SOLUTION: on a trip, palmer s. friedrich drove at a rate of 55 mph. he desided to return by a route that was 5 miles longer. his rate returning was 45 mph. if it took him 1 hour longer retur
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Question 99078: on a trip, palmer s. friedrich drove at a rate of 55 mph. he desided to return by a route that was 5 miles longer. his rate returning was 45 mph. if it took him 1 hour longer returning, how long was the second route?
You can put this solution on YOUR website! s=speed t=time d=distance
s*t=d
Going: d=d s=55 t=h
Returning: d=d+5 s=45 t=h+1
Distance going = distance returning -5. Why -5 because it wouldn't be an equation if we didn't make both sides equal. We could have added 5 to the left instead of subtracting 5 from the right, the answer would have been the same.
55h=45(h+1)-5
55h=45h+45-5
55h=45h+40
subtract 45h from each side: 10h=40
divide 10 into each side: h=4 hours going and 5 hrs returning.
45*5=225 mi (distance of the 2nd route) Ans.
Check:
55*4=220 mi (distance of the 1st route)
225-220=5 mi longer for the second route.
Ed
