- This topic has 2,250 replies, 179 voices, and was last updated 2 years ago by noitallmr.
January 30, 2009 3:40 am at 3:40 am #1068192
goody613: the rule that every rule has an exception?January 30, 2009 3:50 am at 3:50 am #1068193goody613Member
yeahFebruary 4, 2009 4:00 pm at 4:00 pm #1068194
“I always lie” is another one… it cant possibly be true because he must be lying when he says it!February 5, 2009 10:21 pm at 10:21 pm #1068195
A boy drops a rubber ball from a height of 8 feet above a concrete floor. Each time the ball hits the ground it bounces up 75% of its previous maximum height (it reaches 6 feet on its first bounce, 4.5 feet on the second bounce…). What is the total vertical distance the ball will eventually travel?February 5, 2009 10:27 pm at 10:27 pm #1068196
Dr. Pepper?? wow, pass a canFebruary 5, 2009 10:30 pm at 10:30 pm #1068197charlie brownMember
and a can for me too, please.February 5, 2009 10:33 pm at 10:33 pm #1068198
do you know that i wrote that message when i saw your name and i forgot to read your riddle?? had to go back a second timeFebruary 5, 2009 10:38 pm at 10:38 pm #1068199
dr. pepper: well technically if it always went 75% of the distance it was at then it would never stop bouncing (just like if ur 10 feet from a wall and keep walking half the distance u will never reach it) nbot sure if this is the answer…if its not it would depnd on how high the ball is (so that ui could figure out at which point it would stop bouncing
moish01:hahaFebruary 5, 2009 10:46 pm at 10:46 pm #1068200
yeah but gravity, so it’s gotta stop at some point. i don’t know this stuff but i do know common sense.February 5, 2009 10:53 pm at 10:53 pm #1068201
moish01: i know but that is why the question makes no sense…but like i said it could b but u would need to know the height of the ball so that u can no at which point the ball will stop bouncing…once the height of the ball and the amoint it will bounce is the same then it is resting on the ground…im making this all up but i think it makes senseFebruary 5, 2009 10:57 pm at 10:57 pm #1068202
ah. i hear.February 5, 2009 11:03 pm at 11:03 pm #1068203
moish01: haha really? well mayb u could explain it to me bec im a little confused 😛 jkjkFebruary 5, 2009 11:18 pm at 11:18 pm #1068204
I hope the mishpacha is doing well.
The riddle thread missed you.
On to the task at hand…
The denominator of the fractional equivelent of the missing percentage is how many times the original number will be just short of.
In other words, if it bounces 50%, and 1/2 is the missing portion of 100% the total would be just shy of 2 * the original height.
Since it bounced 75%, and 1/4 is the missing portion of 100% the total would be just shy of 4 * the original height.
The answer is: just shy of 24 feet.February 5, 2009 11:37 pm at 11:37 pm #1068205
I Can Only Try, did you factor in the bounce on the way up and down, or only the way down? 8 feet down, 6 feet up, 6 feet down, 4.5 feet up, 4.5 feet down. That is 29 feet already, and it is still bouncing. That was the “chap” of the riddleFebruary 5, 2009 11:43 pm at 11:43 pm #1068206
… unless the up and down are both counted, in which case I guess it’s the six original feet, plus 2 * (just shy of 18 feet) which would equal just shy of 42′ altogether. (I’d better add the caveat “I think”). 🙂February 5, 2009 11:44 pm at 11:44 pm #1068207
I think you’re right.
Please see my prior post, which I just submitted before seeing your answer.February 5, 2009 11:46 pm at 11:46 pm #1068208
…and my remedial reading teacher pointed out that the original drop said 8′, not 6′
I’m going down to the minors, to work on my swing. ;pFebruary 6, 2009 1:23 am at 1:23 am #1068209
I can only try- It’s great to be back here. The family is great thanks for asking. This morning I saw the baby kicking her feet while deep in concentration- it was the cutest thing- I thought she was trying to figure out where her center of gravity was so she could try to roll over. My wife figured out exactly what she was doing and made me change the diaper…
It’s really touching to see that I was missed. I’m afraid to log on at home when I should be helping my wife and at work our department was assigned some challenging projects so I’m a little reluctant to log on over there (I’m still trying to explain to facilities why I had 3,121 coconuts, 4 colleagues and a monkey in my 5′ X 7′ cubicle…).
