<PRE><B>Question 1106300</B>
speed*time=distance
x k/h*t1=15; t1=15/x
(x-27)*t2=6; t2=6/(x-27)
40 minutes =2/3 hour, and t1+t2=2/3
15/x+6/(x-27)=2/3
multiply everything by 3x(x-27) to clear fractions
15*(3*(x-27))+6(3x)=2x(x-27)
15(3x-81)+18x=2x^2-54x
2x^2-54x-45x+1215-18x=0
2x^2-117x+1215=0
(2x-27)(x-45)=0
x=13.5
x=45
only the second can remain positive after subtracting 27
started at 45 k/h and took 20 minutes to go 15 km (1/3 of an hour)
then drove at 18 k/h for 20 minutes, and that is 6 km (1/3 of an hour)
Times are equal.



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