Two pipes can fill a tank in 12 minutes and 20 minutes respectively. Both pipes are opened together and after some time the first pipe is closed and the tank is full in totally 10 minutes. For how many minutes was first pipe open?
35 minutes
36 minutes
8 minutes
6 minutes
Explanation:
It is given that two pipes can fill a tank in 12 minutes and 20 minutes respectively.
Tank filled by first pipe in 1 minute =1/12
Tank filled by second pipe in 1 minute =1/20
Tank filled by both pipes in 1 minute = 1/12 + 1/20 =1/10
Let first pipe was open for (10 -x) minutes.
$\dfrac{1}{12} \times (10-x) + \dfrac{1}{20} \times 10 $=1
$\dfrac{10}{12}-\dfrac{x}{12}+\dfrac{1}{2} $ =1
$\dfrac{16}{12} -1 =\dfrac{x}{12} $
$\dfrac{4}{12} =\dfrac{x}{12}$
x=4
Since the first pipe was open for (10-x) minutes,
therefore first pipe was open for 6 minutes.
It is given that two pipes can fill a tank in 12 minutes and 20 minutes respectively.
Tank filled by first pipe in 1 minute =1/12
Tank filled by second pipe in 1 minute =1/20
Tank filled by both pipes in 1 minute = 1/12 + 1/20 =1/10
Let first pipe was open for (10 -x) minutes.
$\dfrac{1}{12} \times (10-x) + \dfrac{1}{20} \times 10 $=1
$\dfrac{10}{12}-\dfrac{x}{12}+\dfrac{1}{2} $ =1
$\dfrac{16}{12} -1 =\dfrac{x}{12} $
$\dfrac{4}{12} =\dfrac{x}{12}$
x=4
Since the first pipe was open for (10-x) minutes,
therefore first pipe was open for 6 minutes.