SOLUTION: 2.mr. ong makes regular business trips to kaulalumpur a distance of 420 km from singapore (a)on his way up to kaulalumpur he travels along the trunk road at an average speed of x k

Question 388273: 2.mr. ong makes regular business trips to kaulalumpur a distance of 420 km from singapore (a)on his way up to kaulalumpur he travels along the trunk road at an average speed of x km/h. write down the expression in trm of x for the time taken in hours to travel from singapore to kaulalumpur
(b)on his return journey to singapore he travels along the north south highway and increases the average speed by 15 km/h.write down the expression in term of x for the time taken in the hours to travel from kaulalumpur to singapore.
(c)if the time difference between the two journey is 40 min from an equation in x and show that it reduces x^2+15x-9540=0
(d)solve the equation and find the time taken for the trip from singapore to kaulalampur
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
2.mr. ong makes regular business trips to kaulalumpur a distance of 420 km from singapore (a)on his way up to kaulalumpur he travels along the trunk road at an average speed of x km/h. write down the expression in trm of x for the time taken in hours to travel from singapore to kaulalumpur
(b)on his return journey to singapore he travels along the north south highway and increases the average speed by 15 km/h.write down the expression in term of x for the time taken in the hours to travel from kaulalumpur to singapore.
(c)if the time difference between the two journey is 40 min from an equation in x and show that it reduces x^2+15x-9540=0
(d)solve the equation and find the time taken for the trip from singapore to kaulalampur
...
FORWARD
speed x km/h
distance = 420 km
...
t=d/r
t= 420/x hours
...
RETURN
-------
speed = x+15 km/h
distance = 420 km
..
time = 420/(x+15) hours.
...
40 minutes = 40/60 = 2/3
time going - time returning = 2/3 hours
(420/x)-420/(x+15)=3/4
...
LCD = x(x+15)
multiply equation by x(x+15)
420(x+15)-420x=2/3* x(x+15)
420x+6300-420x=2/3*x(x+15)
6300=2/3*x(x+15)
multiply by3
18900=2x^2+30x
2x^2+30x-18,900=0
/2

...
x^2+15x-9450=0
x^2+105x-90x-9450=0
x(x+105)-90(x+105)=0
(x+105)(x-90)=0
x= 90 km/h
...
distance = 420 km
speed = 90 km/h
time = 420/90
hours Time to Kaula lumpur
...
m.ananth@hotmail.ca

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