Math Problems!........share It Here!

[black cat]

Bat ganun.. when I was still studying, I won twice sa Math Wizard/Math Quiz.. but reading the math problems here.. ewan.. mga joke ba to? O napurol na lang ang utak ko? hehehe

[Killer5]

You are in bar... chillin out with your 7 friends... you table is 10 meters away from the bar. if you can carry 3 bottles of beer with one hand, how many trips do you have to make from your table to the bar if you and your friends want to drink up 14 bottles of beer?

Edited September 20, 2004 by Killer5

[boy男孩]

i move that this thread be closed!

wala naman pakialam ang bilang ng taong tumataya sa probability of winning sa lotto.

[quintix]

How about 2, 3 or 4 persons winning the same draw?

Kung ang tanong eh probability na dalawang tao manalo sa isang draw, kailangan natin malaman kung ilan ang tumataya at kung ilang number ang tinataya nung bawat isa.

 

For example kung isa lang ang tumataya, kahit ano gawin mo hindi dalawa ang mananalo. Kung dalawa ang tumataya, silang dalawa ang dapat manalo. Kung tatlo ang tumataya, you have to add the possibilities na kung sinong dalawa sa tatlo ang mananalo. Kasama pa sa model mo yung kung ilang number ang tinataya nung bawat isa... there are two levels of complexity here, yung personal odds of winning nung bawat individual and the number of winners... there are two random variables at work here.

 

Anyway, mahabang usapan talaga ang probability atsaka yung mga 'in how many ways'. The most crucial part is kung nagkakaintindihan kung ano ba ang tanong.

 

This is not the right place for questions like this.. patulan na lang nating yung beer bottle problem, ok?

[boy男孩]

Kung ang tanong eh probability na dalawang tao manalo sa isang draw, kailangan natin malaman kung ilan ang tumataya at kung ilang number ang tinataya nung bawat isa.

 

For example kung isa lang ang tumataya, kahit ano gawin mo hindi dalawa ang mananalo. Kung dalawa ang tumataya, silang dalawa ang dapat manalo. Kung tatlo ang tumataya, you have to add the possibilities na kung sinong dalawa sa tatlo ang mananalo. Kasama pa sa model mo yung kung ilang number ang tinataya nung bawat isa... there are two levels of complexity here, yung personal odds of winning nung bawat individual and the number of winners... there are two random variables at work here.

 

Anyway, mahabang usapan talaga ang probability atsaka yung mga 'in how many ways'. The most crucial part is kung nagkakaintindihan kung ano ba ang tanong.

 

This is not the right place for questions like this.. patulan na lang nating yung beer bottle problem, ok?

salamat sa advice mo. nasagot na yung tanong ko. ibig sabihin naintindihan nung sumagot. mag aral ka muna nng statistics bago ka kwento ng kwento. sa UP ka mag aral ha.

 

yung beer problems para yun sa puzzles doon sa JOKES SECTION.

[quintix]

boy, salamat sa sagot mo...

 

di ko na pipilitin ang pananaw ko, halata namang nalalaman mo din ang sinasabi mo. pero kung may panahon ka itanong mo sa isa mong kakilala na bihasa sa statistics kung ano tingin niya.

 

peace be with you.

[Killer5]

ano to? extension ng fight club? para kayong mga gago..

lalo ka na boy chekwa

Edited September 22, 2004 by Killer5

[boy男孩]

quintix: bago ka kasi sabat ng sabat basahin mo muna ang history ng tanong. basically, ang buod ng tanong ko eh. paano ma-derive ang 1:5.3million chance na lumabas ang isang 6- number combination from a 42 balls lotto draw. yung ibang tanong any nag sanga sanga nalang. nasagot naman yung general formula. ewan ko sayo bakit napunta sa number of bettors and pinagsasabi mo. ang punto ko d2 kung ikaw ang tumaya sa lotto (assuming 1 bet lang) ang chance mo manalo at 1:5.3million. mag isa kaman tumaya o isang million kayong tumaya. totally independent yun.

[miguel2]

I cant believe it ... What's the probability na magkita at magtalo ang dalawang gunggong na sina " BOY HAPON" at "KILLER5" dito sa thread na to ...?

 

Take it to the Fight club and leave this thread alone if you have nothing to contribute or solve.

Edited September 22, 2004 by chunkyhunk

[Guest chunky]

You are in bar... chillin out with your 7 friends... you table is 10 meters away from the bar. if you can carry 3 bottles of beer with one hand, how many trips do you have to make from your table to the bar if you and your friends want to drink up 14 bottles of beer?

Two probable answers:

 

1. If onlly you are to carry the beer, you'll need a total of three trips. 6 on the first, 6 on the second, and 2 on the last.

 

2. If this is a trick question, why would I carry all of the beer ten meters away and make three trips when I can ask a couple of my friends to help me carry the rest of the bottles?

 

On a serious note: Please take it easy boys. Take your Fight to the Club...

