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Something for your mind
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| Cosmo |
These were translated, so don't spank me 2 hard! :toothless
Easy ones:
#1:Imagine there's 100 businessmen. 80 of them have mobiles, 70 - pagers and 60 - tamagochies :)
How many of them definetly have all three things?
#2:Imagine: There's an abyss with small bridge over it. It's night. Four men on one side of the bridge need to get to the other side. One of them crosses it in 1 min, second - 2 min, third - 5 min, last one - in 10 min. They have lantern and can cross the bridge only in two or less, and one of them had to carry lantern. Also if two of them crosses it, they walk with the speed of the slowest. I.e. if first and last cross it they'll get to the other side in 10 min.
In what minimal time they all get to the other side? ;) |
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| Sir. Lunchalot |
1. 40
2. 12 min
I hope....;) |
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| Fir3start3r |
I think #2 is 19 min.
#1? I'm workin' on it. :) |
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| Cosmo |
| Both wrong :toothless |
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| OrZonE |
| the first one is 30 |
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| Hell_Copter |
#1 is 30...
i think... |
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| Sir. Lunchalot |
Argh, forgot the ones with the pagers on the first one. Ok, here we go:
1. We have 100 people, 20 of these do no own a mobile. These 20 people could be 20 people who own a pager, that leaves 50. These 50 people could hold the 40 people who don`t won a tamagochi, which leaves 10 people as the correct answer.
2. I stay with my 12 min. Every possible way has to be done with pairs, since pairs are always faster as if one went seperate. One has to go with the slowest (10min). You`d probably take the one with the second slowest time (5min) making it 10 min for the first group. The second group (1min+2min) is as slow as the slowest (2min) making it 2min for the second group. 10+2=12 minutes minimum. |
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| Hell_Copter |
i believe #2 is 20 minuetes
but i'm not sure... i'll think it through tomorrow
i have to go now... |
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| Hell_Copter |
| quote: | Originally posted by Sir. Lunchalot
Argh, forgot the ones with the pagers on the first one. Ok, here we go:
2. I stay with my 12 min. Every possible way has to be done with pairs, since pairs are always faster as if one went seperate. One has to go with the slowest (10min). You`d probably take the one with the second slowest time (5min) making it 10 min for the first group. The second group (1min+2min) is as slow as the slowest (2min) making it 2min for the second group. 10+2=12 minutes minimum. |
you don't understand...
if someone is going... he have to come back
that means that is 1+10 are going... 1 has to go back and that's already 11 minuetes |
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| Sir. Lunchalot |
| quote: | Originally posted by Hell_Copter
you don't understand...
if someone is going... he have to come back
that means that is 1+10 are going... 1 has to go back and that's already 11 minuetes |
AHHHH! That`s what he meant with the lanter....Now I understand! Thanks for pointing that out...:) |
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| Fir3start3r |
for #2...Assuming you have the fastest guy always carrying the lantern...
I'm not about to type all of this so I'm going to use a little Algebra... :p
A = 1 min. man, B= 2 min. man, C= 5 min., D= 10 min.
If A always carrys the lantern...
AB = Trip 1 = 2 min.
AC = Trip 2 = 5 min.
AD = Trip 3 = 10 min.
= AB + A(return trip!) + AC + A + AD
= 2 + 1 + 5 + 1 + 10
= 19 min.
:p |
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| Cosmo |
| quote: | | 1. We have 100 people, 20 of these do no own a mobile. These 20 people could be 20 people who own a pager, that leaves 50. These 50 people could hold the 40 people who don`t won a tamagochi, which leaves 10 people as the correct answer. |
Brilliant! :D
The answer is 10
| quote: | for #2...Assuming you have the fastest guy always carrying the lantern...
I'm not about to type all of this so I'm going to use a little Algebra...
A = 1 min. man, B= 2 min. man, C= 5 min., D= 10 min.
If A always carrys the lantern...
AB = Trip 1 = 2 min.
AC = Trip 2 = 5 min.
AD = Trip 3 = 10 min.
= AB + A(return trip!) + AC + A + AD
= 2 + 1 + 5 + 1 + 10
= 19 min.
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Nope! :toothless
Add-on:when u'll done with this one, some really terrible awaits you! :whip: :D |
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