The speed of both the cars are 80km/h and 60km/h
<u>Step-by-step explanation:</u>
Let the speed of one car be 'a' and the speed of other car be 'b'.
The total distance (d) = 480km
It is given that the speed of one car is 20km/h faster than the other.
We can write,
a = b+20
The slower car takes 2 hrs more to reach the suburb than the other car.
Let the time taken by the fastest car be t
Speed = distance/time
So,
a = 480/ t
b = 480/(t+2)
We got the values of a and b.
a = b+20
480/t = (480/(t+2)) + 20
Taking LCM on the right side.
480/t = (480 + 20t + 40) / (t+2)
480(t+2) = (480+20t+40) t
480t + 960 = 480t + 20t(t) + 40t
20t(t) + 40t - 960 = 0
Divide the whole equation by 20 to simplify the equation.
t(t) + 2t - 48 = 0
Solve the quadratic equation by splitting the middle terms.
t(t) + 8t - 6t - 48 = 0
t(t + 8) - 6(t + 8) = 0
(t- 6) (t+8) =0
t = 6 (or) -8
t is time and cannot be negative. So t= 6hrs
a = 480/t
a = 480/ 6 = 80
the speed of the fastest car is 80km/hr
a = b + 20
b = a - 20
b = 80 - 20
b = 60km/h
