Conquer Club

Time yourself. Post your times only, and your solution.

There's a bus. On it are 7 girls. Each girl holds a bag. Inside the bags each are 7 cats who each have 7 kittens. How many legs are on the bus?

Ramned wrote:Time yourself. Post your times only, and your solution.

There's a bus. On it are 7 girls. Each girl holds a bag. Inside the bags each are 7 cats who each have 7 kittens. How many legs are on the bus?

7 girls= 14 legs

49 cats= 196 legs

196 kittens= 604

804 legs.

2 min.

Wrong. Both are. It took me 5 minutes. Think closely. Be cautious.

um buses don't have legs sooo i'm going to go out on a limb here and say zero. zero legs are on the bus.

Wrong still

right

7 girls...14 legs

7 cats...28 legs

forget the kittens they are not on the bus

1 bus driver...2 legs

44 legs in total

wrong, kittens are on the bus. Assume all is on the bus, assume no complications.

HINT: 7 girls..........they each hold 7 bags apiece, and within every single damned bag there are 7 cats and 7 kittens. They all have legs. This should be your thought process.

NOT

7 girls * 7 cats * 7 kittens * 2 legs for the girls * 4 legs for the cats and kittens.

Ramned wrote:HINT: 7 girls..........they each hold 7 bags apiece, and within every single damned bag there are 7 cats and 7 kittens. They all have legs. This should be your thought process.

NOT

7 girls * 7 cats * 7 kittens * 2 legs for the girls * 4 legs for the cats and kittens.

from your first post i thought the girls had 1 bag each lol.

1) 7 girls = 7 girls with 2 legs = 14 legs
2) 7 girls x 7 backpacks = 49 backpacks = 0 legs
3) 49 backpacks x 7 cats = 343 cats (4 legs) = 2,401 legs
4) 343 cats x 7 kittens = 2401 cats (4 legs) = 9,604 legs

Total = 10,990 legs

Unless, of course, there's a bus driver carting around this menagerie which would add another 2 legs, I presume.

Girls can't drive buses.

guylian wrote:

Ramned wrote:HINT: 7 girls..........they each hold 7 bags apiece, and within every single damned bag there are 7 cats and 7 kittens. They all have legs. This should be your thought process.

NOT

7 girls * 7 cats * 7 kittens * 2 legs for the girls * 4 legs for the cats and kittens.

from your first post i thought the girls had 1 bag each lol.

1) 7 girls = 7 girls with 2 legs = 14 legs
2) 7 girls x 7 backpacks = 49 backpacks = 0 legs
3) 49 backpacks x 7 cats = 343 cats (4 legs) = 2,401 legs
4) 343 cats x 7 kittens = 2401 cats (4 legs) = 9,604 legs

Total = 10,990 legs

Unless, of course, there's a bus driver carting around this menagerie which would add another 2 legs, I presume.

CORRECT

i still like my answer.

Ramned wrote:Time yourself. Post your times only, and your solution.

There's a bus. On it are 7 girls. Each girl holds a bag. Inside the bags each are 7 cats who each have 7 kittens. How many legs are on the bus?

You said the girls hold "a" bag each meaning one, misleading question

Also You never said the kittens were in the bags, you merely said the cats had kittens, they could be anywhere.

Anyway here is a more awesome problem, I was well chuffed when i worked it out:

A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.

On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.

The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:

"I can see someone who has blue eyes."

Who leaves the island, and on what night?

I'd like to congratulate everyone on completing this guys homework for him. Well done

This is dumb.

pmchugh wrote:

Ramned wrote:Time yourself. Post your times only, and your solution.

There's a bus. On it are 7 girls. Each girl holds a bag. Inside the bags each are 7 cats who each have 7 kittens. How many legs are on the bus?

You said the girls hold "a" bag each meaning one, misleading question

Also You never said the kittens were in the bags, you merely said the cats had kittens, they could be anywhere.

You also made no reference to how many other people may or may not be on the bus. Therefore, it is not possible to exactly answer your question.

7 girls = 14 legs

7 girls * 7 cats = 49 cats = 196 legs

49 cats * 7 kittens = 343 kittens = 1372 legs

Total = 1582 legs.

