seb--morin
Im high
+152|7077|Montréal, Québec
Hi guys, i have this stupid maths homework and i think im too tired because i just dont get it.... It seem to be easy but im blocked..

prob 1- 
A guy lost 1/3 of his money then played again and lost 1/3 of what remained. He lost again 1/3 a 3rd time and then a 4th time he lost 1/3 again. At the end he have 32$, how much he had at the begining?

prob 2-
A rectangle has a perimeter of 500 meters and the area is 14400 m2 .... i have to find the lenght of the sides. I need to put this in an equation like f(x)= ......

2x+2y=500
x*y=14400    right? well  how come i get stuck at   14400=250-x 

prob 3-
we can cover a floor with 2000 tile of a certain size but if they where longer and larger of 1 cm, we would only need 1280. Find the dimension of the tile.
Surgeons
U shud proabbly f off u fat prik
+3,097|6927|Gogledd Cymru

prob 1 - 3*3*3*3*32 = $2592
Funky_Finny
Banned
+456|6570|Carnoustie, Scotland

0seb--morin0 wrote:

Hi guys, i have this stupid maths homework and i think im too tired because i just dont get it.... It seem to be easy but im blocked..

prob 1- 
A guy lost 1/3 of his money then played again and lost 1/3 of what remained. He lost again 1/3 a 3rd time and then a 4th time he lost 1/3 again. At the end he have 32$, how much he had at the begining?
$32 x 33.3% - Answer x 33.3%, do this four times.

prob 2-
A rectangle has a perimeter of 500 meters and the area is 14400 m2 .... i have to find the lenght of the sides. I need to put this in an equation like
f(x)= ....
2x+2y=500
x*y=14400    right? well  how come i get stuck at   14400=250-x 

f(x)= area
f(500) = 14400

Infact I have no fucking clue, I'm doing this in school as well, but I don't get this one.

prob 3-
we can cover a floor with 2000 tile of a certain size but if they where longer and larger of 1 cm, we would only need 1280. Find the dimension of the tile.
Lost again, hope I helped with the first one. What year are you in?

Last edited by Funky_Finny (2007-11-29 15:08:44)

Brasso
member
+1,549|7068

#1 - $2592

#2 - Hmm, I forgot how to do functions, but these are the two equations you want to use:
2l+2w=500
l*w=14400

#3 - No idea.  Shouldn't you have learned this in class?
"people in ny have a general idea of how to drive. one of the pedals goes forward the other one prevents you from dying"
Hurricane
Banned
+1,153|7068|Washington, DC

I always hated these word problems. They try to reflect real-life situations but honestly, what kind of room planner or architect wouldn't know the measurements of SOMETHING (in reference to the second problem)? Some guy working with dark matter?
Drakef
Cheeseburger Logicist
+117|6799|Vancouver
If the man had $32 remaining after losing a third of his money four times, then he had $162 to begin.

First loss: $162 to $108
Second loss: $108 to $72
Third loss: $72 to $48
Fourth loss: $48 to $32

Unless I misunderstand the question, judging from the other results, but I'd say $162
SenorToenails
Veritas et Scientia
+444|6568|North Tonawanda, NY

0seb--morin0 wrote:

prob 2-
A rectangle has a perimeter of 500 meters and the area is 14400 m2 .... i have to find the lenght of the sides. I need to put this in an equation like f(x)= ......

2x+2y=500
x*y=14400    right? well  how come i get stuck at   14400=250-x
Easy:

2x + 2y = 500

x * y = 14400

so, x = (500 - 2y) / 2

put that into the area equation, and solve for y.  Then use the first equation to get the numerical value of x.  Check to make sure that it works.

prob 3-
we can cover a floor with 2000 tile of a certain size but if they where longer and larger of 1 cm, we would only need 1280. Find the dimension of the tile.
longer and larger?  do you mean longer and wider?  as in the usual tile is x by y, the new tile is x+1 by y+1? (see post further down for answer with assumptions)

Last edited by SenorToenails (2007-11-29 15:31:37)

Ender2309
has joined the GOP
+470|7009|USA
for the area one:

2x+2y=500

xy=14400

x=14400/y

2(14400/y) +2y=500

28800/y +2y=500

28800 +2y^2=500y

2y^2 -500y +28800=0

y-160=0 or y-90=0

160*90=14400

X=160
Y=90
Ender2309
has joined the GOP
+470|7009|USA

Drakef wrote:

If the man had $32 remaining after losing a third of his money four times, then he had $162 to begin.

