Question 999970: 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”?
Answer by addingup(3677)   (Show Source):
You can put this solution on YOUR website! The woman walked up a hill, turned around and walked down the hill. She walked as far uphill as she did downhill (same hill, no?)
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At 3mph, TU (Time Up) using the distance (d) formula:
TU= d/3
And at 6mph the time it took her to come down the hill (TD) is:
TD= d/6
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So we have 2d: d/3 and d/6. Now we can find the rate
2d= [(d/3 + (d/6)]r
2d/[(d/3 + (d/6)]= r
2/[(1/3 + (1/6)]= r
12/3= 4= r
OK, so the woman's average rate up and down the hill was 4mph.
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The woman goes up and down the hill at an average of 4mph and goes on the level road at 4mph. OK, see what we discovered? Since her up/down average is 4mph and on the flat surface is also 4mph, she covers as much ground up and down the hill as she does on the level road.
Except that she spends 2/3 of her time on level ground and 1/3 of her time on the hill. Using d for distance, let's write an equation.
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d= (2/3)(5 hours)(4 miles/hour) + (1/3)(5 hours)(4 miles/hour)
d = 40/3 + 20/3
d = 60/3
or
d = 20 miles
The woman walked a total of 20 miles.
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