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 Maurice's speed be V1
Let Martin's speed be V2
V1 = V2 + 20
elapsed times:
Morris' time = Martin's time + 1
400/V2 = 400/V1 + 1
Susbtitute for V1:
400/V2 = 400/(V2 + 20) + 1
Solve for V2; then V1
so I got up to:
V2 = (V2 + 20) + 1
right? because 400 cancels out?!?!?
I'm confused.
No, you made a mistake when you simplified the equation. Let's go through the steps again to find the correct solution:
1. Let's start with the given information:
- Maurice's speed: V1
- Martin's speed: V2
- Maurice drives 20 km/h faster than Martin: V1 = V2 + 20
- Maurice's time is 1 hour less than Martin's time: 400/V2 = 400/V1 + 1
2. Substitute V1 in terms of V2 into the time equation:
400/V2 = 400/(V2 + 20) + 1
3. To solve for V2, we need to clear the denominators by multiplying both sides of the equation by V2 and (V2 + 20):
V2 * (V2 + 20) * (400/V2) = V2 * (V2 + 20) * (400/(V2 + 20)) + V2 * (V2 + 20)
4. Simplify the equation:
400 * (V2 + 20) = 400 * V2 + V2 * (V2 + 20)
5. Distribute on both sides of the equation:
400V2 + 8000 = 400V2 + V2^2 + 20V2
6. Simplify further and rearrange:
V2^2 + 20V2 - 8000 = 0
7. We now have a quadratic equation. You can solve it by factoring, completing the square, or using the quadratic formula. In this case, let's use factoring:
(V2 - 80)(V2 +100) = 0
8. Set each factor equal to zero and solve for V2:
V2 - 80 = 0 or V2 + 100 = 0
V2 = 80 or V2 = -100
9. Since V2 represents the speed, we can ignore the negative solution. Therefore, V2 = 80 km/h.
10. Substitute V2 back into the equation V1 = V2 + 20:
V1 = 80 + 20 = 100 km/h
So, Maurice drove at a speed of 100 km/h, and Martin drove at a speed of 80 km/h.