I had a question and answered it, but I've been told that my solution is incorrect. What's the mistake here?

The Question

Runner A is running at the speed of x in a triangular path (each side of the triangle is of length a) and Runner B is running in the same track at the speed of x+3. Runner A passed 3a+60 (a>60) while at the same time Runner B passed 6a-60. Express the value of the perimeter (3a, I believe) using the speed x.

My solution

Using this table

        |  time  |  speed  |  location  | 
Rider A |   t    |    x    |     tx     |
Rider B |   t    |   x+3   |   t(x+3)   |

We can understand that tx=3a+60 and t(x+3)=6a-60 so t(x+3)-3a+120=3a+60=tx and then we get tx+3t-3a+120=tx and then 3t=3a-120 => t=a-40. If we put that back in the first equation tx=3a+60 we get

(a-40)x=3a+60 => xa-40x=3a+60 => (x-3)a=40x+60 => a=(40x+60)/(x-3) meaning that the permiteter is


What is incorrect here? (Sadly I'm not sure what the correct answer is so I can't add it here)

  • $\begingroup$ Where did the equation $t(x-3)-3a+120=3a+60=tx$ come from? $t$ and $a$ have different units, so you should not subtract them. $\endgroup$ Jun 14, 2020 at 19:30
  • $\begingroup$ I decided to add −3a+120 inorder to make t(x-3) minus that equal to 3a+60, which is also tx. t and a have different unites, but tx and -3t don't, why is that problematic? did that cause the mistake? $\endgroup$ Jun 14, 2020 at 19:36
  • $\begingroup$ in this comment the x-3 was a typing error, it was meant to be x+3. But I can't find where the sign drop is. $\endgroup$ Jun 14, 2020 at 19:48
  • $\begingroup$ The constants in the equations actually hide units, because $t$ is time, $x$ is speed, and $a$ is length. It is a good troubleshooting technique to keep track of the units because it will highlight algebra errors. Here the hidden units allow you to subtract $a$ from $t$. I agree with your answer. It would be easier to read with the equations in a column and any explanation between them. $\endgroup$ Jun 14, 2020 at 19:55

2 Answers 2


You need to eliminate $t$, so write $$tx=3a+60\\t(x+3)=6a-60\\t=\frac {3a+60}x\\t=\frac {6a-60}{x+3}$$ $$(3a+60)(x+3)=(6a-60)x\\ 3ax+9a+60x+180=6ax-60x\\ 120x+180=3ax-9a\\ \frac{120x+180}{x-3}=3a$$

  • $\begingroup$ Thank you, could you explain what was incorrect in my method? $\endgroup$ Jun 14, 2020 at 19:36
  • $\begingroup$ Where did I do that? $\endgroup$ Jun 14, 2020 at 19:48
  • $\begingroup$ So it appears that my solution was correct? $\endgroup$ Jun 14, 2020 at 19:52
  • $\begingroup$ @snatchysquid : Comments about errors in your shown work would be made to your Question. Comments about errors in this Answer are made to this Answer. Why do you believe that a comment to this Answer is about your work? $\endgroup$ Jun 14, 2020 at 19:52
  • $\begingroup$ @snatchysquid: Eric Towers was complaining about my answer, not yours. You and I now agree $\endgroup$ Jun 14, 2020 at 19:52

There are several defects of the question, so let's start with those and how they affect your answer.

  • Why do you believe both runners started at the same time?
  • You are told the speed of each runner, but not their direction, so why do you believe both runners are running around the track in the same direction?
  • "passed $3a+60$" is not unambiguously interpretable -- there is no place that records completed laps (the "$3a$"). The place is some point labelled "60". It is unclear if that label is on the track so that runners running in either direction pass the place labelled "60" at the same point on the track or if it labelled along a runner's path so that "$3a+60$" means "has completed one lap and proceeded an additional 60 units in whichever direction that runner is running".
  • (There is a potential problem that the distance units for the speeds and the distance units along the track are different. It is typical to change units so that this is not the case, so I do not consider this potentiality further.)

These defects make each column of your table questionable. Putting all that aside by assuming both runners start at the same time, both runners run around the track in the same direction, and "passed" is followed by a total distance specifier, not a track position marker, ...

From "$(x-3)a=40x+60$", you obtain $$ a = \frac{40x+60}{x-3} \text{,} $$ but that manipulation is only valid if $x \neq 3$ (otherwise you are dividing by zero, an operation that is undefined). You have not addressed the case $x = 3$. (That case leads to an impossibility, but you have to actually say so, otherwise your solution is incomplete. I recommend going back to the original pair of equations, substituting $x \mapsto 3$, then seeing how/if a contradiction is obtained.)

OP in comments asserts that the runners run in opposite directions. This means, picking a sign convention, that the velocity of Runner A is always $x$, the speed of Runner B is $x+3$ and the sign of the speed of runner B is the opposite of the sign of $x$. Therefore, the velocity of Runner B is $$ \begin{cases} -(x+3) ,& 0 < x \\ x+3 ,& -3 < x < 0 \\ -(x+3) ,& x < -3 \end{cases} $$ The case $x = 0$ is excluded because Runner A is asserted to run some nonzero distance. The case $x = -3$ is excluded because Runner B is asserted to have run at least 300 units.

Solving the system when $x < -3$ or $0 < x$, we get $a = \frac{-20}{x+1}$. (This assumes $x \neq -1$, which is true since $x$ is positive or less than $-3$ in this case.) This may seem to give a negative length, $a$, but this $a$ is positive when $x < -1$.

Solving the system when $-3 < x < 0$, we get $a = \frac{40x+60}{x-3}$. (This assumes $x \neq 3$, which is true in this case.)

Finally, the question has the constraint, "$a > 60$" which is ignored in the Question's answer and in the above. To enforce $ a > 60$,

  • When $x < -3$ or $0 < x$, $a > 60$ implies $\frac{-20}{x+1} > 60$, which we reduce to $-4/3 < x < -1$, so this case is incompatible with the given constraints. (As above, you must write that you have produced and checked this potential solution, otherwise, your solution is incomplete.)
  • When $-3 < x < 0$, $a > 60$ implies $\frac{40x+60}{x-3} > 60$, which we reduce to $3 < x < 12$, so this case is incompatible with the given constraints. (As above, you must write that you have produced and checked this potential solution, otherwise, your solution is incomplete.)

Having rejected both cases, there is no value of $a$ that satisfies the constraints of the problem.

In fact, the maximum value of $a$ in either case is $10$.

  • $\begingroup$ They are not running in the same direction, I'm interested only in sizes and not directions. This "60" point is given to us and you may assume x is not 3. I assumed people would understand what I mean (as Ross did). I'm only interested in the algebric technique and I'm aware of the importance of units and undefined points. $\endgroup$ Jun 14, 2020 at 20:06
  • $\begingroup$ @snatchysquid : If you take that the two runners are running in opposite directions, then $x = 3$ leads to a consistent solution which you have omitted. $x = 3$ only leads to an impossibility if both runners run in the same direction. $\endgroup$ Jun 14, 2020 at 20:08
  • $\begingroup$ @snatchysquid : Also, you have ignore the constraint $a > 60$. I've added a bit to my answer to capture your revelation "They are not running in the same direction" and the consequences of the constraint on $a$. $\endgroup$ Jun 15, 2020 at 18:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .