Question

one pipe p1 can fill an empty tank in 5hours less than another pipe p2.both the pipes together can fill the tank in 5hours.then how long would it tak pipe p2 alone to fill the tank​

Answer 1

Answer:

Pipe A can billion empty tanks in 5hrs

⇒ billed by pipe A in hr =

5

1

Pipe B can fill an empty tank in 6hrs.

⇒ Part filled by pipe B in 1hr=

6

1

∴ tank filled by pipe AandB together in 1 hr =

5

1

−

6

1

=

30

1

∴ the tank can be filled in 30hrs.

Answer 2

Given:

Time taken by pipe P1 to fill tank = 5 hours less than pipe P2

Time taken by both the pipes to fill tank = 5 hours

To find:

Time taken by pipe P2 to fill tank alone.

Solution:

Let the time taken by pipe P2 to fill the tank alone be $y$. Then, time taken by pipe P1 to fill the tank is $y-5$

The rate at which pipe P2 can fill the tank in 1 hour = $\frac{1}{y}$

The rate at which pipe P1 can fill the tank in 1 hour = $\frac{1}{y-5}$

The rate at which both the pipes can fill the tank in 1 hour = $\frac{1}{y} +\frac{1}{y-5}= \frac{1}{5}$

$\frac{1}{y}+ \frac{1}{y-5} =\frac{1}{5}$

The denominators are not the same. Hence, we cross-multiply.

$\frac{y-5+y}{y(y-5)}=\frac{1}{5}$

$5(2y-5)=y(y-5)$

$10y-25=y^{2} -5y$

$y^{2} -5y-10y+25=0$

$y^{2} -15y=-25$

By completing the squares method,

$y^{2}-15y+(\frac{15}{2})^{2} =-25+(\frac{15}{2})^{2}$

$(y-\frac{15}{2})^{2}=\frac{125}{4}$

$y-\frac{15}{2}=\frac{11.2}{2}$

$y=\frac{15}{2}+\frac{11.2}{2}$

$y=\frac{26.2}{2}=13.1$ ≈ $13$ hours

Hence, pipe P2 alone will take 13 hours to fill the tank.

Pipe P2 alone will take 13 hours to fill the tank.