A pipe can empty a tank in 40 min. A second pipe with diameter twice as much as that of the first is also attached with the tank to empty it. The two together can empty the tank in ?
I am attaching my solution also here but I want a good verification of it because in the textbook which I am following says the other way.
Solution :
Let the Diameter of the 1st pipe be $d_1$ and that of the other pipe be $d_2$, where $d_2 = 2d_1$. So, technically speaking the other pipe can transfer 4 times the amount transferred by the 1st pipe.
So, it would take 1/4 of the time taken by the 1st pipe.
Therefore, Pipe 1 takes 40 mins. Pipe 2 will take 10 mins only.
So, together they will take 8 mins to empty the tank.
Is it correct ?
The "formula" to use here is $W = RT$ where $W$ = work, $R$ = rate in work per minute, and $T$ = time in minutes. The tank is empty when $W=1$.
For the first pipe, $W_1 = 1$ when $T_1 = 40$. So $R_1 = \dfrac{1}{40}$.
For the second pipe, $R_2 = 4R_1 = \dfrac{4}{40}$
The combined rate is $R_{together}=R_1 + R_2 = \dfrac{5}{40} = \dfrac 18$.
So, $T_{together}=8$ minutes.