I'm a high school student, and I have come across a problem that I cannot solve. I feel there must be something obvious that I'm not seeing.
Problem: The distance between two train stations is $96$ km. One train covers this distance in $40$ minutes less time than another one. The second train is $12$ km/h faster than the first one. Find both trains' speeds.
What I have done:
Set $v_1+12 = v_2$ (the speed of train $2$ is $12$km/h more than speed of train $2$), and $96/(v_1) = (96/v_2)-40$ (the time it takes for train 2 to transverse the distance between the stations is $40$ minutes less than the required by train $2$) Now, from here I get to:
$v_1 = v_2-12$.
\begin{align}
&\frac{96}{v_2-12} = \frac{96}{v_2}-40 \\
&\qquad\implies \frac{96}{v_2-12} = \frac{96-40v_2}{v_2} \\
&\qquad\implies v_2\cdot 96 = (v_2-12)\cdot (96-40v_2) \\
&\qquad\implies v_2\cdot 96 = v_2\cdot 96-40v_2^2-1152-380v_2 \\
&\qquad\implies 0 = -40v_2^2-380v_2-1152
\end{align}
Solving this quadratic equation yields no real roots.
Could you please suggest the right way to go?
Best Answer
Let $v_1$ km/hr denote the speed of the faster train, and $v_2$ km/hr denote the speed of the slower train (I seem to have reversed your notation—sorry, $v_1$ feels like faster variable to me than $v_2$). First off, we can relate the amount of time it takes for each train to travel the 96 km to the speed of each train. So, let $$ t_1 \text{ hrs} = \frac{96 \text{ km}}{v_1 \ \frac{\text{km}}{\text{hr}}} = \frac{96}{v_1}\text{ hrs} \qquad\text{and}\qquad t_2 \text{ hrs} = \frac{96 \text{ km}}{v_2 \ \frac{\text{km}}{\text{hr}}} = \frac{96}{v_2}\text{ hrs}\tag{1}$$ denote these two times. We know that the faster train arrives 40 minutes (that is, $\frac{2}{3}$ of an hour—watch the units! (this is an easy mistake to make—I messed it up, too!)) earlier than the slower train, which implies that $$ t_1 \text{ hrs} = t_2 \text{ hrs} - \frac{2}{3} \text{ hrs} = \left( t_2 - \frac{2}{3} \right)\text{ hrs}, $$ and we know that the faster train is 12 kph faster than the slower train, hence $$ v_1 \ \frac{\text{km}}{\text{hr}} = v_2 \ \frac{\text{km}}{\text{hr}} + 12 \ \frac{\text{km}}{\text{hr}} = \left(v_2 + 12\right) \ \frac{\text{km}}{\text{hr}}. $$ It should be noted that the only major mistake that I see in your work is in the above step—in your model the faster train takes more time to cover the distance, which is a problem. Substituting these into the equations at (1) (and eliding units—the units of time are hours, the unit of distance are kilometers, and the units of speed are km/hr), we get the system $$ \begin{cases} t_2 - \dfrac{2}{3} = \dfrac{96}{v_2 + 12} \\ t_2 = \dfrac{96}{v_2}. \end{cases} $$ Can you solve it from here?