Question

A boy started on a trip across a lake by motorboat. After he had traveled 15km, the motor failed and he had to use a banca for the remaining 6 km to his destination. His average speed by motor was 4kph faster than his average speed while rowing. If the entire trip took 5 1/2 hours, what was his average speed while rowing?

Answer 1

Total distance = 21 km
Let x = rowing speed in kph
total trip time = 5.5 hours

15km/(x+4)kph + 6km/x kph = 5.5 hours

Multiply both sides by x(x+4), where x>0,
15x+6(x+4)=5.5x(x+4)

Solve for x and reject the negative root to get x=2 kph.

Answer 2

first, recall that speed, v, is distance traveled over a certain period of time:
v=d/t
thus t=d/v *we'll need this equation later*

represent the unknowns using variables:
let x = average speed of motorboat (in kph)
let y = average speed of rowing (in kph)
set-up the equations:
(1) x = y + 4
(2) (15/x) + (6/y) = 5.5 *this is t=d/v*

substitute equation (1) to (2):
15/(y+4) + 6/y = 5.5 *simplify.
15y + 6(y+4) = 5.5(y)(y+4) *multiply by 2 to make 5.5 whole number
30y + 12y + 48 = 11(y^2 + 4y)
42y + 48 = 11y^2 + 44y
11y^2 + 2y - 48=0
(11y+24)(y-2)=0
y = 2 kph *average speed of rowing

*note: the other root, y=-24/11 is extraneous since speed cannot be negative.

so there,, =)