Aptitude - Pipes and Cistern - Discussion

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.

Nice explanation @Tanu.

Why 1/x-9 Comes to negative?

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.

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.

@Lavanya.

Answer is 3 mins.

@Lavanya.

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

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.

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.

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?