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Ski Lift Mathematics (February 2024 Challenge Question)

Posted January 31, 2024 12:00 AM
Pathfinder Tags: challenge question

Mark Frank is going snow boarding.

To get to the top of the mountain, he must ride the ski lift. Mark Frank gets in lift chair No. 10.

Exactly halfway up the ski lift, he passes chair No. 100.

Assuming all chairs are equally spaced apart, how many chairs did Mark Frank pass on his ascent?

ANSWER (updated 2.26.24)

To know how many chairs on the the ski lift total, first identify how many on each half of the lift cable (100 - 10 = 90). Then double it for the full amount. 180 chairs total on the ski lift.

Mark Frank gets on and off the ski lift at the lowest and highest points of it, respectively. He passes all chairs except his own. 179.

Sure, there are some potential missing variables in the challenge. Where specifically Mark Frank climbs aboard the ski lift. Total length of the ski lift. If the lift operators can pause the lift or engage a clutch on the chairs. If Mark Frank got off early or fell.

Make the necessary assumptions to solve this; state them if you need to.

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#1

Re: Ski Lift Mathematics (February 2024 Challenge Question)

01/31/2024 6:19 AM

180

His "ascent" is all the way to the top.

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#2
In reply to #1

Re: Ski Lift Mathematics (February 2024 Challenge Question)

01/31/2024 7:28 AM

179!

I think I counted the 10/100 crossing twice.

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#32
In reply to #2

Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/26/2024 10:25 AM

I'm equal parts proud and ashamed you got this so quickly.

Nonetheless - congrats!

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#33
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Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/26/2024 10:39 AM

Look at the diagram in the #27 answer. You do not pass the chair in front of you and you do not pass the chair behind you because they are on the end when you get off and on the chair lift. You also do not pass the chair you are on. That makes the correct answer 177

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#12
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Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/01/2024 8:32 PM

I think you are confusing ascent with summit...

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#17
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Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/03/2024 9:17 AM

I think I am using the word ascent like the noun trip.

If you go to A to B via C, your trip is from A to B: especially if the passing of point C is only noticed coincidentally by a third party.

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#3

Re: Ski Lift Mathematics (February 2024 Challenge Question)

01/31/2024 7:49 AM

Chair number 100 must have been at the top when he boarded at the bottom for him to pass it half way up. When he started, #10 - #99 were going up, 90 cars, and 90 cars going down. So he passed 90.

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#5
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Re: Ski Lift Mathematics (February 2024 Challenge Question)

01/31/2024 7:53 PM

I've changed my mind. I believe Randall is correct.

There are 90 chairs going up and 90 coming down for a total of 180.

Why did Mark Frank pass 179 cars? When he got on he was the lowest chair. When he got off he was the highest chair. He passed every chair but his own.

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#34
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Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/26/2024 2:08 PM

Yes, points!

Thank you for stating your logic.

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#4

Re: Ski Lift Mathematics (February 2024 Challenge Question)

01/31/2024 4:36 PM

44....

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#15
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Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/02/2024 4:14 PM

I though the answer was 42.

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#25
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Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/04/2024 8:18 PM

Well we're both coming and going, so it would be * 2 or 84, add to that 5, for being off topic, and we have our answer 89...we must adjust our answer to address the dimensional reality we are currently in...

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#26
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Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/05/2024 1:26 PM

Nicely done.

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#6

Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/01/2024 12:50 AM

177

I had to diagram it.

At the halfway point including 10 and 100 there are 91 chairs to the top total.

That means 45 on each side and one on the end turning around. Because there is one turning around at each end this means there will be two at least two chairs he does not pass. so 180 minus the two on the end turning around and minus the one he is sitting in.

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#7

Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/01/2024 7:23 AM

For the chair 100 and the chair 10 to meet exactly halfway on the skilift, chair 100 must be stopped at the very top while Mark Frank is getting on board on chair 10, which means there are about 98 chairs in between them..

To make the loop, the 98 /2 = means that Frank riding #10 on the way up must pass at most 44 or 45 chairs in order to encounter and/or meet chair #100 midway up the skilift.

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#8
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Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/01/2024 2:51 PM

98?

I think you meant 90.. after all 98/2 doesn't equal 44 or 45

My response to the original question: Too many assumptions within the original question. Equal spacing is good info, but not enough. Chairs (with their numbers) are often removed for maintenance and moved around in position on the lift cabling. No reason to assume the numbers have remained in sequence on the lift.

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#9
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Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/01/2024 3:06 PM

Yes you are right and it was a mistake on my part.

Thank you for clarifying and pointing out this error in my thinking….

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#10

Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/01/2024 8:00 PM

Since you are starting at the bottom of the lift in car #10 and at the halfway point to the top you pass #100, that must mean there are 90 cars coming and 90 going, since we are only passing cars on the adjacent side and only passing half the cars that would mean we passed 44 cars at the halfway point since you can't pass yourself...neither are you passing cars that are travelling with you, the only cars you are passing are going the opposite direction...when you get to the top you will have passed 89 cars in total..but we don't have any information saying we made it to the top, only that we passed car #100 halfway up...at least that's my take...

