Finding the ratio of two persons time spent driving to a meeting Mark and pat drive separately to a meeting. mark's average driving speed is $1/3rd$ greater than pat's and mark drives twice as many miles as pat. 
What is ratio of number of hours mark spends driving to the meeting to the number of hours pat spends driving to meeting  
 A: $m_s$: Mark's driving speed, $\quad p_s$: Pat's driving speed
$$m_s = p_s + \frac 13 p_s = \frac 43 p_s\tag{1}$$
$m_d$: Distance Mark drives, $\quad p_d$: distance Pat drives
$$m_d = 2 p_d\tag{2}$$
$m_t$: Time Mark spends driving, $\quad p_t$: Time Pat spends driving
$$\large\frac{m_{t}}{p_{t}} \ = \ \frac{m_{d}/m_s}{p_d/p_s}\; = \; \frac{m_{d}}{p_d} \cdot \frac{p_s}{m_s} $$ $$= \frac{2p_d}{p_d}\cdot \frac{p_s}{\frac 43 p_s}\tag{substitution (1), (2)}$$ $$ = 2\cdot \frac{1}{4/3} =  \frac 32\tag{cancel units and evaluate}$$
A: Many problems are made simpler by producing a comparison ratio; in fact, this problem specifically asks for one.  Since the time $ \ T \ $ to cover a distance $ \ D \ $ at a constant speed $ \ v \ $ is given by $ \ T = \frac{D}{v} \ $ , we can write
$$ \frac{T_{Mark}}{T_{Pat}} \ = \ \frac{D_{Mark}/v_{Mark}}{D_{Pat}/v_{Pat}} \ = \ \frac{D_{Mark}}{D_{Pat}} \cdot \frac{v_{Pat}}{v_{Mark}}  \ . $$
You would now interpret the information given in the problem in order to use this appropriately.
A: Presumably you are to use $x$'s and $y$'s and $t$'s. This answer is intended to show that (in this case) we can dispense with these things. But you should write up a more conventional solution. 
The planet that Mark and Pat live on was not specified. So we can suppose that in the units used on that planet, Pat's speed is $3$ and Mark's is $4$. We can also suppose that in the units used on that planet, Mark drives a distance $12$, so Pat drives $6$. Mark then takes $3$ units of time, and Pat takes $2$. 
A: We know that distance=average speed x time. 
 Let's put this as $d=v\cdot t \;$
and denote:
$$\begin{align}
& Mark's\; average\; speed: v_M \\
& Pat's\; average\; speed:v_P \\
& distance\; driven\; by Pat\; :d \\
& time\; taken\; by\; Mark:t_M \\
& time\; taken\; by\; Pat:t_P \\
& ratio\; of\; number\; of\; hours\; mark\; spends\; driving\; to\; the\; meeting\; to\; the\; number\; of\; hours\; \\ 
& pat\; spends\; driving\; to\; meeting\;=\frac{t_M}{t_P} \\
\end{align}$$
Hence 
$\;v_M=\frac{2d}{t_M} \; and \;v_P=\frac{d}{t_P} $
$$\begin{align}  
 \frac{2d}{t_M} & =\frac{4}{3} \cdot \frac{d}{t_P}  \\
\end{align}$$
$$\begin{align}  
 \frac{t_M}{t_P} & =\frac{3}{2}  \\
\end{align}$$
