Found 29th Nov 2006
First to get it right can have rep from me :P
Sounds difficult at first but start at one corner n work round..

A professor was conversing with a colleague who told him about a dinner party he had attended with his wife. The colleague told him the following things about the party:

1. There were four married couples present.
2. Each person had a unique hobby.
3. The eight people were seated around a dinner table with the host and hostess at either end and three people seated on either side.
4. Only one married couple were seated beside each other.
5. A man was seated on either side of the hostess.
6. A woman was seated on either side of the host.
7. The hostess likes to ride horses.
8. Donna collects stamps.
9. Carol and her husband were seated on the same side of the table.
10. The piano player was seated next to his brother-in-law.
11. The person who grows roses was seated next to the person who does needlepoint.
12. Frank was seated directly across from the person who builds model planes.
13. Harold is seated to the immediate right of the hostess.
14. George and Betty were seated directly across from each other.
15. Alice is married to Edward.
16. Donna's sister-in-law is seated directly across from Carol.
17. The piano player was seated next to the hostess.
18. The stamp collector's husband was seated across from the model plane builder.
19. Carol was seated immediately to the left of George.
20. Alice is married to the fisherman.
21. The person who does needlepoint was seated across from the actress.
22. The actress was seated immediately to the right of the host.
23. The fisherman was seated across from his sister.
24. Frank was seated next to Edward.

What was each person's hobby, and where did they sit?

  1. Misc
  1. Misc

I like riddles but this ones just way to late for me to give it a go,maybe i'll give it a try in the morning when i know what i'm doing:)
Nice post:thumbsup:

Original Poster

Heh want an easier one?

You traveled three hours at a certain speed. If you had taken 1 minute less to do each mile you would have gone 30 miles farther than you actually did.

How far did you go?

60 miles altogether

If its a simple maths problem and not an actual riddle, I think i'm right!


Original Poster

Yup that's right, here's full explanation for those who can't jump to answer.
If r is the actual rate in miles/minute and R is the hypothetical rate in miles/minute, then, since you would have taken one fewer minute per mile,

1/R=1/r - 1 = (1-r)/r

so R=r/(1-r)

Then, according to the distances

180r + 30 = 180r/(1-r)

Solving for r, we get r = 1/3 mile per minute, so in 180 minutes, you went 60 miles.

Rep will be added in a sec.

Nice one! Good question seancampbell

Original Poster

Ty! how about another then..
In the village of Much-Pedaling-in-the-Marsh, one third of the children can swim, two thirds can ride a bicycle, and one seventh can both swim and ride a bicycle (though not necessarily at the same time). Given that there are fewer than 40 children in Much-Pedaling-in-the-Marsh, how many of them can niether swim nor ride a bicycle?

Oh man, I haven't done sub-sets in aaaaaaaaaaagees. I got as far as 18/21, but I can't be sure that's correct. If it is I guess 3 children. But that's what it is, a guess

zero as 1/3 can swim and 2/3 can ride a all of them can do one or the other.

I'm gonna say 6 children can do neither (or 5 considering "fewer than 40 children")

The reason its not zero fungus is that there are one seventh that can do both (so make up the figure of 1/3 and 2/3). To work this out, I worked out how many kids could ONLY swim or ONLY ride a bike, then minused the total from 40. Not sure if i'm right, I have surgery tomorrow morning and am supposed to be revising!! (Med student!)


BTW, X10 was correct with 18/21, just that that fraction of 40 is 6, not 3


Okay, to answer the initial puzzle (None of the things here are riddles really):

Clockwise starting with the Hostess:
(Hostess)Betty - Horses
Frank - Model Planes
Edward - Fisherman
Alice - Actress
(Host)George - Roses
Carol - Needlepoint
Donna - Stamps
Harold - Paino Player


BTW, X10 was correct with 18/21, just that that fraction of 40 is 6, not … BTW, X10 was correct with 18/21, just that that fraction of 40 is 6, not 3J

LoL, so I can do the hard part, but I can't figure out how many that is from 40, oh my, my noggin.

So what third would the people who can do both be put ???

ok back to the swimming/biking thing......i get 3.....common denominator of 7 and 3 on a number lower than 40 gives 21 children....7 can swim, 14 can bike and 3 can do 3 bikers are also swimmers, so it becomes 7 & 11 which gives 18 and leaves 3 kids that can't do jack.
probably looking at it totally wrong again considering I said zero first time but whatever...

