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?

(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!

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

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 ?

arggh, go with Damon on the first question,

I read it as 1/3 of the time, not 1/3 hour less

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

sorry - I used 15 mph instead of 18

1. To find x in the first question, you can set up a chart with the following information:

Trip Speed (mph) Time (hours)
Outgoing trip x t
Return trip x + 5 t - 1/3

According to the problem, the distance traveled on the outgoing trip is 120 miles, so you can write the equation:

Distance = Speed × Time
120 = x × t

For the return trip, the time is one-third of an hour less than the outgoing trip, which means t - 1/3. Since the return speed is 5 miles per hour faster, you can write the equation:

Distance = Speed × Time
120 = (x + 5) × (t - 1/3)

Now you have a system of equations:

120 = x × t
120 = (x + 5) × (t - 1/3)

To solve this system, you can substitute the first equation into the second equation:

120 = (x + 5) × (x × t - 1/3)

Simplifying this equation will give you a quadratic equation in terms of x. Solving this quadratic equation will give you the value of x.

Regarding the confusion between +one-third and -one-third, it depends on how you set up your variables and equations. In this case, the problem states that the return trip took one-third of an hour less time than the outgoing trip. This means you subtract one-third instead of adding it. If you were to set up your variables and equations differently, the signs might change accordingly.

2. To find the length of the lake in the second question, you can also set up a chart with the following information:

Speed (mph) Time (hours) Distance (miles)
12 x d
18 x - 1/4 d

According to the problem, the time it took to travel the length of the lake at 12 miles per hour is greater by one quarter (1/4) of an hour compared to the time it took at 18 miles per hour.

So you can set up an equation using the given information:

12 × (x + 1/4) = 18 × x

Simplifying and solving this equation will give you the value of x, which represents the time it took in hours to travel the length of the lake at 12 miles per hour.

Once you have the value of x, you can find the length of the lake by substituting it into the equation for either scenario:

Length = Speed × Time
Length = 12 × (x + 1/4)

Again, the x - 1/4 represents the time it took for the slower speed, which is one quarter (1/4) less than the time for the faster speed.

By calculating the length of the lake using this equation, you will find the answer you're looking for.