A motorist makes a journey of 240 km from Singapore to Malacca to visit his in-laws at an average speed of x km/hr. Write down an expression for ...the number of hours taken for the journey. On...

A motorist makes a journey of 240 km from Singapore to Malacca to visit his in-laws at an average speed of x km/hr. Write down an expression for ...

the number of hours taken for the journey.

On his return journey from Malacca to Singapore, his average speed is reduced by 6 km/hr due to heavy traffic at the causeway. Write down an expression for the time taken for the return journey.

If the return journey takes 20 minutes longer, form an equation in x and solve it to find the average speed for each journey, giving your answer correct to two decimal places.

pohnpei397 | College Teacher | (Level 3) Distinguished Educator

Posted on

The above answer must in some way be wrong because the speeds in km/h do not work with the data given.

For the speeds to be correct as given in the above answer, they must yield times that are 20 minutes apart.

To check the times given by the above answer, we divide 240 km by 131.59 km/h and 125.59 km/h.  Doing that gives us times of 1.82 hours (going) and 1.91 hours (returning).  This gives us a difference of only .09 hours which is (60*.09) about 5 minutes.  Therefore the above answer is wrong.

The above answer is correct in reaching the quadratic equation x^2-6x-4320 = 0.

But what one must then do is factor this equation.  Doing so, we get

(x+62.795)(x-68.795) = 0

This tells us that x (the speed on the journey to Malacca) must be 68.795 km/h.  And the speed on the return journey must be 62.795 km/h

So the expression for the number of hours on the way to Malacca is y = 240/x where x is the speed in km/h and y is the time in hours.

The expression for the return journey is y-.33 = 240/(x-6)

This is because the return journey takes 20 more minutes (.33 hours) and the speed is 6 km/h slower.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.