### Discussion :: Pipes and Cistern - General Questions (Q.No.3)

Priyanka Sharma said: (Nov 17, 2010) | |

Please tell how it is (1/2 - 3/7) ? |

Paushali Mukhoty said: (Dec 30, 2010) | |

If the total area of pump=1 part The pumop take 2 hrs to fill 1 part The pumop take1 hour to fill 1/2 portion Due to lickage The pumop take 7/3 hrs to fill 1 part The pumop take1 hour to fill 3/7 portion Now the difference of area = (1/2-3/7)=1/14 This 1/14 part of water drains in 1 hour Total area=1 part of water drains in (1x14/1)hours= 14 hours So the leak can drain all the water of the tank in 14 hours. |

Jad1980 said: (Feb 18, 2011) | |

I'm confused, please check my logic 1. It takes 2 hours for the pump to fill the tank without the leak. 2. The leak adds 1/3hrs onto that time time, which is an increase in time of 1/6th. 3. Then the amount leaked = 1/6 of the amount pumped. 4. When the pumped is turned off the leak will drain the tank at 1/6 of the speed at which the pump fills it 5. If it takes the pump 2hrs to fill it then it should take 12 hours for the leak to drain the tank |

Sandeep said: (May 23, 2011) | |

Paushali Mukhoty Good explanation thank you. |

Vivek said: (Jul 5, 2011) | |

Nice explanation by Paushali Mukhoty. |

Krunal said: (Aug 8, 2011) | |

Very well explaination. I understand very well. |

Amitha said: (Sep 6, 2011) | |

How cum 7/3 hours to fill 1 part for the leakage? |

Radhika Krishna said: (Sep 12, 2011) | |

Their total tank divided in to 7 parts i.e 20 min to fill 1/3min to fill 1 part |

Zakeer said: (Oct 27, 2011) | |

How come 7/3 hours to fill 1 part for the leakage? |

Pgcr ( Sri Lanka) said: (Nov 4, 2011) | |

Just because of the leak it took 2 1/3 hours. Normally it takes only 2 hours. Hence, 1 1 1 - - - = - 2 X (2 1/3) OR 1 1 1 - - - = - 2 (2 1/3) X So, X = 14 hours. |

Chakri said: (Jan 5, 2012) | |

Please tell me why it's 7/3 changed into 3/7? |

Ashu said: (Jan 13, 2012) | |

@chakri If you take 7/3 hrs to complete 1 job Then, what length of job will you take to complete in 1 hour ( what you do here is...use reciprocal value when u shift the value) that is 7/3=>1 1=>3/7 |

Ashu said: (Jan 13, 2012) | |

PGCR great logic. What type of logic you used here? Please explain. |

Deepi said: (Feb 5, 2012) | |

A takes 2 hrs lets take as x Leakage time lets y 2 1/3 hrs=7/3hrs => (x*y)/(y-x)=7/3 => (2y)/(y-2)=7/3 => 2y=(y-2)(7/3) => 2y=(7y/3)-14/3 => -y/3=-14/3 => y=14 |

Raj said: (Mar 1, 2012) | |

It is very simple. Pump takes 2hr to fill & due to leakage it takes 7/3 hrs calculate it in terms of 1 hr. pump will fill 1/2 part of tank & due to leakage 3/7 part of tank so leakage in 1 hr will be 1/2 - 3/7 = 1/14 part, so full tank will be empty in 14 hr. |

Allhad said: (Apr 10, 2012) | |

To all who don't understand how 7/3. Actually time given is 2hrs and 20min i.e. 2hrs+20min (1/3 hrs). |

Vinay Arora said: (Sep 17, 2012) | |

In 120 mints it complete = 1 tank but it take 140 mints now so, in 1 mint it fill =1/120 extra time it take be= 140-120= 20 mints in 20 mint it fill= 20 * (1/120) =1/6 1/6 tank fill in 2 Hr 20 mint that is 7/3 so 1/6 of tank fill= 7/3 1 tank fill in= (7/3)*6= 14 hr ok buddy...... |

Sai Triveni said: (Sep 2, 2013) | |

As per question a pump can fill and due to leakage it took 1/3 of time extra. As we can say {PUMP-(PUMP+LEAKAGE)== IF WE SIMPLEFY THIS == PUMP-PUMP-LEAKAGE THIS GIVE RESULT AS = -LEAKAGE. I CAN'T UNDERSTAND WHY U REVERSED THE VALUE IT MUST BE 7/3 BUT U TOOK 3/7 HOW? |

