Velocity of a train A train travels with a velocity that doubles every hour. In 3 hours, it has travelled 350km. What was the trains velocity in the second hour?
A. 225 km/h
B. 175 km/h
C. 150 km/h
D. 100 km/h
Answer is D. Can someone explain?

Assuming that the intent is to solve this using linear algebra, here are the steps
$$ x =\mbox{distance traveled in the first hour}$$
$$ y =\mbox{distance traveled in the second hour}$$
$$ z =\mbox{distance traveled in the third hour}$$
This problem is expressed mathematically as
$$x+y+z=350$$
$$2x=y$$
$$2y=z$$
Which can be rewritten as
$$x+y+z=350$$
$$2x-y+0z=0$$
$$0x+2y-z=0$$
The augmented matrix of this linear system is
$$\left[\begin{array}{ccc|c}
1 & 1 & 1 & 350 \\
2 & -1 & 0 & 0 \\
0 & 2 & -1 & 0 \\
\end{array}\right]$$
$$=\left[\begin{array}{ccc|c}
1 & 1 & 1 & 350 \\
0 & -3 & -2 & -700 \\
0 & 2 & -1 & 0 \\
\end{array}\right]$$
$$=\left[\begin{array}{ccc|c}
1 & 3 & 0 & 350 \\
0 & -3 & -2 & -700 \\
0 & 2 & -1 & 0 \\
\end{array}\right]$$
$$=\left[\begin{array}{ccc|c}
1 & 3 & 0 & 350 \\
0 & -7 & 0 & -700 \\
0 & 2 & -1 & 0 \\
\end{array}\right]$$
$$=\left[\begin{array}{ccc|c}
1 & 3 & 0 & 350 \\
0 & 1 & 0 & 100 \\
0 & 2 & -1 & 0 \\
\end{array}\right]$$
$$=\left[\begin{array}{ccc|c}
1 & 0 & 0 & 50 \\
0 & 1 & 0 & 100 \\
0 & 2 & -1 & 0 \\
\end{array}\right]$$
$$=\left[\begin{array}{ccc|c}
1 & 0 & 0 & 50 \\
0 & 1 & 0 & 100 \\
0 & 0 & 1 & 200 \\
\end{array}\right]$$
Which implies that
$$ x =50\ \mathrm{km}$$
$$ y =100\ \mathrm{km}$$
$$ z =200\ \mathrm{km}$$
Since $y$ represents the distance that the train traveled in the second hour, the answer then is $100 \frac{\mathrm{km}}{h}$.
