Two cars left the city for a suburb, 120 km away, at the same time. The speed of one of the cars was 20 km/hour greater than the speed of the other, and that is why it arrived at the suburb 1 hour before the other car. What is the speed of both cars.?

1 Answer
Apr 7, 2017

Slower car: 40"km/h"40km/h
Faster car: 60"km/h"60km/h

Explanation:

Let x=x=hours it took the slower car to reach the suburb
The time it took the faster car to reach the suburb is x-1x1

Since distancediv÷time==speed,
The speed of the slower car is 120"km"div(x)"h"=120/x"km/h"120km÷(x)h=120xkm/h
The speed of the faster car is 120"km"div(x+1)"h"=120/(x-1)"km/h"120km÷(x+1)h=120x1km/h

Since the faster car travelled 20"km/h"20km/h faster than the slower car,
120/(x-1)=120/x+20120x1=120x+20

Now multiply both sides by (x-1)(x1) and then xx to start solving.
(120(x-1))/(x-1)=(120(x-1))/x+20(x-1)120(x1)x1=120(x1)x+20(x1)
(120x(x-1))/(x-1)=(120x(x-1))/x+20x(x-1)120x(x1)x1=120x(x1)x+20x(x1)

Simplify, distribute, factor, disregard, etc. Essentially, solve.
(120xcolor(red)cancelcolor(black)(x-1))/color(red)cancelcolor(black)(x-1)=(120color(red)cancelcolor(black)x(x-1))/color(red)cancelcolor(black)x+20x(x-1)
color(red)cancelcolor(black)(120x)=color(red)cancelcolor(black)(120x)-120+20x^2-20x
0=20x^2-20x-120
0=x^2-x-6
0=x^2+2x-3x-6
0=x(x+2)-3(x+2)
0=(x+2)(x-3)
x+2=0orx-3=0
x="-"2orx=3

Since x is the number of hours for the slower car to reach the suburb, a negative solution can be disregarded as extraneous as negative time does not (yet?) exist. x=3

We know that the speed of the slower car is 120/x"km/h" and that of the faster car is 120/(x-1)"km/h" so plugging in x, we get:
Slower car: 120/((3))=120/3=40"km/h"
Faster car: 120/((3)-1)=120/2=60"km/h"