Bunuel
A bus trip of 450 miles would have taken 1 hour less if the average speed S for the trip had been greater by 5 miles per hour. What was the average speed S, in miles per hour, for the trip?
(A) 10
(B) 40
(C) 45
(D) 50
(E) 55
Kudos for a correct solution.
Let's start with a
word equation:
travel time at actual speed =
travel time at faster speed + 1 hour
In other words:
travel time at S mph =
travel time at (S + 5) mph + 1 hour
travel time = distance/speedSo, we get:
450/S=
450/(S + 5) + 1
Multiply both sides by S to get: 450 = 450S/(S+5) + S
Multiply both sides by S+5 to get: 450(S + 5) = 450S + S(S+5)
Expand: 450S + 2250 = 450S + S² + 5S
Subtract 450S from both sides: 2250 = S² + 5S
Rewrite as: S² + 5S - 2250 = 0
Factor: (S + 50)(S - 45) = 0
So, EITHER S = 50, OR S = 45
Since the speed can't be negative, the correct answer must be S = 45
Answer: C
Cheers,
Brent
Hello, why do we have to add 1 in the first equation? Shouldn't we subtract? Isn't the same distance reached while traveling 1 hour less, but 5 mph faster?