SOLUTION: a man drove at a constant speed from his city 180 miles away. On the return trip, he increased his constant speed by 5 mph and made the return trip in 24 minutes less time than had

Question 853754: a man drove at a constant speed from his city 180 miles away. On the return trip, he increased his constant speed by 5 mph and made the return trip in 24 minutes less time than had been required to drive to the city. What was his sped from home to the city?
Found 2 solutions by mananth, lwsshak3:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
Time taken in forward journey = 180/x
x is the speed
return journey speed increased by 5 mph. (x+5)
Time forward - time return = 24/60 = 2/5 hours

Take LCM of denominator
5x(x+5)
180*5*(x+5) -180*5*x=2*x*(x+5)
900x+4500-900x=2x^2+10x
2x^2+10x-4500=0
/2
x^2+5x-2250=0
x^2+50x-45x-2250=0
x(x+50)-45(x+50)=0
(x+50)(x-45)=0
x= -50 OR 45
Speed from home was 45 mph

Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
a man drove at a constant speed from his city 180 miles away. On the return trip, he increased his constant speed by 5 mph and made the return trip in 24 minutes less time than had been required to drive to the city. What was his sped from home to the city?
***
let x=speed from home to city
x+5=speed on return trip
travel time=distance/speed
..
24 min=24/60=2/5 hr

LCD:5x(x+5)
5(180x+900)-900x=2x(x+5)
900x+4500-900x=2x^2+10x
2x^2+10x-4500=0
x^2+5x-2250=0
(x+50)(x-45)=0
x=-50 (reject)
or
x=45
speed from home to city=45 mph
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