Aptitude - Pipes and Cistern - Discussion

5. 

A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which 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:

[A]. 6 hours
[B]. 10 hours
[C]. 15 hours
[D]. 30 hours

Answer: Option C

Explanation:

Suppose, first pipe alone takes x hours to fill the tank .

Then, second and third pipes will take (x -5) and (x - 9) hours respectively to fill the tank.

1 + 1 = 1
x (x - 5) (x - 9)

x - 5 + x = 1
x(x - 5) (x - 9)

(2x - 5)(x - 9) = x(x - 5)

x2 - 18x + 45 = 0

(x - 15)(x - 3) = 0

x = 15.    [neglecting x = 3]


Prasad said: (Dec 31, 2010)  
How you have taken 9 hours?

Vijay said: (Jan 12, 2011)  
As 2nd one is 4 hrs slower than 3rd pipe, 3rd pipe should be 4 hrs faster than 2nd. So, 3rd pipe is 9 hrs faster than 1st one.

Pranesh Kumar said: (May 29, 2011)  
Why you neglect x=3 for what reasons?

Karthikeyan said: (Jun 2, 2011)  
How got this equation 1/x+1/x-5=1/x-9 ?

Manasa said: (Jun 11, 2011)  
@pranesh kumar

In options "3" is not given...so we considered 15
both "3" and "15" are correct.

@karthikeyan

In the question they given

The first two pipes operating simultaneously fill the tank(1/x+1/X-5)... in the same time means ("=") during which the tank is filled by the third pipe alone(1/X-9)
how much water gets filled up by the first two pipes in 1 hour...the same amount of water gets filled up by the third pipe in 1 hour.

Sachin said: (Aug 6, 2011)  
How can a pipe fill in both 3 and 9 hours. Hows it possible?

Only one answer must be true?

which one how to get?

Praneeth said: (Aug 6, 2011)  
We have considered the x-5 and x-9 as time taken by 2nd and 3rd pipes, but after substituting x=3 we will get -2hr and -7hr respectively for the 2nd and third pipes.

So, we cant have negative values for time practically.

Sai said: (Aug 31, 2011)  
Why we have taken 1/x, 1/x-5, 1/x-9. Please can any one say me the reason?

Kasi Srinivas said: (Sep 1, 2011)  
@ sai :

first pipe alone takes x hours to fill the tank => in 1 hour 1/x part of the work is completed.
similar way we get 1/(x-5) and 1/(x-9) part of the work is completed in 1 hour.
it is the standard practice that we consider the work done in an hour.

If you understood the question properly, you can understand this :)

Ghousia said: (Sep 6, 2011)  
Why is x-9 and x-5 taken as the pipe speed is represented faster then it should be x+9 and x+5? Please clarify my doubt.

Sai Krishna A said: (Nov 10, 2011)  
@ghousia faster means in less time. if for example A takes 20hours then B takes(20-5=15)hours not 20+5=25hours. got it

Venky said: (Jan 17, 2012)  
4 hours slower than the third pipe = means how we took 1/X-9 and that also like this 1/X +1/X-5= 1/X-9

third pipe speed we dont know ...the how {x-9}

Deepi said: (Feb 6, 2012)  
@ venky

See the question they have given as second pipe 5 hours faster than the first and 4 hours slower than third.

Lets consider first pipe takes X hrs for ex. 10hrs
Second pipe takes 5 hours faster than first it means 5hrs (X-5)

Third pipe is faster than second by 4hrs. That is 1hour(10-9)=(X-9).

Rofiqul Hoque said: (Jun 27, 2012)  
Let 1st pipe take time = x hours.

2nd pipe is 5 hours faster means its take 5 hours less than 1st pipe.

i.e., 2nd = (x-5) hours.

3rd pipe is 4 hours slower than the 2nd pipe.

i.e., 3rd +4 = (x-5).

=>3rd = (x-5) -4.

=>3rd = x-9.

Now,

1/x + 1/ (x-5) =1/ (x-9).

Dear friends now just solve it. Bye friends meet you again.

Chinna Jgc said: (Aug 22, 2012)  
2nd pipe is 5 hours faster than 1st.so 2nd pipe=5x

1st pipe automatically takes 1 hour to fill the tank.1st pipe=1x.

in question 2nd pipe is 4 hours slower than 3rd pipe.

i.e 5x+4x=9x

3rd pipe =9x

add all 3 pipes i.e 1x+5x+9x=15.

Rohit Sharma said: (Dec 14, 2013)  
Is it wrong if we suppose X time taken by the 2nd pipe and solve as per the problem i.e :

Time taken by 1st pipe : X+5.
Time taken by 2nd pipe : X.
And time taken by 3rd pipe : X-4.