Original Poster

Yup it is 3.
Betty - Horses,
Frank - Piano,
Edward - Fishing,
Alice - Actress,
George - Roses,
Carol - Needlepoint,
Donna - Stamps,
Harold - Planes
was correct too

Original Poster

You have twelve identical-looking coins, one of which is counterfeit. The counterfeit coin is either heavier or lighter than the rest. The only scale you have to use is a simple balance. Using the scale only three times (Note: not loading, but using for balancing), find the counterfeit coin

Do I get my free rep now?

Original Poster

lol sure.. you'll love the little blue user can't affect you block :P

"You have given out too much Reputation in the last 24 hours, try again later."
aww shucks. will have to wait a few min.

How about I give you rep for putting some interesting things on the Misc. forum?

Original Poster

Sure, if ya want :P hardly going to say no lol. roll on block number 5:P


You have twelve identical-looking coins, one of which is counterfeit. The … You have twelve identical-looking coins, one of which is counterfeit. The counterfeit coin is either heavier or lighter than the rest. The only scale you have to use is a simple balance. Using the scale only three times (Note: not loading, but using for balancing), find the counterfeit coin

1. Take 6 of the coins.

2. Put three on one side, three on the other.

3. If the scales balance, discard these six and move on to the other 6.

4. Of the remaining 6, put 2 on one side, 2 on the other. If these balance, discard these 4, meaning of the 2 you have left, one is the odd one.

5. Then, take 2 of the coins you know to be equal on one side of the scale, set one on the other, and add on one of the coins of the 2 you had narrowed it down to. If the scales balance, the odd one is the remaining one you haven't weighed. If the scales don't balance, it's the last one you set on the scales.

If the scales didn't balance in step 2, simply repeat the number 6 process with the 6 coins you set on in point one.

If the scales hadn't balanced in step 4, ah well... i think you know what i means....!

Makes sense to me anyway!!

Nin-JA !


Nin-JA !



LoL - Awesome picture!
I think emmajk42 should give the riddle solvers rep, that would be fair.
Oh, wait, that's me twice and 1 for herself.....I wonder what motivation my suggestion had ;P

Original Poster

Nice one Em!

Another easy one...
There are two concentric circles of radii r and 2r, respectively. of the inner circle 120 degrees are unshaded and the rest shaded, and vice versa for the outer circle, what is the ratio of the total shaded area to the total unshaded area?

1:3 ?

I'm confused, is this what we're meant to be calculating?

If so I concur with ]emmajk42

Original Poster

Heh guess my crib sheet must be wrong as I have it written down as 5:7

[SIZE=2]I get 4:5 or [/SIZE]

[SIZE=2]5:4 can't remember whether i did that as shaded to unshaded or tother way round (did it at work)[/SIZE]

I think we all lost that one then
Next please

Original Poster

Okies, another easy one to get everyone back into it..

You are given 5 bags. There are 10 beads in each of the bags. In four of the bags, the beads each weigh 10 kilograms. In the remaining bag, each bead weighs only 9 kilograms. All the bags and beads look identical. You must find out which bag has the lighter beads. The problem is that all the bags look identical and all the beads look identical. You can use a scale, but it has to be a single-tray scale, not a two-tray balance scale. Also, you may use the scale only once. How can you find out which bag has the lighter beads?

[SIZE=2]Take a bead out and weigh it?[/SIZE]

[SIZE=2]EDIT - ooops - thought there were two bags, now I see there were 5...[/SIZE]

Original Poster

lol you can use the scales once alone so would not work unless you got rather lucky.

getting back to the circles...I now get 3:2...give r a value it works out.

Original Poster

Anyone got an answer to bags & beads or do people want the answer?

I'm still thinking, been working all day so not had a chance to give it some thought. I have to say though this doesn't seem as easy as some of the others.

On a side note I'm listening to Michael Jackson's "Who Is It" and your dancing santa seems to go well with the chorus ]seancampbell

Take different amounts of beads from each bag and put them on the scale e.g. 1 from 1 bag, 2 from another, 3 from another, etc.

Put them all on the scale together, and work out how much it SHOULD weigh if each bead was 10kg. Depending on how much less the weight is, you will know which bag has the lighter beads. e.g. if the weight is 2kgs less than it shoiuld be, the lighter beads are from the bag that you only took 2 from, etc.
(hope we're allowed to take beads out of the bags!!!) :s


"You can only use the scale once"
Put all the bags on the scale together and lift each one off separately (you are still only using the scale once?). The total weight of all the bags will be 490kg? If you take off one bag with the 10kg beads in the weight will drop to 390kg and then the next bag off weight will drop to 290kg. If you take off the bag with the beads weighing 9kg then the balance will go to 200kg/300kg/400kg.
So by taking off the bags separately and noting the remaining balance you should be able to determine which is which.
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