College Algebra

Question

One car goes 8 km/h faster than another. In the time that it takes the faster car to go 238 km, the slower car will go 210 km. What is the speed of the faster car?

Answer 1

Hello, thank you for taking the time to post your question!

The key on this type of question is to find a way to relate the variables to one another algebraically based on the information given in the question. On this one we have

238 / F = 210 / S     since in the same amount of time the faster car goes 238 the slower car goes 219

F = S + 8  since the faster car goes 8 km/h faster than the slower car

From there then I would plug the second equation into the first equation and then solve for the value of S

238 / (S + 8) = 210 / S

238S = 210(S + 8)

238S = 210S + 1680

28S = 1680

S = 1680/28 = 60, meaning that the speed of the slower car is 60 km/h

Then you can go back to the second equation to find the speed of the faster car

F = 60 + 8 = 68, meaning that the speed of the faster car is 68 km/h

I hope that helps you get moving in a better direction on this type of question! Feel free to reach out if you have any additional questions beyond that :)

Answer 2

d= rt

distance = rate of speed times time traveled

first car goes r+8 times t = 238

(r+8)t = 238

2nd car, the slower one, goes r times t = 210

rt = 210

t = 210/r

substitute that into the last equation for the 1st car

(r+8)210/r = 238

multiply by r

(r+8)210 = 238r

210r + 1680 = 238r^2

238r^2 - 210r - 1680 = 0

119r^2 - 105r - 840 =0

either factor, complete the square or use the quadratic formula

r= 105/238 + or - (1/238)sqr(105^2 + 4(119)(840)

ignore a square root that gives a negative speed