Raja said: (Sep 12, 2014) | |

@Sai. Its actually 1/(7/3). |

Abhilash said: (Nov 5, 2014) | |

Actually tank could have filled in 2 hr. Part will be 1/2. Let the time taken by the leakage is y. Therefore, part is 1/y. 3/7 = (1/2-1/y) here 3/7 is the total time taken to fill the tank (in parts). 1/y = ((7-6))/14. y = 14. |

Mohanapriya said: (Nov 14, 2014) | |

2 hrs compose 120 mins. But due to leak it takes 120+20 = 140 mins. 1st 2hr - 20 goes of [the amount has to be filled in 20 mins]. 4nd hr - 40 mins. 6th hr - 60 mins. 8th hr - 80 mins. 10th hr - 100 mins. 12th hr - 120 mins. 14th hr - 140 mins. And so it takes 14 hours to empty a full tank. |

Govindarajan said: (Nov 18, 2014) | |

20 mins extra for 2 hour i.e. 120 mins. 120 mins is first tank. 140 mins is second tank. 120/x = 140/20. 140/20*120 = 840 mins. 1 hour = 60 mins. 14 hours = 14 hours. |

Mgh said: (Nov 19, 2014) | |

@Mohanpriya. Why have we done the calculation till 140 mins? |

Sandeep said: (Nov 19, 2014) | |

Lets us simplify by doing it minute basis:. Let the Volume be V without leak tank is filled in 120 min. So, in 1 min = V/120. With leak, = V/140. Leak is withdrawing water per min = V/120 - V/140 = V/840. i.e 840th part is filled in 1 min. So the tank is filled in 840 min i.e. 14 hours. ----------------------------------------------------------------Hour Basis, In 1 hour V/2 with leak, 3V/7. Work by leak = V/2-3V/7 = V/14. V = 14 h. |

Elizabeth said: (Nov 19, 2014) | |

@PGCR. Very well explained. Thank you :). |

Anjali said: (Apr 13, 2015) | |

A pump can fill a tank with water in 2 hours. Because of a leak, it took 2 hours to fill the tank. No where it is written that its taking 2.20 mints to fill while after leakage. Then why everyone taking (7/3). I am lost. Do help. |

Priya said: (Jul 9, 2015) | |

I had one doubt if suppose they are asking to find when the tank is filled due to its leakage means how to solve it? Please can anyone explain? |

Bunty said: (Jul 9, 2015) | |

For those who didn't got that how 7/3 come from? Here it is 2*1/3 hours means (2*3+1)/3 = 7/3. |

Juned Shah said: (Aug 12, 2015) | |

/*Other method to solve this problem */. A, A-l. Time 120, 140. Time/min 7, 6. Capacity 840, 840 of tank. So due to leakage difference in time/min (7-6) = 1. So 840/1 = 840 min. Convert it in hr. = 840/60 = 14 hr. & This is our require answer. |

Mouni said: (Sep 20, 2015) | |

I can't understand how 7/3 is converts in to 3/7? |

Mouni said: (Sep 21, 2015) | |

A cistern is normally filled in 8 hrs. But takes 2 hrs more to fill because of a leak in its bottom. If the cistern is full, the tank will empty it in? Please tell me with explanation. |

Venkat said: (Oct 23, 2015) | |

@Mouni. Please check this one: Let us take Cistern is filled by pipe A and leak is due to pipe B (In this way it is easy to understand). First point : Cistern filled by pipe A in 8 hrs. Second point : Cistern filled by pipe A and pipe B (it is actually leak) is 10hrs (8+2). Now take LCM of these two LCM of 8, 10 is 40. This implies to fill Cistern, total 40 units of work needed. Then if only pipe A can fill the Cistern in 8 hrs that means 5 units of work for min by pipe A. Both pipe A and pipe B can fill the Cistern in 10 hrs that means 4 units of work by pipe A and pipe B. So if you see, in second case 4 units of work per min where as in first case it is 5 units of work. This means if you include leak it reduces the work units by 1 unit ---> leakage per hr is 1 unit. So to drain all the water in cistern, leak has to do 40 units of work. So if it is doing 1 unit of work to drain the tank. For 40 units it has to 40 hrs of work. So answer : 40 hrs. |

Narendra said: (Feb 14, 2016) | |

For inlet it takes one hour for 1/2 part. For leakage, let's consider it takes 1 hour for 1/x parts. For 2 hours and 20 minutes the tank will be filled as question says: (1/2-1/X)*7/3 = 1 (the full tank having 1 part). |

