USSR Exam problem

Question

I obtained this problem from here.

A car starts from point $A$ towards $B$ at the same time as a motorcycle starts from $B$ to $A$ (but with a lesser speed). At the moment they meet, a second motorcycle starts from $B$ and also meets the car. The distance between the two meeting places is $\frac{2}{9}$ the distance between $A$ and $B$. Had the car's speed been less by $20$kmh, the distance between the two points would have been $72$km and the first encounter would have taken place $3$ hours after the car started from $A$. Assuming the two motorcycles have the same speed, find the distance between $A$ and $B$.

I started off using this method. First I know that the car and the first motorcycle will travel a distance $i$ in $3$ hours. Let $v_m$ denote the speed of the motorcycle and $v_c$ denote the velocity of the car. Then, $i=(v_c-20)(3)$ and $i=(v_m)(3)$. We know that the distance between the two meeting points is $72$ and after the first meeting, we have two more equations: $i+72=(v_c-20)(3+t)$ and $d-i-72=(v_m)(3+t)$ where $d$ is the distance between $A$ and $B$.

I'm wondering if I'm even on the right track with this problem. Or, is the answer as obvious as noting that $72$ is $2/9$ of the distance.

Answer 1

$S$ - path length
$v_m$ - motorcycle speed
$v_c$ - car speed (in the 1st case)

Case 1: Car speed is $v_c$
$t_1$ - time to 1st meeting
$t_2$ - time between 1st and 2nd meeting

Case 2: Car speed is $v_c-20$
$t_3$ - time to 1st meeting
$t_4$ - time between 1st and 2nd meeting

Just express all statements with equations
(and verify their logic 3-4 times):

$$v_c.t_1 + v_m.t_1 = S$$

$$v_c.t_2 + v_m.t_2 = S - v_c.t_1$$

$$v_c.t_2 = \frac{2}{9} S$$

$$(v_c-20).t_3 + v_m.t_3 = S$$

$$(v_c-20).t_4 + v_m.t_4 = S - (v_c-20).t_3$$

$$(v_c-20).t_4 = 72$$

$$t_3 = 3$$

Now we have 7 equations and 7 variables, right?
So the rest should be more or less routine.

EDIT:

The rest is not routine at all. One may easily get swamped in calculations (if one is solving it manually, as I did) and thus lose too much time to say the least.

OK, I got 4 different solutions/paths of which 3 turned out impossible/invalid due to facts are which are implicitly implied or explicitly given in the problem statement (the statement implicitly implies that $v_c-20 \geq0$, and also it says explicitly "the motorcycle has lesser speed").
The only valid solution turns out to be the following:
$v_c = 80, v_m=40, t_4 = 6/5, t_3 = 3, S = 300$.

Btw, this problem doesn't seem too recreational :) which is expected taking into account its source. Based on my modest experience the idea behind such problems is that all but the best candidates would get swamped in calculations, and would waste too much time on this problem, and thus will be left with too little time for the other problems. So the idea is to test the candidates' math techniques too (any 6th grader can write the above equations but the rest is not so easy at all). But on the other hand, from all 5 problems this one seems the easiest maybe :) It's not an easy exam to say the least.

Answer 2

Let $x$ be the distance between $A$ and $B$, in kilometers. Let $h$ be the number of hours from when the car starts to when it meets the first motorcycle. Let $bx$ be the distance traveled by the motorcycle in $h$ hours. During the first $h$ hours, then, the car travels a distance $(1-b)x$. The distance between the car and the second motorcycle is therefore $bx$ at the instant the second motorcycle starts; it takes an additional $bh$ hours for them to meet, during which the second motorcycle travels a distance $b^2x$ and the car travels $(1-b)bx$.

But the problem statement says the car travels a distance $\frac29 x$ between meeting the two motorcycles, so $(1-b)bx = \frac29 x.$ Dividing by $x$, $$(1-b)b = \frac29.$$

The two possible solutions of this quadratic equation are $b = \frac13$ and $b = \frac23.$ We reject $b = \frac13$ because the motorcycle travels faster than the car. (Why do I say that? See my comments at the end.)

