First, if you enjoy the Friday Puzzle, I have just posted 101 of them here.
OK, to this week’s puzzle. On Friday I posted this puzzle…..
John is a gardener. The other day he bought a stack of posts and decided to put them around his garden. He planted the posts 2 feet apart and found that he needed 200 extra posts. Annoyed, he pulled up his posts and replaced them, but this time placing them 3 feet apart. Now he had 25 posts over. How many posts did he buy?
If you have not tried to solve it, have a go now. For everyone else, the answer is after the break.
The equation is..
2(x+200) = 3(x-25), which resolves to x=275.
UPDATE: Sorry, that was a typo, answer should be 475!
Did you solve it? Any other ways of getting to the answer?
If you enjoy the Friday Puzzle, remember that I have produced a kindle ebook containing 101 of the previous Friday Puzzles! It is called PUZZLED and is available in the UK here and USA here.
April 25, 2011 at 5:37 am |
I guess u mean 475 not 275!
April 25, 2011 at 6:02 am
Yes, Richard seems to have made a math error. I get 475 as well. I blew this one, however, by not reading the problem properly. I missed that the 25 posts were surplus rather than additionally necessary, like the 200. I need to read more carefully!
April 26, 2011 at 8:46 am
2(x+200)=3(X-25)
2x+ 400 =3x-75
475=x
Accidents (and errors) never happen in a perfect world…
Seems somone wasn’t able to design a perfect world…
April 26, 2011 at 6:09 pm
Faulty logic never happens in a perfect world …
Seems someone wasn’t interested in being coherent …
April 25, 2011 at 5:37 am |
Ummm….I make that x=475. Am I being a div?
April 25, 2011 at 5:42 am |
Yes, it should be 475 (had me going for a moment!)
April 25, 2011 at 5:54 am |
Does this puzzle also prove that you cannot do proper maths in 59 seconds?
April 25, 2011 at 6:29 am |
I came up with 470 because I didn’t think of using the original post twice. I assumed there were two endpoints to the fence which would have left the equations being:
(x+199)2=(x-24)3
This leaves you with x=470.
I suppose I had the idea right but was just a little off.
April 25, 2011 at 1:10 pm
I started out the same way as you did but then realized that the first pole is indeed the last pole as well.
April 25, 2011 at 7:12 am |
How can that be correct, it doesn’t work if you relate back. I had 225, 2x 200 = 3x – 25 = 225 which works if you think about the question. Any thoughts?
April 25, 2011 at 7:38 am
I think you need to check your maths! The equation expands to 2x 400 = 3x – 75
April 25, 2011 at 7:40 am
Typos thick and fast this morning! Should be a plus sign between 2x and 200
April 25, 2011 at 7:50 am |
2x 200 = 3x – 25.
April 25, 2011 at 7:53 am |
I meant to type 2x 200
April 25, 2011 at 7:54 am |
Twitter bug, the plus sign doesn’t come out!!
April 25, 2011 at 8:06 am |
Surely if you are laying a fence, the number of panels is one less than the number of posts, ie if you had three posts, you would have two panels, so in the first instance three posts would mean four feet (3 * 2 – 2) not six feet.
So, equation should be 2(x 200)-2=3(x-25)-3
Which resolves to x=476, one more than the answer above, given that I assume the final fence panel is finished by a final post.
April 25, 2011 at 8:28 am
Twitter bug….2(x plus 200)-2=3(x-25)-3
April 25, 2011 at 8:32 am
Only the question doesn’t mention fences, fence posts or fence panels at all. Just posts. Maybe they’re decorative, or he plans to run wire round them or something. The answer stands, I think.
April 25, 2011 at 8:54 am
Hmm. Even if it’s just posts and no fence, surely the distance covered by them is the distance within the posts? Ie if he puts down only two posts, there is only one space between those posts. So two posts equals two feet not four.
April 25, 2011 at 1:42 pm
I, too, thought this warrants some care with the number of posts vs. fields. As a side note, there is a programming term named exactly for this issue – fencepost error.
However, it turns out that the error does not arise here because its effect is masked. If you denote c as the garden’s circumference in feet, and you consider the number of fields based on x posts, then the equations come out as:
c/2 = x – 1 + 201 = x + 200
c/3 = x – 25
In the first equation, x posts leaving a gap give you x-1 fields covered, and an additional n posts in the gap will hold n + 1 fields — the gap’s end posts are already there.
In the second equation, no correction is needed, since the number of fields and posts along a cyclic arrangement is the same.
April 26, 2011 at 11:36 am
Being a little fussy, I’d say the garden would have a gate of some sort. If this is the case, it is indeed an example of the fencepost error, thus the answer would be 476 posts, not 475.
April 25, 2011 at 8:28 am |
I misread the puzzle as him having 25 extra posts after having bought the 200 extra. If only I’d read it right, I wouldn’t have kept on coming up with x=-175.
April 25, 2011 at 8:30 am |
Carly, only if you are doing a straight fence, if you are doing an enclosing fence the last panel goes back to the first post.
