Answer:
Train P :
![t_(P) =(330)/(55)=6\ \text{hours}](https://img.qammunity.org/2021/formulas/mathematics/high-school/i0uhcb2b100akiwu2frig7jb0gg0m7kdee.png)
Train Q :
![t_(Q) =(330)/(55)+(1)/(2)=6\ \text{hours}\ 30\ \text{minutes}](https://img.qammunity.org/2021/formulas/mathematics/high-school/n8alnsa82d1f7db4lkid09fupyiyi8usgy.png)
Explanation:
The formula to compute the distance traveled by a train is:
![d=s* t](https://img.qammunity.org/2021/formulas/mathematics/high-school/ew1r2nthk3h5payl208qv2n9uw91u7liyh.png)
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)](https://img.qammunity.org/2021/formulas/mathematics/high-school/ngkt33nou3o1f1kotz2aoy0rjjrd8he0jm.png)
Train Q :
d = 330 km
s = (x - 5) km/h
![t=(330)/(x)+(1)/(2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/yfkugvbb2993u54sa7g0wd8oe29b0u5ldk.png)
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](https://img.qammunity.org/2021/formulas/mathematics/high-school/dypcy11wfwqujkuvwnnu4fvud09lgaoutq.png)
Compute the time taken by the two trains as follows:
Train P :
![t_(P) =(330)/(55)=6\ \text{hours}](https://img.qammunity.org/2021/formulas/mathematics/high-school/i0uhcb2b100akiwu2frig7jb0gg0m7kdee.png)
Train Q :
![t_(Q) =(330)/(55)+(1)/(2)=6\ \text{hours}\ 30\ \text{minutes}](https://img.qammunity.org/2021/formulas/mathematics/high-school/n8alnsa82d1f7db4lkid09fupyiyi8usgy.png)