<PRE><B>Question 55445</B>
15. Two cars with average speeds of 40 and 50 km/h respectively moved toward each other at 4:00 AM. If the square of the distance traveled by the second car is 250 times the distance traveled by the first car, at what time will they meet? how far has each traveled by then?

LET THEIR TIME OF TRAVEL FROM START TO TILL THEY MEET = T HRS
DISTANCE TRAVELLED BY I CAR = 40T
DISTANCE TRAVELLED BY SECOND CAR = 50T
SQUARE OF DISTANCE TRAVELLED BY II CAR = 50T*50T=2500T^2
250 TIMES DISTANCE TRAVELLED BY I CAR = 250*40T=10000T
HENCE
10000T = 2500T^2
4T=T^2
T^2-4T=0
T(T-4)=0
T=4 HRS.....THAT IS THEY MEET AT 4:00 AM+4 = 8:00 AM
BY THIS TIME 
I CAR TAVELLED = 4*40  = 160 KM
II CAR TRAVELLED = 4*50 = 200 KM.
HENCE THEIR INTIAL DISTANCE OF SEPERATION = 160+200 = 360 KM.</PRE>