### Discussion :: Pipes and Cistern - General Questions (Q.No.4)

Swetha said: (Oct 7, 2010) | |

How Part filled by (A + B) in x min + Part filled by A in (30 -x) min = 1? |

Gopal said: (Oct 29, 2010) | |

Please explain clearly. |

Ashwin said: (Nov 7, 2010) | |

If pipe B is turned off after x min, which means it is opened for x mins along with pipe A. therefore for x mins both pipes A & B is filling the tank ie : x*(A+B)----(1) As after x mins pipe B closed, pipe A alone filling the tank. In question it asked in total 30 min tank should be filled. therefore pipe A alone fills the remaining tank in (30-x) mins ie : A*(30-x)-----(2) Hence the total work is obtained as equ (1) + (2) = 1 |

Priya said: (Jun 19, 2011) | |

Sorry I didn't understood. Please explain me in detail. |

Ricky said: (Aug 8, 2011) | |

Excellent. |

Vishnupriya said: (Aug 17, 2011) | |

I can't understand still. Can any one say in detail. |

Mahesh Patil said: (Aug 23, 2011) | |

Shortcut (75/2)-30 ----------- * 45 = 9 ....after solving (75/2) |

Kasi Srinivas said: (Aug 31, 2011) | |

@ Gopal , Priya , Vishnupriya : Both pipes A and B are opened initially. Now these both pipes (A and B) fill the cistern together for some time...let us say they are filling the cistern together for x mins. Which implies pipe B is turned off after x min (according to the question). If u understood this u can follow the problem.. Pipes A & B are filling the tank for x mins => x*(A+B) --- (1) As after x mins pipe B closed, pipe A alone filling the tank. He said cistern will be filled in half an hour ie 30 mins. => pipe A alone is filling the remaining tank in (30-x) mins => A*(30-x)---(2) So, part filled by (A + B) in x minutes + Part filled by A in (30 -x) minutes = 1. here 1 indicates the tank is full. by substituting we get, x*(2/75 + 1/45) + (30-x)* 2/75 = 1 on solving we get x = 9. - K @ $ ! (kasi srinivas) |

Venkatesh said: (Sep 1, 2011) | |

Excellent explanation Kasi Srinivas |

Vishnupriya said: (Sep 1, 2011) | |

Ya I got it now. Thank you Kasi Srinivas. |

Amitha said: (Sep 6, 2011) | |

Good explanation -K @ $! |

S.Meenakshi said: (Nov 14, 2011) | |

Excellent explanation KASI. |

Sandip said: (Mar 14, 2012) | |

Here " 1 " means the complete part of the cistern....we learnt such theory in early school days. |

Rupinder said: (Jun 9, 2012) | |

@Sandip: That's what aptitude is meant for, i.e., How much knowledge we've dragged with us from our primary schooling, which indicated how consistent the candidate was throughout his academic period. Aptitude marks tells this apparently. And that's what recruiter is looking for, a consistency. Plus it tells your mental ability to grasp the things and to sort it out |

Neha Sharma said: (Jun 10, 2012) | |

@Mahesh Patil. What's the logic behind the short cut? |

Pintu Francis said: (Aug 8, 2012) | |

Y going complicated? be simple. A work in 1 min 2/75. B work in 1 min 1/45. So, (2/75) 30+ (1/45) x=1. Since A work for 30 min and B for x min. Solving X=9. |

Nags said: (Aug 11, 2013) | |

In (2/75)30+(1/45)x = 1. Where how 30 will come? I can't understand. |

Meraj Husain said: (Aug 16, 2013) | |

I can get you understood here: Since cistern is one which is to be filled by A & B. Hence 1 = work of 1 min of A *30+work in 1 min of B*x. 1 = (2/75)*30+(1/45)*x. x/45 = 1-60/75 = 15/75 = 1/5. x = (1/5)*45 = 9 mins. |

Arshad1605 said: (Oct 31, 2013) | |

In 1 minute, both together will fill 2/75 + 1/45 = 11/225 of the cistern. Cistern should be thus filled in 225/11 minutes. But cistern filled in 30 minutes. Therefore, B turned after 30 - 225/11 = (330 - 225)/11. = 105/11. = 9(6/11). Hence B turned off after 9 minutes. |