About a minute.



EDIT: Wait, each girl has 7 bags? You never said that in the original question...

e_i_pi wrote:I'd like to congratulate everyone on completing this guys homework for him. Well done

yeah that is what astrospace engineering majors are assigned for homework.


Sorry for my imperfect recital of the problem.

pmchugh wrote:

Ramned wrote:Time yourself. Post your times only, and your solution.

There's a bus. On it are 7 girls. Each girl holds a bag. Inside the bags each are 7 cats who each have 7 kittens. How many legs are on the bus?

You said the girls hold "a" bag each meaning one, misleading question

Also You never said the kittens were in the bags, you merely said the cats had kittens, they could be anywhere.

Anyway here is a more awesome problem, I was well chuffed when i worked it out:

A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.

On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.

The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:

"I can see someone who has blue eyes."

Who leaves the island, and on what night?

Okay.

If there was 1 blue eyed person, they would know instantly and leave on the first night. If there were two, then they each would expect the other to leave. When the other person didn't, they would know that they had blue eyes, and both leave the second night. Three people, likewise, would all leave on the third night.

So, if there are 100 blue eyed people, they will all leave on the 100th night.

As for the brown eyed people, they will never know their eye colors unless the Guru tells them something else.

I think the seats the girls are on all have legs as well, work that out

If this problem is on an island, couldn't they just look into the water and see their reflections?

Ramned wrote:

e_i_pi wrote:I'd like to congratulate everyone on completing this guys homework for him. Well done

yeah that is what astrospace engineering majors are assigned for homework.


Sorry for my imperfect recital of the problem.

Also you didn't say the bus was moving - thus there may not be a driver. And, if there was, it might be one of the girls.
If this is a few years ago, the bus will also have a ticket collector.
Maybe an Inspector.
And, wait, each of these girls is carrying 7 bags full of cats/kittens? What about the (inset name of appropriate animal-rights group for your country) protesters?
How do we know all the girls/cats/kittens have all their limbs?
What about the fleas on the cats?
What about mutant girls/cats/fleas with extra limbs.
Are any of the girls Siamese twins?
Is there any food aboard? Chicken legs?
What about the dogs chasing the cats? have any of them managed to catch the bus? (This will depend on whether the bus does indeed have a driver).
And, um, so on.

Incidentally, in answer to a very common logic problem, I currently have 9 odd white socks. Just because they are the same colour doesn't mean they are in pairs.

At Mickey D's, you can get McNuggets in boxes of 6, 9, and 20. What's the largest number of nuggets that you can't order?

john9blue wrote:Okay.

If there was 1 blue eyed person, they would know instantly and leave on the first night. If there were two, then they each would expect the other to leave. When the other person didn't, they would know that they had blue eyes, and both leave the second night. Three people, likewise, would all leave on the third night.

So, if there are 100 blue eyed people, they will all leave on the 100th night.

As for the brown eyed people, they will never know their eye colors unless the Guru tells them something else.

correct!

Blinkadyblink wrote:If this problem is on an island, couldn't they just look into the water and see their reflections?

Sorry i forgot this bit:

There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."

And lastly, the answer is not "no one leaves."

Get round that!

john9blue wrote:At Mickey D's, you can get McNuggets in boxes of 6, 9, and 20. What's the largest number of nuggets that you can't order?

43 is the answer.

44 = 20 + 6*4
45 = 9*5
46 = 20*2 + 6
47 = 20 + 9*3
48 = 6*8
49 = 20*2 + 9

With this list of 6 in a row, you just need to add a box of 6 McNuggets to each of them to get the next 6 numbers. And this repeats forever. So the first time you find 6 in a row, is when you know the answer.

So to know there is no solution to 43...well, that involves knowing that both 6 and 9 are multiples of 3...so if the number is not divisible by three (as 43 is not), then you subtract 20 (because there will be no solution for that number by only using 6 and 9)...43-20 = 23....23 is not divisible by 3, so subtract 20 giving you 3....3 is divisible by 3...but, alas, it is the only number divisible by 3 that cannot be attained by adding the values 6 and 9.

Good one.