help me solve this =P |
Please respect our community and follow the rules. There are many types of humor so we can do without those that aim to hurt/offend individuals and groups of people alike.
The community guidelines are addressed to ALL forums, which means the humor forum is undoubtedly included. However, we stress that these rules are especially observed in this forum:
NO OBSCENITY
This includes, but is not limited to excessive swearing, flaming, posting of pornographic images Racism, Homophobic, sexist remarks or bigotry of any sort.
PICTURES: No nudity of any type is allowed on the boards.
NO DUPLICATE TOPICS
If a topic exists a couple of pages away covering the same issues then the new one will be deleted or merged. Look through the pages to see if it has already been posted, if not then it should be okay to post.
Please do not violate the guidelines. It is here for a reason and is not to be ignored.
Thank you.
![]() ![]() |
help me solve this =P |
Aug 1 2004, 01:31 AM
Post
#1
|
|
![]() doot doot doot ![]() ![]() ![]() ![]() ![]() ![]() Group: Member Posts: 1,803 Joined: Jan 2004 Member No: 1,954 |
Part I - there are 12 balls same size same weight same color, everything the same except one all that is heavier than the others... u have a balance scale (the one for the libra sign) how do u tell which is the heavier ball? u can only weigh 3times.
Part II - Same scenario, except this time u dont knoe if the ball is heavier or not, u can weigh 3times. how u do it? |
|
|
|
Aug 1 2004, 01:43 AM
Post
#2
|
|
![]() doot doot doot ![]() ![]() ![]() ![]() ![]() ![]() Group: Member Posts: 1,803 Joined: Jan 2004 Member No: 1,954 |
ok.. i figured out the first part...
first u put 6 and 6 on each scale, since the "odd" ball is heavier, that means its gonna be on the heavier side.. so u take the 6balls that are heavier and split into 3... and then take the heavier 3 and put one on each scale, if they're both equal then the one u didnt put is the "odd" one.. heh i need help! on the second one |
|
|
|
Aug 1 2004, 01:16 PM
Post
#3
|
|
![]() doot doot doot ![]() ![]() ![]() ![]() ![]() ![]() Group: Member Posts: 1,803 Joined: Jan 2004 Member No: 1,954 |
anyone?
|
|
|
|
Aug 1 2004, 01:18 PM
Post
#4
|
|
![]() Word. ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Member Posts: 3,004 Joined: Jul 2004 Member No: 34,673 |
i don't understand the question..... grr.
|
|
|
|
Aug 1 2004, 02:25 PM
Post
#5
|
|
![]() s a r a h r o s e <3 ![]() ![]() ![]() ![]() ![]() Group: Member Posts: 575 Joined: Apr 2004 Member No: 12,944 |
lol im totally confused sry
|
|
|
|
Aug 2 2004, 01:31 AM
Post
#6
|
|
![]() Your love is a razorblade kiss ♥ ![]() ![]() ![]() ![]() ![]() ![]() Group: Member Posts: 1,794 Joined: Apr 2004 Member No: 9,959 |
*confuzed*
|
|
|
|
Aug 2 2004, 05:21 AM
Post
#7
|
|
![]() bang bang! my baby shot me down! ![]() ![]() ![]() ![]() ![]() Group: Member Posts: 754 Joined: Jun 2004 Member No: 23,848 |
ya i got the first one. i heard this question before and remember how i did it hehe... i will b thinking about the asnwer to part 2...
|
|
|
|
Aug 2 2004, 08:13 AM
Post
#8
|
|
![]() Dark Lord of McCandless ![]() ![]() ![]() ![]() ![]() ![]() Group: Member Posts: 2,226 Joined: May 2004 Member No: 16,761 |
The second scenario works exactly the same as the first one.
Stage one, you put 6 on each. If no ball is heavier, tyhey are equal. If one is heavier, than your solution stands. |
|
|
|
Aug 2 2004, 08:21 PM
Post
#9
|
|
![]() dizzy me up. ![]() ![]() ![]() ![]() ![]() ![]() ![]() Group: Member Posts: 3,191 Joined: Apr 2004 Member No: 11,139 |
UM. could u repeat dat in english plz
|
|
|
|
Aug 2 2004, 08:25 PM
Post
#10
|
|
![]() Senior Member ![]() ![]() ![]() ![]() ![]() Group: Member Posts: 875 Joined: Apr 2004 Member No: 13,841 |
uh lol, seems like youre better at this
|
|
|
|
Aug 2 2004, 09:29 PM
Post
#11
|
|
![]() unify and defeat... divide and crumble ![]() ![]() ![]() ![]() ![]() ![]() Group: Member Posts: 2,759 Joined: Mar 2004 Member No: 6,379 |
ur problem confuzzling, but if I understand correctly, what ComradeRed said should be exactly right
|
|
|
|
Aug 2 2004, 09:30 PM
Post
#12
|
|
![]() I Luv Yooh! ![]() ![]() ![]() ![]() ![]() Group: Member Posts: 874 Joined: Jul 2004 Member No: 29,183 |
ackk...kunfused....very, very kunfused!
