Horsing Around!
Cam and Chris left the same town, at the same time, to attend a TOM conference in the nearby metropolis of Puzzletown. One of the puzzlers travelled via horse, and the other drove a vehicle.
It became apparent:
That if Chris had travelled three times as far as he had, he would have half as far to ride as he had ridden;
and
That if Cam had travelled half as far as he had, he would have three times as far to ride as he already had.
Who rode the horse? ... and what color was the horse? (oops ... darn pain meds ... forget the color question).
It became apparent:
That if Chris had travelled three times as far as he had, he would have half as far to ride as he had ridden;
and
That if Cam had travelled half as far as he had, he would have three times as far to ride as he already had.
Who rode the horse? ... and what color was the horse? (oops ... darn pain meds ... forget the color question).
posted by Zaux at
5:58 AM
16 Comments:
It appears as though both men have travelled two-sevenths of the distance and the language sounds as though they may both being travelling together, so I will guess that the horse in question is some kind of vehicle as opposed to the four legged mamal variety.
If that is the case a quick Google search has revealed one such vehicle which is green in color and both are riding.
On the other hand my maths is probably worng and massively overthinking this question, what do you guys think?
Hi Antn'y ...
There is an actual horse and an actual vehicle involved.
Another English comprehension test! I certainly sympathise with Antn'y. Both seem to have travelled 2/7 ths of the distance. So there is no way to tell who rode what.
OR if the reference meant the distance that they would then have then travelled, then Chris is 2/9 (0.222...) ths of the way, and Cam 1/4 th (0.25) of the way. Cam used the faster transport.
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Okay, I solved this independent of the published solution. Got the same thing you guys got ... Cam and Chris both traveled 2/7 of the distance, and therefore no solution is apparent as the problem is stated.
I went back and checked the statements to make sure the intended logic was in tact ... no problems there ...
I'm fairly disgusted that the sources seem fairly unreliable. I will not post as many problems, but the ones I do post will have been solved and verified.
I apologize for this problem.
Hi Zaux. My only real problem with this sort of problem is decoding the English (that's not your fault, it's a limitation of mine). I should have put a ;) after my English comprehension utterances.
At worst it's irritating when a problem is flawed, but as you can see, I thoroughly enjoyed trying to solve "The Robbery". I'm still not sure if I actually have done it (B is the robber) or not, my sophistry is so good that I have started to believe it myself.
I also know that it is very hard to try to keep up the problem posting. I think you have been doing a fantastic job, so don't crucify yourself about flawed problems. Obviously you're not obliged to keep it up, especially at the astonishing rate that you have been doing.
i solved it!-then looked at the comments and saw i was way off :[
Chris travelled 2/9 ths
Cam 1/6
and the color of Cam's horse? pale
Inspired by Knightmare's answer, I had another try:
Let d be the distance we're talking about having to cover.
Let Chris have travelled a distance s. Then I now get:
d = 3s + (1/2)3s = 9s/2 => s = (2/9)d
Let Cam have travelled a distance c, then:
d = c/2 + 3(c/2) = 4(c/2) => c = d/2
So c/s = (d/2)/((2/9)d) = 9/4 = 1.75.
Now I suspect that Chris rode the horse (assuming the vehicle is faster). I can't imagine how I got the results I posted earlier. Hmmm, now I come to think of it, it has been 40 years since I got my cerstifitake in English.
So it's the repliers who are flawed. Knightmare, wow!
Zaux...you know what that means-you ARE obliged to keep it up-HA
LOL. Poor Zaux, getting all chagrined because of our incompetence.
In my previous solution, I read it as: if Chris had travelled three times as far as he had actually travelled, then he'd have to go half as much again as the distance that he would have travelled having travelled three times as far as he'd acually travelled. The alternative solution (which is no good) is: if Chris had travelled 3 times as far as he'd actually travelled, then he'd have to travel half as much again as the distance that he'd actually travelled.
I'll spare you (and me) the similar interpretations of Cam's progress.
The last interpretation gives:
d = 3s + (1/2)s = (7/2)s
d = (1/2)c + 3c = (7/2)c
Blame the lack of meds for the other solution in my first post.
Here's the gist of the published solution:
x = total distance from town to Puzzletown.
y = distance Chris rode
(x-y) = remaining distance
If he had ridden 3y distance, (x-3y) would be remaining distance.
He would have had half as far left: 1/2(x-y). Then:
x-3y = 1/2(x-y)
2x-6y = x-y
x = 5y
y = 1/5 x
Therefore, Chris had ridden 1/5 of the total distance.
Total distance = x
z = distance Cam rode.
(x-z) = distance remaining.
If he had ridden only 1/2 z distance, he would have (x - 1/2 z) distance remaining.
He would have had 3 times as far remaining:
x - 1/2 z = 3(x-z)
2x-z = 6x-6z
5z = 4x
z = 4/5 x
Therefore, Cam had ridden 4/5 of the total distance.
Chris - 1/5
Cam - 4/5
Being that the vehicle should be faster than the horse, Chris rode the horse.
Thanks for that Zaux. I think the original setter has a serious problem with English.
Interpretations (bear in mind I've missed the 'would have had') that immediately precedes what I've written:
Half as far to ride as he had [actually] ridden =>
x-3y = y/2 => y = 2x/7.
I think this is the most natural interpretation.
or
Half as far to ride as he had [hypothetically] ridden =>
x-3y = (1/2)3y => y = 2x/9
This is a reasonable interpretation.
or
Half as far [as the hypothetical remaining distance] to ride as he had [actually] ridden =>
(1/2)(x-3y) = y => y = (1/5)x (strangely enough).
I'm not happy with that one.
or
Half as far [as the hypothetical remaining distance] to ride as he had [hypothetically] ridden =>
(1/2)(x-3y) = 3y => y = (1/9)x
I'm not happy with that one.
But to equate half the actual remaining distance to the hypothetical remaining distance is taking the mick.
Still didn't obtain Knighmares's 1/6 th LOL.
I could probably milk this one more, but won't.
... I've re-read the question. It's phrased slighlty differently for Cam:
That if Cam had travelled half as far as he had, he would have three times as far to ride as he already had.
Could be read as:
That if Cam had travelled half as far as he [actually] had [ridden], he would [then] have three times as far to ride as he [*******] had [left to ride].
[*******] needs to be 'actually' or 'currently'. Then get:
x-z/2 = 3(x-z) => z = (4/5)x as in the official answer.
So I still can't accept the official interpretation, I can bend the statements regarding Cam into line with it.
OK I'm now moving on.
No problems Zaux, I still liked the problem even though the answer I got wasn't what was looking for.
Chris rode a green horse with purple stripes while Cam rode a red VW Bug with black polka dots on it. That is the only clean answer i can give you. have fun reteiving the data that i gave you!!!!!
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