Brain Teasers are Back!

Thanks. I’ve still the addition one to fix, but it’s time for sleep. We’re on holiday but heading back tomorrow and I need a clear head.

The big news is, I visited the dealer for a cup of tea and didn’t buy anything! Thinking about a Super-Lumina interconnect for next year though…

I’m glad you enjoyed your (free !) tea at the dealer’s. But next year and those Super-Luminas don’t seem to be too far away !!!

Enjoy the rest of your holiday !

Just to tidy things up, and to confirm Mike’s solution above.

The last half dozen or so were completed as 1-2; 1-6; 0-5. Then a leap down to 5-5; 3-6; 1-5; and 1-4.

If anybody other than Mike gave it a shot, I hope you enjoyed it, whether you finished it or not.

Cheers
Don

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Gifts for those in need

A generous man near us sets aside a certain sum of money each week for equal distribution as gifts among the needy of his acquaintance.

Last week he said, “If next week there a five fewer applying for assistance than this week, those receiving a gift will receive £2 more than this week.”

However, this week there were actually four more applying for a gift than last week, and he pointed out “In which case, each will receive £1 less than last week.”

How much did each person receive this week ?

£6 for 20 people.

I’m going to give you 10/10 Eoink.

However, That was last week’s distribution. This week there were 24 recipients, each receiving £5 ie £1 less than last week when 20 recipients each received £6.

If there had been 5 fewer this week than last week, ie only 15 people, they would each have received £8, ie £2 more than last week.

The sum available for distribution each week is £120

Perhaps I should have put this one into the Use of English thread !

Well done Eoink

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Ah, good point Don. I used “last week”’s numbers as the base for my algebra, so used them for the answer having not read the question again to realise I’d answered the wrong thing.

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No problem Eoink,

I knew you had cracked it and what you had done.

Getting the words for last week, this week, next week was a brain teaser in itself !

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Eoink,

This time round, I had to resort to a bit of trial and error. I think I’n the past I avoided this technique.

I set up a formula with “n” as the number of recipients and kept increasing the number from 1 until the amount of money for distribution was constant for the three scenarios.

Hi Don, my solution was as follows:
Let N be the number of recipients, and G the size of each person’s gift last week. The total given is T.
T = N*G
We also know that there are two other cases for this week:
T = (N-5) * (G+2) = NG - 5G +2N -10
T = (N+4) * (G-1) = NG + 4G - N - 4
Thus NG = NG - 5G +2N - 10
=> 2N - 5G = 10
Also NG = NG +4G -N - 4
=> 4G - N = 4 => 8G - 2N = 8
Add together yields 3G = 18 so G = 6
4G - N = 4 so N = 20

So a total of £120 last week shared between 20 people at £6 each.

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Yes ! Thank you Eoink, that’s great.

I shall have to search out my old hand-written solutions, most were on that 5mm squared graph paper that helped me keep numbers and writing in neat lines and columns. It was failing to correctly multiply the 4G - N = 4 line by two that forced me to resort to the trial and error route. (I didn’t use T; G and N - I used P; x and n - but you know what I mean !)

Just in passing, that penultimate line has a typo - the “46” (I have the same problem, typing up my manuscripts generates typos no matter how much care I take !!!) You might still have time to correct it ?

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Thanks Don, typo fixed.

Neat !

A Wheel Fallacy

I noticed that nobody has tackled the Wheel Fallacy teaser.

OK, the distance AB = π x D so where D = 28 then AB = 28π

Because π is only ever an approximation likewise AB

Now the curved path of a point on the wheel starting at A will pass through X (see diagram below) and end at B after one revolution of the wheel. The length of the curve AXB can be calculated precisely, ie it doesn’t depend upon π

Does anybody know or can anybody guess what that distance AXB is for a 28” dia wheel ?

Just to be clear, the curve AXB is the locus of
Point A on the wheel, on its way to B.

Hope that makes sense ?

…and its called a cycloid …

Ok, the length of the cycloid AXB is 4x28=112" precisely.

That just leaves the area between the curve AXB and the line AB …

The area of the wheel A = πR^2.

The area beneath the curve AXB is 3A

The area beneath the curve AXB, either side of the wheel in its central position is A

To summarise, the area beneath the curve is therefore

A to the left of the central wheel

A = area of the central wheel

A to the right of the central wheel

I thought these mathematical outcomes were rather neat.

Iane Dabbot and the Police !

Iane Dabbot quoted a number (the weekly pay in pounds of an average police constable) that needed to be multiplied by 409 (the number of such constables that would be dismissed each week on the road to total anarchy). No time for the calculator, this was a John Humphries interview, live on air, and the necessary homework was somewhat lacking !

In the time-honoured fashion of many of us in a hurry, Iane placed the first figure of the product by 4 below the second figure from the right instead of below the third. Well, this often happens when there is a 0 in the multiplier. And the subsequent figures were likewise displaced.

The result of Iane’s mistake was that the anticipated weekly cost saving was wrong by £328,320.

What was the multiplicand, ie the weekly pay of an average police constable that Iane initially quoted ?

Drift …… yes, I know. Many of our threads “drift” …. But….

The airplane is flying form A to B at 120kts True Airspeed. The desired track is 270⁰.

The wind is blowing from 225⁰ at a steady 30kts.

What heading should the airplane adopt in order to make good the desired track and what ground speed will be achieved. ?

What will be the resultant drift ? (you can use the terms Left/Right or Port/Starboard)