Two cars left the city for a suburb, 480 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 2 hour earlier than the other car. Find the speeds of both cars.

The speed of both the cars are 80km/h and 60km/h

Step-by-step explanation:

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

80 and 60

Step-by-step explanation:

480/t = 480/t+2   +20

t=6

480/6=80

80-20=60

1 x=22.52

2: 65/39=604+39x

65x-39x=604

26x=604

x=604/26

×=23.23