Please explain me.

Tanu said: (Jan 19, 2014)  
1st can fill tank in x hrs.

2nd can fill 5 hrs faster than 1st = x - 5.
3rd is 4 hr faster than 1st = (x - 5) -4 = x-9.

Hence,
1/x +1/(x-5) = 1/(x - 9).

Sabyasachi said: (Mar 13, 2014)  
Is there any short tricks for this type of questions?

Sabyasachi said: (Mar 13, 2014)  
How it became (x-5) and (x-9) ?

Thekingoftheworld said: (Apr 14, 2014)  
The resolution here presented lies on the assumption that the volume is 1 cubic meter.

In fact, this problem would have infinite solution as no volume is specified in the question.

Pulkit said: (Jun 6, 2014)  
Can anybody answer my question with correct explanation that why we have neglected x=3 and answered x=5. I know the explanation but I want to now that does anybody else knows it.

Inu said: (Jul 1, 2014)  
Time cannot be negative. Thats why 3 is neglected.

Sameeera said: (Aug 10, 2014)  
Why can't we solve it in this way:
Let c=3rd pipe, b=2nd pipe, a=1st pipe.

Let us fix c's time = x.
c=x ; a=x+9; b=x+4.

(a+b)'s time =c's time.
1/(x+9) + 1/(x+4) = 1/x.

By solving we get x=6.

Now a=9+6=15.
Ok I think you have understood.

Chandni said: (Apr 23, 2015)  
I solved it by taking 2nd pipe's speed as x.

Then 1st pipe speed would be x-5 and 3rd pipe = x+4.

If we resolve now, we get y = -10 and y = 2.

So, I was finding answer according to 2.

Lakshmi said: (May 13, 2015)  
Why neglecting 3 in the answer please can any one say?

Vicky said: (Jul 19, 2015)  
Can you explain that 1/x-9 how come to right hand side positive.

Its positive on left side so it may be come to the right side than the equation -1/x-9.

Please any body explain.

Deepak Pareek said: (Jul 28, 2015)  
Solve through options. It will be easier.

Lakshmichandra said: (Aug 21, 2015)  
Here why we can not consider x+5? I did not understand this problem can any please explain?

Dipika Sinha Bhaduri said: (Oct 5, 2015)  
This question is a very easy one.

Let 2nd pipe's speed be x.

So 1st pipe's speed will be x+5.

And 3rd pipe's will be x-4 (as per the ques).

Now, it is given in the question that The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone.

Hence 1/x + 1/x+5 = 1/x-4.

Solving this we get x = -2 and 10. Obviously x = 5 therefore speed of 1st pipe is 10+5 = 15.

Vaisakh said: (Oct 10, 2015)  
Let first pipe alone takes x hours to fill the tank. Then if we take (x+5), it means 2nd pipe fills the tank 5 hours slower than the first pipe. So it's not possible. Just think about it.

Amol Phad said: (Nov 5, 2015)  
Please somebody tell me in easier way.

Sagar said: (Jan 2, 2016)  
Looking at the question, it is quite evident that the first two pipes simultaneously fills the tank in same time.

Assuming the time taken by pipe A to fill the tank is X hours, then for every hour, it will fill 1/X of the tank.

Now, "second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe" --> So the second pipe will require lesser time than pipe A to fill tank i.e. (X-5) hours.

Second tank is also 4 hours slower than the third pipe, so third pipe will be X-5-4 = X-9 hours.

Summing up, the equation goes like:

1/X + 1/X-5 = 1/X-9.

Santosh H said: (Mar 8, 2016)  
Just remember these:

1. Faster fil = lesser time.

2. If total time=x, then for 1 hr it is 1/x.

3. When you get two different answers after solving the equation just substitute the answers (like x=15) in the original equation and check whether it is feasible.

Brryne said: (Mar 31, 2016)  
Dear Friends. The solution according to me is correct.

The reason as to why X = 3 is neglected is because there is no negative time.

For example: 3 hrs. If you substitute X with 3, then,

1/(x-5) will be 1/(3-5).

Which are 1/-2 hours (negative thirty minutes: never in real life!)

Minion said: (May 6, 2016)  
@Sameeera.

Your solution seems to be apt to this problem. Thank you.

Robot said: (Jul 2, 2016)  
If you assume the value corresponding to the second pipe to be x,

The other two factors would be x + 5, x - 4 with respect to the 1st and the 3rd pipe.

