- Two taps fill a tank in 6 hours. Alone, one of them spends 5 hours more than the other. Determine the time it takes one of them to fill this tank in isolation.
**Solution**:

Consider x the time taken for the 1st tap to fill the tank and x + 5 the time taken for the 2nd tap to fill the tank.

Within an hour, each tap fills the following fraction of the tank:

In an hour, the two taps together will fill of the tank; look at the corresponding equation:

Solving it, we have:

6 (x + 5) + 6x = x (x + 5)

6x + 30 + 6x = x^{2 }+ 5x

x^{2}- 7x - 30 = 0

x '= - 3 and x "= 10

Since the negative root is not used, we will have x = 10 as a solution.**Answer:**The first tap fills the tank in 10 hours and the second tap in 15 hours.

- At a celebration dinner, a prize of R $ 24,000.00 would be distributed equally among the guests. As 5 people were missing, each of the gifts received an increase of $ 400.00 in your prize. How many people were present at this dinner?
**Solution**:

We can represent by:

Solving it:**Answer:**At this dinner 20 people should be present. As there were 5, so 15 people were present at dinner.