Find the distance between two towns given train timings

Question

While practicing maths and starting to learning it, I found question this question:

A train running between two towns arrives at its destination 10 minutes late when it goes 40 miles per hour and 16 minutes late when it goes 30 miles per hour. The distance between the two towns is:

A. 720 miles
B. 12 miles
C. 8-6/7 miles
D. 12-7/7 miles
E. None of these

I am trying to solve it like this: 40 miles per hour, hours have 60 minutes, so 40 miles in 60 minutes, and it is ten minutes late. So one mile per minute is $40/60 = 0.66$, so miles in 10 minute is 6. So in 70 minutes it covers about 46 miles.

But how can I solve it further, to find the distance?

Answer 1

Let $d$ be the distance in miles, and $v$ the speed in mph of the train at which it arrives in time.

The time is the distance divided by speed. Now, use the fact that the train is $10$ minutes late (or $10/60$ hours late) when it goes $40$ mph: $$\frac{d}{40}=\frac dv+\frac{10}{60}$$

The first member is the time that the train spends in arriving at $40$ mph, and the second member is the time that it spends at speed $v$ plus $10$ minutes. They are equal, of course.

Now, write a similar equation for the other data, and solve the system.

Answer 2

Your approach might work, but the power of algebra is pretty helpful here.

Let $D = $ unknown distance between two towns, in miles

Then $D/40$ is the time, in hours, to travel between the towns at 40 mph, so

$T_1 = 60 * D/40$ is that time, in minutes.

Similarly,

$T_2 = 60 * D/30$ is the time, in minutes, for travelling at 30 mph.

The problem tells us that $T_2 = T_1 + 6$.

So $$ 60*D/40 + 6 = 60*D/30 \\ 60D + 40*6 = 40 * 60 * D/30 \\ 30*60*D + 30 * 40 * 6= 40 * 60 * D\\ 30*40*6 = 10*60*D \\ 1200*6 = 600 * D \\ 2 * 6 = D $$ so $D = 12$.

Answer 3

The basic equation is $d=v\cdot t$, where $d$ is the distance, $v$ the velocity and $t$ the time needed.

Let's say the train is scheduled to arrive after a duration $t_0$. Then, $d=40\mathrm{mph}(t_0+\frac{1}{6}\mathrm{h})=30\mathrm{mph}(t_0+\frac{16}{60}\mathrm{h})$. Solve this for $t_0$ and plug it in one of the $t_0$-equations and you're good.

Answer 4

Here's a fast method (in fact, it can be done mentally)

$60$ mph = $1$ minute/mi. and $40$ mph $= 1.5$ minute /mi.

A difference of $0.5$ min./mi increases delay by $6$ minutes,

thus distance $= 6/0.5 = 12$ mi.