Now instead of changing the speed of the car, I'll simply send out a second car at the same time, traveling $20$ kph slower. The second car meets the first motorcycle after $3$ hours, at which time the first car is $60$ km ($20$ kph times $3$ hours) ahead of the second car. Recall that the motorcycle meets the first car at $h$ hours and the second car at $3$ hours, and that the motorcycle travels twice as fast as the first car. Between $h$ hours and $3$ hours, therefore, the motorcycle travels $40$ km while the first car travels $20$ km.

Let $ax$ be the distance traveled by the second car in $3$ hours, that is, the distance from $A$ to the meeting of the second car and the first motorcycle. The motorcycle meets the first car $40$ km farther from $A$, at $\frac13 x$ km from $A$, so $ax + 40 = \frac13 x;$ equivalently, $$x = \frac{40}{\frac13 - a}.$$

At this point I assume the second motorcycle has already traveled $40$ km, the same as the first motorcycle traveled since meeting the first car. That is, I assume the second motorcycle always starts $h$ hours after the first one, regardless of the speed of the car. (See my comments below for further explanation.) At the instant when the second car meets the first motorcycle, therefore, the distance between motorcycles is $\frac23 x$, just as it was when the first motorcycle met the first car and the second motorcycle started.

The second car covered $ax$ km in the time the first motorcycle traveled $(1-a)x$ km, so its speed is $\frac{a}{1-a}$ of either motorcycle's speed. Between the instant when it meets the first motorcycle and when it meets the second motorcycle, the second car travels $\frac23 ax$ km while the second motorcycle travels $\frac23(1-a)x$ km. But we are also told the car travels $72$ km between these same two events, so $\frac23 ax = 72,$ and $$x = \frac{108}{a}.$$

We now have two different ways of expressing $x$ in terms of $a$; set them equal: $$\frac{40}{\frac13 - a} = \frac{108}{a}.$$

Cross-multiplying, we get $40a = 108\left(\frac13 - a\right),$ and solving this linear equation in $a$ yields $a = \frac{9}{37}.$ Therefore, $\frac13 - a = \frac{10}{111}.$ Since we previously found that $x = \frac{40}{\frac13 - a},$ we have

$$x = 444.$$

---

Checking this answer: from $x = 444$ we have $ax = 108$, so the second car travels $36 kph$, the first car $56$ kph, and the motorcycles $112 kph$. We also find $\frac13 x = 148$; check that $148 - 108 = 40$, the distance between where the first motorcycle meets the first car and where it meets the second car. At $3$ hours the distance between the second car and second motorcycle is $444 - 108 - 40 = 296$ km, which is the combined distance they travel in $2$ hours: the motorcycle travels $224$ km and the car travels $296 - 224 = 72$ km.

---

Some remarks on the interpretation of the problem, which I found less clearly stated than I would like. (Something lost in translation from Russian, perhaps?)

For the relative speeds of the car and motorcycle, I read the problem by paraphrasing it as, "A car starts at the same time as a motorcycle, but with lesser speed." That is, a parenthetical phrase is equivalent to a phrase set off by commas, and the car is still the subject of the main clause of the sentence. The motorcycle would be slower if the problem had said, "A car starts from point A towards B. At the same time, a motorcycle starts from B to A (but with a lesser speed)."

For the case where the car traveled $20$ kph slower, I had difficulty deciding whether the second motorcycle should start $3$ hours after the first motorcycle (at the instant when that motorcycle meets the car) or whether the second motorcycle's starting time is independent of the car's speed and only coincidentally occurs $h$ hours after the first motorcycle starts. Initially, I took the first interpretation, leading to the answer $x = \frac{600}{3\sqrt{34} - 16} \approx 401.914.$ I wrote up the second interpretation above because the problem statement did not say the second motorcycle would leave later if the car was slower, and also because the second interpretation gives a neater answer that one can easily check without use of a calculator.