### Discussion :: Pipes and Cistern - General Questions (Q.No.11)

Pranathi said: (Jan 22, 2011) | |

Why cant we take 3x for faster pipe and x for slower pipe? if I take in that way im getting answer as 9. Is it right or wrong?. |

Pri said: (Jan 26, 2011) | |

Please expain this solution more clearly. |

Reshma said: (Feb 22, 2011) | |

@Pranathi. We are discussing about d time. If we take the time taken by slower pipe as x and faster time pipe as 3x. It means that faster pipe takes more time. Which means our assumption is wrong. |

Karthiga said: (Feb 24, 2011) | |

Why we take 1/x+3/x instead of x + x/3 =36 ? |

Miss Mathphobia said: (Mar 30, 2011) | |

Why not x+x/3=36? It gives me answer as 27... |

Anusha said: (May 15, 2011) | |

its simple ya , one pipe =1/x; anther pipe =3/x: =4x36 sec=144. comment pls. |

Manasa said: (Jun 11, 2011) | |

Thanks RESHMA. |

Sonu Choudhary said: (Jun 25, 2011) | |

I am still not understand why we take 1/x+3/x=1/36. Please reply. |

Vinoth Kumar said: (Jun 29, 2011) | |

@sonu: if A takes x mins given tat B works thrice as fast as A i.e, B takes 1/3 of A's time=x/3 we know tat 1/A+1/b=1/36 1/x+1/(x/3)=1/36 1/x+3/x=1/36 4/x=36 x=4*36 x=144 |

Rishab said: (Aug 3, 2011) | |

How do we know that 1/a + 1/b = 1/36 ? |

Bhushan said: (Aug 8, 2011) | |

Above ans is right. Thing this way.....we want to find the time that the slower pipe alone can fill the tank not the time from total 36mins hence we take 1/A+1/b=1/36 so that we are finding the time from 100% and not from 36 mins. We are basically generalizing by taking the numerator as 1. |

Poornima said: (Aug 8, 2011) | |

Why do we need to take time as 1/time? |

Kailash said: (Aug 15, 2011) | |

Whenever we say ram works twice as fast as shyam in a given timeframe. it means it takes two shyams to complete that work in that given timeframe because single shyam can only do half of ram's capacity. kindly always keep this common sense in mind while attempting these kind of ques independent of their form. So solution: one pipe works 3 times faster than other..let name of this pipe A let name of other pipe (slower)B B is equivalent to shyam here..so if both pipes can fill one tank in 36 mins. both means A+B and as per our common sense A is equivalent to 3 B's. therefore (B+B+B)+B = 36 B = 36/4? NOT POSSIBLE as if 4 people can do a work in 1 min then 1 person will do it in 4*1 mins. coz more people less time and less people more time. What we understand by this is timse is inversely proportional to work done. therefore if four B's doing it in 36 mins then single B will do it in 4*36 = 144. |

Dolo said: (Aug 25, 2011) | |

Great kailash. |

Jamloki said: (Aug 26, 2011) | |

Kailash you are genious. |

Saurav said: (Sep 12, 2011) | |

If i take A =x and B=3x then I am getting 27 min as answer is it wrong? If I assume x as time then A = 9 and B =27 after calculating x+3x=36 is it wrong? please explain. |

Laxman said: (Sep 18, 2011) | |

The simplest way is. Together the two pipes can fill in 36 min. That means the fastest one can fill 36*3=108. And the slowest one is =108+36=144. |

Shanti Lal Dhaker said: (Oct 18, 2011) | |

We may also solve in this manner i.e. x + x/3 =36 => 4x/3 = 36 or => x=27 |

Teju M.B said: (Nov 13, 2011) | |

@Laxman:. Your way of approaching to the problem is good, simple and understandable. |

Mahe said: (Jan 6, 2012) | |

Slower pipe x,faster 3x.... 1/x+1/3x=1/36 =>x=48 then 1/36-1/48 we got 1/144 |

Indrajit said: (Feb 13, 2012) | |

Slower pipe fills in x min ,faster fills x/3. Let for complete the tank use symbol C. For slower pipe: in x min it fills C in 1 min it fills C/x Similarly in 1min faster pipe fills 3C/x Both fills in 36min so,in 36min both fills C in 1 min both fills C/36 Now, C/x+3C/x=C/36 or 1/x+3/x=1/36 Hope you got it! |