Saurav said: (Mar 5, 2014) | |

Why x*(A+B)+A*(30-x) = 1. It should be x*(A+B)+A*(30-x) = 1/2, because after 30 mins the tank should be half filled? |

Nins said: (Jul 13, 2014) | |

What is the meaning of "1" here? |

Shwetha Cr said: (Jul 17, 2014) | |

1 indicates complete cistern. |

Sameera said: (Aug 10, 2014) | |

1 indicates complete work, a+b=1 means a and b have done a complete work. |

Silvi said: (Nov 3, 2014) | |

This is not logical reasoning. By multiplying them with minutes, you can not estimates their unit to be a part. Hence, the "=1" here is unaccepted. You must provide with minutes unit too. |

Ratih said: (Nov 3, 2014) | |

I still don't get the logic. Why you talk about minutes needed and suddenly talk about part. You can't write it "=1". |

Silvi said: (Nov 3, 2014) | |

If we reverse the answer to the common formula: 9x11/225+21x2/75. We will get answer = 1. But we actually want to get answer = 30 minutes, right? This explanation is clearly not logical and misleading. We can't assume minutes unit is same with part unit. You can't multiply them with minutes and suddenly make the equation "=1" because it's clearly not same unit. |

Raj said: (Nov 17, 2014) | |

Please tell me how we get 2/75? |

Asif said: (Mar 22, 2015) | |

Simply. (1-Given time/A's time)*B's time. = (1-30/75/2)45. = (1-60/75)*45. = (1-0.8)*45. = 0.2*45 = 9. |

Nehal said: (Mar 24, 2015) | |

Hey, After first step just open the brackets i.e, 2x/75 + x/45 + 60/45 - 2x/75 = 1. (2x/75 - 2x/75) + x/45 + 60/45 = 1. 0 + x/45 + 4/5 = 1. Take 1/5 as common, x/9 + 4 = 5. x/9 = 1. x = 9. No need of complex calculations. |

Prakash Kumar said: (Jul 19, 2015) | |

Let us understand the problem first, we can easily analyze that Here Pipe A is run for HALF an hour. Total part filled by part A alone in half an hour is {2/75}*30 = (4/5). Now (1- (4/5) ) , that is (1/5) part is still remaining. And it is filled by Pipe B and after that it is made off. So (1/5) part filled by Pipe B in, = 45*(1/5) = 9 minute. Hence after 9 minute pipe B will made off. |

Rohit said: (Aug 5, 2015) | |

Pipe A fills tank in 6 hours, pipe B in 8 hours. Both pipes opened simultaneously, then after how many hours should pipe B should be closed so that tank is filled in 4 hours. I am getting a negative value and answer I am getting is 16/3 which is wrong. Please help. |

Abhishek said: (Aug 13, 2015) | |

It is easy. It is given that the cistern will be filled in just 30 minutes. And we don't know that who between A and B takes how much time from 30 minutes to fill the tank. Suppose A takes x minutes. Then B will take (30-x) minutes. But here is given that B is turned off in some time. We don't know when. So take it as Y minute. After Y minute only A will fill remaining part in 30-Y minute. So first A+B will fill for Y minute and then B is turned off. So only A will fill in 30-Y minute. So the equation is (A+B)Y + A(30-Y) = 30 minute. |

Abhishek said: (Aug 13, 2015) | |

@Rohit are you kidding bro? If A and B take more than 4 hours to fill the tank (they takes 6 and 8 hours). Then how could anyone of them fill it in 4 hours. Use some mind. |

Unik said: (Aug 14, 2015) | |

@Rohit. Is the answer 3? |

Ashu said: (Aug 25, 2015) | |

Why x is multiplied by total work? |

Bryan said: (Sep 7, 2015) | |

I'm here with an easy solution. In 1 min A fill 2/75 of tank. In 30 min A fill 30*2/75 = 4/5 of tank since A is turned on for 30 min while B is shut after few mins that we have to find: Now tank which B will fill in x min = 1-4/5 = 1/5. Now work/work in 1 min = Total no. of min (x). (1/5)/(1/45) = x. So x = 9 min. |

Shekhar Ssc said: (Sep 14, 2015) | |

Let A one min work = 1/37.5 = 2/75. Let B one min work = 1/45. Let B turn off after x min. And so tank filled in 30 min. So Tank full = Part filled by A and B + Part filled by only A. 1 = (A+B)x+(30-x)A. 1 = 11x/225+(30-x)2/75. That's way x = 9. |

