Carcass wrote:
The time it took car A to travel 400 miles was 2 hours less than the time it took car B to travel the same distance. If car A's average speed was 10 miles per hour greater than that of car B, what was car B's average speed in miles per hour?
Car A's average speed was 10 miles per hour greater than that of car BLet
x = Car B's average speed
So,
x + 10 = Car A's average speed
The time it took car A to travel 400 miles was 2 hours less than the time it took car B to travel the same distanceIn other words:
(car A's travel time) = (car B's travel time) - 2time = distance/speedSo, we get:
400/(x + 10)) = 400/x - 2Multiply both sides of the equation by (x + 10) to get: 400 = 400(x + 10)/x - (x + 10)(2)
Multiply both sides of the equation by x to get: 400x = 400(x + 10) - (x)(x + 10)(2)
Expand: 400x = 400x + 4000 - 2x² - 20x
Subtract 400x from both sides to get: 0 = 4000 - 2x² - 20x
Multiply both sides by -1 to get: 0 = -4000 + 2x² + 20x
Rewrite as follows: 2x² + 20x - 4000 = 0
Divide both sides by 2 to get: x² + 10x - 2000 = 0
Factor: (x + 50)(x - 40) = 0
So, EITHER x = -50 OR x = 40
Since the speed can't be negative, it must be the case that x = 40
Answer: 40
Cheers,
Brent