April 25, 2011 at 9:01 am
Ian, who puts a fence round four sides of their garden? Most gardens I know have them just round three sides, hence the two ends don’t join. Is this a good example of a different answer when it comes to practical application of the problem? I guess if you read the question in it’s strictest sense (“around his garden”) I am wrong. But if you assume in the real world that’s very unlikely to happen in that way then I’m right. C.
April 25, 2011 at 9:14 am |
Nice and straighforward, if he doesn’t need a gate or corners or planning permission.
April 25, 2011 at 9:28 am |
It took me several unsuccessful attempts until I suddenly realized that the answer is blindingly obvious. The difference between 2X and 3X is X. In other words the answer is the difference between -400 (2 x -200) and +75 (3 x 25).
April 25, 2011 at 9:32 am |
I see I’m not the only one who managed to get the right equation but the wrong answer due to bad Maths skills…
April 25, 2011 at 1:41 pm |
After about 10 minutes, I somehow stumbled upon the correct answer while avoiding the dreaded algebra. Good for me, I guess.
April 25, 2011 at 3:07 pm |
I was frightened for a time, and then I saw the correction to the typo… phew!… (and although it gave me sweat, it is better to have things these way, wrong answer first, correction later, or else, it would have been puzzling to understand half the comments).
I reach the same result, but as I said Friday, this implied some assumptions, proper for lazy. The posts are to be put around a garden so they must define a closed path, and gaps in number equal to the number of posts. But the closed path cannot be of any kind. I think that’s where the puzzle becomes interesting.
Could the garden have be a rectangle? Likely, yes. But is difficult to see how, and most of all, the basic assumption to solve the problem, to equal 2(x+200) to 3(x-25), is likely forfeit too. The easiest way to puts those posts is to put one in each corner and fill between. This is possible if the number of posts is even (the case where they are 3 ft apart if the solution given is right), and in that case, the sum of distances between posts will give the perimeter of the rectangle. But the assumption of the problem is that, it is possible to do it also with x+200=675 posts, 2 ft apart. Now, it is easy to see that if we start with one post in a corner, the other corners cannot have posts. That means there will be 3 distances that will cut inside the rectangle implying that 2(x+200)<3(x-25), and the problem cannot be solved the way it was. Forgetting the fact that that distance will perhaps cut inside the rectangle and the garden, it is conceivable to dispose the posts at the rectangle border with the same total sum of distances between them, but at least for four small regions at the corners, the path will not coincide. If the path doesn't coincide, and the sum of distances is forced by us, it is a question of a non-verified assumption from our part to solve 2(x+200)=3(x-25).
No problem of this kind would have happened if Richard had said the garden was a pentagon, with a post in each corner…
April 26, 2011 at 12:14 am |
[...] Answer to the Friday Puzzle! First, if you enjoy the Friday Puzzle, I have just posted 101 of them here. [...]
April 26, 2011 at 6:33 am |
I got two answers:
1) same as Richard’s, with a garden completely surrounding the house: 2(x+200) = 3(x-25): x = 475.
2) If I would do this at home, where the houses are build contiguous, I would need an additional 2 or 3 feet after the last post, giving:
2(x+200)+2 = 3(x-25)+3: x = 474.
April 27, 2011 at 4:02 pm |
I was stumped for awhile — not by the math — but looking for the puzzle … a simple algebra word problem is not particularly puzzling, so I assumed there must be something more to it. Thanks, Richard, for the Friday puzzle. It is terrific, and it’s quite ok with me if you phone it in now and again.
April 30, 2011 at 5:35 pm |
I am one week late but I have an interesting error :
as a french I didn’t figured what a “post” meant,
so I thought it was a seed that richard was about to plant across the garden (yes you said “around” the garden, well, that is not so clear).
When you plant 2 feet apart or 3 feet apart, on a surface, you resolve
n+200=S*1/2²
n-25=S*1/3²
n=(9*25+4*200)/(9-4)=(225+800)/5=205
which is closer to richard’s first answer
May 4, 2011 at 3:20 pm |
What I found intriguing is this… Ordinarily your garden abuts your house. If you plant the posts all the way around so that the first post is the last then you just fenced yourself out of your house!
Therefore I assumed one would only fence (or post) along three sides, giving the other answer that some came up with. So the length being posted is:
2(X – 1 +200)
3(X – 1 – 25)
2(X+199)=3(X-26)
2X+398=3X-78
X=476
Mathematically sound and consistent with the general house/garden layout together with the normal human instinct not to ‘fence oneself into a corner’!
May 27, 2011 at 4:10 am |
I came to the answer in headspace without the math… In going from every 2 feet to every three, he increased the length of the fence (over the original 2 foot increment length) by the missing 200 plus the extra 25, or 225. This was a 50% increase in length over the first layout, so the first fence was 3 times that in length, or 675. But he was 200 short that time, so he must have had 475. And since we know 675 at 2 foot increments was the target length, then the length of the garden’s perimeter is 1350 feet.