Brain Teasers ? or 50 Years On........... ?

Hopefully, before then, somebody else will step in........................

Perhaps this has refreshed a few minds ?

Anyone want to Resolve Vertically at X to find M1 ?

I'm working on the theory that all the Applied Mathematicians are of summer vacation..........

Anyway, I hope it brought back fond memories, rather than nightmares to a few people

Three Sisters............

Three sisters, Susie, Sally and Sheila, had 24 boyfriends between them (don't ask !), each having the number of boyfriends equal to their age three years before. The youngest one, Sally, proposed a swap as follows!!

“I will keep only half the boyfriends I currently have and divide the rest between you two equally. But then you, Susie (being my middle sister), keeping half your accumulated boyfriends, must divide the rest equally between Sheila and me. And then you, Sheila, must do the same.”

They agreed. The result was that each ended up with eight boyfriends. How old is each sister at present ?

Until the Australian Government sort out their tectonic plate movement, this might be of interest to anybody contemplating a driverless car................

PS you can assume no slipping between wheel and kerb.

Don Atkinson posted:

Three Sisters............

Three sisters, Susie, Sally and Sheila, had 24 boyfriends between them (don't ask !), each having the number of boyfriends equal to their age three years before. The youngest one, Sally, proposed a swap as follows!!

“I will keep only half the boyfriends I currently have and divide the rest between you two equally. But then you, Susie (being my middle sister), keeping half your accumulated boyfriends, must divide the rest equally between Sheila and me. And then you, Sheila, must do the same.”

They agreed. The result was that each ended up with eight boyfriends. How old is each sister at present ?

Thought it looked easy enough for a quick go, and it was: They are all the same age: 11.

24 boyfriends 3 years earlier, equal to sum of their ages then, gives away that they are children, even if apparent lack of fidelity didn't. Let a, u, h be sarah's susie's and shiela's ages-3 respectively.

a/2 + u/4 + h/4 = a/4 + u/2 +h/4 = a/4 + u/4 + h/2 =8

solving any first pair gives a=u, similarly h=a or u.  =8, +3

Don Atkinson posted:

Until the Australian Government sort out their tectonic plate movement, this might be of interest to anybody contemplating a driverless car................

PS you can assume no slipping between wheel and kerb.

 

Its a slow day...

My guess:

P = W * sqrt(2Rh - h^2)/(R-h)

Best wishes

Matthew

Innocent Bystander posted:

Don Atkinson posted:

Three Sisters............

Three sisters, Susie, Sally and Sheila, had 24 boyfriends between them (don't ask !), each having the number of boyfriends equal to their age three years before. The youngest one, Sally, proposed a swap as follows!!

“I will keep only half the boyfriends I currently have and divide the rest between you two equally. But then you, Susie (being my middle sister), keeping half your accumulated boyfriends, must divide the rest equally between Sheila and me. And then you, Sheila, must do the same.”

They agreed. The result was that each ended up with eight boyfriends. How old is each sister at present ?

Thought it looked easy enough for a quick go, and it was: They are all the same age: 11.

24 boyfriends 3 years earlier, equal to sum of their ages then, gives away that they are children, even if apparent lack of fidelity didn't. Let a, u, h be sarah's susie's and shiela's ages-3 respectively.

a/2 + u/4 + h/4 = a/4 + u/2 +h/4 = a/4 + u/4 + h/2 =8

solving any first pair gives a=u, similarly h=a or u.  =8, +3

Ah ! Good start Innocent

If they were each 8 years old three years before, then each would start out with 8 boyfriends.

Sally's proposal, would first involve her giving two boyfriends to Susie and two boyfriends to Sheila. Susie and Sheila would each now have 10 boyfriends - so far so good. Now, Susie has to give half her accumulation of boyfriends away. That is five of them - equally divided between Sheila and Sally. Two and a half each..............whoops !!

Matthew T posted:

Don Atkinson posted:

Until the Australian Government sort out their tectonic plate movement, this might be of interest to anybody contemplating a driverless car................

PS you can assume no slipping between wheel and kerb.

 

Its a slow day...

My guess:

P = W * sqrt(2Rh - h^2)/(R-h)

Best wishes

Matthew

A Guess Matthew ?? Well done that man !

Same answer as mine, only in trying to "simplify" I moved an "h" outside of the first set of (brackets) viz:-

P = W * sqrt[h(2R - h)]/(R-h)

Not sure which I prefer really.

Sally is 7, Susie is 10 and Sheila is 16.

It's easiest to find the solution if you work backwards i.e. Sally must have had 16 boyfriends before sharing to end up with 8 and the other girls would have had 4. Susie must have had 8 bfs before sharing (giving 2 each to her sisters who would have had 2 [Sally] and 14 [Sheila]). Hence Sally started with 4, Susie with 7 and Sheila with 13.

Adding 3 years gives their ages as 7, 10 and 16.

Oops, the first line of the explanation should have read "Sheila must have had 16 boyfriends....." 

sjbabbey posted:

Sally is 7, Susie is 10 and Sheila is 16.

It's easiest to find the solution if you work backwards i.e.Sally  must have had 16 boyfriends before sharing to end up with 8 and the other girls would have had 4. Susie must have had 8 bfs before sharing (giving 2 each to her sisters who would have had 2 [Sally] and 14 [Sheila]). Hence Sally started with 4, Susie with 7 and Sheila with 13.

Adding 3 years gives their ages as 7, 10 and 16.