Hardik said: (Oct 8, 2015) | |

Both together can do a work in 225/11 minute. Now pipe A can do a work in 75/2 minute. He alone did work for 30 minutes in last. 30*2/75=4/5. Now 1/5 had been filled by a and b together. 1 cistern in 225/11 minutes then 1/5 in how many minutes ? |

Amit said: (Dec 6, 2015) | |

Simply put A can fill in 75/2 minutes. Therefore in 1 min it fills 2/75 part. Therefore in 30 min it will fill 2*30/75 = 4/5th part. Now after 30 min amount remaining to be filled will be 1-4/5 = 1/5. (In above subtraction 1 means whole cistern), so now B has to fill 1/5 i.e. remaining part. Question is how many minute does B takes to fill 1/5 part of the cistern. We know from question that B takes 45 min to fill the cistern. Therefore in 1 minute fill 1/45th part. Therefore x *1/45 = 1/5 (where x= minute it actually takes to fill the cistern). Solving X = 9 minutes. |

Azharudhin said: (Apr 9, 2016) | |

Nice explanation @Mahesh Patil. Thank you. |

Azharudhin said: (Apr 9, 2016) | |

Hi @Meraj Husain, please explain how 60 came? Your explanation is nice, please explain clearly once again. Thank you for your comments. |

Ashok said: (Jul 8, 2016) | |

Go from the option. Choose option 9 minutes and use it in B 1 minute work = 1/45 and find it for 9 minutes it's 1/5 then go for remaining works it is 1-1/5 = 4/5. This remaining work will be done by A then. 4/5 work in 75/2 minutes 4/5 * 75/2 = 30 that is the given minutes taken to fill the tank. |

Aka said: (Aug 12, 2016) | |

Yet simpler approach: A's 1 minute work = 2/75, B's 1 minute work = 1/45, Let B be opened for first x minutes while A is opened for entire 30 mins. => 2/75 * 30 + 1/45 * x = 1. By solving it, we get the answer. |

Kouser said: (Sep 24, 2016) | |

Good explanation @Kasi Srinivas. |

Kumar said: (Sep 29, 2016) | |

@Arshad1605, Please tell me how 9 came after 105/11? |

Ranjit said: (Sep 29, 2016) | |

@Arshad 1605. How the answer 9 came after the step 105/11? |

Naidu said: (Oct 6, 2016) | |

How 30 came after 2/75? |

Sudhakar P said: (Oct 25, 2016) | |

I can't understand please explain clearly. |

Satendra Singh said: (Nov 5, 2016) | |

(30-X) how came? Please explain me. |

Hemant said: (Nov 20, 2016) | |

In 1min=2/75 part of the cistern is filled by pipe A. And In 1 min=1/45 part of the cistern is filled by pipe B. Now, Pipe A worked for 30 min and pipe B work for x min we suppose, To fill complete cistern = 1 part. 2/75 * 30 + 1/45 * x = 1, 12/15 + x/45 = 1, (36 + x)/45 = 1, x = 9 min. |

Kunal Aryan said: (Jan 26, 2017) | |

For this type of question use Short Trick : [Y*(1 - t / x)] Here X= 75/2 ; Y=45 and t= 30mint. (ie. half an hour). Soln: 45*(1 - 30 * 2 / 75) = 9 mints. answer. |

Madhubala said: (Jul 2, 2017) | |

For 1min it takes( 2/75). For 30 min (1/2 hour) it takes 2/75*30=4/5 part. So remaining only 1 part out of 5 which is going to fill by B. So (1/5*45) = 9 min. |

Sandhiya said: (Aug 2, 2017) | |

Can anyone explain how 2/75 comes? please. |

Manikant said: (Sep 9, 2017) | |

2/75 + 1/45 = 8/225. How it can be 11/225? |

Abhijeet Pol said: (Sep 20, 2017) | |

Much simple way is the LCM method. LCM of (75/2 and 45) = 225, assume this is total unit of work. Thus, efficiency of A = [(225)/(75/2)] = 6 units/min and B = 225/45 = 5 units/min. Since A works for 30 mins, he will finish = 6 x 30 = 180 units. Remaining = 225 - 180 = 45 units to be completed by B in = (45/5) = 9 mins. |

