংশ পূর্ণ করতে 12 ঘন্টা সময় লাগবে।

Question: Two taps, P and Q, can fill a tank in 6 and 8 minutes respectively. A leak (outlet pipe) R can empty 20 liters of water per minute. If all three are opened together, the tank is filled in 12 minutes. What is the capacity of the tank in liters?

সমাধান:
প্রথম নল P এর 1 মিনিটে পূর্ণ করে = 1/6 অংশ
দ্বিতীয় নল Q এর 1 মিনিটে পূর্ণ করে = 1/8 অংশ
তিনটি নল একত্রে 1 মিনিটে পূর্ণ করে = 1/12 অংশ।

ছিদ্র নল R এর 1 মিনিটে খালি করার অংশ = (P + Q এর কাজ - সম্মিলিত কাজ)
= (1/6 + 1/8) - 1/12 অংশ
= (4 + 3)/24 - 1/12 অংশ
= 7/24 - 1/12 অংশ
= (7 - 2)/24 অংশ
= 5/24 অংশ।

অর্থাৎ, ছিদ্র নল R একা 5/24 অংশ খালি করে 1 মিনিটে।

ছিদ্র নল R একা সম্পূর্ণ ট্যাঙ্কটি খালি করতে সময় নেয় = 1/(5/24) মিনিট
= 4.8 মিনিট।

ছিদ্র নল R 1 মিনিটে খালি করে 20 লিটার।
∴ ট্যাঙ্কটির মোট ধারণ ক্ষমতা = (মোট সময় × প্রতি মিনিটের নির্গমনের হার)
= (4.8 × 20) লিটার
= 96 লিটার।

Question: A pipe can fill an empty tank in 15 minutes. Another pipe drains water at a rate of 10 liters per minute. If both pipes are opened simultaneously, the tank gets filled in 60 minutes. What is the capacity of the tank (in liters)?

Solution:
মনে করি, ট্যাঙ্কটির মোট ধারণক্ষমতা V লিটার।

প্রথম পাইপটি 15 মিনিটে খালি ট্যাঙ্কটি পূর্ণ করে।
∴ প্রথম পাইপ দিয়ে প্রতি মিনিটে পানি পূর্ণ হয় = V/15 লিটার।

দ্বিতীয় পাইপ দিয়ে প্রতি মিনিটে পানি বের হয়ে যায় = 10 লিটার।

দুটি পাইপ একসাথে খোলা থাকলে ট্যাঙ্কটি 60 মিনিটে পূর্ণ হয়।

∴ প্রতি মিনিটে ট্যাঙ্কটি কার্যকরভাবে পূর্ণ হয় = V/60 লিটার।

প্রশ্নমতে,
(V/15) - 10 = V/60
⇒ (V/15) - (V/60) = 10
⇒ (4V - V)/60 = 10
⇒ 3V/60 = 10
⇒ 3V = 10 × 60
⇒ 3V = 600
⇒ V = 600/3
∴ V = 200

সুতরাং, ট্যাঙ্কটিতে 200 লিটার পানি ধরে।

Let,
The pipe Z alone takes A minutes to fill the tank.
Given that,
Y is thrice as fast as Z.
Then, Y takes A/3 minutes to fill the tank.
And, X is thrice as fast as Y.
X takes (A/3)/3 = A/9 minutes to fill the tank.
Now,
Part filled by X in 1 minute = 9/A
Part filled by Y in 1 minute = 3/A
Part filled by Z in 1 minute = 1/A
Net part filled by (X+Y+Z) in 1 minute = 9/A + 3/A + 1/A
= 13/A
(X+Y+Z) take 10 minutes to fill the cistern.
Part filled by (X+Y+Z) in 1 minute = 1/10
Thus,
We have,
1/10 = 13/A
⇒ A = 13 × 10
⇒ A = 130
Therefore, Z alone takes 130 minutes
So, X can fill the cistern in 130/9 minutes.
Hence, the correct answer is - ঘ) none of these

Time taken by one tap to fill half of the tank = 3 hrs.
Part filled by the four taps in 1 hour = 4 × (1/6) = 2/3
Remaining part = (1 - 1/2) = 1/2
2/3 : 1/2 :: 1 : x
=> x = (1/2 × 1 × 3/2) = 3/4
So, total time taken = 3 hrs. 45 mins.

a = 8; b = 9; C = 12
Capacity of a tank
= a × b × c/(c-a)
= (8 × 9 × 12)/4
= 216 Litre.

