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Train P travels for 330 km at an average speed of x km/h from Kuala Lumpur to Kota Bahru. Train Q travels from Kota Bahru at an average speed of (x-5) km/h and arrives at Kuala Lumpur 30 minutes earlier than train P. Calculate the total time, in hours taken by the 2 trains. ​

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Answer:

Train P :
t_(P) =(330)/(55)=6\ \text{hours}

Train Q :
t_(Q) =(330)/(55)+(1)/(2)=6\ \text{hours}\ 30\ \text{minutes}

Explanation:

The formula to compute the distance traveled by a train is:


d=s* t

Here,

d = distance traveled

s = speed of the train

t = time taken

From the provided information:

Train P :

d = 330 km

s = x km/h

Then time taken is:


t=(d)/(s)=(330)/(x)

Train Q :

d = 330 km

s = (x - 5) km/h


t=(330)/(x)+(1)/(2)

Both the trains traveled the same distance.


x* (330)/(x)=(x-5)* [(330)/(x)+(1)/(2)]\\\\330=(330(x-5))/(x)+(x-5)/(2)\\\\330* 2x=[660(x-5)+x(x-5)]\\\\660x=660x-3300+x^(2)-5x\\\\x^(2)-5x-3300=0\\\\x^(2)+60x-55x-3300=0\\\\x(x+60)-55(x+60)=0\\\\(x-55)(x+60)=0\\\\x=55

Compute the time taken by the two trains as follows:

Train P :
t_(P) =(330)/(55)=6\ \text{hours}

Train Q :
t_(Q) =(330)/(55)+(1)/(2)=6\ \text{hours}\ 30\ \text{minutes}

User Erdem Akkas
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