The speed of the journey is expressed as the rate of journey distance per hour and the speed obtained by dividing the distance by the numberof hours taken for the journey.
If t is the time in hours for the journey of a distance d , the speed , x is given by x= d/t. Therefore, if the speed x and the distance d of journey are known , then time t could be given by the expression:
t = d/x
The speed of the motorist is already assumed by data as x km/hour
The time taken by the motorist taken from Singapore to Malacca= d/x = 240/x..................(1), as d=240km, by data.
The speed while coming back, reduced by 6 kms, is x-6 km/hour.Threfore the time taken for the return journey (from Malcca to Singapore) = 240/(x-6)..................(2),The time as given is more by 20 minutes, or 20/60hours = (1/3) hour.
Therefore the required equation is:
Time taken for return journey - time taken for onward journey =1/3 hours or
240/(x-6) - 240/x = (1/3).
To solve the equation, let us remove the denominator, by multiplyin the LCM of the denominators, i.e, 3(x-6)x. Then,
240*3x-240*3(x-6) = (x-6)*x
720x-720x+4320 = x^2-6x. Simplifying this, we get:
x^2-6x-4320 = 0 is a quadratic equation .
Therefore, roots are given by:
x = [6 +or-sqrt(6^2+4*1*4320)]/(2*1) or
x = 131.5902732 km/h is the speed of the onward journy.
x-6 = (131.5902732-6)km/h = 125.5902732kms is the speed of the return journey.
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