Aptitude - Pipes and Cistern - Discussion

Why it take x-5?

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

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.

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.

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).

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).

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?

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?

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.