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Hi Bobby,
You got it!
What method did you use?
I'm just starting to type out a method that basically follows the spreadsheet one, but expresses it differently. I'll post it soon.
Hi Bobby,
Hi phrontister;
If I got this right then I have a different method.
Sorry, but that can't work because:
They're good!
I also reckon that before anyone gets to X-out the first 5 words listed there they'd have Xed-out a full 5-by-5 grid just filled with "in terms of"...a good escape clause by which one can avoid having to think about how to properly phrase their exact intention.
Good for politicians, media presenters, lecturers...and such-like, etc, and so forth, and so fifth.
Another "How old am I?" puzzle like the one for Edna's age that I posted a couple of days ago:
Rita is twice as old as Ann will be when Rita is half as old as Ann will be when Ann is 3 times as old as Rita was when Rita was 6 times as old as Ann.
The sum of the ages of Rita and Ann is 44.
What is Rita's age?
Hi Bobby,
Thanks!
I've got another one of these that can be solved in much the same way (algebra and spreadsheet). I'll post it later on today.
Here are the equations for the algebraic solution I mentioned in my previous post:
Excel spreadsheet image below (for the solution, just change Edna's bogus age in D11 as per the yellow highlighted instruction). Goal Seek aid included.
The formulas in column E were copied from the coloured cells to their left...ie, from columns C & D. Simply type them into those coloured cells to get your spreadsheet to run.
Btw, I used Conditional Formatting to change the fill colour in C20 to green once a correct solution is found...and "WRONG!" changes to "SOLVED!" because of the formula.
Just ignore the second image, as it's replaced by the first. It's playing stubborn and refuses to be deleted.
Hi Avon,
You're quite right! And I use that right-to-left strategy too.
I have a spreadsheet (Excel) solution that I'll post an image of when I get home from work.
I've included formulas and explanations in it, and also an option for using Goal Seek.
I also have an algebraic solution that I'll post too.
Glad you liked it. I've got a couple more of these that I'll post later. One's a new one that I found yesterday and haven't tackled yet, and I'd like to solve it first before posting.
Gotta rush...:(
Hi 1a2b3c2212,
"so confusing". Yes...good for giving one an eye-ache!
Sorry again, but both those equations are incorrect. Understanding the puzzle's logic is at least as difficult as working out a solution strategy!
And if you want a really good headache, try the "Self-Referential Aptitude Test" puzzle I posted about 5 weeks ago (it's just about to fall off page 1 on this forum). It's just a logic puzzle...but very testing!!
but the qn didn't say how big the circle was. could be so big that in 1 day the horse already travel 1 mile leg difference.
Impossible! The horse would have to be 840 feet wide! See my reasoning in post #6 re the single-lap scenario.
Hi 1a2b3c2212,
size of the route is not essential right?
"two of its legs travel 30 miles each day and the other two legs travel nearly 31 miles".
Hi Bobby,
I think that the circle is the best shape for getting the maximum distance between the horses hoofs.
It maintains a consistent increase in the distance difference.
To achieve that 1-mile hoof difference over 30.5 miles (centre of horse's torso), would travelling in a spiral give the same result as for travelling in circles? Too hard for me.
Hi Bobby,
That prog can get all these types of problems.
Eventually...it took 15 minutes.
Had time to wash the dishes, dry them, put them away, fill the kettle, boil the kettle, make a cuppa...came back to the computer...it was still thinking.
Henry is 2 years old and Edna is 6 years old.
Sorry, 1a2b3c2212, but that is incorrect.
If Henry is 2 now, in 10 years' time he will be 12. Half of 12 is 6 - which is older than he is now - and so that doesn't agree with the tense of "was half the age he will be 10 years hence."
Hi 1a2b3c2212,
was bout to discover the ambiguity of the bonus qn until bobbym said it
After I saw Bobby's post I edited my original post yesterday to try to remove any ambiguity. Maybe there are other loopholes I haven't plugged, but my idea was that the horse goes round and round in circles...like one might do at a mill. And so I introduced my bonus question with "Assuming the horse travels in a circle:".
it can be any shape right?? just have to switch direction and 0 displacement.
The shape in the original (non-bonus) question isn't specified...it just has to be such that results in the 1 mile distance difference within one day.
Hi Bobby,
I used Janet's (LibertyBasic) program from shiwaji's "Sum of fractions which equals 1" thread - with some adjustments for your problem - and it found only the solutions you posted above.
Edna is as old as Henry will be when Edna is twice as old as Henry was when Edna's age was half the sum of their present ages.
Henry is as old as Edna was when Henry was half the age he will be 10 years hence.
How old is Edna?
Hi 1a2b3c2212,
You got it...nice!
I've just got a couple of comments about your answers to the bonus questions:
Hi Bobby,
If 1a2b3c2212 is right, that curve is not a circle.
It isn't necessarily a circle, but it can be. I've reworded the bonus question to remove any doubt about it.
Thanks for alerting me to that.
Hi 1a2b3c2212,
that's practically the so-called strategy
Ah...thanks for that. You rescued me from trying to draw the solution in a series of steps, which wouldn't have been easy!
Your time of 2 min 02 sec is pretty impressive! Is that just for the physical crossing bit, or did that also include working out the strategy? If it's both, that sounds mighty incredible!!
A horse travels the same distance every day. Oddly, two of its legs travel 30 miles each day and the other two legs travel nearly 31 miles.
How is this possible?
This is just your everyday, common garden-variety, Joe Average kind of adult horse, of medium build and normal conformation.
Bonus questions once you've solved it:-
Hi 1a2b3c2212,
If the max number of persons on a plank at any one time is 6, then I reckon the minimum number of plank placements is as per my drawing B (ie, 11).
I'd split the 13 people into 2 groups of 6 + a single. So that the max limit of 6 per plank isn't exceeded, people from those groups, and the single, would at times have to occupy stepping stones (also occupied by one end of a plank) while the planks are moved into position (only one person per stepping stone is allowed, I presume). The planks would be passed from one person to the next as best suits the occasion and positioned in the order shown in the drawing.
I think the foremost planks and stepping stones should be populated as soon as possible.
I kept the Word file for my drawing B and moved objects around to find the optimum strategy, but I don't think I could illustrate that well in a single drawing...I'd have to do a series of them.
in reality the plank could squeeze up to 6 people.
I take it that doesn't include the part of a plank that rests on a stepping stone or another plank, otherwise I'm pretty sure it can't be done.
Hi Bobby,
Oops...muffed it! Got lost in my own scribble, and didn't copy down the right answers for #14 & #15. I've corrected my post.
I think I posted the answer to Question #14 from my workings for Bonus Question #4.
Hi Bobby,
Notice how instead of getting 0 he gets 0.20539126e-14, but at least the for next loop terminates.
Yes - I'll have to remember that for situations where that is important (or maybe I should make it a habit anyway).
Some day this quirk won't exist and we won't have to dream up workarounds...and we'll laugh at the clunky ways of the old days.
Hi quittyqat,
Unlike Bobby I could only do a few in my head...but I didn't use a calculator.
More left! (Q13?)