The Official IQ Questions Thread
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Originally posted by skythewood:
divide the 27 balls into 3 groups, A, B, C
Step 1: compare group A & group B
results/conclusion: if A is heavier. the heavier ballis in set A.
if B is heavier. the heavier ballis in group B.
If A and B same weight, ball is in group C.
divide the group with heavier ball into 3 sets, A, B, C
Step 2: compare set A & set B
results/conclusion: if A is heavier. the heavier ballis in
group A.
if B is heavier. the heavier ballis in set B.
If A and B same weight, ball is in set C.
divide the set with heavier ball into 3 balls, A, B, C
Step 3: compare ball A & ball B
results/conclusion: if A is heavier. the heavier ballis in ball A.
if B is heavier. the heavier ballis in ball B.
If A and B same weight, ball is in ball C.
I don't see how this question is 'up-ed'When you stated the odd ball is heavier.. it already made things easier
Anyway, back to my original question. Any takers? I'll be back in an hour to reveal the answer!
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Originally posted by skythewood:
divide the 27 balls into 3 groups, A, B, C
Step 1: compare group A & group B
results/conclusion: if A is heavier. the heavier ballis in set A.
if B is heavier. the heavier ballis in group B.
If A and B same weight, ball is in group C.
divide the group with heavier ball into 3 sets, A, B, C
Step 2: compare set A & set B
results/conclusion: if A is heavier. the heavier ballis in
group A.
if B is heavier. the heavier ballis in set B.
If A and B same weight, ball is in set C.
divide the set with heavier ball into 3 balls, A, B, C
Step 3: compare ball A & ball B
results/conclusion: if A is heavier. the heavier ballis in ball A.
if B is heavier. the heavier ballis in ball B.
If A and B same weight, ball is in ball C.
eh but 1 gram leh.
how u noe the scale is accurate?
hw u noe u can see a difference?
haha
would've solved it nt for the 1 gram.
lol
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the question is, you don't know if the odd ball is lighter or heavier
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so easy lar.
u do 3 steps, den u ask another person to do another 3 steps.
keep the chain going until u get the answer.
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There is a weighing balance and 12 balls. The 12 balls all look the same from the outside, but 1 weighs a little different from the other 11.
Using the weighing balance not more than 3 times, find out which is the odd ball and whether it is lighter or heavier :P
-UNSOLVED-
Note: You do not know if the odd ball is lighter or heavier. TS will be back in an hour to reveal the answer!
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hmm.. i kind of forgot the answer already. but still remember that the first steps is divide the 12 balls into 3 groups of 4 and then take any 2 groups to have their first reading taken. the rest forgot liao.
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Originally posted by Detached:
There is a weighing balance and 12 balls. The 12 balls all look the same from the outside, but 1 weighs a little different from the other 11.
Using the weighing balance not more than 3 times, find out which is the odd ball and whether it is lighter or heavier :P
-UNSOLVED-
Note: You do not know if the odd ball is lighter or heavier. TS will be back in an hour to reveal the answer!
not so fast leh. i still trying to recall.
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For TS question,
divide the 12 balls into 3 groups, A, B, C
Step 1: compare group A & group B
results/conclusion: if A and B is not same weight. conclusion (I)
If A and B same weight, ball is in group C. conclusion(II)
From conclusion (II) the diff weight ball is in C,
Step 2: compare 3 balls from A & 3 balls from C
results/conclusion: if A and C is same, the odd ball is unweighed ball in C.
if A and C is different, the odd ball is in weighed ball in C
take note whether the ball is heavier or lighter
Step 3: compare 2 of the 3 ball in C
results/conclusion: if unequal, one of them is the ball
if equal, the unmeasured is the ball
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Originally posted by Detached:
By weighing balance, I meant a non electronic scale (imagine the see-saw)
And also, by removing any ball from the balance, it automatically counted as another usage on the balance.
You very ma-fan one.....Put six balls on each side, remove 3 balls from each side first, then followed by 2 balls if they are still balance, then you know which ball is unbalance.
