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Originally posted by EarlNeo:
I assume itis a balance weight that u are talking abt. as in those with 2 tray on each side.
sperate the ball into group of 3, 3 and 2.
weight the two group of 3 balls on each tray. If balance, weight the other 2.
else not blance, take the group of 3 balls (heavier gp) and put one on each tray and leave the remaining one out.
If not blance, u got ur answer. Else if balance, the remaining one is the heavy one
Yup, this is the answer! Well done!
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Hi all,
More IQ questions for you to solve!
Question:
1. You have 8 balls. 7 balls are of the same weight, shape, colour, and texture except...
2. One ball is heavier, everything else cetera peribus as above.
3. You have 2 weighs. Calibrated and everything is identical to the dot.
4. You only have 2 tries with the weighs to tell find out the heavier ball.
How?
Note:
1. Some guys might be overly distressed knowing that they have eight balls. Please treat this hypothetically.
2. You cannot use other tools except for those mentioned above.Thank you.
Edited by nehpyh 11 Nov `08, 11:05AM
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Originally posted by FireIce:
Man 1 Man 2
C1 C2
Pro 1 Pro 2
Man 1 wear C1, and C2 over C1
Man 1 bibaboom Pro 1
(Pro 1 in contact with C2 only)
Man 1 gives C2 to Man 2
Man 2 wear C2 to bibaboom Pro 1
Man 1 proceed to bibaboom Pro 2 with C1
Man 1 finish and pass C1 to Man 2
Man 2 wear C1 over C2 and bibaboom Pro 2
but then, since the Pros oredi got STDs, then last step is no need lah
Man 2 can wear C2 and bibaboom Pro 2
Sorry for the hiatus....busy with budgeting :(Yes, and the winner is our very own Vanila FireIce-the-one-with-the-crooked-ciggie-and-a-cuppa mod - the coolest mod in SGF! She got every details right.
I must confess I should have clarified that bonkers cannot cross-infect the pros too. So more points to Vanila-... mod for being a true CSI!
Actually, LifeIsBeautiful's answer is the easiest (based on Occam's Razor) and 'morally correct' tyres - Dun do it. But I guess we cannot answer that way. Imagine you go to the exam hall and you simply write "you should know better" in all answer scripts!!! LOL.
Well done!
Edited by nehpyh 11 Nov `08, 10:44AM
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Originally posted by Mostwanted5125:
Let exterior of outer condom be x
let interior of outer condom be y
let exterior of inner condom be z
let interior of inner condom be f
Guy A slides in the combine condom and boom girl1,
Then slide out inner condom to boom girl 2
And y and z are both fluid free.
So guy B slide into y and boom girl 1 without guy A fluid.
Then B continue to boom girl 2.
Is the assumption that both guys cannot touch each other fluid or all the 4 of them?
Guys cannot touch the fluid such that they may contract STDs but they can touch, fondle, feel, etc...the ladies...if not no fun.Still 70/100.
Missing links at the end. Also, denotations are not used effectively. Can be clearer. ;)
Edited by nehpyh 07 Nov `08, 12:02PM
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Originally posted by Mostwanted5125:
Hmm. guy A use 2 condoms one on top of the other boom girl 1. Then take it off.
The inner condom exterior no fluid..so boom girl 2.
While Guy B use the outer condom of boom girl 1 cos its interior is clean. After boom girl 1 then he cna go boom girl 2.
This way, Both guys not touching each other fluid!
Close but no, sorry. 70/100." After boom girl 1 then he cna go boom girl 2."
No illustration here.
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See if you can solve this...
- 2 desperate men with clean bill of health
- 2 condoms
- 2 street walkers with STDs
How can each man boom-boom both girls without getting infected with STDs or touching each other's fluid?
- only normal intercourse (no rear, ear, nose, mouth, or hand, armpit, foot, back of knee etc...)
- no washing of condoms
Answers must be illustrated clearly without bypassing steps and no assumptions.
Dun worry, I have real no-nonsense answers.
Edited by nehpyh 07 Nov `08, 11:44AM
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