Maurice drove 400 km from Edmonton to Battleford in 1 hour less time than it took Martin to drive the same route from Battleford to Edmonton. If Maurice drove 20 km/h faster than Martin, at what speed did each of them drive?
Show a complete algebraic solution.
Let V = Maurice's speed and v = Martin's speed, measured in km/h.
V = v + 20
Maurice's time = 400/V = Martin's time - 1 = 400/v - 1
400/V = 400/v - 1
You now have two equations in two unknowns, V and v. Sove them by standard algebraic methods. One way would be to substitute v+20 for V in the second equation, and solve for the remaining unknown v.
To solve the equations, substitute V = v + 20 into the second equation:
400/(v + 20) = 400/v - 1
To eliminate the fractions, multiply both sides of the equation by v(v + 20):
400v = 400(v + 20) - v(v + 20)
Expand and simplify:
400v = 400v + 8000 - v^2 - 20v
Rearrange the equation to bring all terms to one side:
0 = -v^2 - 20v + 8000
Now, to solve this quadratic equation, factorize or use the quadratic formula. In this case, it is easier to use the quadratic formula:
v = (-b ± √(b^2 - 4ac))/(2a)
In this quadratic equation, a = -1, b = -20, and c = 8000. Substituting these values into the quadratic formula:
v = (20 ± √((-20)^2 - 4(-1)(8000)))/(2(-1))
Simplifying further:
v = (20 ± √(400 + 32000))/(-2)
v = (20 ± √32400)/(-2)
v = (20 ± 180)/(-2)
Now solve for the two possible values of v:
v = (20 + 180)/(-2) = 200/(-2) = -100
v = (20 - 180)/(-2) = -160/(-2) = 80
Since speed cannot be negative, we ignore the value v = -100.
Therefore, Martin's speed is 80 km/h.
To find Maurice's speed, substitute this value of v back into the equation V = v + 20:
V = 80 + 20 = 100
Therefore, Maurice's speed is 100 km/h.