Confusing Math Problem: Solving for the Length of a Ski Slope in 30 Minutes

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Jake ascends a ski slope at 5 miles per hour and descends at 25 miles per hour, completing the round trip in 30 minutes. The discussion revolves around determining the length of the slope using algebraic equations for time taken to ascend and descend. One participant suggests using different variables for the ascent and descent times to simplify the calculations. After some calculations, the proposed answer of 25/6 miles is questioned for accuracy. The conversation emphasizes the importance of clear variable usage in solving the problem.
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1. Jake rides a lift to the top of a maountain at a speed of 5 miles per hour. He then skis straight back down the slope to the bottom of the ski lift at an average speed five times faster than he went up the mountain. If the whole trip took 30 minutes, how long is the ski slope?

a. 1 mile
b. 15/7 miles
c. 17/6 miles
d. 25/6 miles
e. 5 miles


my attempt at a solution:

d=5t
d = 25t/5

From here on I tried plugging in the answers for the distance but they all seem to work. Then I found another way

at 5mph he's traveling at 1/12 miles per minute and at 25mph on the way down its 5/12 mpm. The I multiplied 1/12 by 5 to get 25/12 and did the same for 5/12 and got 25/12. I added up the distance and got 25/6 but I am not sure at all if its correct.

Im pretty confused with this one so any algebraic solution would really help. Thanks.
 
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hi nando94! :smile:
nando94 said:
d=5t
d = 25t/5

you'll never solve this if you use the same letter for two different things

use t and T (or t1 and t2), with t + T = 30 :wink:
 
haha yeah. I realized that later but I forgot to put that here. I was wondering if 25/6 was correct?
 
(just got up :zzz: …)

hmm …

let's see some algebra :wink:

if d is the length of the slope, what are the equations for t and T ? :smile:

(btw, no need to change to minutes, it's easier to use 30 min = 0.5 hour)
 
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