Aptitude - Pipes and Cistern - Discussion

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.

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.

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.

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.

@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.

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.

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).

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.

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.