How old?

HybridTech · 06-02-2005 2:32am

ixoy, I wish everybody in his country had your brains,especially our politicians!!!! Now that's fast for the permutations and combinations involved!

ixoy · 06-02-2005 2:39am

fade2black wrote:

A bag of potatoes weighs fifty kilograms divided by half the weight of the bag of potatoes. What is the weight of the bag of potatoes in kilos?

Come on mate! Give me a challenge

The answer is 10kg.

Let x = the weight of the bag of potatoes.

50/(x/2) = x
---> 50.2 = x
----
x

---> 100 = x^2

x=10.

Half of 10 is 5. 50/5=10... So yeah I'm right

fade2black · 06-02-2005 2:40am

Bingo. You want a challenge do ya....ok..give me a couple of seconds...

HybridTech · 06-02-2005 2:41am

Thanks for the inclusion, but gotta go horizontal and recharge batts for another day. Good luck to Ireland v. Italy. So, good night from the European Capital of Culture for another night. God Bless y'all!

fade2black · 06-02-2005 2:42am

In a household, there are some boys and girls. Every girl has as many sisters as she has brothers. Every boy has twice as many sisters as brothers. How many boys and girls are in this household?

ixoy · 06-02-2005 3:00am

fade2black wrote:

In a household, there are some boys and girls. Every girl has as many sisters as she has brothers. Every boy has twice as many sisters as brothers. How many boys and girls are in this household?

4 girls, 3 boys.

Let x = no. of girls
Let y = no. of boys
Every girl has (x-1) sisters and y brothers (as they can't include themself)
Every boy has x susters and y-1 brothers (as they can't include themself)

Therefore -->
1) x-1 = y
2) y-1 = 2x

Solving simultaenously we get values of x=4 and y=3.

Then we can sub it back in. Each boy has 2 brothers and 4 sisters. That's correct according to Line 2.
And each sister has 3 brothers and sisters. As specified.

YAY ME!

fade2black · 06-02-2005 3:04am

ixoy wrote:

4 girls, 3 boys.

Let x = no. of girls
Let y = no. of boys
Every girl has (x-1) sisters and y brothers (as they can't include themself)
Every boy has x susters and y-1 brothers (as they can't include themself)

Therefore -->
1) x-1 = y
2) y-1 = 2x

Solving simultaenously we get values of x=4 and y=3.

Then we can sub it back in. Each boy has 2 brothers and 4 sisters. That's correct according to Line 2.
And each sister has 3 brothers and sisters. As specified.

YAY ME!

Excellent. You're welcome on my quiz team anytime!

ixoy · 06-02-2005 3:12am

fade2black wrote:

Excellent. You're welcome on my quiz team anytime!

Thanks - that was a bit of fun to challenge the old brain. I'm off to toddle to bed now. Ta ta.

fade2black · 06-02-2005 3:15am

Later smart arse

StupidLikeAFox · 19-02-2005 1:40am

haha, i remember getting those types of questions in the junior cert honours, used to love working them out!

grimsbymatt · 21-02-2005 1:04pm

Shrimp wrote:

ok.. I'm on to something..

72 1 1 = 74, 36 2 1 = 39, 24 3 1 = 28, 18 4 1 = 23, 18 2 2 = 22, 12 6 1 = 19, 12 3 2 = 17, 9 4 2= 15, 9 8 1 = 18, 8 3 3 = 14, 6 6 2 = 14, 6 4 3 = 13.

Both 6+6+2 and 8+3+3 equal 14, which means that It has to be one of those two? Cos otherwise it'd be unsolvable..

His youngest likes ice cream or somthing, meaning that there's one thats the youngest.. this would lead us to belive that the combination in 6+6+2.. simple really?

Harking back to this one: there is no indication in the original post that the sum of the ages is 14! Neither is it necessary for the two older children to be the same age - the youngest is still the youngest! Therefore, 6, 4 and 3 could be the ages of the children, as could 6, 6 and 2; 9, 4 and 2; 12, 3 and 2!

Crumbs · 22-02-2005 3:00pm

grimsbymatt wrote:

Harking back to this one: there is no indication in the original post that the sum of the ages is 14!

There is.
As you quoted from Shrimp above, there are 12 groups of 3 numbers with a product of 72. The bartender then tells the customer the sum of the 3 numbers but the customer still can't deduce what the 3 numbers are. This implies that more than one group must have the same sum. The only two groups from the twelve to have the same sum are (2,6,6) and (3,3,8). That's how we know the sum is 14.

Then the third piece of information implies that (2,6,6) is the answer.

grimsbymatt · 25-02-2005 1:09pm

Oh yes. I should have looked more carefully before opening my big gob!

grimsbymatt · 25-02-2005 1:14pm

Ok, here's a little one.

You have a hotel with an infinite number of rooms, all currently empty. One morning, a party of an infinite number of people arrive, each requesting a room. 'No problem,' you say. However, there then arrives another infinitely sized party of people, each requesting a room also.

How are you going to accommodate them all?

[don't Google - that would be too easy!]

Flashling · 04-03-2005 9:20pm

Double up.

Six friends are playing a word game. To see who will start, they each draw a letter tile from a bag containg all the letters of the alphabet. Anna pulls out a J, Simon pulls out an R, Lauren pulls out and H, Joe pulls out a P and Marie Claire pulls out a C. Jaqueline's letter completes the sequence perfectly. What is it?

grimsbymatt · 07-03-2005 11:38am

Flashling wrote:

Double up.

You can't have them double up, they want a room each.

Blisterman · 07-03-2005 8:08pm

They'll all have a room, since there's infinate rooms.

grimsbymatt · 08-03-2005 11:47am

But you have two infinite sets of people, so you won't be able to put the second set in a room, as you will still be putting the first set in for an infinite amount of time (i.e. there can be no point (in terms of room numbers) at which one set ends and the other starts).

Blisterman · 08-03-2005 3:02pm

But infinity X 2 = Infinity, since you have infinity rooms, they each have a room.

Crumbs · 10-03-2005 3:48am

Put one set in the odd numbered rooms and the other set in the even numbered rooms ???

grimsbymatt · 11-03-2005 5:17pm

Crumbs wrote:

Put one set in the odd numbered rooms and the other set in the even numbered rooms ???

bingo

Blisterman · 12-03-2005 10:21pm

Well my answer make sense too.

grimsbymatt · 14-03-2005 1:17pm

Blisterman wrote:

Well my answer make sense too.

It's not possible to give them all a room without employing some sort of strategy. As the groups are infinite in size, you will never get to the end of the first group to be able to assign rooms to the second group, therefore an alternating strategy is necessary.

Blisterman · 16-03-2005 2:06pm

Oh, I get you.

pentesh · 22-03-2005 12:36pm

Flashling wrote:

Six friends are playing a word game. To see who will start, they each draw a letter tile from a bag containg all the letters of the alphabet. Anna pulls out a J, Simon pulls out an R, Lauren pulls out and H, Joe pulls out a P and Marie Claire pulls out a C. Jaqueline's letter completes the sequence perfectly. What is it?

I imagine the answer to this is K. am I right.... Well Flashling??

How old?

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