Here,
1/ (x) + 1/ (x + 5) = 1/ (x - 4).
In this case, the answer turns out to be 10.

Help me.

Lalitha said: (Aug 2, 2016)  
Why you neglect 3?

Debasish said: (Aug 11, 2016)  
Is there any shortcut method for solving it?

Virajgouda said: (Aug 11, 2016)  
Why do you neglect 3? What is the reason?

Prawesh Pradhan said: (Sep 26, 2016)  
@Virajgouda, if take X as 3 the answer will be negative. And time can't be negative. X - 5 and x - 9 would be -2 and -6.

Bhagyasree said: (Mar 15, 2017)  
Suppose A takes x hours then B takes x-5 hours.

Ex: It means that A completed in 10 hours where B Completed only in 5 hours
Now C take y hours where B Completes in y+4 hours(slower),
So x-5 =y+4..............y=x-9,
Here capability of A+B = C,
1/x+1/x-5 = 1/y,
Now, You can easily solve this eq 1/x+1/x-5 = 1/x-9.

Reshma said: (Mar 17, 2017)  
Nice explanation @Tanu.

Basavaraj said: (Mar 30, 2017)  
Why 1/x-9 Comes to negative?

Devendar said: (May 9, 2017)  
Guys, please read the question properly their it is given as.

2nd pipe 5hrs father then 1st pipe.
2nd pipe 4 hrs slower then 3rd pipe.
Now consider 2nd pipe as x.

So,
1st pipe 5hrs slower then 2nd pipe =x-5.
3rd pipe 4hrs faster then 2nd pipe =x+4.

Now solve the equation.
1/(x-5) + 1/(x)= 1/(x+4).
x-5+x/(x (x-5))=1/(x+4).
2x-5 = x (x-5)/(x+4).
(x+4)(2x-5)=x (x-5).
2x^2 + 8x -5x - 20 = x^2 -5x.
x^2 + 8x -20 = 0.
x^2 - 2x + 10x - 20 =0.
x (x-2) + 10 (x-2) =0.
(x-2)(x+10) =0.
x =2 or.
x = -10.

This is correct.

Lavanya said: (Jun 29, 2017)  
Two pipes M and N can fill a cistern in 24 min and 32 min, repectively. If both the pipes are opened together, then after hoe many minutes N should be closed so that the tank is full in 18 minutes?

Please answer it.

Ronnie said: (Jul 26, 2017)  
@Lavanya.

Answer is 3 mins.

Monu Dhaka said: (Aug 4, 2017)  
@Lavanya.

Answer is 10 minutes {(18/24)+(8/32)}=1.
18-8 = 10minutes.

Sana said: (Sep 2, 2017)  
Suppose, second pipe alone takes x hours to fill the tank.

Then, first pipe and third pipe will take (x + 5) and (x - 4) hours respectively to fill the tank.

1/(x+5) + 1/x = 1/(x-4)
( x + x + 5 ) / ( x^2 + 5x ) = 1/(x-4)
Srossmultiply ( 2x + 5 ) * ( x - 4 ) = 1 * (x^2 + 5x )
Solve, you get
(x - 10)(x + 2) = 0
x = 10. [neglecting x = -2]
1st pipe = x + 5 = 15 hours.

Rabindra said: (Sep 16, 2017)  
The ratio of performance is A:B;C = 1:5:9 if we take lcm then we get the tank capacity i.e 45litre that means,
1+5+9=45,
15=45,
per hour litre=45/15=3 litre.
A can fill 3 lite per hour and B can 15 lit/hour, C can fill in 27 lit/hour,
so A will take time,
45/3=15 hour.

Pradhyumna said: (Oct 15, 2017)  
Can't we pick the 'B' to be as x and do the same things since When I did I got four first just by generally calculating it?

Nehal said: (Feb 4, 2018)  
By taking pipe b=x.
Then,
For others, it will be ...(x+5) for a.
And (x-4) for c.
Eq: 1/(x+5) +1/(x)=1/(x-4).
After solving ans comes as x=10.

Is it correct?

Nagarjuna said: (Jun 28, 2018)  
Can anyone help me?

If suppose A and B are filling a tank started at same time.
A alone can fill it in 'a' minutes B alone can fill it in 'b' minutes.
So we are taking (1/a)+(1/b)=1 to find the time.

But when they are filling it simultaneously why are we taking same (1/x+1/x-5=1/x-9)?
I think after completion of a ,b is started it means we have to take a+b right? But not (1/a)+(1/b).

Nagarjuna said: (Jun 28, 2018)  
Can anyone help me?

