SOLUTION: Mary and Richard leave Edmonton to go to Jasper, a distance of 400 km. Mary drives at an average speed of 15 km/h faster than Richard. If Mary arrives in Jasper 40 minutes before

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Mary and Richard leave Edmonton to go to Jasper, a distance of 400 km. Mary drives at an average speed of 15 km/h faster than Richard. If Mary arrives in Jasper 40 minutes before      Log On


   

Question 1030814: Mary and Richard leave Edmonton to go to Jasper, a distance of 400 km. Mary drives at an average speed of 15 km/h faster than Richard. If Mary arrives in Jasper 40 minutes before Richard, find Richard's average speed, to the nearest km/h.
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
Mary and Richard leave Edmonton to go to Jasper, a distance of 400 km. Mary drives at an average speed of 15 km/h faster than Richard. If Mary arrives in Jasper 40 minutes before Richard, find Richard's average speed, to the nearest km/h.

Let Richard,s speed be x
Mary's speed = x+15
Time taken by Richard - time taken by Mary = 40 minutes = 2/3 hours
400/x - 400/(x+15) = 2/3
400(x+15)-400x = (2/3) *x(x+15)
6000 = (2/3) *x(x+15)
18000 = 2x^2 +30x
/2 and rearrange
x^2 +15x -9000 =0
a= 1 , b= 15 c= -9000

b^2-4ac= 225 + 36000
b^2-4ac= 36225
%09sqrt%28%0936225%09%29=%09190.33%09
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29
x1=( -15 + 190.33 )/ 2
x1= 87.66
x2=%28-b-sqrt%28b%5E2-4ac%29%29%2F%282a%29
x2=( -15 -190.33 ) / 2
x2= -102.66
Ignore negative value
Richard's speed =87.66 mph