Karthik said: (May 7, 2012) | |

First pipe fills 3 times faster than second one so x=3x If together the two pipes can fill the tank in 36 minutes so x*3x=36 x=36/3x x=12/x x=144 |

Kant said: (Aug 27, 2012) | |

Let the volume b unit meter cube so rate of individual filling = rate of combine fillin so 1/X +1/(X/3) = 1/36 so x=144 |

Ali said: (Aug 27, 2012) | |

@Karthik how it is x square=12 and x =144??? |

Yasin said: (Sep 8, 2012) | |

I AM TRYING LIKE THIS but not getting. First pipe x to fill tank then second one is 3x. now both 1/x+1/3x=36 4/3x=36 1/x=9??? What's the problem? |

Mahima said: (Sep 23, 2012) | |

To all those who have not understood yet. Try this. It is given that- One pipe can fill a tank three times as fast as another pipe. Not let 1st pipe is A (faster) and 2nd pipe is B (slower). Now let pipe A fills the tank in x/3 minute. Therefore in 1 minute pipe A will fill 1/ (x/3) part of the tank. Similarly let pipe B fills the tank in x minute (because pipe B is slower therefore it will take more time). Therefore in 1 minute pipe B will fill 1/x part of the tank. Therefore in 1 minute both the pipe (A and B) fill 1/36 part of the tank. Now it is given that- Together the two pipes can fill the tank in 36 minutes. Therefore A+B=36 mins. Substitute the value of A and B according to 1 min we get. => 1/ (x/3) + 1/x =1/36. => 3/x +1/x= 1/36. => 4/x =1/36. => 4*36=x. =>x=144 (ans). |

Manoj said: (Jan 31, 2013) | |

So simple, Time=work/capacity, So total capacity=1/x+3/x. And total work=1. So total time taken by both of them=36=1/(1/x+3/x). |

Harshdeep said: (May 19, 2013) | |

Another simpler approach: (slower) pipe A takes 'x' min to fill complete tank alone. Then pipe A in 1 min fills (1/x) portion of tank alone. (faster) pipe B takes (x/3) min to fill the complete tank alone. then pipe B in 1 min fills (3/x) portion of tank alone. When both pipes together fills the tank then in 1 min they fill (1/x + 3/x) portion of tank. This means complete tank can be filled in (x + x/3) minutes. (4x/3) = 36.... so x = 144 min . Answer. |

Ravikanth said: (Jun 2, 2013) | |

The best approach to these kinda questions are the logic of replacing the variables with the numbers but be careful in doing so because one should be aware of the numbers which are being replaced to the variables should be satisfying the condition of the variables. So guys the easiest approach to the above problem is to: P1 is three times faster than P2 right? So I will write down a statement as follows, P1=3P2. So am gonna replace these with numbers as below, P1=3. If p1=3 the above statement becomes 3=3P2. By the cancellation we get the value of P2 as 1, P2=1. If a tank is to be filled by the two pipes together in 36 minutes then the values P1=3 becomes 3L/M P2 = 1L/M. So my next statement is, P1+P2 = 3+1. P1+P2 = 4L/M. If both the pipes are working together then the tank is being filled at a rate of 4L/M. So the time taken by the to pipes to fill the tank is 36M. Then we need to multiply the combined rate of pipes with the time to get the capacity of the tank. 4L/M*36M=144L which the capacity of the tank. Now coming to the last part of the question. "slower pipe alone will be able to fill the tank in:" I already mentioned the slower pipe as P2 and the rate of P2=1L/M so the time taken by P2 to fill the tank of 144L capacity is (tank capacity/P2)=144L/1L/M=144M. And finally sorry for the big explanation and guys please pardon me if there are any grammatical errors. In brief, P1 = faster pipe P2 = slower pipe. P1=3P2. If P1 = 3 =>3 = 3P2 =>P2 = 1. Time = 36M. P1+P2 = 1+3 =>P1+P2 = 4. As the units of the pipes are liters/minute L/M Find tank capacity, (P1+P2)*Time=4*36=>Tank capacity = 144L. 144L capacity tank will be filled by slower pipe at its slower rate right? rate of slower pipe P2? P2 = 1L/M. Time=Tank capacity/rate of pipe P2 => Time = 144/1 = 144. Which is the required answer. |

