March 16, 2010

A math puzzle for you

Since I am a math teacher, I offer you a math puzzle. You can listen to the puzzle, then pause the video, think it over, and then resume to see my solution and see whether you agree with me. If for any reason you cannot access the video here, you can find it at YouTube, search edward tarte. (The YouTube version includes a bunch of interesting comments:))
video

10 comments:

asda said...

Hello Edward! This is Alex (vescovifan) from ICC, I hope youre treatment is going well and you get better soon. Maybe we can schedule a game sometime. About the math problem, 60 was my original answer, I noticed you subtracted the 5 intersections, from 10-10 etc. Here's my reasoning: At time 0, 0 sheep have gone across. At time 10min (not included, or just before 10min) 10 sheep have gone across. At time 20min (again not included), 20 sheep have gone across and so on. I think the trick is in those intersections, and the way you phrase the rate, 10min/10sheep and not the only way around. I'll think about it some more.

Edward Tarte said...

Hello, Alex. At time 0, 0 sheep have gone across? No. The problem says it takes exactly 10 minutes for 10 sheep to go over. The word 'exactly' is important. That requires that at time 0, sheep #1 is in mid-air going across; and at time 10, sheep #10 is in mid-air going across. Therefore, at a steady rate, the time interval between the instant that one sheep goes over and the instant that the next sheep goes over is 10/9 minutes. This is not a rate of one sheep every minute. Since you are a university engineering student, I trust that this will make sense to you. Take another look at my two-column table.

jawala said...

"if it takes exactly 10 minutes for ten sheep to jump over a fence,
at that rate, how many sheep can jump over the fence in an hour ?"


rate is defined as 'a ratio between two measurements, often with different units.'


let x represent one unit of time. (10 minutes, in this puzzle)
let h represent six units of time. (1 hour, in this puzzle)
let n represent the total number of sheep.

the ratio between the number of sheep which jump the fence, and the time elapsed is ::
10 sheep per x
which can be represented as :: 10/x

(the representation is not x/10, because time is not dependent on the act(s) of hypothetical sheep -
if it were, there would be far more concerning issues in this math puzzle)


number of sheep per unit h = (rate)(the elapsed time)

n / h = (10 / x)(h)
n / h = (10 / x)(6x)
n / h = 60x / x
n / h = 60

therefore, the number of sheep per unit h is equal to 60.



further speculation ::

there are two factors not addressed within the original question :
the time it takes for one sheep to jump over the fence, and
the time between each jump.

both are of some interest, because if there were NO time between each jump, and
the act of one sheep jumping a fence were instantaneous,
the rate would be an infinite number of sheep per unit time.


implementing those two factors /may/ change our perception of the question.

so, picture the movement of the sheep as a sine or cosine function.
each crest represents the height of a jump; the x-axis represents the ground.
everything below the x-axis represents the time between jumps.


now imagine we were to time the sheep.

at 00:00 - the timer has not yet started.
if we assume that one sheep is already exactly half-way over the fence,
then it would be dishonest to represent the rate as anything other than --.5,
UNLESS, after 10 minutes, the last sheep is also half-way over the fence.

associating the first sheep to the first 10 minutes would mean
we would have to associate the last "half-sheep" to the next ten minutes.


"... at time 0, sheep #1 is in mid-air going across; and at time 10, sheep #10 is in mid-air going across."

is that accurate ?

short answer : no.

0 - sheep #1
1 - sheep #2
2 - sheep #3
3 - sheep #4
...
8 - sheep #9
9 - sheep #10


your error lies in a simple counting mistake.

"... and at time 10, [sheep #11] is in mid-air going across."



as a math puzzle, the question falls short.
as a thought experiment, the question is great.


you are a good man, mr. tarte;
do not allow pride to hinder your sound judgement.


regards,
jawala

bananayoghurt said...

The description of the problem just says it takes 10 minutes for 10 sheep to jump over, It doesn't say anything about them jumping one by one.

