Question
Solve the problem. A man rode a bicycle for 12 miles and then hiked an additional 8 miles. The total time for the trip was 5 hours. If his rate when he was riding a bicycle was 10 miles per hour faster than his rate walking, what was each rate?
Answers
let walking rate be x mph
let biking rate be x+10
time for walking = 12/x
time biking = 8/(x+10)
isn't 12/x + 8(x+10) = 5 ?
multiply each term by x(x+10)
you will get quadratic, solve for x
let biking rate be x+10
time for walking = 12/x
time biking = 8/(x+10)
isn't 12/x + 8(x+10) = 5 ?
multiply each term by x(x+10)
you will get quadratic, solve for x
Let the walking speed be v and the bike speed V = v + 10.
12/(v+10) + 8/v = 5 is the sum of the times spent bidking and walking.
Solve that equation for v. You will have to start with a common denominator.
[12v + 8(v+10)]/[v(v+10)] = 5
5 v^2 + 50 v = 20 v + 80
v^2 + 6v - 16 = 0
(v + 8)(v -2) = 0
Choose the positive root, v = 2 miles/hr
The bike speed was then v + 10 = 12 mph.
12/(v+10) + 8/v = 5 is the sum of the times spent bidking and walking.
Solve that equation for v. You will have to start with a common denominator.
[12v + 8(v+10)]/[v(v+10)] = 5
5 v^2 + 50 v = 20 v + 80
v^2 + 6v - 16 = 0
(v + 8)(v -2) = 0
Choose the positive root, v = 2 miles/hr
The bike speed was then v + 10 = 12 mph.
go with drwls solution, I mixed up the two rates.
Solve. V^2-6v-16=0
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