Deepak said: (Nov 26, 2017) | |

Best solution: Parts filled by A in 1 min = 2/75 Parts filled by B in 1 min = 1/45 Now; In 30 min tank is full so, Parts filled by A in 30 min = 30*2/75 =60/75 Parts filled by B in X min = X/45 So, 60/75 + X/45 = 1 Therefore ,X = 9 min. |

Tapz Patel said: (Nov 26, 2017) | |

A*(A'work)+B*(B'work)=1, It means 30(A)+x(B)=1, Then, x=9. |

Khagendra said: (Dec 4, 2017) | |

@Kasi Srinivas. Nice explanation, anyone can easily understand this, Thanks. |

Arjun said: (Dec 7, 2017) | |

Nice work @Kasi Srinivas. |

Leela said: (Jan 12, 2018) | |

Thanks @Deepak. |

Nikhil Jha said: (Feb 23, 2018) | |

pipe ----part filling----lcm of A & B----multiplying factor(cal for how much will fill in 1 unit time) A ----75/2 ----225 ----225%(75/2)= 6 i.e A can fill 6 ltr in 1 min B ---- 45 ---- 225 ---- 225/45=5 i.e B can fill 5 ltr in 1 min. * NOTE : if fraction comes then while calculating LCM take lcm of numerators only. now given that tank will fill in half-hour means A will fill till half hour ie for 30 min. in 1 min A is filling 6 ltr so in 30 min =30 *6 =180 ltr fill by pipe A. Means remeaning will be filled by B. i.e 225-180=45 ltr will be filled by PIPE B. But pipe B is filling 5 ltr in 1 min so for 45 ltr = 45/5 = 9 min it will take. so B will open for 9 min. (here A will continuously open, we only have to stop B. so 1st calculate how much water will A can fill in given time and after that calculate how much time it will take to fill remeaning by closing pipe i.e B ) |

Sarika said: (Apr 7, 2018) | |

I didn't get this sum please give me some tricks to understand the problem. |

Unnimaya said: (Jul 6, 2018) | |

Well said @Kasi Srinivas. |

Hari said: (Aug 14, 2018) | |

@Sarika. Just take LCM of 37.5 and 45=450(total units). You need 30mins to fully fill so(450/30)=15mins u have in total if Both worked non stop now A works for a full period so 450/75 (convert 37.5 into the whole num as used for LCM)=6 mins. So, B has worked 15-6=9mins. |

Narula said: (Oct 24, 2018) | |

The much simple way is the LCM method. LCM of (75/2 and 45) = 225, assume this is a total unit of work. Thus, efficiency of A = [(225)/(75/2)] = 6 units/min and B = 225/45 = 5 units/min. Since A works for 30 mins, he will finish = 6 x 30 = 180 units. Remaining = 225 - 180 = 45 units to be completed by B in = (45/5) = 9 mins. |

Swapna said: (Dec 10, 2018) | |

Thank you @Pintu Francis. |

Swapnil said: (Jan 21, 2019) | |

Pipe A fills the tank in 75/2 Min. As pipe A runs 30 mins it fills, 75/2 min = 1 tank. 30 min = 1*2/75*30 = 4/5, Pipe A fills 4/5 in 30 mins, remaining 1/5. Pipe B fills 1 full tank in 45 mins,so. 1 tank = 45 min. 1/5 tank = 45/5 = 9 min. |

Hithma said: (Feb 16, 2019) | |

I Can't understand this. Please, someone explain to me. clearly. |

Devi said: (Jul 5, 2019) | |

For this type of question, we can approach like; Here pipe A fill-in 75/2min(let it be 'x'), pipe B in 45 min(let it be 'y'). At what time B pipe should be close to fill a total cistern in 30 min (let it be 't'min)is, y(1-(t/x))= 45(1-((30*2)/75)) = 9min. |

Abey said: (Sep 8, 2019) | |

Thanks for explaining the answer @Kasi. |

Vini said: (Sep 24, 2019) | |

How will A become 2/75, can anyone explain? |

Nilesh12 said: (Oct 10, 2019) | |

Formula(1 - total time/2nd pipe time)*1st pipe time. then, (1-30/75/2)*45. Means, (1-60/75)*45. (75-60/75)*45. (1/5)*45. Answer 9. |

