Two taps are filling a tank Two taps are filling a tank together in 60 minutes (each tap has a different filling speed). 
If we open only one tap until we fill 1/3 of the tank, then close it, and then let the other tap to fill the rest of the tank - then it will take 120 minutes.
How long would it take to any of the taps to fill the tank on its own? 
I tried to solve it by 3 equations with 3 parameters but I'm not sure I'm in the right way...
 A: Let $r_1$ and $r_2$ be the different rates of the taps. From the first condition, we know that if they are filling together, namely at the rate of $r_1 + r_2$, the time will be $60$ minutes = $1$ hour. Without loss of generality, assume that the tank has capacity $1$. Then we have the equation (using capacity / rate = time) $$\frac{1}{r_1 + r_2} = 1 \implies r_1+r_2 = 1$$ From the second condition, we can derive a similar equation using the information given: $$\frac{1/3}{r_1} + \frac{2/3}{r_2} = 2$$Can you continue from here? 
NOTE: As Deepak pointed out, there is an easily observable solution to these equations, namely $(r_1, r_2) = (1/2, 1/2)$. Why? It happens that if we assume the two taps are identical, then the first condition tells us that at twice the rate, we fill it in an hour. The second condition tells us that at the normal rate, we can fill it in two hours.
A: Here's a quick "mental" solution:
The second condition implies that with $\frac 23$ of the tank left to fill, the second tap takes $120$ minutes. Which means that the second tap working alone would take $\frac 32 \times 120 = 180$ minutes to fill the entire tank.
Now consider the first condition. The two taps working together take $60$ minutes. Within this time, the second tap would've filled $\frac 13$ the tank on its own, meaning the first tap would've filled $\frac 23$ the tank on its own. So the first tap is twice as fast flowing as the second.
So first tap would take $90$ minutes and second tap $180$ minutes, each on its own, to fill the tank.
Note that the above solution interpreted the $120$ minutes in the second condition as the time taken purely by the second tap to complete filling the tank. There is some ambiguity in the question statement. The following edit addresses this.

A helpful comment by @JohnDouma mentioned that I might have misunderstood the second condition - that the $120$ minutes refers to the total time taken for both taps acting sequentially (and not in parallel). Taking this possibility into account, here is a more formal algebraic solution:
Let the rates of the first and second taps be $r_1$ and $r_2$, where the rates are measured in the convenience unit "tank per minute".
The first condition gives: $\frac 1{r_1 + r_2} = 60$, which can be rearranged to $r_1 + r_2 = \frac 1{60}$
The second condition gives $\frac{\frac 13}{r_1} + \frac{\frac 23}{r_2} = 120$
Solving the two simultaneously results in the quadratic $21600r_1^2 - 300r_1+1 = 0$, which has the solutions $r_1 = \frac {1}{120}$ or $r_1 = \frac{1}{180}$.
The first solution for $r_1$ gives $r_1 = r_2$, so this (each tap taking $120$ minutes on its own) should be dismissed since the problem explicitly disallows equal rates (however, if this was not disallowed, it is a viable solution, as pointed out by @paulinho).
So the correct solutions are $r_1 = \frac 1{180}, r_2 = \frac 1{90}$, which means that the first tap takes $180$ minutes on its own while the second takes $90$ minutes.
Note that, interestingly, this seems to be exactly the reverse of the solution with the alternative interpretation that I used at first.
