Madelaine88 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?
A. 20
B. 30
C. 40
D. 50
E. 80
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 - 2NOTE: At this point, we can either test the five answer choices to see which one satisfies the above equation, or we can solve the equation algebraically. Although testing the answer choices is probably faster, let's solve the equation algebraically
Multiply 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: C
Cheers,
Brent