Riddle
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You have 10 bottles of pills... one of these bottles contains poisonous pills.
The poisonous pills are EXACTLY the same as the normal pills in size, color, etc except that they are 0.01g heavier each. Each normal pill weighs 1g (therefore the poisonous pill weighs 1.01g)
You have an electronic weighing scale with a 0.01g resolution.
How will you USING ONE READING only (ie, you cannot drop more and more pills on top to see how the readings change), to determine which bottle contains the poisonous pills?
If you have heard this one b4, pls don't post the answer till sometime later, let the others have a chance...
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haha storm dun quite understand the weighing part lehz...erm since the poisonous one each pill is heavier den whole bottle shd oso be heavier den weight the bottle one by one lor...see which bottle the heaviest...
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haha storm dun quite understand the weighing part lehz...erm since the poisonous one each pill is heavier den whole bottle shd oso be heavier den weight the bottle one by one lor...see which bottle the heaviest... pass or not?
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haha storm dun quite understand the weighing part lehz...erm since the poisonous one each pill is heavier den whole bottle shd oso be heavier den weight the bottle one by one lor...see which bottle the heaviest... pass or not?
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ya, but you only have one weighing wor... that is, you can only use the weighing machine ONCE... how are you going to weigh all the bottles and still tell which bottle among the 10 is heavier?
OK sorry, you don't have to weigh the BOTTLE... CLUE #1: you weigh the pills. But how are you going to do it? -
OK this is not a riddle. It's more of an IQ question....
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Wat the difference b/w Riddle n IQ question then ??
Isnt it the same ?? Both testing on IQ ??[/quote] -
One need to use brain, the other don't need. -
Which needs ? Which nt needs ?Originally posted by Parka: One need to use brain, the other don't need.
Wat is IQ ??
ME totally blurRed -
using probability to calculate?
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hmmm label each pill u took frm which bottle one den all at once put on the weighing machine. den take out one at a time to see the decreasing rate. u say cnnt drop more n more den i take out one by one can???Originally posted by Storm:How will you USING ONE READING only (ie, you cannot drop more and more pills on top to see how the readings change), to determine which bottle contains the poisonous pills? -
scarli the bottles there got label
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no snowflake, you cannot remove the pills also.
and there are no labels on the bottle. you cannot tell other than the weight of the pill itself.
how? anyone? -
hee. nope...Originally posted by lyonen3:using probability to calculate?
there IS an answer, and it's perfectly logical. No tricks, nothing. Just using pure logic... -
my answer is that , all u have to do is to say ' i gave , i dun know ' then the person who test u will tell u the answer lah...
if he dun know which one is the poisonous one...then u anyhow pick one lor..... -
Highlight the below portion if you want to know the answer.
To simplify the answer, let's use four bottles instead, having one containing the poisonous pills.
Remove 1 pill from Bottle 1.
Remove 2 pills from Bottle 2.
Remove 3 pills from Bottle 3.
Remove 4 pills from Bottle 4.
Combine the 10 pills and weight them together.
If Bottle 1 has the poisoned pills, the 10 pills will weigh 10.1 grams.
If Bottle 2 has the poisoned pills, the 10 pills will weigh 10.2 grams.
If Bottle 3 has the poisoned pills, the 10 pills will weigh 10.3 grams.
If Bottle 4 has the poisoned pills, the 10 pills will weigh 10.4 grams. -
steady lah www!
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Hi Storm,
this is an extension of the qtn u posted earlier:
same scenerio:
You have 10 bottles of pills, but u do not know WHICH and HOW MANY of these bottles contains poisonous pills.
The poisonous pills are EXACTLY the same as the normal pills in size, color, etc except that they are 0.01g heavier each. Each normal pill weighs 1g (therefore the poisonous pill weighs 1.01g)
You have an electronic weighing scale with a 0.01g resolution.
How will you USING ONE READING only (ie, you cannot drop more and more pills on top to see how the readings change), to determine which bottle contains the poisonous pills? -
steady lar!Originally posted by Storm:steady lah www!
honestly speaking, i got the answer somewhere else. -
POWER LAH!!! a beri cleber riddle n answer mann... c",)
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Originally posted by hanniball:
well, thinking about it with my half-sleepy head, I think the principle behind is still the same lor... highlight if u want to see the answer...
Hi Storm,
this is an extension of the qtn u posted earlier:
same scenerio:
You have 10 bottles of pills, but u do not know [b]WHICH and HOW MANY of these bottles contains poisonous pills.
The poisonous pills are EXACTLY the same as the normal pills in size, color, etc except that they are 0.01g heavier each. Each normal pill weighs 1g (therefore the poisonous pill weighs 1.01g)
You have an electronic weighing scale with a 0.01g resolution.
How will you USING ONE READING only (ie, you cannot drop more and more pills on top to see how the readings change), to determine which bottle contains the poisonous pills?
[/b]
if we take powers of 2 number of pills out (1,2,4,8,16,32,64,124,256,512) and weigh them and find out the sum total, we should be able to determine WHICH and HOW MANY bottles are poisonous. I chose powers of two because no matter how you add the bottles, you will always get a unique number... (tinkled with prime numbers for a while, but it didn't work too well)
example: if bottles 1 through 9 are poisonous, you should have an excess weight of 1+2+4+...+256 = 511*0.01g = 5.11g... which is still 0.01g less than if ONLY bottle 10 was poisonous (and therefore with an excess weight of 5.12g)
OK in summary, find the excess weight, convert it to binary format, then where you have the digit 1 is where the poison is. (eg in the above 511 in binary will be 0111111111 so the poisonous pills are in bottles 9,8,7,6...,2 and 1).
Another e.g: if the excess weight was 0.75g, then in binary will be 0001001011 meaning bottles 7,4,2 and 1 are poisonous.
I'm sure there are other combinations of numbers that make up unique sums, but this is my first cut answer... -
lol
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ya correct man. sleepy head can think so well like dat IQ sure pass 130.
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Originally posted by hanniball:ya correct man. sleepy head can think so well like dat IQ sure pass 130.
no lah, initially was sleepy after finishing talking to J'da... then when I saw your post I was awake again. Went back to sleep after posting the answer and checking to see if it holds water lor...
besides... it helps to be trained in electrical engineering! -
OK here's another one. This one should be quite easy lah... prob most ppl heard of this b4 liao.
You're trapped in a room 5m by 5m wide with no doors and no windows. The ceiling is 3m high. In the ceiling there's a small hole, only 1cm across. Water is flowing rapidly from a tap and its flooding the room at a rate of 10cm per minute. That means that you will have only 30 minutes before water fills the whole room and you are drowned. You have only 3 items in the room: a 5m-long rope, a chair and a 1m long metal bar.
Your friends will arrive in 45 minutes to save you from the room. How will you survive that long?
Clue #1: The answer is not to put your nose at the hole. The hole is there for the air to flow out of the room when water displaces it.