Riddle anyone? (bahah)

[Naughtiest_Maximus]

^^^

3000?

 

[astrosmurf]

5000

 

[~::Trip'nPuSSy::~]

4100, not counting for any abiguity implied by the first word.

 

[smileyfish]

4100 - the 10 makes you think it's 5000

 

[astrosmurf]

4100 - the 10 makes you think it's 5000

yeah, i worked that out after i'd posted my answer, but i wasn't going to be so mean as to spoil the fun for others smiley

 

[BopGirl]

Moe, you can't over take the person coming last in the race If they are coming last, there is no one after them :D But yeah ... you can argue that you are lapping them ... bah! hahah

*************************************

At a recent Pets Anonymous reunion, the attendees were discussing which pets they had recently owned. James used to have a dog. The person who used to own a mouse now owns a cat, but the person who used to have a cat does not have a mouse. Kevin has now or used to have a dog, I can't remember which. Becky has never owned a mouse. Only one person now owns the pet they previously had. Rebecca kept quiet throughout the meeting and nobody mentioned the hamster. Can you determine who owns which pet and what they used to own?

 

[che_melbourne]

james had dog has mouse
rebecca had mouse has cat
becky had cat has hamster
kevin had hamster has dog

it works

 

[BopGirl]

Nup! Cos someone still has the same animal they started with

 

[Sllip]

3 guys go for dinner and the bill comes to $25 dollars. They each hand over a $10 note. The waiter decides that it will be too hard for the men to break up $5 each so keeps $2 and returns $1 to each person.
So this means that each person pays $9 for their dinner.

But 9x3+2(for the tip)=29. What happened to the other dollar?

 

[muzby]

dinner hasn't come to $9 for each person plus tip.

dinner has come to $9 for each person INCLUDING the tip.



here's one for ya.

spell out each number, starting at one, then two, three etc etc...

what is the first number you will come to that has the letter "a" in it??

 

[SupaDiscoBreaka]

The three guys were paying for a $25 bill, not a $30 bill. The guys are paying $9 each which is $27. Take away $2 for the tip = $25 (the price of the meal)

and...

2) You have 8 balls, one ball is slightly heavier than the others.

There is no possible way of telling which one is the heaviest without weighing the balls.

You have a tip scale (the old fashion ones with two see-saw plates which balance when both weights are the same).

You can only use the tip scale twice to find out which is the heavist ball.

pft.... thats the girls version...

Try this one out for size:

You have 12 balls and one of them is *either* lighter or heavier. In three turns on the tip scales you must discover which one is the odd one out and whether it is lighter or heavier.

Anyone who can get this without using pencil and paper deserves a gold star.

 

[De Spook]

one thousand

 

[ez_555]

um... i think this may be wrong... but one hundred and one?

 

[ez_555]

ooh shit damn i was wrong *runs and hides*

 

[NickyJ]

Honestly, who goes to a pets anonymous meeting?

How bout this-
James had dog has mouse
Kevin had cat has dog
Becky has always had a hamster
Rebecca had mouse has cat

 

[che_melbourne]

ok... its a long one tho

place 4 balls on each side, and leave 4 balls behind
the scale will be either even (a) or unballanced (b)

FOR (a):
for (a) remove balls on scale and place 3 of the 4 remaining on one side and 3 of the previously used balls on the other they will either be even (a1) or unballanced (a2)

for (a1) remove balls on scale and place the remaining ball (odd one out) on one side and one of the other balls on the other side to find if it is heavier or lighter

for (a2) with the 3 balls just placed (one of which is the odd one out), place 1 one each side of the scales (giving one ball left over) they will either be even (a21) or unballanced (a22)

for (a21) the remaining ball is the odd one out, place it on the scales with another ball to find if it is heavier or lighter

for (a22) one of those balls is the odd one. remove the heavier ball and replace it with the ball left over, if the scales are still uneven then the lighter of the two balls is the odd one and it is lighter than the rest, if they are even then the ball previously removed is the odd one and is heavier than the rest of them

FOR (b):
for (b) remove 3 balls from each plate. place the 3 removed from the heavy plate onto the light plate and place 3 normal balls (from step 1 the ones left over) and place them on the light plate. the plates will either remain the same (b1), become even (b2) or tip the other way (b3)

for (b1) remove the balls that have just been placed as the odd ball is one of the ones that remained in its place, remove the heavier ball and comparethe lighter one with a normal (one that is of equal weight to the others) ball, if it is even then the ball removed is the odd one and it is heavier. if it is still unbalanced then the odd ball on the scales is the lighter one

for (b2) one of the 3 balls removed is the odd one out. from the previous weigh-in it can be seen that this group contains a lighter ball. place 1 of these 3 on each side of the scale, if it is even then the odd ball out is the one remaining and it is lighter, if it is unbalanced then the lighter of the 2 balls is the odd one out

for (b3) the odd ball is amongst the 3 balls moved from one plate to the other. it can also be seen from the previous weigh-in that the odd ball is heavier than the rest. place 1 of these 3 on each side of the scale, if it is even then the odd ball out is the one remaining and it is heavier, if it is unbalanced then the heavier of the 2 balls is the odd one out.

sorry i joined steps 2 and 3 for part b... it was just dragging onn tooooooo much for me.

now... that better be one fuking big star and it better be 21ct

-dee

 

[KANDY-d(-_-)b-RAVER]

man too many drugs for this tread :S

 

[SupaDiscoBreaka]

for (a1) remove balls on scale and place the remaining ball (odd one out) on one side and one of the other balls on the other side to find if it is heavier or lighter

for (a21) the remaining ball is the odd one out, place it on the scales with another ball to find if it is heavier or lighter

for (a22) one of those balls is the odd one. remove the heavier ball and replace it with the ball left over, i

place 1 of these 3 on each side of the scale, if it is even then the odd ball out is the one remaining and it is heavier, if it is unbalanced then the heavier of the 2 balls is the odd one out.

I think you keep forgetting that if the scale is unbalanced it doesn't mean that one is heavier, it could also mean the other one is lighter

 

[che_melbourne]

^^^ SDB i know that but once you have established the odd ball out it is compared to a ball that has been compared to other balls and it is of equal weight...

trust me it works... just read through it carefully or even act it out

 

[BopGirl]

Someone give me another challenge??