### Math Notes

Subjects

#### Algebra Solutions

##### Topics || Problems

It takes a motorboat $$1\frac{1}{3}$$ hours to go 20 miles downstream and $$2\frac{2}{9}$$ hours to return. Find the rate of the current and rate of the boat in still water. Solution:

$$V_d = V_b + V_c$$

$$\frac{distance}{time} = V_b + V_c$$

$$\frac{20}{1\frac{1}{3}} = V_b + V_c$$

$$15 = V_b + V_c$$

$$V_b = 15 - V_c$$

$$V_u = V_b - V_c$$

$$\frac{distance}{time} = V_b - V_c$$

$$\frac{20}{2\frac{2}{9}} = V_b - V_c$$

$$V_b = 9 + V_c$$

$$15-V_c = 9 +V_c$$

$$V_c = 3$$

$$V_b = 9 + 3 = 12$$

Thus the rate of the boat on still water is 12 mph and the current is 3 mph