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While working on the following word problem, I could not solve it by making the 'time' a variable, but rather only by making the distance a variable, and I'm wondering why that is. Here is the problem:

The distance between stations A and B is 148 km. An express train left station A towards station B with the speed of 80 km/hr. At the same time, a freight train left station B towards station A with the speed of 36 km/hr. They met at station C, and by that time the express train stopped at an intermediate station for 10 min and the freight train stopped for 5 min. Find:
The distance between stations C and B.

My first attempt for a solution was to mark the time between stations A and C with the variable 'a', and similarly for variable 'b', thus:
80a = distance from A to C
36b = distance from B to C

But then I can come up with only this equation:
148 - 80a = 36b

But I can't get to the other equation needed to solve this.

However, if I mark the distance from A to C with a variable, then it becomes possible to solve it:
x = distance between B to C, thus:
(148-x)/80 +10/60 = x/36 +5/60

Why is it that I can't find a second equation when I mark the time as being unknown and then represent the distance with it?
After all, I don't know the time nor the distance...

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  • $\begingroup$ I think I see at least one problem with my first attempt. The way I defined 'a' as "the time it takes to get from A to C", while there's actually no such thing as "time it takes to get to.." because "time to get to" always depends on speed. So if I said "the time it takes to get from A to C at 80 km/hr" then I wouldn't use 'a', but rather a term that includes speed in it. Specifically, distance divided by speed. Did I spot my error correctly? $\endgroup$ – Pineapple29 Dec 7 '16 at 8:18
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Just write down and translate what is given in the problem statement and introduce unknowns as they come along:

Let $t$ be the time span between the beginning and the meeting at $C$. We know that it takes $t-10\,\text{min}$ to get from $A$ to $C$ at $80\,\text{km/h}$, so if we denote this distance with $a$, we have a first equation $$a = (t-10\,\text{min})\cdot 80\,\text{km/h}.$$ We have similar data for the other train, just with different constants, so $$b = (t-5\,\text{min})\cdot 36\,\text{km/h}.$$ That's two equations for three unknowns, so did we forget something? Yes, of course! We also have $$a+b = 148\,\text{km}. $$ Now that the problem has been translated, you can through all your solving skill at it to find the desired $b$.

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  • $\begingroup$ Thanks for the reply, but I'm not sure how this answers my question. If I mark the time as being unknown, then I must introduce a third variable to solve it? What's the reasoning behind this? $\endgroup$ – Pineapple29 Dec 7 '16 at 7:50
  • $\begingroup$ No, you don't add more variables, you define the two you used for time by showing how they related to each other. $\endgroup$ – user304051 Dec 7 '16 at 7:51
  • $\begingroup$ So the mistake I made was giving the times from A to C and B to C different variables? When I think of it now, that time should be equal, right? $\endgroup$ – Pineapple29 Dec 7 '16 at 8:11
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When you added two variables for time you changed from using one variable to show distance. Instead of $(a,b)$ you could say something more meaningful like $(a, a+2)$.

The difference between your distance equation and time equation is that with distance you only used $(x)$ but with your time you used two variables $(a,b)$ without defining their relationship beforehand.

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