[boy男孩]

Two probable answers:

 

1. If onlly you are to carry the beer, you'll need a total of three trips. 6 on the first, 6 on the second, and 2 on the last.

 

2. If this is a trick question, why would I carry all of the beer ten meters away and make three trips when I can ask a couple of my friends to help me carry the rest of the bottles?

 

On a serious note: Please take it easy boys. Take your Fight to the Club...

more so, you can put the 14bottles of beer in a bucket and ask the waiter to carry it for you. hehehe sabi ko na nga ba pang jokes and more tong tanong eh.

[quintix]

I lost my cool when a reference was dropped regarding my miseducation, and for this I apologize.

 

I shouldn't have to defend myself, knowing my own credentials, accepting my own fallibility, and realizing the limitations of this medium for relevant mathematical discourse.

 

To the cooler heads who have intervened, thanks for pointing out this was getting out of hand.

 

--------------------------------------------------

 

Here's my contribution, observe that:

2 + 2 = 2 x 2

1 + 2 + 3 = 1 x 2 x 3

 

Is there a similar equation with the sum of four numbers being equal to their product?

 

(I don't have any answer to this, but if there is a correct answer, everyone will know without any further discussion necessary)

[boy男孩]

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i move that this thread be closed!

at first I thought your question was meant to put an end to this thread. but wait, to answer your question, yes there are.

 

obvious one, 0 + 0 + 0 + 0 = 0 * 0 * 0 * 0

 

another : 1 + 1 + 2 + 4 = 1 * 1 * 2 * 4 = 8

 

 

and this continues....

 

 

for 5 numbers:

 

1+1+1+2+5=1*1*1*2*5 = 10

 

 

for 6 numbers,

 

1+1+1+1+2+6 = 1*1*1*1*2*6 = 12

 

so on and so forth:

 

the general eqn would look like this:

 

(n-2)+2+n = 1^(n-2)*n

[floppydrive]

on dividing by zero:

 

to divide by a number x is to multiply by its multiplicative inverse, 1/x

 

the reciprocal/inverse of a number is that which when multiplied by the number is 1

 

so we need the value of 1/x when x=0. however, any number multiplied by zero is zero (zero property of multiplication).

 

and so zero has no inverse, and we can't divide anything by it. ganun na yun, parang langis at tubig na di maghahalo kahit ano pa ang bate mo.

Here's another way of looking at the problem:

 

Formula: A/x = ? when X = 0

 

There are three solutions:

if the number A is already zero, then the answer is zero.

if the number A is NOT equal to zero, then:

 

The formula A * 1/x = ?

As x approaches zero, the answer increases:

 

x=1, 1/x = 1/1 = 1

x=0.5, 1/x = 1/0.5 = 2

x=0.1 1/x = 1/0.1 = 10

x=0.01 1/x = 1/0.01 = 100

x=0.0001 1/x = 1/0.0001 = 10,000

x=0.00001 1/x = 1/0.00001 = 100,000

and so on.

 

As X approaches 0 (i.e. x=0.00000000001), the answer approaches infinity, Positive or negative, depending on the numerator.

-----------------------

 

I think this is a thread that allows us to exercise our thinking skills, not our debating nor fighting skills. Generating solutions is the goal, not determining who has a bigger ego.

 

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Regarding "langis at tubig na binate", add an emulsifier and it will mix! Example: the surfactants in detergent permit the mixing of oil with water.

 

Life is not always black and white. The enjoyment is in the search for DIFFERENT solutions.

 

Peace People!

[boy男孩]

I think this is a thread that allows us to exercise our thinking skills, not our debating nor fighting skills. Generating solutions is the goal, not determining who has a bigger ego.

 

------------------------

 

Regarding "langis at tubig na binate", add an emulsifier and it will mix! Example: the surfactants in detergent permit the mixing of oil with water.

 

Life is not always black and white. The enjoyment is in the search for DIFFERENT solutions.

 

Peace People!

I agree with floppydrive.

 

And before somebody will MOVE to close this thread again, i would like to correct and elaborate my last solution.

 

whereby:

 

the general eqn would look like this:

 

(n-2)+2+n = 1^(n-2)*n

 

it should read like this:

 

 

(n-2) + 2 + n = 1^(n-2) * 2 * n

 

 

where (n-2) should be expanded as: 1 + 1 + ... + 1

 

wherefore: 1^(n-2) = 1^(1+1+1+...+1) = 1*1*1*...*1 where the # of term is equal to (n-2).

 

to illustrate: (example problem) : find the the solution where the sum = product of a say, 12 numbers.

 

solution: let n=12

 

(n-2) + 2 + n = 1^(n-2) * 2 * n

 

(10) + 2 + 12 = 1^10 * 2 * 12

 

1+1+1+1+1+1+1+1+1+1+2+10 = 1*1*1*1*1*1*1*1*1*1*2*10 = 24