SOLUTION: Jessica drove to her cabin on the lake and back. it took one hour longer to go there then it did to come back. The average speed on the trip was 50 km/h. The average speed on the w

Algebra ->  Rate-of-work-word-problems -> SOLUTION: Jessica drove to her cabin on the lake and back. it took one hour longer to go there then it did to come back. The average speed on the trip was 50 km/h. The average speed on the w      Log On


   



Question 1044407: Jessica drove to her cabin on the lake and back. it took one hour longer to go there then it did to come back. The average speed on the trip was 50 km/h. The average speed on the way back was 75 km/h. how many hours to the trip there take?
Found 3 solutions by josgarithmetic, MathTherapy, ikleyn:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
                  SPEED        TIME        DISTANCE

Going there         r           t+1          d

Coming back         75           t           d

Total                         (t+1)+t        2d

AVERAGE            50

Three unknown variables. Basic rule RT=D for travel rates.

system%28r%28t%2B1%29=d%2C75t=d%2C2d%2F%28t%2B%28t%2B1%29%29=50%29

You can solve for all three variables, but the question asks for solution and value of %28t%2B1%29%2Bt, the time for the whole trip , as round-trip.

Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!
Jessica drove to her cabin on the lake and back. it took one hour longer to go there then it did to come back. The average speed on the trip was 50 km/h. The average speed on the way back was 75 km/h. how many hours to the trip there take?
This is done based on the assumption that the speed to the cabin was 50 km/h
Let time taken to get to the cabin be T
Then time taken on return trip = T – 1
We then get the following DISTANCE equation: 50T = 75(T – 1)
50T = 75T – 75
50T – 75T = - 75
– 25T = - 75
T, or time taken to get to the cabin = %28-+75%29%2F%28-+25%29, or highlight_green%28matrix%281%2C2%2C+3%2C+hours%29%29

Answer by ikleyn(52810) About Me  (Show Source):
You can put this solution on YOUR website!
.
Jessica drove to her cabin on the lake and back. it took one hour longer to go there then it did to come back.
The average speed on the trip was 50 km/h. The average speed on the way back was 75 km/h. how many hours to the trip there take?
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Let D be the distance from Jessica home to her cabin on the lake.
Let x be the Jessica' speed on the way from home to the cabin.

Then the condition says

2D%2F50 = D%2Fx+%2B+D%2F75.   (1)  

The left side is the total time for the trip from home to the cabin and back. 
Surely, it is the sum of the time in one direction (first addend) and in the opposite direction (second addend).

The condition also says

D%2Fx+-+D%2F75 = 1.      (2)   ("it took one hour longer to go there then it did to come back. ")

In the first equation we can cancel D in both side. Then you get

2%2F50 = 1%2Fx+%2B+1%2F75,  or  1%2Fx = 2%2F50+-+1%2F75 = 6%2F150+-+2%2F150 = 4%2F150 = 2%2F75.

Hence, x = 75%2F2 = 37.5 km%2Fh is Jessica' speed on the way from home to the cabin.

Now, substitute this value of x into equation (2). You will get

2D%2F75+-+D%2F75 = 1,  or  D%2F75 = 1,  or  D= 75 km is the distance in one (in each) direction.

Having this, you can estimate the time from home to the cabin. It is

D%2Fx = 75%2F37.5 = 2 hours.

The time from the cabin to home is 75%2F75 = 1 hour.

Check. Let us check the average speed for the entire trip.


       It is %2875+%2B+75%29%2F%282%2B1%29 = 150%2F3 = 50 km%2Fh.

Teaching students to solve Travel and Distance problems using tables is the same
as teaching the healthy children to walk using crutches.

Same for using tables to solve any other word problems.

It is my personal view.