A driver covers the first 70% of a distance at an average speed of 67 mph. At what speed must she drive the remainder of the distance so as to average 55 mph for the entire trip?
-
Let D = the length of the total trip.
Let r be the rate of speed for the remaining 30% of the trip.
D*55 = ((7D/10) * 67) + ((3D/10) * r).............Divide both sides by D
55 = 46.9 + 3r/10....................Multiply both sides by 10
550 = 469 + 3r
3r = 550 - 469 = 81
Divide both sides by 3.
r = 27 <----------- Answer (rate of speed for last 30%)
Let's check:
(0.7 * 67) + (0.3 * 27) = 46.9 + 8.1 = 55
The check is valid and the answer is correct.
.
Let r be the rate of speed for the remaining 30% of the trip.
D*55 = ((7D/10) * 67) + ((3D/10) * r).............Divide both sides by D
55 = 46.9 + 3r/10....................Multiply both sides by 10
550 = 469 + 3r
3r = 550 - 469 = 81
Divide both sides by 3.
r = 27 <----------- Answer (rate of speed for last 30%)
Let's check:
(0.7 * 67) + (0.3 * 27) = 46.9 + 8.1 = 55
The check is valid and the answer is correct.
.