If suppose A and B are filling a tank started at same time.
A alone can fill it in 'a' minutes B alone can fill it in 'b' minutes.
So we are taking (1/a)+(1/b)=1 to find the time.

But when they are filling it simultaneously why are we taking same (1/x+1/x-5=1/x-9)?
I think after completion of a ,b is started it means we have to take a+b right? But not (1/a)+(1/b).

Seenam Goel said: (Jul 3, 2018)  
Let us suppose, The 2nd pipe can fill the tank in X hrs.

Hence, 1st pipe will take (X-5)hrs.
& , 3rd pipe will take (X+4)hrs.

Now, acc to the question.
work done by 1st and 2nd pipe together = work done by 3rd pipe alone.

Hence, 1/(x-5)+1/x =1/(x+4) ;
after solving the above part we will get a quadratic equation,
i.e. x^2+8x-20=0 ;

After solving the quadratic equation we get,
x=2 & x= -10 as the result and we can see that both of them are not adequate.
Hence, acc to me there is a problem with the data given in the question.

So, DATA IS INADEQUATE.

Swapnil said: (Aug 7, 2018)  
1/x + 1/ (x-5) = 1/(x-9).

Why 1/(x-9) equate with 1/x + 1/ (x-5)?

Pankaj said: (Nov 1, 2018)  
A+B = C.

Now,

Let A pipe fill the tank in =X hr.
2nd pipe 5 hr slower than A,
B = (X-5) hr
3rd pipe 4 hr slower than B,
C= X- (5+4) hr = (X-9) hr.
LCM Of (A+B) is;

A= X
X(X-5) LCM.
B=X-5.

A's 1hr work= X(X-5)/X = X-5.
B's 1hr work= X(X-5)/(X-5) = X.
_______________________________
A+B. = 2X-5
_________________________________

A+B =C.

X(x-5)/2x-5 = x-9,
X^2-5x/2x-5 = x-9 cross multiple,
2x^2-5x-18x+45 = x^2-5x (5x cancel out),
2x^2-18x+45,x^2 = 0,
X^2-18x+45 = 0,
X^2-(15+3)x+45= 0,
X^2-15x-3x+45 = 0,
X(x-15) - 3(x-15) = 0,
(X-15) (x-3) = 0,
Ans= 15 hr.

Nita said: (May 4, 2019)  
If we subject pipe 3 (P3) = x.

Then pipe 2 (P2) = x+4 (since P2 is 4 hours slower than P3 )
And Pipe 1 ( P1) = ( x+4)+5= x+9 (since P1 is 5 hours slower than P2; indirectly implied in the question).

So, the question becomes;

1/(x+9) + 1/ (x +4) = 1/ x.
Solving the equation we get x =+6,-6;
Taking x=+6.
Then pipe1 = (x +9) = 6+9=> 15.

Trinesh said: (Jul 3, 2019)  
Why it take x-5?

The second pipe fills the tank 5 hours faster than the first pipe so there will be X+5.

Harshvardhan said: (Oct 14, 2019)  
Simple if we take X time taken by pipe 2 then,

Time is taken by first pipe 1 = X+5.
Time is taken by second pipe 2 = X.
Time is taken by third pipe 3 = X-4.
So as per the question,
Pipe 1 +pipe 2 = pipe 3.
1/(X+5) + X = 1/(X-4).
After solving this equation we will get X=10.

Then, we know that time taken by pipe 1 = X+5.
Simply put the X=10, so pipe 1 will take =15 hours.

Saimanasa said: (Jun 3, 2020)  
Thank you @Manasa.

Amaterasu said: (Sep 8, 2020)  
Consider 1st, 2nd and 3rd pipe as a,b,c.

Given 1/a+1/b=1/c or a+b/ab=1/c because the time taken to fill by a and b is same is c
b=a-5 can be written as a=b+5.
b=c+4.
1st put value of a in a+b/ab=1/c then u get 2b+5/(b+5)b = 1/c
Then put the value of b you will get c=6 so b will 10 and a will be 15.
And the time taken by a is 15hr.

Sarbajit Rai said: (Oct 21, 2020)  
@Ghousia,

Here the question says to find the time and thus it is logical that lesser the time it takes faster will be the work in filling the tank. For instance, car A is moving faster than car B by 10 kmph means car A is taking less time to cover the same as that of car B. That is the reason why 1/x+1/ (x-5) =1/(x-9).

Shubhadip Das said: (Jan 30, 2021)  
In how much time 2nd and 3rd pipe together fo the work? Please explain the answer.

Anonymous said: (Sep 6, 2021)  
Thanks @Sai Krishna.

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