Laxmikant said: (Apr 25, 2016) | |

@Paushali Mukhtoy is very well explained. Thank you. |

Mahendra said: (Jul 29, 2016) | |

First, let's find the rate of filling up with and without leakage. Assume that the volume of the tank is 2 liters (for easy calculation). Then the rate of filling without leak = 1 liter/hour. Rate of filling with leakage = 2 liters/ (7/3) hours = 6/7 liters per hour. Now we get the leakage rate ie = 1 - 6/7 = 1/7 liters/hour. So, time taken = total volume/rate of leakage = 2/ (1/7) = 14 hours. |

Saiprakash Holi said: (Jul 31, 2016) | |

Let A take 2hrs to fill. B take 'x' hrs to empty the tank. (A-B) per = 2 1/3hrs =7/3. (A-B) per= Aper -Bper. 1/ (3/7) =1/2 - 1/x. 1/x = 1/2 - 7/3. x = 14. |

Rajeev said: (Sep 15, 2016) | |

Here it is mentioned as 2 hrs for the tank and for leak tank time also given 2 hrs instead of 2hrs 20 min. SO I am also confused that how 3/7 comes? |

Jishnu C V said: (Nov 12, 2016) | |

@Mouni. Work done for 1 hr without leak = 1/2, Time taken to fill the tank with leak = 2 1/3hrs = 7/3 hrs, Work done with leak in 1 hr = 3/7, Work done by leak in 1 hr = 1/2 - 3/7 = 1/14, So, tank will be empty by the leak in 14 hrs. |

Praveen said: (Jan 10, 2017) | |

Good explanation, Thank you @Mohanapriya. |

Sudheer said: (Feb 23, 2017) | |

How to substitute this on for answer 1/2-3/7 =1/14? |

Champion said: (Mar 3, 2017) | |

How can we say that 1/14 part of tank fills in 1 hour and it takes 14 hrs to drain all water? |

Murali said: (Mar 12, 2017) | |

A pipe can fill a tank in 3 hours. Because of the leak in the bottom, it is filled in 4 hours. If the tank is full, how much time will the leak take to empty it? Can anyone solve it? |

Anjali Tengli said: (May 1, 2017) | |

Without leakage it takes 2 hrs. In 1 hr it take 1/2. With leakage, it takes 7/3 In 1 hr 1/7/3 I.e. 3/7. 1/2 - 3/7= 11/14. |

Ankit said: (May 6, 2017) | |

Good explanation. Thank you @Deepi. |

Sahid said: (Aug 20, 2017) | |

First, it takes 120 min for leakage it lake (120+1/3*60=120+20)=140 min then LCM(120 and 140 it. It give 840min. 840/60 = 14hours. |

Sandhya said: (Sep 6, 2017) | |

How 1/3 got over there? please explain. |

Anuj said: (Sep 8, 2017) | |

Without leakage a takes 2 hours with, leakage 7/3 hour take lcm of 7/3 *3=7and 2*3=6. lcm of 6 and 7=42 it takes more 1/3 hour so divide 42 by 3 =42/3 =14. |

Pranav said: (Dec 27, 2017) | |

Ans : D Tank fill without leak = 2 hours Tank fill with leak = 2 1/3 hours = 2/1 * 1/3 = 7/3 hours (imp) The time taken to drain a tank is = Part filled in 1 hour (without leak) - partly filled in 1 hour (with leak). = 1/2 - 1/ 7/3 => 1/ 7/3 = 3/7 = 1/2 - 3/7 = 1/14. So, time taken to drain the tank is 14 hours. |

Swati said: (Jan 14, 2018) | |

Why 42/3? @Anuj. |

Mrunal said: (Jan 19, 2018) | |

Here is a different solution. Let tank capacity is 42 liters. Pump A filling rate is = 42 liters/ 2 hours= 21 lph. Pump A filling rate due to the leak is = 42 liters/ 7/3 hours = 18 lph. Now the leaking rate is 21-18= 3 lph. Leak can empty tank in = 42 liters/ 3 lph= 14 hrs. Hope this is simple and easy to understand. |

Jamsheed said: (Mar 4, 2018) | |

Please tell me if a small pump drained a tank in x hours. In 3 hours how much it will drained? Anyone solve this. |

