SOLUTION: Jessie recently drove to visit her parents who live 552 miles away. On her way there her average speed was 14 miles per hour faster than on her way home (she ran into some bad weat
Algebra ->
Percentage-and-ratio-word-problems
-> SOLUTION: Jessie recently drove to visit her parents who live 552 miles away. On her way there her average speed was 14 miles per hour faster than on her way home (she ran into some bad weat
Log On
Question 1157117: Jessie recently drove to visit her parents who live 552 miles away. On her way there her average speed was 14 miles per hour faster than on her way home (she ran into some bad weather). If Jessie spent a total of 23 hours driving, find the two rates. Found 2 solutions by ikleyn, MathTherapy:Answer by ikleyn(52851) (Show Source):
Let x be the average rate moving to parents (in miles per hour).
Then the average rate moving back is x-14 mph.
is the time going to.
is the time going back.
The time equation is
+ = 23 hours.
Cancel the factor 23 in both sides
+ = 1.
Reduce to the standard form quadratic equation and solve it.
Then evaluate x-14.
You can put this solution on YOUR website! Jessie recently drove to visit her parents who live 552 miles away. On her way there her average speed was 14 miles per hour faster than on her way home (she ran into some bad weather). If Jessie spent a total of 23 hours driving, find the two rates.
Let outbound-speed be S
Then return-speed = S - 14
We then get the following TIME equation: ------ Dividing by numerator-GCF, 23
24(S - 14) + 24S = S(S - 14) ------ Multiplying by LCD, S(S - 14)
(S - 56)(S - 6) = 0
S - 56 = 0 OR S - 6 = 0
Outbound-speed = OR S = 6 (ignore)
Do you think you can find the return-speed?