Anil Kumar said: (Jan 27, 2014) | |

If slower pipe is x. Than faster 3x. Take both 36 min. means x = 9 let 9+3 = 12 and 12 square = 144. |

Aditi said: (Feb 9, 2014) | |

If we take faster one as x, Then d slower one will become 3x, Thus the work done by them is equal to 1/36. Therefore, 1/x + 1/(3x) = 1/36. (3x + x)/(3x * x) = 1/36. 4/3x = 1/36. x = 48. Thus the slower one will become 144, thats the solution for taking x and 3x as the time. |

Raj said: (Jul 9, 2014) | |

First pipe x. Second pipe 3 times faster then it is 3x. Together will fill it in 36 min. 1/x+1/3x=1/36. Then we will get x=48 minutes and 3x= 144 minutes faster pipe is 48 minutes Which takes less time and slower pipe is 144 minutes which takes more time. |

Ahammed Riyas said: (Jul 15, 2014) | |

If we take x+x/3 instead of 1/x+3/x we cannot get answer because the method of solving will be more complicated. |

Vaibhaw said: (Aug 18, 2014) | |

I have another solution, Suppose slower pipe is taking X min, then in that time faster pipe will fill 3 tanks. Now both start filling tanks,then when slower pipe will finishing filling till that time faster also done with filling 3 tanks. Total number of tanks filled is Slower= 1, and faster = 3, sum total is 4 tanks, NOw time taken for filling 1 tank is 36, Therefore time taken for filling 4 tanks is 36*4 = 144. Therefore it indicates same time taken by slower pipe also to complete fill the tanks. |

Senthil said: (Sep 7, 2014) | |

Why can't we take x+3x=36 and why we are taking 1/x+3/x=1/36 explain this solution in detail? |

Senthil said: (Sep 7, 2014) | |

Why can't we take x+3x=36 and how 1/x+3/x is taken? |

Anurag said: (Sep 21, 2014) | |

@Senthil. Lets start afresh, If you think in terms of speed/rate of filling the tank of 'l' Litres capacity, then the speed of slower pipe will be x litre/min and the speed of faster pipe will be 3x litre/min. But, if you think in terms of time, then the slower pipe will take x min(or 3x min) and the faster one will take x/3 min(or x min) to fill the tank. Now, if you consider the second case i.e.,consider in terms of time, then you will only need to find out how much portion of the tank's capacity each pipe can fill in 36 min. Since, in 1 min the slower pipe can fill 'l'/x litres of the tank's capacity(In x min 'l' litre), and In 1 min the faster one can fill 'l'/(x/3) litres of the tank's capacity(In x/3 min 'l' litre). Therefore, in 36 min slower pipe will fill 36*('l'/x) liters, And In 36 min faster pipe will fill 36*('l'/(x/3)) liters. Since, both the pipes are contributing in filling the tank of capacity 'l' Litres. Therefore, 36*('l'/x) + 36*('l'/(x/3)) = 'l'. ('l'/x) + ('l'/(x/3)) = 'l'/36. (1/x) + (1/(x/3)) = 1/36. (1/x) + (3/x) = 1/36. Hope, it will be clear now. |

Sameer Sharma said: (Nov 2, 2014) | |

If a is times faster then b and takes less minutes then in how many time they will fill tank. |

L.Vignesh said: (Mar 29, 2015) | |

Above answer is right 144 mins. How when the tank was empty by one slow pipe (B) & 3 times faster pipe then slow its called pipe (A). Know imagine the tank contain 4 liters of water and time taken to empty is 36 min. As per above condition the B pipe can take 1 liter of water is 36 min mean the A pipe will take 3 liters at the time (3 times more than B. So 3x1=3). If the pipe A as been removed the work as gone for B. But the B is 3 times slower. Now A pipe as take 36 min for 3 liter means B will take 3 times more time 108 min (36x3=108). Already the pipe B was have taken 36 min (for 1 liter) and adding the 108 min for balance 3 liter after adding the answer was 144 min. (or). For 1 liter it take 36 min means for 4 liter its 144 min as per above calculation. |

Madhu Reddy said: (Apr 3, 2015) | |

Let's A = Fast pipe. B = Slow pipe. A takes x time. And B takes 3x time. So, in 1 min. 1/x + 1/3x = 1/36. By solving. 4x/3x^2 = 1/36. Finally x = 4*36/3. x = 48. Time taken by slow pipe = 3x = 3*48 = 144 min. |