Manuj said: (Mar 5, 2018) | |

From, where 2 1/3 came? |

Stella Brown said: (Mar 31, 2018) | |

How is 1/2-3/7 possible? Please explain indetail. |

Rakesh said: (Apr 3, 2018) | |

It takes 2 hours to fill the tank. It takes 2 1/3 (7/3 hrs)hours to fill the tank including the leak. Now, Tank fill in 1 hour=1/2, Tank fill in 1 hour including leak = 3/7, Tank leak for 1 hour= tank fill in 1 hour- tank fill including tank leak in 1 hour = 1/2 - 3/7=1/14, So, 1/14 of the tank will be empty in 1 hour. Now, 1/14*14 of the tank will be empty in 1*14 hour. (Multiply 14 on both sides), The Complete tank will be empty by leak in 14 hours. |

Anil said: (Jun 9, 2018) | |

Let consider 12-litre tank and pump fill in 2 hours(120 mins) w/o leakage. with leakage, it takes 2 hours 20 mins, means its take 20 mins extra, so pump fill 2 litres extra in leakage condition, total 14 litres fill by the pump but in tank 12 litres left, means in 2hrs 20mins leakage leaks only 2 litres, the total is 12 litres, 2 litres x6=12 litres. 7/3 hours(2 hrs 20 mins)x6 = 14 hours. |

Ashish said: (Jun 18, 2018) | |

Good explanation. Thanks @Paushali Mukhoty. |

Yamini Dhanasekar said: (Jul 19, 2018) | |

It can also solved by; Let the Filling pump be 'A' and leakage be 'B'. A takes 2 hrs to fill, In 1 hr -it will complete 1/2 of the work, ie half of the tank. In addition of B, A-B=2 1/3 hrs, In 1 hr they complete 1/21/3 which is; A-B=3/7. We can substitute A, 1/2 -B=3/7 B=1/14. So, In hr, B drains 1/14 part of the tank. Total work-1. Hr -- work 1 -- 1/14 ? -- 1. 1/1/14= 14 hours. |

Yamini Dhanasekar said: (Jul 19, 2018) | |

@Jamsheed. Always, we can consider total work be 1. So Hour ---- Work completed. x ---- 1 3 ---- ? By, cross multiplication, 3/X of the work. Hope you understand. |

Salman said: (Aug 7, 2018) | |

Thanks @Ash. |

Hari said: (Aug 14, 2018) | |

Assume they are 2 different pipes. So total units of water to be filled is lcm of 2 and 7/3 =42units. The leak takes 1/3time extra than 1st pipe. So work done by the leak is 1/3 * 42 = 14. |

Rahul said: (Sep 19, 2018) | |

Pipe A could fill an empty cistern in 18 hours while Pipe B can drain a filled cistern in 30 hours. When the cistern is empty, Pipe A is turned on for an hour and then turned off. Now Pipe B is allowed to drain out water from the cistern for an hour and then turned off. The pipes were alternately left open for an hour each time till the cistern was full. How much time did it take for the cistern to be full? Can anyone help me to solve this? |

Rachel Vaz said: (Feb 17, 2019) | |

Through LCM method: Pump A fills in = 2hrs. Leaking pump = 7/3 hrs. Take LCM of 2 and 3 = 6 units, Therefore, Pump A = 3 u/h, Leaking pump = 14 u/h, Hence, the leaking pump can drain all the water in = 14 hours. |

Lavanya said: (Mar 26, 2019) | |

Explain how come 2 1/3 = 3/7. |

Zahid said: (Apr 15, 2019) | |

Let total units to be filled be 1 unit. Then work done by the pump in 1h=1/2 now let x be the time taken by leakage to empty the tank i.e. 1 unit. The work done by leakage in one is 1/x units. Now, as it takes 2 1/3 (or 7/3) to fill tank then work done in one hour with leakage will be =1/ (7/3) =3/7. Therefore. Work is done by pump without leakage per hour-work done by leakage in one-hour =work done by pumy with leakage. i.e 1/2-1/x = 3/7. i.e x = 14hours. |

Lulu Fathima said: (Apr 22, 2019) | |

How 7/3 becomes 3/7? |

Behara Satish Kumar said: (May 21, 2019) | |

Let v take two pipes A&B, Let time of A=2hrs = 120months Therefore the capacity of A(per mnt) = 1/120. Now, The time is taken by B=2 1/3 hrs = 140months. Therefore the capacity of B(per mnt) = 1/140. The difference between two pipes be the leakage capacity i.e A-B = (1/120)-(1/140) = 1/840. Therefore time was taken of leak = 840mnts = 14h. |

Siddu said: (Sep 7, 2019) | |

A initially fills in 2 hrs. Hence in one hr it fills 1/2 units, With leak it takes 7/3 hrs, Hence in one hr it fills 3/7 units (with leak). (1/2)- leak per unit = 3/7, Leak per unit = (1/2)-(3/7), =1/14. Per hr, it leaks units 1/14. Total time taken to empty is 14 hrs. |