|
|
|
|
Aug 2 2004, 11:24 PM
Post
#13
|
|
![]() Wow, i dont know whats going on... ![]() ![]() ![]() ![]() ![]() ![]() Group: Member Posts: 1,439 Joined: Apr 2004 Member No: 10,977 |
Holy shit TRIPLE POSTING, thats just crazy
|
|
|
|
Aug 3 2004, 11:19 AM
Post
#14
|
|
|
I am Sandy. Hear me roar. ![]() ![]() ![]() ![]() ![]() ![]() Group: Member Posts: 1,152 Joined: May 2004 Member No: 15,896 |
yerp, comradered is right, i think
|
|
|
|
Aug 3 2004, 12:58 PM
Post
#15
|
|
|
Member ![]() ![]() Group: Member Posts: 13 Joined: May 2004 Member No: 17,124 |
stage one : split the group in half. the heavier six has the heavier ball
stage two: split the heavier group into 3. heavier 3 has the heavier ball stage three: this has two senarios: you weigh only two of the balls -if one ball is heavier, than you got the heaviest ball -if both balls balance, than the extra ball is the heaviest |
|
|
|
Aug 3 2004, 01:19 PM
Post
#16
|
|
|
tag! you're it! ;D ![]() ![]() ![]() ![]() ![]() Group: Member Posts: 555 Joined: May 2004 Member No: 17,572 |
whoa S0 *confuseing ..
|
|
|
|
Aug 3 2004, 01:33 PM
Post
#17
|
|
![]() doot doot doot ![]() ![]() ![]() ![]() ![]() ![]() Group: Member Posts: 1,803 Joined: Jan 2004 Member No: 1,954 |
yea co0nster421 got part 1
part2 cant be done like part 1 cuz u dont knoe if the "odd ball" is heavier or not.. in part 1 i told u its gonna be heavier thats why when u weigh 6 and 6, u take the 6 that are heavier.. in part 2, u dont knoe if the odd ball is heavier or lighter tho lol lets say the odd one is lighter, then u choose the lighter side of course but u dont knoe it, thats thing ... basically part2 asks how do u find the "odd" ball by only weighing the 12balls 3times |
|
|
|
Aug 5 2004, 08:43 PM
Post
#18
|
|
![]() Senior Member ![]() ![]() ![]() ![]() ![]() ![]() Group: Member Posts: 1,384 Joined: Aug 2004 Member No: 37,337 |
alright alright thsi is how u do it:
Number the balls 1 to 12. Weigh 1, 2, 3, and 4 against 5, 6, 7, and 8. If (1, 2, 3, 4) and (5, 6, 7, 8) balance: Weigh 9 and 10 against 11 and 8 (we know 8 is not the odd ball). If (9, 10) and (11, 8) balance: then 12 is the odd one. Weigh 12 against any other to find out if it is heavy or light. If (9, 10) and (11, 8) do not balance: suppose 11 and 8 are heavier, than 9 and 10; then either 11 is heavy, or 9 is light, or 10 is light. Weigh 9 against 10; if they balance, 11 is heavy; if they do not, the lighter of 9 and 10 is the odd ball. (Similar argument if 11 and 8 are lighter than 9 and 10). If (1, 2, 3, 4) and (5, 6, 7, 8) do not balance: Suppose 5, 6, 7, and 8 are heavier than 1, 2, 3, & 4. Then: one of (1, 2, 3, or 4) is light, or else one of (5, 6, 7, or 8) is heavy. Weigh 1, 2, and 5 against 3, 6, and 9. If they balance: then either 7 is heavy, or 8 is heavy, or 4 is light. Weigh 7 against 8; if they balance, 4 is the odd ball, otherwise the heavier of 7 and 8 is the odd ball. If (1, 2, 5) and (3, 6, 9) do not balance: suppose 1, 2, and 5 are lighter than 3, 6, and 9; then either 6 is heavy, or 1 is light, or 2 is light. Weigh 1 against 2 to find out which one of the three choices is true. Otherwise, suppose 1, 2, and 5 are heavier than 3, 6, and 9; then either 3 is light, or 5 is heavy. Weigh 3 against (say) 2 to find out which of the two choices is true. (Similar argument if 1, 2, and 5 are lighter than 3, 6, and 9). |
|
|
|
Aug 6 2004, 12:01 AM
Post
#19
|
|
![]() Senior Member ![]() ![]() ![]() ![]() Group: Member Posts: 121 Joined: Aug 2004 Member No: 37,813 |
oh wow.. i am so bad a logic problems, i would never have thought of the solution.. hmm i will ask some of my friends i they can solve it ^-^
|
|
|
|
![]() ![]() |