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#11
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Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/01/2024 8:09 PM

I concur and shares your analysis...

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#13
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Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/02/2024 1:22 PM

90 cars going up (#10-#99) and 90 going down (#100-#180, #1-#9) - correct

If the cars going down were not moving, you would pass 90 cars on your trip up.

But as you are going up, the cars on the other side are going down. At the halfway point, you pass #100, the top car coming down when your started. For the rest of your trip you are passing the cars on the way down that were in front of you when you started.

90 of the original cars going down + the 89 cars in front of you = 179 cars passed

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#14
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Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/02/2024 2:14 PM

I think to enable the meeting of the two chairs, #10 and #100 exactly halfway in their travel, chair #10 should be stopped at the bottom to enable Frank to get on and board the chair #10.. Now in order for the two chairs to meet halfway, chair #100 must also be stopped at the top end of the skilift... which means that the length of the trip must be divided into 2 equal parts..

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#18
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Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/03/2024 8:36 PM

Yes, my mental picture is one car at the bottom and one car at the top where, at each, passengers exit and enter. Between these two cars are two rows of 89 cars, one ascending and the other descending. Think of a very vertically elongated, 180 sided polygon, with the entry/exit vertices at the top and bottom.

When #10 is at the bottom, #100 is at the top, and, by symmetry, #10 passes #100 at the halfway point.

When ascending, #10 passes the 89 cars in front of #100, #100, and the 89 cars in front of himself.

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#19
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Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/04/2024 8:20 AM

I'm sorry, I got lost when you added one additional row to make the 2 rows ....

I was under the impression that the ski lift is just one continuous string of chairs.. with chair #100 serving as the lead... and the 2nd loop was created when chair#100 started its descent to looped back downward simultaneously pulling chair#10 to start its accent..

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#21
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Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/04/2024 10:35 AM

It is very possible for the skilift to have strung only 2 chairs, one on each end of the skilift trip that arbitrarily were labeled as chair#10 on the opposite end of the loop that was labeled as chair #100...

As it is also very possible for the skilift to have a finite number chairs to be equally spaced along one continuous string that was looping around.. In which case the chairs can be sequentially labeled as chair#..... #s... The limiting factor would only be the length of the looped string!

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#16
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Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/02/2024 7:15 PM

Ah yes of course relative speed, if you are travelling at 3mph and they are travelling at 3mph in the opposite direction then your relative speed would be 6mph, and doubling the number of chairs passed....then at the halfway point it would be 89 chairs passed...how could I have missed that....well, that's why we have peer review init...

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Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/04/2024 9:30 AM

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#22

Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/04/2024 11:01 AM

Apart from being bad spellings of two defunct European currencies:-

So maybe there's more to this than meets the eye?

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#23
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Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/04/2024 4:28 PM

I'm thinking it's probably this guy...

https://en.wikipedia.org/wiki/Marc_Frank_Montoya

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#24

Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/04/2024 5:16 PM

Not counting the bull whips I get 90 on top half and 90 on bottom half. When ascending one would pass half of the total. I think he sees 90 cars go down while ascending.

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#27

Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/10/2024 3:44 PM

Below is a map of chair numbers, which may help. Numbers 55 and 145 are going around the end pulleys. The orientation is symmetrical, whether left/right, up/down, or increasing/decreasing chair numbers.

You could pass all other chairs but your own, but if the emkark/debark points are a bit distant from the ends of the loop, you might miss a chair or two.

...145
146-144
147-143
|
|
178-112
179-111
180-110
001-109
002-108
|
010-100
011-099
|
|
053-057
054-056
...055

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#28
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Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/10/2024 8:02 PM

This is what I mentioned earlier. You don't pass the chair behind you and you don't pass the chair in front of you and of course you don't pass the chair you are sitting in so 177 chairs are passed.

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#29
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Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/11/2024 8:08 AM

Why do you think you wouldn't pass the chair in front of you?

Passing the car in front at the getting on point would be similar.

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#30
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Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/11/2024 11:09 AM

Look at the diagram, Unlike the picture there is a chair at the end of the system when you get off and on. According to the original post, chairs 10 and 100 are right beside each other. Ten to 100 is 91 chairs. That is 45 on the down side, 45 on the upside and one on the end.

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#31

Re: Ski Lift Mathematics (February 2024 Challenge Question)

02/20/2024 5:07 PM

Frank is in chair 10, I posit that chair 1 is ahead of Frank, chair 11 is trailing Frank, and chair 100 is at the top opposite Frank. Therefore, the answer is: Frank passes one chair. Clue to my thinking: You might think my answer is off, but if you think about it, you will see my answer is spot on.

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