Hopefully we should be finished the projects by early July and I’ll be able to return to this thread full time.February 6, 2009 1:24 am at 1:24 am #1068210
moish01- The soda with the same name as me is Dr Pepper (not Dr. since it’s not a real doctor), I hope this removes any confusion.
teen- The problem says “What is the total vertical distance the ball will eventually travel?” not “What is the total vertical distance traveled by the ball when it comes to a complete stop?”. The ball will never come to a complete stop but the vertical distance travelled will eventually reach a limit as the amount of time elapsed approaches infinity, what is this limit?
The answer is 56 feet.
I can only try- 24 feet is the upward distance, multiply that by two for the distance it travelled down each time it bounced up plus the original 8 foot drop gives the correct answer of 56 feet.February 6, 2009 5:19 am at 5:19 am #1068211February 8, 2009 7:57 pm at 7:57 pm #1068212gleegMember
Dr. Pepper – you seem like a wise man but the correct answer is -8 feet. The total distance traveled can be easily calculated since it is a telescoping sum:
Distance = – 8 + 6.5 – 6.5 + 4.875 – 4.875 + …..
‘. = – 8(+ 6.5 – 6.5)+(4.875 – 4.875)+(…..
= – 8February 9, 2009 1:37 am at 1:37 am #1068214
gleeg- if you studied physics at all, you would realize that negative/positive signs are there only to signify direction (left or right from the starting place). He asked for the “total vertical distance” which would be the absolute value of the distances… or scalar vs vector quantities…
am I right Dr. Pepper?February 9, 2009 5:53 am at 5:53 am #1068215
does anyone have a riddle that doesnt require such…uhhh…mathematical knowledge?February 11, 2009 1:02 am at 1:02 am #106821622OldGoldParticipant
what is greater than G-d
more meaner than the devil
a poor man has it
a rich man needs it and when you eat it you die?February 11, 2009 3:11 am at 3:11 am #1068217
nothing is greater than god
nothing is meaner than the devil
a poor man has nothing
a rich man needs nothing (else)
and when you eat nothing you die
😀 make sense?February 11, 2009 3:32 am at 3:32 am #1068218
more meaner? english, dude!February 15, 2009 6:06 am at 6:06 am #1068219
The distance traveled can not be a negative number. If the question was how much higher is the ball when it stops bouncing then -8 would be correct.February 15, 2009 6:12 am at 6:12 am #1068220
Special for valentines day;
You have a regular deck of 52 cards with two cards missing (you don’t know which 2).
A card is selected at random, what is the probability that the suit of the card picked is Hearts?February 15, 2009 5:39 pm at 5:39 pm #1068221
If I asked the other guy which road not to take, which one would he tell me?
They would both lead you to the ideal pathFebruary 15, 2009 5:40 pm at 5:40 pm #1068222
What 4 Brachos are said once a year, every year?
(Please note, this is geared towards Jews in the diaspora, therefore, brachos like “al achilas Marror would be incorrect)February 15, 2009 6:23 pm at 6:23 pm #1068223
is it a 24% chance?February 15, 2009 9:16 pm at 9:16 pm #1068224
I’ll go with “Teen”s answer. Unlike the “Monte Hall” puzzle, if the missing cards are truly random I don’t think the odds are affected (four suites plus two wildcards, I think – I don’t really know cards).
1) Tal 2) Geshem 3) Biur Chometz 4) I’ll (probably) kick myself for missing this one.February 15, 2009 9:21 pm at 9:21 pm #1068225
There is no Bracha for Tal, nor for geshem. They are t’fillas. I am referring to brachos!
Biur chametz is correct thoug (3 left!)February 15, 2009 9:49 pm at 9:49 pm #1068226
jaymatt what is the diaspora? and whats it got to do with marror?February 15, 2009 9:50 pm at 9:50 pm #1068227
jaymatt: how about the one where u find the chametz (forgot the name of it)February 15, 2009 10:01 pm at 10:01 pm #1068229squeakParticipant
1) Neiros Erev Yom Kippur
2) Biur Chometz
3) Bircas Ilanos
4) The chasimah for Nachem (on tisha b’av by mincha – though some sefardim say it 3x)
Note: For those of litvish origins, the brocho of “Haposeach lonu shaarei rachamim” is also said once a year (instead of “Yotzer ohr u’voray choshech” on Yom Kippur). But some other nusachos never say it, and some others say it on Rosh Hashono as well. So litvaks could count it as a fifth.February 15, 2009 10:09 pm at 10:09 pm #1068230squeakParticipant
Here is a riddle of my own:
In what situation might one not be allowed to daven shemona esrai – even though one is in a perfectly kosher shul – because of a bowing-down problem?February 15, 2009 10:36 pm at 10:36 pm #1068231
Lehadlik nair shel Yom haKipurim (not sure if you’re counting this since men usually don’t make this brocha)February 15, 2009 10:38 pm at 10:38 pm #1068232
Eruz Chatzeiros? dont people do that once a year?
hmm… what about one that not everyone says, but that one normally only says once a year… ie: the bracha a father says at his son’s bris? it is possible though to have two sons born within a year who were not twins but that is not the norm… ok, not everyone has a son every year but I do know some families where it goes almost without fail Bli Ayin Hara…
the bracha for megila? wait, you do that though in the morning and the evening…
ok, I am not sure if any of these are what you are looking for, but it was a try…February 15, 2009 10:47 pm at 10:47 pm #1068233
I gave up and googled it.
