Question 1123723: A pair of hikers, 18 miles apart, begin at the same time to hike toward each pther. , and if they meet 2 hours later, how fast is each one walking?
Found 3 solutions by Theo, Alex.33, josgarithmetic:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! rate * time = distance.
for the first hiker:
time = 2
distance = d
rate = x
for the second hiker:
time = 2
distance = 18 - d
rate = x + 1
rate * time = distance for the first hiker becomes 2x = d
rate * time = distance for the second hiker becomes 2(x + 1) = 18 - d
since 2x = d from the first equation, replace d in the second equation with 2x to get:
2(x + 1) = 18 - 2x
simplify to get 2x + 2 = 18 - 2x
subtract 2 from both sides of this equation and add 2x to both sides of this equation to get 4x = 16
solve for x to get x = 16/4 = 4
the first hiker travels at 4 miles per hour.
the second hiker travels at 5 miles per hour.
in 2 hours, the first hiker has traveled 8 miles.
in 2 hours, the second hiker has traveled 10 miles.
they meet at the 8 mile mark from the first hiker, which is the same as the 10 mile mark from the second hiker.
the question was how fast is each walking.
the first hiker walks at 4 miles per hour.
the second hiker walks at 5 miles per hour.
that's your solution.
Answer by Alex.33(110)  (Show Source):
You can put this solution on YOUR website!
we assume the faster one's speed to be v1. The other's will be v2.
we have 2v1+2v2=18, v1=v2+1
Solve for v1, v2 and you're done.
Answer by josgarithmetic(39620)  (Show Source):
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