Algebra Distance Problem?
Here it is:
Every weekend you ride your bike on a forest preserve path. The path is 20 miles long and ends at a waterfall, at which point you relax and then make the trip back to the starting point. One weekend, you find that in the same time it takes you to travel to the waterfall,you are only able to return 10 miles. Your average speed going to the waterfall is 5 miles per hour faster than the return trip. What was your average speed going to the waterfall?
In calculating this, my answer came out to be 10 miles per hour. Is that correct?
Comments
who cares? that waterfall is soooo relaxing. i am soooooo sleepy...
if we must...let's see...dist = rate * time and...
let x = rate on return trip. then the whole problem becomes:
10 miles = x * t and
20 miles = (x+5)*t let's solve for t, we're not doing anything else...
10 miles/x mph = t and
20 miles/(x+5)mph = t set them equal to each other gives:
10/x = 20/(x+5) and solve for x
10(x+5) = 20x and
10x + 50 = 20x variables on one side, constants on the other (how many times have you heard that?)
10x = 50 and
x = 5mph (since x is a rate)
Ans: the average speed going to the waterfall is 10mph (i.e. 5+5mph)
oh look...a bunny rabbit...
rt=d
to: r=f+5
(f+5)t=20
ft=10
f=10/t
(10/t +5)t=20
10 +5t=20
5t=10
t=2
(f+5)t=20
(f+5)2=20
f+5=10
f=5
f+5 = 10
Yes, I think 10 is the answer