It could be a very large fence and all 10 sheep could eat grass for 9 minutes 59 seconds then all jump over at once.

I don't believe your initial description was definitive enough to have a correct or wrong solution, thx

duffry said...

This was an interesting variation of the fencepost puzzle. Thanks.

If you don't know it I remember it from early on in my school days and was delighted to have to answer it for my GCSE oral exam:

"Ten fence posts are spaced in a straight line, one meter apart. What is the total length of the fence?"

I'm reciting that from memory so it may be a little scruffy.

Thanks again.

rafaeldcortes said...

As a math teacher you should know better.... if it takes exaclty 10 minutes for 10 sheep to go over the fence that means that your table is wrong because from 12:00 to 1:00 there are 61 minutes...

12:00 - 12:59 = 60 minutes = 1 hour
or more precisely 12:00:00 - 12:59:59

1 12:00
2 12:01
3 12:02
4 12:03
5 12:04
6 12:05
7 12:06
8 12:07
9 12:08
10 12:09
11 12:10
12 12:11
13 12:12
14 12:13
15 12:14
16 12:15
17 12:16
18 12:17
19 12:18
20 12:19
21 12:20
22 12:21
23 12:22
24 12:23
25 12:24
26 12:25
27 12:26
28 12:27
29 12:28
30 12:29
31 12:30
32 12:31
33 12:32
34 12:33
35 12:34
36 12:35
37 12:36
38 12:37
39 12:38
40 12:39
41 12:40
42 12:41
43 12:42
44 12:43
45 12:44
46 12:45
47 12:46
48 12:47
49 12:48
50 12:49
51 12:50
52 12:51
53 12:52
54 12:53
55 12:54
56 12:55
57 12:56
58 12:57
59 12:58
60 12:59

Your logic is flawed.
Assuming sheep 1 was crossing at 12:00:00 sheep 2 should have crossed at 12:01:00 thus sheep 60 would have crossed at 12:59:00

You may want to rethink your table.

Michael G Montague said...

At the rate of sheep you are counting, it is only 9 sheep per 10 minutes, not ten. Your statement of ten sheep in ten minutes is incorrect. If ten sheep go over the fence in a ten minute time period, then that is one per minute and the answer would be sixty, NOT 55. With your scale, the only correct number of sheep according to your proposal would be between 12;00 and 12:10.

edward said...

This is fascinating. -Edward Woltin

Paul AIA said...

A Math Puzzle for Mr. Tarte

Dear Mr. Tarte:

I enjoy watching and listening to your videos concerning religion, piano and math.

Here is a math puzzle that you may find interesting:

Twenty years ago, I was pondering the song "The Twelve Days of Christmas." I tried to come up with a formula for determining the cumulative number of gifts the singer received on any day since the gift-giving began.

Please note: The singer does NOT receive one gift on the first day, two on the second day, three on the third day, etc. If the sum were simply 1+2+3 . . . +12, the formula would simply be [(n+1)n]/2.

A literal following of the song reveals the singer receives one gift on day 1, three gifts on day 2, six gifts on day 3, . . . and 78 gifts on day 12. The total number of gifts received is 364. I could only arrive at this total by adding the total number of gifts received each day. I tried to work out a formula per the above series, but I was stumped.

I presented the problem to my father, who is a retired engineer. (Incidentally, he was, like you, born in the 1st half of the 1930s. Also, my dad is a devout Catholic deacon.)

My dad got to work on the problem, and arrived at a solution. But, he found the problem to be much more complex than he had anticipated.

The specific question is thus: What is the formula for determining the cumulative number of gifts received as of any given day per the song The Twelve Days of Christmas, where n equals the ordinal number of days that have elapsed since the beginning of the series?

If you like, I can send you my dad's solution, along with two scanned pages of calculations he wrote.

Best regards,
Paul Arquette

Dan Shaw said...
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