I never would’ve gotten 3 or 4, although I should’ve gotten 3.
Good question(s).February 16, 2009 2:46 am at 2:46 am #1068234
Yes ames you are correct, the answer is 1/4.
Here’s how I solved the problem when I had it on an exam.
There are 1,326 unique ways to choose 2 cards from a standard 52 card deck. (52 x 51 / 2 = 1,326. We divide by two since order doesn’t matter i.e. a seven of spades and a king of clubs is the same group as a king of clubs and seven of spades.)
Of the 1,326 possibilities 78 include 2 hearts (13 * 12 / 2= 78), 507 include one heart and one non-heart ( 2 * (13 * 39 / 2) = 507) and 741 consist of two non-hearts (39 * 38 / 2 = 741). (Please note that 78 + 507 + 741 = 1,326.)
If two hearts are missing from the deck the probability of picking a heart is 11/50.
If one heart is missing from the deck the probability of picking a heart is 12/50.
If no hearts are missing from the deck the probability of picking a heart is 13/50.
So the probability of two hearts missing (78/1,326) x the probability of picking a heart (11/50) = (78/1,326)*(11/50).
Summing up the probabilities from all three scenarios gives us (78/1,326)*(11/50) + (507/1,326)*(12/50) + (741/1,326)*(13/50) = 1/4.February 16, 2009 3:28 am at 3:28 am #1068235
when can you not bow down?? if there is an untzniyus lady (for some reason) on the men’s side… You said in a perfectly kosher shul, so I thought that refered to no A”Z there, but maybe there was a person there that should not be there. (Or what about on a stone floor on yom kippur when the chazzen bows down??)February 16, 2009 3:47 am at 3:47 am #1068236
ames: i dont think u had all that in mind haha
dr pepper: uhhh sure haha if u say so lol
what i did was find the chance of 2 hearts beings picked out first then the chance that if those 2 hearts were picked of picking another heart but i see where i went wrong i automatically assumed 2 hearts were picked right away but i realize now its not for sure like that thatFebruary 16, 2009 4:11 am at 4:11 am #1068237qwertyuiopMember
dr. pepper: i wonder if anyone besides you(and probably Icot) understood that.$February 16, 2009 5:07 am at 5:07 am #1068238
I doubt this is the answer you’re looking for, but are you referring to one who may have difficulty controlling their bodily emissions?
Your explanation was detailed, technical, correct and way over my head.
I would simply say that if any number of cards are removed from a deck at random, the odds that the next card in the deck is a heart are 1/4 (a 52 card deck with no wild-cards are included).February 16, 2009 1:49 pm at 1:49 pm #1068239xeroxMember
well squeek- watz the answer? :o)February 16, 2009 8:06 pm at 8:06 pm #1068240
I can only try
That’s correct, no matter how many cards are removed (up to 51 and assumning that it is not known which cards are removed) the probability of the picking a heart will be 1/4. Let’s see you prove it for where 32 cards are removed.
(Hint- there are 125,994,627,894,135 different ways that 32 cards can be chosen from a deck of 52 where order doesn’t make a difference. Hope this helps!)February 17, 2009 12:32 am at 12:32 am #1068241The Big OneParticipant
Dr. Pepper, I saw your post in the Funny Shidduchim Stories thread, about your ’89 camp days. I gotta say you’re a lot younger than I imagined. Embarassingly enough, when you hinted at your mazal tov a few months back (without saying exactly for what), I was about to say that you should have a lot of nachas with this einekel and all you einekelach! Thankfully, by time I got back to post you already indicated it was your child.February 17, 2009 1:43 am at 1:43 am #1068242
The Big One-
My wife can’t stop laughing at your post. What did I say that made you think I am older than I really am?February 17, 2009 2:00 am at 2:00 am #1068243The Big OneParticipant
Dr. Pepper, I’m glad to have brought laughter to your family 🙂
I can’t pinpoint anything in particular, but probably your vast sum of mathematical genius played a role, as well as the sophistication of your other postings.
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