Question: One pipe can fill a tank four times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the faster 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) = 1/36
⇒ 5/x = 1/36 
∴ x = 180

The slower pipe alone fill the tank in 180 minutes.
the faster pipe alon will be able to fill the tank in = (180 ÷ 4) minutes.
= 45 minutes

Let the time taken to fill the cistern by 3 taps A, B and C be X, Y, and Z minutes respectively.
Then the short cut formula for,
Time taken to fill the tank when all the pipes are opened = XYZ/(XY + YZ + ZX) minutes
Here, X = 9 minutes, Y = 6 minutes and Z = 3 minutes.
Now the required time = {(9) × (6) × (3)}/{(9 × 6) + (6 × 3) + (3 × 9)} minutes
= (9 × 6 × 3)/(54 + 18 + 27)
= (9 × 6 × 3)/9{6 + 2 + 3}
= 6 × 3/(6 + 2 + 3)
= 18/11
= 1(7/11) minutes
Hence the answer is 1(7/11) minutes.

Let, Capacity larger jar = x
Say 20 litters of water poured into each jar, because the same amount of water stored for a while in smaller jar
So x×1/4 = 20
Or, x = 80
Total water of larger jar after pouring the water of smaller jar = 20 + 20 = 40
So, the fraction is = 40/80 = 1/2

Question: Three taps A, B, C can fill an overhead tank in 4, 6 and 12 hours respectively. How long would the three taps take to fill the tank if all of them are opened together?

Solution:
Work with rates (fraction of tank per hour).
A fills 1/4 per hour.
B fills 1/6 per hour.
C fills 1/12 per hour.
∴ Combined rate = 1/4 + 1/6 + 1/12
= (3 + 2 + 1)/12 
= 6/12
= 1/2 (tank per hour).

∴ Time to fill one tank = 1 ÷ (1/2) = 2 hours.

∴ The three taps would take 2 hours to fill the tank if all of them are opened together.

For 6 pipes, it takes 1 hour 20 minutes
1 hour 20 minutes = 60 + 20 = 80 minutes

For 5 pipes, let the time taken be x.

This is inverse proportion case:
80 × 6 = x × 5
x = 480/5
= 96

Question: Pipe B takes 12 hours to fill a tank by itself, while pipe A works three times faster. If both pipes are opened together, how many hours will they need to fill the tank?

Solution:
B নল দ্বারা চৌবাচ্চা পূর্ণ হয় = 12 ঘণ্টায় 
∴ 1 ঘণ্টায় পূর্ণ হয় = 1/12 অংশ

A নল দ্বারা পূর্ণ হয় = 12/3 = 4 ঘণ্টায় 
∴ 1 ঘণ্টায় পূর্ণ হয় = 1/4 অংশ

দুইটি নল দ্বারা একত্রে 1 ঘণ্টায় পূর্ণ হয় = (1/12) + (1/4)
= (1 + 3)/12
= 4/12
= 1/3 অংশ 

দুইটি নল দ্বারা একত্রে,
1/3 অংশ পূর্ণ হয় = 1 ঘণ্টায়
∴ 1 অংশ পূর্ণ হয় = 1 × 3 = 3 ঘণ্টায় 

Question: A reservoir can be filled in 4 hours by three taps, P, Q, and R. Tap R is three times as fast as Q, and tap Q is twice as fast as P. How long will tap Q alone take to fill the reservoir?

Solution:
ধরি, নল P একা চৌবাচ্চাটি পূর্ণ করতে x ঘন্টা সময় নেয়।

যেহেতু নল Q, P এর দ্বিগুণ দ্রুত গতিতে পানি সরবরাহ করে, তাই Q এর সময় লাগবে x/2 ঘন্টা।
যেহেতু নল R, Q এর তিনগুণ দ্রুত গতিতে পানি সরবরাহ করে, তাই R এর সময় লাগবে (x/2)/3 = x/6 ঘন্টা।

প্রশ্নমতে, তারা একসাথে 4 ঘন্টায় চৌবাচ্চাটি পূর্ণ করে।
অতএব,
1/x + 1/(x/2) + 1/(x/6) = 1/4
⇒ 1/x + 2/x + 6/x = 1/4
⇒ (1+2+6)/x = 1/4
⇒ 9/x = 1/4
⇒ x = 9 × 4
⇒ x = 36 ঘন্টা।

অতএব, নল P একা চৌবাচ্চাটি পূর্ণ করতে 36 ঘন্টা সময় নেয়।
∴ নল Q একা চৌবাচ্চাটি পূর্ণ করতে সময় নেবে 36/2 = 18 ঘন্টা।

 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: 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.

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.

Question: Two pipes X and Y can fill a cistern in 10 and 15 hours respectively. Both pipes are opened together. After how many hours should pipe X be turned off so that the cistern is filled in 9 hours?

