SOLUTION: First tourist walked a distance of 30 km in a time 10 minutes shorter than the second tourist. Calculate their speeds if the speed of the first one was 3 km/h greater than that of

Question 164547: First tourist walked a distance of 30 km in a time 10 minutes shorter than the second tourist. Calculate their speeds if the speed of the first one was 3 km/h greater than that of the second one.
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
First tourist walked a distance of 30 km in a time 10 minutes shorter than the second tourist. Calculate their speeds if the speed of the first one was 3 km/h greater than that of the second one.
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Since the problem is dealing in both hours and minutes, we'll just keep it simple and deal in hours
Let t=time it took second tourist to walk 30km
Then t-(1/6)=time it took first tourist to walk 30 km
Rate (speed) of second tourist=30/t
Rate of first tourist=30/(t-(1/6)) and we are told that this is 3 km/hr greater than the speed of the second tourist, so our equation to solve is
(30/(t-(1/6)))-3=30/t multiply each term by t(t-(1/6))
30t-3t(t-(1/6))=30(t-(1/6)) get rid of parens (distributive law)
30t-3t^2+0.5t=30t-5 subtract 30t from and also add 5 to both sides
30t-30t-3t^2+0.5t+5=30t-30t-5+5 collect like terms
-3t^2+0.5t+5=0 multiply each term by -1
3t^2-0.5t-5=0 Quadratic in standard form; solve using the quadratic formula

hrs----time it took second tourist
hrs-----time it took first tourist
rate of second tourist=30/1.377=21.8 km/hr--------------------ans
rate of first tourist=30/1.21=24.8 km/hr------------------------ans
--------------disregard the negative value for t
Hope this helps----ptaylor

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