Hi I a newbie.....got a riddle for u..... challenge anyone?
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there are 10 trees
how do u plant the trees so that they form 5 straight rows
riddle for thought... -
zhuo zhimple
use our hand loh....
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wah....
u win liao lor......
sabbo me like that...
now the whole world know who am i liao.....
U #$%@?!&*
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wah.... sound so simple yet so not simple...Originally posted by cloudysky:there are 10 trees
how do u plant the trees so that they form 5 straight rows
riddle for thought...
errr..... u plant it 2 in a row??? is it??? -
oops miss out somethingOriginally posted by LxBxHeV:wah.... sound so simple yet so not simple...
errr..... u plant it 2 in a row??? is it???
u need to have 5 rows of 4 trees
using only 10 trees -
chop every tree into half??Originally posted by cloudysky:oops miss out something
u need to have 5 rows of 4 trees
using only 10 trees -
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plant them in a star shape, with a tree at every tip of the star and intersection of the lines.
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that's the correct answer manOriginally posted by Storm:plant them in a star shape, with a tree at every tip of the star and intersection of the lines.
congrats
for that u'll get a star
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i am confuse.... pls explain or draw out.......Originally posted by cloudysky:that's the correct answer man
congrats
for that u'll get a star
thanks..... -
oic... i got the ans already for the trees riddle....
if u guys got any riddle, feel free to post it here, but it must be something logical and not all those "qian pian wen da ti"
well... now i got a new riddle for u guys... a simple one...
Three men checked into a hotel room for which they paid $30. The next day, the manager realized that the men had been overcharged. She gave the bellhop $5 to return to the three men. On the way to their room the bellhop decided to keep $2 for himself, and give each of the three men one dollar. The three men had now paid $9 each, or a total of $27. This plus the $2 the bellhop kept makes a total of $29. What happened to the other dollar? -
well.... here's another interesting and simple riddle....
U have a knief and a round cake. How are u going split it into 8 pieces by cutting it only 3 times? -
hello, another riddle.... a challenging one....
u accidently drop 1 of your new coin into a bag 80 old coins.
they all look exactly the same. The only different is that the new coin is 1 gram heavier than the old coins. (eg. the old coin weight 10g, the new coin weight 11g).
By using a balance machine (those type which when u put equal weight on both side, it will be stable. If the right side of the weight is slightly heavier, it will fall to the right), how are u going to find the new coin out of the old coins when u are allow to use the balance machine only 4 times? -
A though riddle.... specially for those who is very good in maths..... challenge anyone?
Is 1=1+1 ?
a=x
ax=x^2
ax-a^2=x^2-a^2
a(x-a)=(x-a)(x+a)
a=x+a
1=1+1
What did it went wrong ?? -
u cut the cake into 4 pieces using 2 cuts from the topOriginally posted by LxBxHeV:well.... here's another interesting and simple riddle....
U have a knief and a round cake. How are u going split it into 8 pieces by cutting it only 3 times?
for the 3rd cut, u cut the cake from the middle -
highlight to see the answerOriginally posted by LxBxHeV:oic... i got the ans already for the trees riddle....
if u guys got any riddle, feel free to post it here, but it must be something logical and not all those "qian pian wen da ti"
well... now i got a new riddle for u guys... a simple one...
Three men checked into a hotel room for which they paid $30. The next day, the manager realized that the men had been overcharged. She gave the bellhop $5 to return to the three men. On the way to their room the bellhop decided to keep $2 for himself, and give each of the three men one dollar. The three men had now paid $9 each, or a total of $27. This plus the $2 the bellhop kept makes a total of $29. What happened to the other dollar?
The men paid a total of $27. the actual room cost is $25. Plus the $2 the bellhop took makes the $27. You double-counted the bellhop's $2 by adding it to the $27 again to make $29... -
highlight to see the answer:Originally posted by LxBxHeV:hello, another riddle.... a challenging one....
u accidently drop 1 of your new coin into a bag 80 old coins.
they all look exactly the same. The only different is that the new coin is 1 gram heavier than the old coins. (eg. the old coin weight 10g, the new coin weight 11g).