সমাধান:
ধরি, মোট সময় 9 ঘন্টা পর চৌবাচ্চাটি পূর্ণ হয়। এই সম্পূর্ণ সময়ে কেবল নল Y খোলা ছিল।

নল Y, 15 ঘন্টায় চৌবাচ্চাটি পূর্ণ করতে পারে।
∴ 1 ঘন্টায় Y পূর্ণ করে 1/15 অংশ।
∴ 9 ঘন্টায় Y পূর্ণ করে = 9/15 অংশ
= 3/5 অংশ।

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

নল X, 10 ঘন্টায় পূর্ণ করে 1 অংশ।
∴ 1 অংশ পূর্ণ করে 10 ঘন্টায়।
∴ 2/5 অংশ পূর্ণ করে = (10 × 2/5) ঘন্টা
= 4 ঘন্টা।

∴ নল X কে 4 ঘন্টা পর বন্ধ করতে হবে।

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: Three pipes A, B, and C can fill a tank in 5, 10, and 30 hours respectively. Pipe A was opened at 8 a.m., Pipe B at 9 a.m., and Pipe C at 10 a.m. When will the tank be completely full?

Solution:
ধরি, চৌবাচ্চাটি 8 a.m. এর x ঘন্টা পর পূর্ণ হবে।

তাহলে, পাইপগুলির কাজের সময়কাল নিম্নরূপ:
A কাজ করেছে x ঘন্টা
B কাজ করেছে (x - 1) ঘন্টা
C কাজ করেছে (x - 2) ঘন্টা

প্রশ্নমতে:
x/5 + (x - 1)/10 + (x - 2)/30 = 1
⇒ (6x + 3(x - 1) + 1(x - 2))/30 = 1
⇒ 6x + 3x - 3 + x - 2 = 30
⇒ (6x + 3x + x) - (3 + 2) = 30
⇒ 10x - 5 = 30
⇒ 10x = 30 + 5
⇒ 10x = 35
⇒ x = 35/10
⇒ x = 3.5 ঘন্টা।

অতএব, চৌবাচ্চাটি 8 a.m. এর 3.5 ঘন্টা পর পূর্ণ হবে।
8:00 a.m. + 3 ঘন্টা 30 মিনিট = 11 : 30 a.m.
∴ চৌবাচ্চাটি 11 : 30 a.m. এ পূর্ণ হবে।

Let the tank get empty in T hours counting from 8 am.
A is on for T hours and work is done by A = Work in 1-hour × T hours = T/1.5 = 2T/3

Similarly, B starts at 9 am i.e. it's on for (T-1) hours & work done is = (T - 1)/2

Similarly, C starts at 10 am i.e. it's on for (T-2) hours & work done is = (T - 2)/(1/2) = 2(T - 2)

Initially, the tank is empty and after T hours too, it is empty. So, the total work done is 0.

According to the question,
2T/3 + (T - 1)/2 - 2(T - 2) = 0
⇒ (4T + 3T - 3 - 12 T + 24)/6 = 0
⇒ -5T + 21 = 0
⇒ 5T = 21
⇒ T = 21/5
= 4.2 hours = 4 hours 12 minutes5
This time is needed for the tank to get empty.
The exact time will be 4 hours 12 min from 8 am = 12.12 pm

Question: A tank can be filled by a tap in 8 hours. After one-third of the tank is filled, two more identical taps are opened. How long will it take to fill the tank completely?

Solution:
One tap fills the tank in 8 hours 
rate of one tap = 1/8 tank/hour.

After one-third of the tank is filled, 2 more taps are opened.

∴ Time to fill one-third of the tank = (1/3)/(1/8) = 8/3 hours

∴ Remaining = 1 - (1/3)
= (3 - 1)/3
= 2/3 of the tank

Now 3 taps are working
∴ combined rate = 3× (1/8)=3/8 tank/hour

∴ Time to fill remaining tank = (2/3)/(3/8)
= (2/3) × (8/3)
= 16/9 hours

∴ Total time = (8/3) + (16/9) hours
= 40/9 hours
= 4.44 hours

Tank filled by A alone in 1 hr = 1/12
Tank filled by B alone in 1 hr = 1/24
Tank filled by C alone in 1 hr = 1/48
D empty the tank at the rate of 6m/hr

So,
Tank empty by D in 1hr = 6/96 = 1/16
Now, tan is to be filled up to 72m i.e., 72/96 =3/4 of tank
So,
Let the 3/4^th of the tank to be filled in 't' hours time

For 1 hr all are opened then B closed

So, for (t - 1) hr A, C and D opened
(1/12 + 1/24+ 1/48 -  1/16) + (t - 1) (1/12 + 1/48 - 1/16) = 3/4
⇒ (1/12) + (t - 1) (1/24) = 3/4
⇒ (t - 1)/24 = (3/4) - (1/12)
⇒ (t - 1)/24 = 2/3
⇒ 3t - 3 = 48
⇒ 3t = 51
⇒ t = 17 hours.