Hi sj, I have highlighted a name that made it difficult for me to follow your logic at first.

But the under-lying logic is sound. I'm guessing it is just a typo.................

.........because you have all the correct ages

Oh! ho! sj, you beat me to it. it took me a while to figure out what you had intended to type and I didn't re-check before posting !!

Soldiers on Parade !

Three soldiers out on patrol crossed a river. Two of them got their ammunition wet. They sorted and shared the one lot of good ammunition equally between them. A short while later they were involved in a brief fire-fight during which they each fired four rounds of ammunition. Their total supply of good ammunition was now down to the number of rounds each had after the division. How many rounds were divided ?

Don Atkinson posted:

Don Atkinson posted:

Not my best diagram, I know. And it seemed to get left behind on the previous page.

Usual conditions. Frictionless pinned joints and a well greased sliding roller, again no friction !

g = 9.8 m/sec²

The system is at rest. what force F is required to keep it static. The connecting rods are "light" ie you can ignore their mass.

You could give the force in Kg, but just multiplying by 9.8 gives it in Newtons and makes it look so much more difficult to a non-engineer or non-scientist

Yes, I know this has been seen before, and fatcat got the answer with a "guess" ..........

This time, the sliding pin or roller is moving left to right at a steady speed of 2m/s. In other words it's  a machine, in which a piston pushes the sliding roller L --> R then pulls it back at a steady speed. The 2kg weight moves up and down.

When the moving machine is in the configuration shown, what is the force F ?

Ok, perhaps a little difficult to cast our minds back to late-school/early university days

The answer is 4.425 Newtons.

I derived a formula for the height "y" of the weight with respect to time "t" (not too difficult, given that the horizontal movement of the weight is a uniform 1m/sec)

I then differentiated this twice to get the vertical acceleration of the weight. (once to get the speed, then again to get the acceleration)

This led me to the vertical forces involved. (gravity opposing the vertical acceleration)

Then, using the concepts from the static version of this "teaser" I derived the horizontal force.

There are other ways of solving this problem, for example you can find the acceleration using geometry or trigonometry.

Oh, for the acceleration I got 3.90625 m/s/s

Don Atkinson posted:

Last winter we were in Canada and in between skiing, trekking and sleigh rides, we enjoyed snowball fights with the kids !

We soon got into the habit of stacking snowballs in a pyramid ready to repel the next attack. I made a tetrahedral pyramid with 10 snowballs and my grandson did the same. So we had 20 snowballs ready to hand. However, this was only just enough to fend off the next attack by Mrs D and our granddaughter, so we decided to double our stockpile.

We found that two tetrahedral pyramids, each of ten snowballs, could be re-stacked as a single tetrahedral pyramid of twenty snowballs. We built two of them and that worked ok to repel the subsequent two or three attacks.

My grandson then suggested that because I could make snowballs faster than him, I should make a “big” stack and he a “little” stack which we could then combine into a single “massive” stack using all the snowballs from the “big” and “little” stacks – with no shortages or remainders……………

We lost the next attack because we took too long to figure out how many snowballs were needed for the “little” and “big” stacks etc.

We also lost the next attack, because we were too busy making all the snowballs needed for the two stacks………….never mind combining them into a single stack……

So, what is the minimum number of snowballs we needed, so as to make one large tetrahedral pyramid from two unequal tetrahedral pyramids. ?

ie “little stack” X snowballs + “Big stack” Y snowballs = “Massive stack” Z snowballs

PS, you can  assume that all the snowballs are perfect spheres, each of the same size.

Let me help..............

Z = 680 snowballs

But how many in each stack ?

123 and 557; completed using a sledgehammer approach

I count a split of 120 and 560 snow balls for pyramid sides of 8 and 14 balls, combined a pyramid side length 15.

Still slow here...

Matthew

Nick from Suffolk posted:

123 and 557; completed using a sledgehammer approach

I also used a "sledgehammer" ie Excel...........

...........but, got a different answer, although it was very close

Matthew T posted:

I count a split of 120 and 560 snow balls for pyramid sides of 8 and 14 balls, combined a pyramid side length 15.

Still slow here...

Matthew

Well done Matthew, my figures exactly

the Soldiers on Parade one, a few posts above, is quite straight forward................

Ok, then 18.

x - 12 = X/3

3x - 36 = x

2x = 36

x = 18

Nicely done sj. Nicely done.

Don Atkinson posted:

Hungryhalibut posted:

Don Atkinson posted:

On a happier note for all those “Remainers” (like me) who are depressed or disgusted with the way things seem to be going at present…………

Winter is approaching at a small village in Siberia. The soil in the top two metres above the perma-frost will soon refreeze so they need to dig enough graves in the village cemetery in anticipation of the  number of deaths ( I told you this was on a happier note than Brexit, didn’t I ?)

The village population is 1000 and it’s known that each person has a one percent chance of dying during the winter period.

What is the least number of graves they should dig so the probability of having enough is at least ninety percent ?

I'm informed that the answer is 14.

Hi HH.

You are well informed. Tell Henry he is well up to the mark.

I hope his A Level results get him into his University Course of choice.

Cheers, Don

Hi Don

As you've been so kind in your support of our Henry, I thought I'd let you know that he has got A*AA in maths, further maths and economics, and is off to Loughborough to study maths. And all from Warblington School and Havant College, with not a penny spent on private schooling or tutoring. We are really proud of him.