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.


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							
{{{	sqrt(	36225	)=	190.33	}}}			
{{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}								
{{{x1=(-b+sqrt(b^2-4ac))/(2a)}}}								
x1=(	-15	+	190.33	)/	2			
x1=	87.66							
{{{x2=(-b-sqrt(b^2-4ac))/(2a)}}}								
x2=(	-15	-190.33	) /	2				
x2=	-102.66							
Ignore negative value								
	Richard's speed	=87.66	mph