1. Cate went on a trip of 120 miles at x miles per hour. Returned over same route at 5miles/hour faster than previous rate.If time for return trip was one-third of an hour less than time for the outgoing trip,how do you find x?

Question

(Set up the chart right, and used the time column. But why is the answer +one-third and not -one-third? I am confused)

2. A motor boat traveling at 18miles/hour traveled the length of a lake in one quarter of an hour less time than it took when traveling at 12 miles/hour. What was the length in miles of the lake?
(I got this question, but I had to try again since I did not set up the time column correctly. Why is it x-(1/4) and not (3/4)x?)

Thank you!

Answer 1

120 = x t so t = 120/x

120 = (x+5)(t-1/3)
or
120 = (x+5)(120/x - 1/3)

120 (3)(x) = (x+5)(360 - x)

360 x = 360 x -x^2 + 1800 -5 x

x^2 + 5 x - 1800 = 0
(x-40)(x+45) = 0
x = 40

Answer 2

rate at first part --- x mph
rate for return --- x + 5 mph
time on first part = 120/x
time on 2nd part = 120/(x+5)
(these should be entries in your chart)

now it says:
time on return = (1/3) time of first part
120/(x+5) = (1/3)(120/x)
divide by 120
1/(x+5) = 1/(3x)
3x = x+5
2x = 5
x = 2.5

2. depends what you define as x

I will let the length of the lake be x miles
If you went at 18 mph , time would be x/18
If you went at 12 mph, time wuld be x/12

It said:
difference in the times = 1/4 hour

x/12 - x/18 = 1/4
LCD is 36 , so multiply each term by 36
3x - 2x = 9
x = 9 miles

what was your definition for x ?

Answer 3

arggh, go with Damon on the first question,
I read it as 1/3 of the time, not 1/3 hour less

Answer 4

at 12 it tskes t
so
d = 12 t
at 15 it will take .25 leas time
d = 15 (t-.25)
so
12 t = 15 t - 15/4

3 t = 15/4
t = 5/4
d = 12(5/4) = 15 miles

Answer 5

sorry - I used 15 mph instead of 18