# Two pipes fill a tank

Question-

Two pipes can fill a tank in 20 min and 30 min respectively. If both are opened, after what time should first be closed, so that tank will be filled in 10 min?

My attempt-

Let x be the time after which first pipe should be closed.

x(1/20 + 1/30)+ (10-x)/30 = 1

(5x)/60 + (20-2x)/60 = 1

3x + 20 = 60

x=(40/3) min

Can't use these tags- "pipes" "cisterns" "tanks"

The book answer is wrong. It takes $12$ minutes for the two pipes to fill the tank together, so you can't fill them in $10$ minutes. Note that your value is greater than $10$ minutes, so you are filling the tank for $10$ minutes with both pipes and $\frac {10}3$ minutes of the first. That does fill the tank.

First pipe fills $\frac{(1)}{30}$ of the tank each minute. Second pipe fills $\frac{(1)}{20}$ of the tank each minute. $\frac{(1)}{30} + \frac{(1)}{20} = \frac{(1)}{12}$ Therefore it would take 12 minutes to fill the tank and it cannot be done in 10.

Lets assume the book has the right answer for the wrong pipe 1 rate.

in ten minutes, a 30 min pipe will fill $\frac{10}{30} = \frac{1}{3}$ tanks. This leaves $\frac{2}{3}$ of the tank to be filled by eight minutes of a $p_{1}$ flow rate:
$\frac{2}{3} = \frac{8}{p_{1}} \implies$
$2p_{1} = 24 \implies$
$p_{1} = 12$
So pipe 1 actually fills the tank in twelve minutes and must be shut off after eight.

What if the pipe 2 rate is wrong?

pipe 1 fills in 8 minutes $\frac{p_2 - 10}{p_2}$ of a tank.
$\frac{8}{20} = \frac{p_2 - 10}{p_2}$
$12p_2 = 200$
$p_2 = \frac{50}{3}$ minutes

What if both rates are wrong?

Then pipe 1 must fill $\frac{p_2 - 10}{p_2}$ of a tank in 8 minutes:
$\frac{p_2 - 10}{p_2} = \frac{8}{p_{1}} \implies$
$p_1p_2 - 10p_1 = 8p_2$