If a tank has 4x liters of total capacity and it holds 3x liters of water and if 30 liters of water is taken out, the tank becomes empty.
It means 3x liters of water is taken out
3x = 30 liters
x = 10 liters
Capacity of tank
= 4x = 4 × 10 = 40 liters

Part filled by tap in 1 h = 1/12
Part emptied by leak in 1 h = 1/20

Net part filled in 1h when both (tap and leakage) work.
= 1/12 - 1/20
= (5 - 3)/60
= 2/60
= 1/30
Therefore, Required time to fill the tank = 30 h

Question: A petrol tank that is 1/2 full has 8 gallons petrol removed. The tank is then 1/10 full. What is the capacity, in gallons of the tank?

Solution:
Let the capacity of the tank be x gallons.

According to the question,
(x/2) - 8 = x/10
⇒ (x - 16)/2 = x/10
⇒ 10(x - 16) = 2x
⇒ 10x - 160 = 2x
⇒ 10x - 2x = 160
⇒ 8x = 160
⇒ x = 160/8 = 20 gallons

Work done by the C pipe in 1 minute
= 1/15 − (1/20 + 1/24)
= (1/15 − 11/120)
= −1/40 [−ve means emptying]
∴ Volume of 1/ 40 part = 3 gallons.
Volume of whole = (3 × 40) gallons = 120 gallons.

Question: One pipe can fill a tank in 20 minutes and the other in 30 minutes. If both pipes are opened at the same time, how many minutes will it take to fill two-thirds of the tank?

Solution:

প্রথম নল দ্বারা,
20 মিনিটে পূর্ণ হয় = 1 অংশ
∴ 1 মিনিটে পূর্ণ হয় = 1/20 অংশ

দ্বিতীয় নল দ্বারা,
30 মিনিটে পূর্ণ হয় = 1 অংশ
∴ 1 মিনিটে পূর্ণ হয় = 1/30 অংশ

দুইটি নল একসঙ্গে খুলে দিলে 1 মিনিটে পূর্ণ হয় = (1/20) + (1/30)
= (3 + 2)/60
= 5/60
= 1/12 অংশ

এখন,
1/12 অংশ পূর্ণ হয় = 1 মিনিটে
∴ 1 অংশ পূর্ণ হয় = 12 মিনিটে 
∴ দুই-তৃতীয়াংশ বা 2/3 অংশ পূর্ণ হয় = 12 × (2/3) = 8 মিনিটে

Question: There are two filling pipes taking 20 min and 24 min to fill a tank and a drain pipe removing 3 gallons per minute. When used simultaneously, the tank fills in 15 minutes. What is the tank’s volume?

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

According to the question,
(1/20 + 1/24) - 1/x = 1/15
⇒ 1/x = (1/20 + 1/24) - 1/15
⇒ 1/x = 1/40
∴ x = 40 minutes

A waste pipe can empty 3 gallons per minute
In 40 minutes it can empty = 3 × 40 = 120 gallons.

∴ Capacity of the tank = 120 gallons.

Suppose pipe A can fill the tank in x hours. Then,
pipe B can fill the tank in x/2 hours,
pipe C can fill the tank in x/4 hours.
Part filled by pipe A in 1 hour = 1/x
Part filled by pipe B in 1 hour = 2/x
Part filled by pipe C in 1 hour = 4/x
Therefore, partly filled by pipe A, pipe B, and pipe C together in 1 hour
= 1/x + 2/x + 4/x
= 7/x
i.e., pipe A, pipe B, and pipe C together can fill the tank in x/7 hours.
Given that pipe A, pipe B, and pipe C together can fill the tank in 10 hours
x/7 = 10
⇒ x = 70.

Question: Two pipes, A and B, can fill a tank in 9 minutes and 18 minutes, respectively. If both pipes are opened together, after how many minutes should pipe B be closed to fill the tank in 7 minutes?

Solution:
ধরা যাক, নল B, x মিনিট চলার পর বন্ধ করা হয়।

ট্যাংকটি সম্পূর্ণ পূর্ণ হতে মোট সময় লাগে 7 মিনিট।
সুতরাং, নল A মোট 7 মিনিট চালু থাকে।
এবং নল B মোট x মিনিট চালু থাকে।

নল A দ্বারা 1 মিনিটে পূরণ হয় = 1/9 অংশ
নল B দ্বারা 1 মিনিটে পূরণ হয় = 1/18 অংশ

প্রশ্নানুসারে,
(নল A দ্বারা 7 মিনিটে পূরণ করা অংশ) + (নল B দ্বারা x মিনিটে পূরণ করা অংশ) = 1 (সম্পূর্ণ ট্যাংক)
⇒ 7/9 + x/18 = 1
⇒ (14 + x)/18 = 1
⇒ 14 + x = 18
⇒ x = 18 - 14
∴ x = 4

অতএব, 4 মিনিট পর নল B বন্ধ করলে ট্যাংকটি 7 মিনিটে পূর্ণ হবে।