# 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(15746)   (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