By using a balance machine (those type which when u put equal weight on both side, it will be stable. If the right side of the weight is slightly heavier, it will fall to the right), how are u going to find the new coin out of the old coins when u are allow to use the balance machine only 4 times?
1) split the pile of 81 coins into 3 piles of 27.
2) weigh any two piles. u should be able to find out which pile of 27 contains the heavier coin.
3) split the heavier pile into 3 piles of 9
4) weigh any two piles. select heavier pile
5) split the heavier pile of 9 into 3 piles of 3
6) weigh any two piles. select heavier pile (should contain only 3 coins now)
7) weigh two coins -
highlight to see the answer: (sorry har, in a puzzle-solving mood today)Originally posted by LxBxHeV:A though riddle.... specially for those who is very good in maths..... challenge anyone?
Is 1=1+1 ?
a=x
ax=x^2
ax-a^2=x^2-a^2
a(x-a)=(x-a)(x+a)
a=x+a
1=1+1
What did it went wrong ??
your fourth line a(x-a)=(x-a)+(x+a): basically, if x=a, x-a equals 0, therefore your fourth line is 0=0, and you divided both sides by zero to get your fifth line, which renders your fifth line invalid. -
here's my riddle... was posted in another thread in this forum, so some of you may already know the answer. It comes in two iterations, the second one harder than the first:
Iteration 1:
You have ten bottles of pills. ONE bottle contains poisonous pills. the poisonous pills are totally identical in shape, size, color, smell etc to the normal pills, except that they are 0.1g heavier. A normal pill weighs 1g (therefore a poison pill weighs 1.1g).
You have an electronic weighing machine capable of 0.1g accuracy. How do you weigh the pills in ONE READING (ie, u cannot put the pills on and remove one at a time) and determine which bottle contains the poisonous pills?
Clue: Assume u have unlimited number of pills in each bottle.
Iteration 2:
You still have ten bottles of pills. Now a random number of bottles contain poisonous pills. The poisonous pills have the properties listed above. How do you use ONE WEIGHING again, to determine which bottles contain the poisonous pills?? -
Originally posted by Storm:
hello... thanks for ur hint below..... i got the ans below.....
here's my riddle... was posted in another thread in this forum, so some of you may already know the answer. It comes in two iterations, the second one harder than the first:
Iteration 1:
You have ten bottles of pills. [b]ONE bottle contains poisonous pills. the poisonous pills are totally identical in shape, size, color, smell etc to the normal pills, except that they are 0.1g heavier. A normal pill weighs 1g (therefore a poison pill weighs 1.1g).
You have an electronic weighing machine capable of 0.1g accuracy. How do you weigh the pills in ONE READING (ie, u cannot put the pills on and remove one at a time) and determine which bottle contains the poisonous pills?
Clue: Assume u have unlimited number of pills in each bottle.
Iteration 2:
You still have ten bottles of pills. Now a random number of bottles contain poisonous pills. The poisonous pills have the properties listed above. How do you use ONE WEIGHING again, to determine which bottles contain the poisonous pills??[/b]
take 1 pill from bottel 1, 2 pills from bottel 2, 3 pills from bottel 3 and so on.....
if the weight result is 100.4g, we will know that the 4th bottel is poisionous......
as for the 2nd question.... I am confuse about the question.....
are we giving unlimited pills again? -
yes, still unlimited pillsOriginally posted by LxBxHeV:hello... thanks for ur hint below..... i got the ans below.....
take 1 pill from bottel 1, 2 pills from bottel 2, 3 pills from bottel 3 and so on.....
if the weight result is 100.4g, we will know that the 4th bottel is poisionous......
as for the 2nd question.... I am confuse about the question.....
are we giving unlimited pills again?