The key is to keep weighing the unbalance balls, be it 3 on each side or 2 on each side. Remember to keep replacing the balance balls with the unbalance balls and stuff them into Detached's mouth and then continue until you get 1 ball on each side.
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Originally posted by parn:
You very ma-fan one.....Put six balls on each side, remove 3 balls from each side first, then followed by 2 balls if they are still balance, then you know which ball is unbalance.
The key is to keep weighing the unbalance balls, be it 3 on each side or 2 on each side. Remember to keep replacing the balance balls with the unbalance balls and stuff them into Detached's mouth and then continue until you get 1 ball on each side.
the very impt thing is 3 tries nia leh -
i know the answer...just have to number the balls from 1 to 12...very tiring if post the whole scenarios..
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i know the answer riao i think. but quite long winded
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From conclusion (I)
mark the heavier side as H1-H4, light side as L1-4, group C ball as S
Step 2: compare H1, H2, L1 against H3, L2, S
results/conclusion = if all the same, the problem ball is H4 is heavy, L3 or L4 is light (A)
= If H1. H2, L1 is heavier, means H1, H2 heavy or L2 is light (B)
= if H3, L2, S is heavier, means H3 is heavy or L1 is light (C)
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From conclusion (A)
Step 3: compare L3 and L4
results/conclusion = if same, H4 is heavy
= if not same, lighter one is lighter
From conclusion (B)
Step3: compare H1 and H2
results/conclusion = if same, L2 is light
-heavier one is heavy
From conclusion (C)
Step 3: compare H3 with S
results/conclusion = if same, L1 is light
= if not, H3 is heavy
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not going to check working le, but you guys get the idea.
How difficult is it to weight the balls like 5, 6 times? KNN......
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group the balls into 3 groups and name the balls ABCD EFGH IJKL.
compare ABCD with EFGH.
if ABCD = EFGH,
then compare AI with JK.
if AI = JK
L is the odd one outif AI > JK or AI < JK
compare J with K.the rest i not sure how. i only know if the first reading balance.
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oh. no wonder.
at least i got the 1st step correct.
haha
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i post some easier one.
3 guys are buried in the desert to the neck.
they are buried in a row, so guy1 see nothing, guy 2 can see guy 1, guy 3 can see guy 1 and 2
if they shout the colour of their hat, they will be saved.
they know the colour of the hat is either black or white, and that all 3 hats is not the same colour.who should guess the hat colour?
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Originally posted by skythewood:
i post some easier one.
3 guys are buried in the desert to the neck.
they are buried in a row, so guy1 see nothing, guy 2 can see guy 1, guy 3 can see guy 1 and 2
if they shout the colour of their hat, they will be saved.
they know the colour of the hat is either black or white, and that all 3 hats is not the same colour.who should guess the hat colour?
if guy 3 nvr shout, guy 2 will know what hat he is wearing.
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Originally posted by skythewood:
i post some easier one.
3 guys are buried in the desert to the neck.
they are buried in a row, so guy1 see nothing, guy 2 can see guy 1, guy 3 can see guy 1 and 2
if they shout the colour of their hat, they will be saved.
they know the colour of the hat is either black or white, and that all 3 hats is not the same colour.who should guess the hat colour?
onli 1 of em can guess and if guess correctly den all 3 will be saved?or all 3 must guess correctly?
or 3 can guess, but just need 1 correct answer?
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freedom is correct....
just one guy shout.
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Originally posted by skythewood:
freedom is correct....
just one guy shout.
explain the answer leh.haha i dun rly understand it.
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scenario 1:
guy 3 see guy 1 and guy2 same hat colour
solution: guy 3 shout the answer.
scenario2:
guy 3 see guy 1 and guy 2 diff hat colour.
solution: if guy3 never shout, guy 2 knows guy1 and guy2 diff hat colour. guy 2 shout answer.
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oh haha i see i see.
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maybe i post a question too. this one quite easy.
a mosquito coil can burn for 1 hour. by giving you 2 identical mosquito coils and a lighter, how can u find out a time of 45 mins. you are only allow to use the given items to find the time.