[SOLVED] Problem Solving help

zp3929 · #1 $Old$ May 2nd, 2008, 12:30 AM

Hi,

Can you give me some help with my yr 10 problem solving?

i just need help with 2 questions

the first one is "a gambler bets half of all he owns on the toss of a coin. pleased is he when he wins, that he bets another 19 times, always betting half of the total amount he has. Happily, he says 'i won as often as i lost. So i presume i've come out even.' is he correct" i have to use algebra to obtain my result
so far i have been able to figure out the when he wins the equation is
x + x/2 (x being the total amount he has)
and when he loses the equation is
x/2
but that is all i have been able to figure out.

the next one is "Each person in a room at a new year's eve party kissed every other person in the room once. If by the end of the night ther had been 190 kisses, how many people were in the room" for this i must "by first finding a rule for the number of kisses ther would be for n people, use algebra to obtain your result"
i have hit a brick wall with this and cant think of anything what so ever

so any help would be greatlt appreciated

thanks

Zac

Kai · #2 $Old$ May 2nd, 2008, 03:47 AM

Erm, for the first one, somehow the question asked itself is eluding me, never mind the answer,..

For the second one, if theres 2 person there is 1+1 kisses, for 3 person theres 2+2+2 kisses ...for n people there will be n(n-1) kisses

To find for 190 kisses, u equate n(n-1)= 190

u get n = 14.293

.....u sure u copied the question correctly ??

zp3929 · #3 $Old$ May 2nd, 2008, 03:58 AM

Quote:

Originally Posted by Kai $View Post$

u sure u copied the question correctly ??

down to every letter, see why i'm having trouble?

Zac

Isomorphism · #4 $Old$ May 2nd, 2008, 04:11 AM

Quote:

Originally Posted by zp3929 $View Post$

Hi,

the next one is "Each person in a room at a new year's eve party kissed every other person in the room once. If by the end of the night ther had been 190 kisses, how many people were in the room" for this i must "by first finding a rule for the number of kisses ther would be for n people, use algebra to obtain your result"
i have hit a brick wall with this and cant think of anything what so ever

so any help would be greatlt appreciated

thanks

Zac

Quote:

Originally Posted by Kai $View Post$

Erm, for the first one, somehow the question asked itself is eluding me, never mind the answer,..

For the second one, if theres 2 person there is 1+1 kisses, for 3 person theres 2+2+2 kisses ...for n people there will be n(n-1) kisses

To find for 190 kisses, u equate n(n-1)= 190

u get n = 14.293

.....u sure u copied the question correctly ??

No.. wrong

A kissed B is same as B kissed A. So you should have divided n(n-1) by 2, since while counting n(n-1) you have counted 2 kisses between a pair of people..,

So n(n-1)/2 = 190 and hence n(n-1) = 380. So n=20

Kai · #5 $Old$ May 2nd, 2008, 04:13 AM

Quote:

Originally Posted by Isomorphism $View Post$

No.. wrong

A kissed B is same as B kissed A. So you should have divided n(n-1) by 2, since while counting n(n-1) you have counted 2 kisses between a pair of people..,

So n(n-1)/2 = 190 and hence n(n-1) = 380. So n=20

Yes, i was thinking so, but couldn't 100% sure though.

zp3929 · #6 $Old$ May 2nd, 2008, 04:15 AM

Quote:

Originally Posted by Isomorphism $View Post$

So n(n-1)/2 = 190 and hence n(n-1) = 380. So n=20

thanks for that, does anyone have any idea for the first questions, i have been trying at it for like an hour but for some reason i just cant see the answer, must be having one of those days...

Zac

Moo · #7 $Old$ May 2nd, 2008, 04:30 AM

Hello,

For the first one, good equations

Do you agree that when he wins, he multiplies his possession by 3/2, and when he loses, he multiplies his possession by 1/2 ?
Here is the mistake of the gambler, it's multiplying, not adding.

Now, we know that he wins 10 times, and loses 10 times.

Let x be the initial amount.

Is this equality correct : $\left(\frac 32 \right)^{10} \left(\frac 12 \right)^{10} x=x$ ?

Do you understand where this formula comes from ?

zp3929 · #8 $Old$ May 2nd, 2008, 04:50 AM

Quote:

Originally Posted by Moo $View Post$

Now, we know that he wins 10 times, and loses 10 times.

Let x be the initial amount.

Is this equality correct : $\left(\frac 32 \right)^{10} \left(\frac 12 \right)^{10} x=x$ ?

Do you understand where this formula comes from ?

i do understand where the equation comes from, evertime he wins he increases the amount he owns by 3/2, whereas when he loses 1/2 of all he owns is lost, because this is done 10 times each that is why the power of 10 is there, is that right?
there must be some flaw in my understanding though as i just tried to put it into my calculator and it cane no where near the value of x i had put in
i put x as 1000 and it came out with 56.313514...
where did i go wrong?

thanks

Zac

Isomorphism · #9 $Old$ May 2nd, 2008, 04:56 AM

Quote:

Originally Posted by zp3929 $View Post$

i do understand where the equation comes from, evertime he wins he increases the amount he owns by 3/2, whereas when he loses 1/2 of all he owns is lost, because this is done 10 times each that is why the power of 10 is there, is that right?
there must be some flaw in my understanding though as i just tried to put it into my calculator and it cane no where near the value of x i had put in
i put x as 1000 and it came out with 56.313514...
where did i go wrong?

thanks

Zac

Moo is trying tell you $\left(\frac 32 \right)^{10} \left(\frac 12 \right)^{10} x \neq x$ . So the gambler is wrong... thats all.

zp3929 · #10 $Old$ May 2nd, 2008, 04:59 AM

Quote:

Originally Posted by Isomorphism $View Post$

Moo is trying tell you $\left(\frac 32 \right)^{10} \left(\frac 12 \right)^{10} x \neq x$ . So the gambler is wrong... thats all.

ahh right thanks, sorry a bit slow today, as you can probably tell

thanks

Zac

Moo · #11 $Old$ May 2nd, 2008, 05:03 AM

Quote:

Originally Posted by zp3929 $View Post$

i do understand where the equation comes from, evertime he wins he increases the amount he owns by 3/2, whereas when he loses 1/2 of all he owns is lost, because this is done 10 times each that is why the power of 10 is there, is that right?
there must be some flaw in my understanding though as i just tried to put it into my calculator and it cane no where near the value of x i had put in
i put x as 1000 and it came out with 56.313514...
where did i go wrong?

thanks

Zac

Good reasoning

For the latter part, Isomorphism told you.