Kuldhoj Bamjan said: (Mar 9, 2020) | |

The problem is exactly useful. But the tank when fulfill and quarter fill, there must be change the speed or press of water in the leak. So the water cannot throughout the same or regular behind to end. |

Anjum said: (May 16, 2020) | |

X pipe can fill in tank. Y tank filled with leak. Time taken by leak to empty tank is (xy/y-x). 2*(7/3)/(7/3-2)=14. |

Shubham Nale said: (Jul 11, 2020) | |

Let take a leak as an outlet named B. The capacity of the tank is 14/3 litr.e A=7/3 litre/hr and, B=-2. A+B=-1/3. Therefore. 14/3 / -1/3 = 14. |

Karthik said: (Jul 26, 2020) | |

There is a simple formula. a(1+(a/x)). a= 2 hours. x= 2 1/3- 2= 1/3. Solution: 2(1+(2/ (1/3) )= 14. |

Anonymous said: (Oct 18, 2020) | |

Work done by pipe A in 1 hour = 1/2 Work done by pipe B in 1 hour = 1/x Total work done is 7/3. Therefore, => 1/2 + 1/x = 1/7/3. => 1/2 + 1/x = 1*3/7. => 1/2 - 3/7 = 1/x. => 1/x = 1/14. Hence the pipe will empty the tank in 14hrs. |

Hika said: (Oct 19, 2020) | |

Can anyone explain the answer in LCM method? |

Mr Pai said: (Oct 23, 2020) | |

Time taken by the pump to fill the tank = 2 hours Work done by the pump in 1 hour = 1/2 Time taken by the pump to fill with leakage = 7/3 Work done by the pump with leakage in 1 hour = 3/7 Time taken by the leakage to empty the tank = x hours Work done by the leakage in 1 hour = 1/x. Therefore, (1/2)-(1/x) = 3/7. -1/x = (3/7)-(1/2). -1/x = (6-7)/14. x = 14 hours. |

Jothigonzi said: (Jan 19, 2021) | |

@Deepi. Nice explanation and easier to understand. Thanks. |

Reddy said: (Jan 27, 2021) | |

How 3/7 with leakage? Explain, please. |

Sudharshan Naidu said: (Feb 19, 2021) | |

The efficiency of a pipe in 1 hour is 1/2 = 0.5lit/h. The efficiency of a pipe with leakage 1/2.33 = 0.429, The difference between 0.5 - 0.429 is 0.07. The efficiency of a leak is 0.07lits/ hour. So, check with an option in how many hours the 1-litre tank empty we get the answer. |

Dhivyanisan said: (Mar 15, 2021) | |

Time is taken by pump without leak = 2hrs. Its discharge capacity says a= 1/2. Time is taken by a pump with leak = 2 (1/3)hrs or 7/3 hrs. It's discharge capacity say b= 1/(7/3) = 3/7. So discharge capacity of the leak is a-b = 1/14. Time taken by the leak to empty tank = 1/(1/14) = 14 hrs. |

R.Jana said: (Apr 1, 2021) | |

How to do by using the ratio process? Please explain. |

Suresh said: (Apr 11, 2021) | |

Convert Hours into minutes. A can fill in 2hrs i.e 120mins. Leakage i.e a-b 2 1/3 is equal to 140mins. Then capacity (LCM) is 840. Efficiency : A is 7 and a-b is 6 and the difference is 1. It can empty one part per minute out of 840 minutes. Then 840 parts can be emptied in 840 minutes. Now convert 840 minutes to hours. 840/60 is 14 hours. |

Messi said: (Apr 28, 2021) | |

Thanks for explaining @Suresh. |

Appu said: (Jun 6, 2021) | |

Shortcut; x = 2. y = 7/3 =====>(2 * 3) + 1. x*y/y-x =====>14. |

Vivek Anand said: (Jun 8, 2021) | |

Guys, it's simple. First thing; A pump can fill water in 2 hours. Due to the leak, it takes 2 1/3 hours. They are asking in how many hours it can empty the tank through leak only. So, 1. Tap ---> 2 hours. 2. Tap+leak ---> 2 1/3 hours (i.e) 7/3 hours. 3. Leak ---> ? Leak = ( tap + leak ) - tap = 1/7/3 - 1/2. = 3/7 - 1/2, = 6-7/14, = -1/14. Total leak = -14 - indicates leak or outlet etc. |

Kalaiselvan E said: (Jul 30, 2021) | |

@Jad. Thanks for explaining. |

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