SOLUTION: Samuel started to walk to school at 6:00 A.M., his brother started to walk at a rate 2 kilometers per hour faster. He overtook him at 6:45 A.M. Find the rate at which each boy walk
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Question 324802: Samuel started to walk to school at 6:00 A.M., his brother started to walk at a rate 2 kilometers per hour faster. He overtook him at 6:45 A.M. Find the rate at which each boy walked.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Samuel started to walk to school at 6:00 A.M.,
his brother started to walk at a rate 2 kilometers per hour faster.
He overtook him at 6:45 A.M.
Find the rate at which each boy walked.
:
I don't think we have enough information to find a unique solution to this
:
From the given information we know Sam walked .75 hrs
Let's assume Bro left at 6:15, then his walking time is .5 hrs
:
we know they walked the same dist
:
Let r = Sam's rate
then
(r+2) = Bro's rate
:
Dist = rate * time
:
Sam's dist = Bro's dist
.75r = .5(r+2)
.75r = .5r + 1
.75r - .5r = 1
.25r = 1
mult both sides by 4
r = 4 km/h is Sam's rate
and
4 + 2 = 6 km/hr is Bro's rate
:
Another solution is possible if Bro leaves at 6:30
Then Bro's walking time is .25 hr
Sam's dist = Bro's dist
.75r = .25(r+2)
.75r = .25r + .5
.75r - .25r = .5
.5r = .5
r = 1 km/h is Sam's rat
and
1 + 2 = 3 km/hr is Bro's rate
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