can somebody help me with these puzzles please?

part of an xmas quiz, I have got the other ones. I am sure the farmer one is quite easy I just can't remember my trigonometry.

I don't think so.

1. Gold
2. 60m squared
3.Baked Beans
4. Not on a tuesday

The cow one is probably based on 3:4:5 right-angled triangles. HTH.

2. 3200
4. 20

maybe

lets say expenses =100

c = amount claimedper day by tory
l = amount claimed per day by labour

8c + 8l = 100

l = 12.5 - c

but 12l = 100 = 150 - 12c

12c= 50, therefore c = 4 and one sixth

100 divided by 4 and one sixth = 24 days. I think??

1 = Terry Hall
2 = the field's in Norfolk
3 = Kill them all
4 = sorry don't know that one

can't even work out how to draw the field one in CAD

field one is probably 3200 but there are many possible combinations.

infact 3200 is wrong

3200 works - square with 56.56m sides

3600m^2 is the correct answer

3600 is correct

the 30m and 50m at opposite corners, 40m at other corner.
Haven't used trigonometry for years...

Hyp is 80m, angle 45.

opp = sin 45*80
= 56.56 squared

= 3200? No?

arses, how you get 3600?

as in, why is it 60m squared?

A bulls neck is about 1 metre wide and this should be taken into consideration

that's what I got first time but I think there is a second solution

it isn't 3600 imo

I just drew it out and now I'm trying to prove it. Gimme a minute.

I worked it out as per Jimbo

ok, if the field is 60m each side, then the hypetnuse must be

square root of 60*60 + 60*60

dunno how you do square root on windows calc...

but its more than 90, then I can't see how the answer is 3600.

what jimbo and jbha says does seem to work. but you were assuming that the 3 and the 5 are in a straight line. giving a hypotenuse of 8. This is correct, though only by chance. I think.

3600 defo wrong.

have drawn both out on AutoCAD and 3200 is defo right

Don't think you can. I'm fairly sure that there is another solution and I tought I could use the sine rule (non right andgled triangle.) but the I realise that I don't have any of the angles in my diagram so I cannot prove it. The easiest answer is of course 3200. I'm just annoyed that I can't work out another solution which I'm fairly sure exists.

If the rope is in the corners, and tight, then it must have 50m one end, and either 30m the other (creating a right angle triangle) or 50m one end, 40m the other, and 30 m pulling it to a corner. Therefor creating an isosles triangle.. will work out that solution now.

This creates a triangle of length 40m * 30m * xm

Can't work out x as I dont have a feckin calcualtor!

once you know x, square it and see if is 56 or 60.

I think the 3200 answer is right, but the derivation flawed.

i haven't seen any convincing proof yet.

how would you work it out Jimbo? I can't do it without at least one angle and I can't seem to find a way to calculate an angle.

Haven't done any trig for many years.

Question 2 is based on a right angled triangle (as pointed out above)

3,4,5 RA triangle is quite common

3 squared + 4 sqaured = 5 squared

9+16 = 25 - thus we know it is a right angled triangle

in this case 30 squared + 40 squared = 50 squared

given that the 30 and 40 side are at right angles we can use the area of a triangle formula which is 1/2 base x height

1/2 x 40 = 20 x 30 = 600 metres squared

btw it looks as though one or two of you have missed the point of the question

2. A farmer has a square field. There is a bull somewhere in the field. The bull is secured by tight ropes to three of the corner posts. The ropes are 30m, 40m and 50m long. What is the area of the field ?

The fact that it is a square field is irrelevant

I thank you ;)

FFT: none of that makes sense to me. the 30,40 and 50 are not forming a triangle

well done FFT.

only 3 of the 4 corner posts are in use, so therefore a triangle has been created

the 4th corner post is doing the square root of feck all

i agree with pos, the 30 40 50 aren't sides of a triangle, they are distances from a single point

the 30 and the 50 combine, the 40 meets the 2 - they dont form a triangle

actually yes, sorry, think I get it now :^0 :(

Good, you were making me confused there fft ;)

We all agreeing on 3200 then?

3200 is at least one correct answer, however this was derived on the "assumption" that te 5 and the 3 were in a straight diagonal line from corner to corner, and the 4 effectvely goes from one of the other corners to the direct centre of the square. then with pythagoras using 8 as the diagonal of the square, 3200 is easily reached. BUT the assumption is only correct by coincidence. I would like to see a proper proof.

FFT still not making any sense to me.

Yes I am agreeing on 3200m, no probs :)

I know where you're coming from now pos4th, my mistake :|

Still haven't seen a satisfactory proof though ;)

if the 3 & 5 go along the diagonal the 4 would not reach to bull or have some slack in it and therefore not be tight to the bull. The answer cannot be 3200.

pos, even if the 50m and 40m were joined, and the 30m is at the corner, it still makes 3200.

