# Algebraic word problem [closed]

A woman walked five hours (total), first along a level road, then up a hill, then she turned around and walked back to her starting point along the same route. She walks 4 miles per hour on the level, 3 uphill, and 6 downhill. (i) Find the total distance she walked. (ii) Can this problem always be solved regardless of the speeds she walks, or were the numbers 4, 3, 6 a “lucky choice”?

Im m not sure how to go about starting this problem. Can anyone help me start it and explain the reasoning behind it? Any help would be appreciated! Thank you!

## closed as off-topic by Travis, graydad, Empty, user91500, user99914 Oct 30 '15 at 9:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Travis, graydad, Empty, user91500, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

• Level distance plus non-level distance equals 10 miles so, since she walked the two legs one direction and then back the opposite direction, she walked 20 miles. – Roy Dec 31 '15 at 6:52
• (i) Level distance plus non-level distance equals 10 miles so, since she walked the two legs one direction and then back the opposite direction, she walked 20 miles. On the level ground she takes x/4 hours twice and on the non-level ground, she takes y/3 hours up and y/6 hours down. Add those four and they equal 5 hours. The resulting equation is x + y = 10 miles. 10 miles one way and 10 miles back equals 20 miles. (ii) Given values are immaterial. For speeds a, b, c, one gets 2bcx + aby + acy = 5abc. After substituting whatever speeds one wants one has mx + ny = 5k, equivalent to the above. – Roy Dec 31 '15 at 7:02

This one is interesting. We have four speeds to worry about.

First level: 4 mph

Uphill: 3 mph

Downhill: 6 mph

Second level: 4 mph.

The average speed on the levels is 4mph. If you don't believe this, you can use the table method below with both levels. To find the average speed on the hills, let's make a table.

Notice that the uphill and downhill distances are the same, so we can set $3t_1 = 6t_2$. Solving for $t_1$ gives us $t_1 = 2t_2$.

Now, the average rate of the uphill and downhill speeds will be:

\begin{align} r & = \frac{d}{t} \\ & = \frac{\mbox{sum of hill distances}}{\mbox{sum of hill times}} \\ & = \frac{3t_1 + 6t_2}{t_1 + t_2} \\ & = \frac{3(2t_2) + 6t_2}{2t_2 + t_2} \\ & = \frac{12t^2}{3t^2} \\ & = 4 \end{align}

It turns out the the total average hill speed will be 4 mph. This means the average speed of the entire trip will be 4 mph. So... if this woman walks for 5 hours, how many miles will she travel?

For the second part, notice that it was pretty nice to end up with an average hill rate that was the same as the level rate. This allowed us to find an average rate for the entire trip and calculate the distance. What would happen if the average hill rate gave us something other than 4 mph? Try picking another pair of rates for uphill and downhill, say 2 uphill and 7 downhill. Find the average hill rate using the table method we did above. Would you be able to calculate the total distance?

• Hi @Alti. If you found this answer useful, I would really appreciate a "check" next to it. If not, is there more explanation or information you need? Thanks! – josh Oct 30 '15 at 13:51