Vik said: (May 28, 2015) | |

Work done by A in 1 min + Work done by B in 1 min = Work done by both in 1 min. |

Pratyush Mishra said: (Jun 19, 2015) | |

Any one tell me simplest way to solve this question? |

Hafee said: (Jul 10, 2015) | |

Let it be 3x, 1x. So, 4x = 1/36. Then x = 144. |

Prakash said: (Jul 24, 2015) | |

Let A be faster pipe & B be slower pipe. We know that [b] Capacity = Work/Time [/b]. Assume the capacity of A as 3 unit/min (i.e. A pipe can fill three times more capacity than B as given in the question). Then the capacity of B is 1 unit/min. Total capacity (A+B) = 3+1= 4 unit/min. If the pipes together (A+B) fill the tank in 36 min, then net work = capacity*time. i.e Net work = 4*36 = 144 units. Now the time taken by slower pipe B = Network/Capacity of B. Time (B) = 144/1 = 144 min. Similarly for A pipe = 144/3 = 48 min. Ans: 144 min. |

Suman Chabri said: (Sep 16, 2015) | |

Thanks now I am understood. |

Pranav Kumar said: (Oct 26, 2015) | |

I didn't understand the problem could you explain once. |

Prashi said: (Oct 28, 2015) | |

Assume A = Fast pipe. B = Slow pipe. A takes x time. And B takes 3x time. So, in 1 min 1/x + 1/3x = 1/36. By solving 4x/3x^2 = 1/36. Finally x = 4*36/3. x = 48. Time taken by slow pipe = 3x = 3*48 = 144 min. |

Shan said: (Jan 28, 2016) | |

I am getting a different answer, Two pipe fills the tank in 36 min. Let's take slower one = A. Faster one as B. So, B = A*3. If B+A = 36. Then B = 27 and A = 9. If B alone has to fill the tank then it will fill in. 27min (of B) + 9/3 of A = 27+3 = 30 min. And if go like this and A alone has to fill the tank then it would be: 30*3 = 90 min. Why I this is not right someone Please explain? |

Yogi M said: (Sep 29, 2016) | |

P1 = 3P2 According to the question. [P1 + P2] = 1/36. [P2] = ? P1 + P2 = 1/36. 3P2 + P2 = 1/36. 4P2 = 1/36. P2 = 1/144 {1/36 * 4 = 144}. ==>P2 = 144min. |

Pavan said: (Mar 20, 2017) | |

Actually A=3B because A pipe fill d tank 3 times fast as B pipe. So A=3B but in time. Then B = A/3 Let, A= x-----(1) assume i.e B=x/3---------(2) Formula 1/A+1/B=1/36 substitute both in 1/x+1/x/3 = 1/36, 1/x+3/x = 1/36, 4/x = 1/36. =>x=4*36 =>144. |

Jayaprakash said: (Jun 25, 2017) | |

Great @Indrajit. |

Khagendra said: (Oct 16, 2017) | |

Most effective solution, Thanks @Laxman. |

Pranathi said: (Dec 6, 2017) | |

Thank you @Reshma. |

Akshatha N said: (Dec 31, 2017) | |

Why cannot we take faster pipe=3x, instead of x/3? |

Akhil Galla said: (Jun 27, 2018) | |

Why can't we take 3x for faster one instead of x/3 for faster one? |

Dhilber said: (Jul 13, 2018) | |

Let take x is the time taken by a faster pipe and 3x be the time taken by the slowest pipe. hence 3x+x=36, x=9. which means in total 36 min. 9 min taken by fastest pipe and remaining 27 min taken by the slowest pipe. So instead of using the fastest pipe, we use the slowest pipe then it takes 3 times more time than the slowest pipe. Hence, 9*3=27. So total time taken will be 27+27=54. Please correct me, if I am wrong. |

Rishabh said: (Feb 4, 2019) | |

So as per ques.(simple trick). Pipe A (slower)complete its work to 1unit/min. Then Pipe B (faster)complete its work to 3unit/min. Then total unit in 1 min=4units/min. Then total units in total time=4*36=144units. NOW B(Slower pipe) time=(1min/1unit)x144units = 144units. |

Santosh said: (Feb 25, 2019) | |

Thanks for explaining @Mahima. |

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