Oh. How did you work that out jimbo?

yes you are right. the only way the 40m rope could exactly hit that diagonal line would be if it went to the middle of the square ie half way along the total (5+3) rope.

You're right, still work to do here I think.

Q 3. Give half of them nothing, give the rest a bit more than they would each indivually receive if you jumped the plank and they divided it equally among themselves (dont know how many pirates there are so cant be exact and too lazy to work out a formula), keep the rest for yourself.

3. You are Captain Jack Sparrow and about to distribute your plunder to your scurvy crew. If more than half of them object to your distribution they will vote to make you walk the plank and keep all the treasure themselves. How can you maximise your own health, safety and take ?

say there are 1000 gold pieces. and 100 crew (includin yourself)

Give 50 of them 11 gold pieces = 550, Give 49 of them nothing, and keep 450 for yourself? Only 49 would vote you out!!! I think that is what jbha is saying. good work, hadn't reallly got a clue what that question was about tbh.

Expenses: does the Tory take the same as the Labour MP: 12 days?

Capt Sparrow: divvy it up between himself and half the crew, leaving the other half of the crew with 0?

Thanks for all the effort guys. I think I will go with 3200 at the moment until I see a more convincing argument.

What do you think about the other questions? I think the pirate one maybe give half the loot to half the crew so they all get an equal share, and keep half the loot yourself.

Here are the other questions from that section (I think I already have the answers to the but wont post the yet so if you want you can have a go)

What is more likely to fall on a Friday than any other day ?

A normal duck has two legs. A lame duck has one leg. A sitting duck has no legs. Ninety nine ducks have a total of 100 legs. There are half as many sitting ducks as normal ducks and lame ducks put together. How many lame ducks are there ?

Where might you get rejected rather then selected ?

What are the 10 types of mathematician ?

1 1 1 + 1 1 + 1 = 4
Imagine the incorrect equation above is

Right.

Draw a square, put a line from one corner to the other. This represents the ropes 50m and 30m
Angle created is 45 deg

Need to prove that from one corner to the middle is 40m, so the rope is tight.

Split one of the triangles in two...

So we have two angles of 45, a side of length 40m, and two unknown and a right angle.

Therefore the line joining to the corner must be 40m.

40m squared + 40m squared = 3200

3200 rooted = 56ish.

how can there be half of 99 ducks?

Jimbo747 02 Dec 11:45


pos, even if the 50m and 40m were joined, and the 30m is at the corner, it still makes 3200.




sorry mate, but I am not convinced.

I am pretty sure the answer to the tory/labour one is 24 days. Best way of thinking of it is this:

imagine they have 120quid to spend. labour do that in 12 days, so they are spending 10 qui a day. but with the tories they did it in 8 days during which time labour would've spent 80 quid meaning the tories spent 40 quid in 8 days = 5/day. so it would take him 24 days to spend 120 quid. Maths is easy using 120quid, but could use any figure if you wanted.

These are some of the other questions I am struggling with:


Financial Section:

What went up 8.3% in April ?

What was called the rate at which banks dont lend to each other ?

What is both a safeguard and an ornament ?

May we help you, your company or your parents with financial affairs in any way ?

Qui etait le père ?

What is unusual about car numberplates in TV adverts ?

How many times did Jonny Quid die this week ?

What 3 items are banned from grounds for football matches but permitted at rugby matches ?

first 4 are financial section, rest are from news and culture

1 1 1 + 1 1 + 1 = 4
becomes

1+1+1+1 =4

wd andy

[i]Right.

Draw a square, put a line from one corner to the other. This represents the ropes 50m and 30m
Angle created is 45 deg

Need to prove that from one corner to the middle is 40m, so the rope is tight.

Split one of the triangles in two...

So we have two angles of 45, a side of length 40m, and two unknown and a right angle.

Therefore the line joining to the corner must be 40m.

40m squared + 40m squared = 3200

3200 rooted = 56ish.[/]

but you are already assuming that the 50 and 30 go in a straight line from corner to corner - you are saying that the diagonal of the square is 80m before you've even started. i Dont think you can make that assumption.

my basic theory on the expenses was, together do it in 8 days, individually therefore should do it in 16 days (or at least the average should be 16 days), one does it in 12 therefore other should do it in 20...rob's looks better though

3200 cannot be right. The 40 rope is not joined to the bull!!!

ornament and safeguard is a pound coin

evski is correct. If you assume the 30 and 50 are along the diagonal, then the 40 rope will go to the centre of the square, when the bull is 10m away from the centre of the square.