Karansingh said: (Nov 20, 2019) | |

There is a shorcut trick here. Find lcm of 75 and 45. Lcm is 225. Divide by 75 and 45 to lcm. Hereafter dividing lcm you will get 3 and 5. Now given is that cistern will be filled in half hour means 30 min if b turned off. They do 30/5 = 6. Now add 3+6 = 9 (why to add this because we asked that if it is closed then we have to add that 3+6). |

Gunjan said: (Feb 21, 2020) | |

Y(1-t/x). X = 75/2. Y =45. T = 30. This is the formula for this question. |

Omkarr said: (May 19, 2020) | |

@All. According to me, the solution is; The cistern will be filled in just half an hour, if the B is turned off after: i.e A work 30 min so, A total work 75/2 minutes. 30 /75/2=4/5. Remaining work is 1-4/5=1/5, So, B can fill in 45 minute. Then 45/5 =9. |

Ruby said: (Jul 16, 2020) | |

Another solution in term of common sense just for MCQ purpose: The time that A takes to fill - 371/2= 37.5 min The time that B takes more than A to fill - 45/37.5=1.2 times After 30 min extra time would have been needed by A to fill - 7.5 minutes. The time that B needs to compensate for - 7.5*1.2= 9 minutes. |

Arun said: (Jul 25, 2020) | |

Let 'B' be turned off after 'x' minutes. Then: Part filled by A and B in 1 minute: [(2/75) + (1/x) ] = [1/30] minutes. [(2x + 75) ] 30 = 75x. x = 225/9 = 9 minutes(Ans). |

Shruti said: (Sep 10, 2020) | |

We get total amount to be filled by taking the LCM of (45,75/2) = 225litres. Efficiency of A=225/(75/2)= 6litres/min Efficiency of B=225/(45) = 5litres/min So A in 30mins =30* 6 = 180litres Remaining amount = 225 - 180 = 45L Now B fills 5L in 1min, 45L it requires 9mins. |

Clinton said: (Sep 19, 2020) | |

((A - Time of closing)/A) * B = Answer. Here, A=75/2. Time of Closing the pipe=30. B = 45. So, Apply in the formula. (((75/2)-30)/(75/2)) * 45 = 9 (Answer). |

Rohini said: (Oct 31, 2020) | |

@Kasi. Nice explanation. |

Jothigonzi said: (Jan 20, 2021) | |

@Abhijeet pol. Nice and easiest explanation. Thanks. |

Rekha said: (May 19, 2021) | |

Well explained, Thanks @Ashwin. |

Joel said: (May 23, 2021) | |

Shortcut: 2/75 * 30 + x * 1/45 = 1, x = 9 min. |

Nazmul Hasan Apu said: (Jul 7, 2021) | |

A's work in 1 hour = 2/75. A's work in 30 minutes = (2/75) * 30 = 4/5. Remaining work (1- 4/5) = 1/5. B's work = (1/5) * 45 = 9 min. |

Subramanya said: (Aug 16, 2021) | |

Well explained, Thanks @Nazmul. |

Saradha said: (Sep 3, 2021) | |

For this kind of question, Use formula (1 - full tank time / 2nd pipe time)* 1st pipe time. Depends on the question what they are asking the 1st and 2nd times will be swapped. if 1st pipe time is asked the 1st pipe time should be outside like the above formula, or else 2nd one time should be outside. Here full time is half an hr should be converted to min 1/2*60 = 30min. Given data: Full time = 30min. 1st pipe time= 37 * 1/2min => 37*2=74, 74+1= 75/2min. 2nd pipe time= 45min. Apply formula (1 - 30/75/2) * 45, 30 and 75 cancelled (1 - 6/15/2) * 45. 1 * 15/2 = 15/2 => 15/2 - 6 = 15-12/2 => 3/2. (3/2 / 15/2)*45 remainders gets cancelled and remaining 3/15*45= 15 and 45 gets cancelled by 3 so 3 * 3 = 9min. So, if u follow the above formula u can get the answer. You must be sure which is asked for if 1st one then outside the bracket if 2nd one then it should be outside. Here b's time is asked so b's time is outside in the formula. |

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