Question

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. ​

Answer 1

Answer:

Train P : $t_{P} =\frac{330}{55}=6\ \text{hours}$

Train Q : $t_{Q} =\frac{330}{55}+\frac{1}{2}=6\ \text{hours}\ 30\ \text{minutes}$

Step-by-step explanation:

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

$d=s\times t$

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:

$t=\frac{d}{s}=\frac{330}{x}$

Train Q :

d = 330 km

s = (x - 5) km/h

$t=\frac{330}{x}+\frac{1}{2}$

Both the trains traveled the same distance.

$x\times \frac{330}{x}=(x-5)\times [\frac{330}{x}+\frac{1}{2}]\\\\330=\frac{330(x-5)}{x}+\frac{x-5}{2}\\\\330\times 2x=[660(x-5)+x(x-5)]\\\\660x=660x-3300+x^{2}-5x\\\\x^{2}-5x-3300=0\\\\x^{2}+60x-55x-3300=0\\\\x(x+60)-55(x+60)=0\\\\(x-55)(x+60)=0\\\\x=55$

Compute the time taken by the two trains as follows:

Train P : $t_{P} =\frac{330}{55}=6\ \text{hours}$

Train Q : $t_{Q} =\frac{330}{55}+\frac{1}{2}=6\ \text{hours}\ 30\ \text{minutes}$