What is more likely to fall on a Friday than any other day ? Work attendance

JamesBlakesHugeArse 02 Dec 12:09


What is more likely to fall on a Friday than any other day ? Work attendance

not monday then?

i think monday would be the answer to that jbha, you might be along the right track though.

no, not monday

sorry no not work attendance

does the FTSE tend to fall on a friday? (i dno im not a financial person)

good friday?

the 13th

friday answer is at the bottom of this page:

http://mathforum.org/dr.math/faq/faq.calendar.html

james has it

1. What is special about 1/7th ?

You find the pattern of 142857 recurs when you divide any number by 7, if it isn't perfectly divisible by 7?

Handy to know when working in a betting shop. £100000 on a 4/7 wins you £57,142.86. Not that see you that bet too often. ;)

bloody hell DK, very good!

1. What is special about 1/7th ?

You find the pattern of 142857 recurs when you divide any number by 7, if it isn't perfectly divisible by 7?

Handy to know when working in a betting shop. £100000 on a 4/7 wins you £57,142.86. Not that see you that bet too often

I picked up on that one too dk. For instance if someone wants to know what 200 divided by 14 is you know it is 14.2857142857etc etc. 400/14 = 28.57142857etc. A couple of times I have had to work something out which involves dividing by 7, and I can reel it off to say 10 decimal places and then they put it in a calculator and see that you were right and they think you are some sort of genius, but it is just a trick.

:)

That's it pos. You know you're getting 14 lots of 7 into 100, and then you carry on going in the right order. You look like a genius to others, when really you just know that. Have to then keep quiet in front of them to keep the veneer up. :D

there are so many tricks like that. working in a drawing office similar to a bookies in that there are alwys numbers flying about and people shouting out sums cos they can't find their calculator. I've managed to work out loads of little tricks, that make you seem a lot smarter than you really are :)

i think Q2 is a trick. If there is rope attached to 3 of the corners, and all 3 lengths of the triangle are different, the field cannot possibly be square as stated in the question. ???????

Why not?

because in a square, all the sides have to be the same length, surely by using 3 corner posts, the enclosed shape has to be a right angle triangle and would have to include 2 edges of the field. so all 3 sides of the triangle cannot be different.

I think you've misunderstood the question.

Q4 is 24 days NAP
i dont think so, i think youve misunderstood me
anyway back to work........

"the enclosed shape has to be a right angle triangle"

No. The shape enclosed by two edges and two separate ropes (the third rope being from the "middle corner" of the three to the junction of the two other ropes) does not have to be a triangle at all.

Draw 4 dots making a square. Connect the left side and the top side. You have a backwards 7. Now draw another dot near the bottom right dot inside the square and think of that as the bull's location. Now draw a line to that"bull" dot from bottom left dot. Draw another Line to that dot from top left dot to "bull" dot. Draw a third line to that "bull" dot from top right dot. You have created a wonky kite shape inside the field.

One of those lines is 30m, one is 40, one is 50. Two sides of the kite are equal lengths. Still no idea how to work it out from that though.

I think you need to break either the shape or the other part of the square up into right-angled triangles (two right-angled triangles and a rectangle for each part) then use some combination of trigonometry and simultaneous equations. However, having said that, I've not actually tried to do it beyond drawing a picture similar to the one you've just described.

this is "cecil rhodes" all over again

I think the bull is a red herring.....

........and I think the answer is 2500m sq. The 30 and 40m ropes are irrelevant

Bul question has me stumped.

Unless you can prove mathematically, that its possible for 3 corners af 3,4 and 5 to be joined.

If its possible to have 50m pinned in one corner, 40m pinned in opposite, and then 30m pinned between them, it will create a triangle with 50*40*64ish

64/ sqrt(2) = 45.25 (length of one side)

which equals an area of 2048


But I have no idea how you would prove that the 30m would reach the bull... Perhaps someone more clever than me could tell ;)

i'm sad, I actually went home from the office just to try and work this out! Can't get any work done from thinking about it!

i think you might need to combine the sin rule and the cosine rule for non right angled triangles and then solve simultaneously to try and find the angles

Ive got

I like your thinking, boggle.

Does no one know the 'true' answer? This really bugging me now :)

i dont like it cause it uses more than 3 of the corner posts. cant better it myself though

if it loops back on itself, fk, its possible to only use 3 posts.

It looks like thats the winner, and I can finally start some work today.

i don't think that is the winner. There is a situation v.similar to the 3200 answer given earlier that will work. just don't now how to prove it

okthis is the last im committing my mind to this now. if all 3 ropes arejoined, making 1 rope 120m long, and said rope is placed tight around 3 posts making a right angle triangle, isosceles triangle, then the area of the square (ie twice the area of the triangle) is 1235.32 sqm to 2dp. this may not be the answer on the paper but it fits the criteria of the question and the maths is right so i rest on that.

Great band - sorry don't know any answers.