Aptitude - Pipes and Cistern - Discussion

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.

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.

Thank you @Manasa.

In how much time 2nd and 3rd pipe together fo the work? Please explain the answer.

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.

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

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

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.

@All.

We can solve by using a single condition.

Here which is efficiency of p1 +p2 =efficiency of p3--->CONDITION
To find individual efficiencies, the traditional x method = time consuming.
So opt for option verification;
On verifying option C, we meet our condition.

Option C says;
p1 takes time---------> 15 hours,
then p2 would be ---->10 hrs(15-5 as from Q 5 hours faster=earlier than p1 so subract),
p3 would be----------> 6hrs (10-4 as from question 4hrs earlier than p2).
Now, for the times of p1, p2, p3 (15,10,6) find the LCM.

LCM = >30.
Derive individual efficiencies from it(use allegation method as follows).
eff of p1=30/15 =>2.
eff of p2=30/10 =>3.
eff of p3=30/6 =>5.

From our condition
verify eff of p1+p2 =eff of p3.
2 + 3 = 5.
So, the assumption option C, 15 hours, would be correct.

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.

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.