First loss: $162 to $108
Second loss: $108 to $72
Third loss: $72 to $48
Fourth loss: $48 to $32

Unless I misunderstand the question, judging from the other results, but I'd say $162
1/3 will increase with every loss. if he ends with 32, thats one third of 96, which is one third of 288, etc. up to 2592
SenorToenails
Veritas et Scientia
+444|6568|North Tonawanda, NY

0seb--morin0 wrote:

we can cover a floor with 2000 tile of a certain size but if they where longer and larger of 1 cm, we would only need 1280. Find the dimension of the tile.
Assume a square tile:

2000 x^2 = 1280 (x+1)^2

solve for x, get 4.

The original tile is 4 cm x 4cm, the new tile is 5 x 5 cm.
Drakef
Cheeseburger Logicist
+117|6799|Vancouver
A guy lost 1/3 of his money then played again and lost 1/3 of what remained. He lost again 1/3 a 3rd time and then a 4th time he lost 1/3 again. At the end he have 32$, how much he had at the begining?
Does this not mean that he lost one-third of his money each time? Or, is it as it seems everyone has said, he was left with one-third each time? If the former, my answer of 162 is correct.
BlackKoala
Member
+215|6763

Drakef wrote:

A guy lost 1/3 of his money then played again and lost 1/3 of what remained. He lost again 1/3 a 3rd time and then a 4th time he lost 1/3 again. At the end he have 32$, how much he had at the begining?
Does this not mean that he lost one-third of his money each time? Or, is it as it seems everyone has said, he was left with one-third each time? If the former, my answer of 162 is correct.
That's how I understood it.  He 'LOST' 1/3, not left with 1/3 each bet.
SenorToenails
Veritas et Scientia
+444|6568|North Tonawanda, NY

Drakef wrote:

A guy lost 1/3 of his money then played again and lost 1/3 of what remained. He lost again 1/3 a 3rd time and then a 4th time he lost 1/3 again. At the end he have 32$, how much he had at the begining?
Does this not mean that he lost one-third of his money each time? Or, is it as it seems everyone has said, he was left with one-third each time? If the former, my answer of 162 is correct.
I believe that you are correct.  At the end of each round, he had 2/3 of what he started with.

The work is simple:

32 = ((2/3)^4) x, where x is the original amount of money.

x = 162
Brasso
member
+1,549|7068

Drakef wrote:

If the man had $32 remaining after losing a third of his money four times, then he had $162 to begin.

First loss: $162 to $108
Second loss: $108 to $72
Third loss: $72 to $48
Fourth loss: $48 to $32

Unless I misunderstand the question, judging from the other results, but I'd say $162
Shit, sry, I calculated that he lost 2/3 his money rather than 1/3.  Drakef is right.
"people in ny have a general idea of how to drive. one of the pedals goes forward the other one prevents you from dying"
l41e
Member
+677|7086

1. It doesn't matter. He'll be broke soon, because he fucking fails at gambling.
2. I don't know.
3. I don't think it's solvable with the information given?
Surgeons
U shud proabbly f off u fat prik
+3,097|6927|Gogledd Cymru

No guys, its not $162

example

A guy has $82 he loses a third of it so he's left with $27.
He then loses a third of his remaining money ($27) which is $9.
He then loses a third of his remaining money ($9) which is $3.
He then loses a third of his remaining money ($3) which is $1 left.

So $32 * 3.3333333 * 3.3333333 * 3.3333333 = $1185.19
BlackKoala
Member
+215|6763

Mr.Surgeons wrote:

A guy has $82 he loses a third of it so he's left with $27.
He then loses a third of his remaining money ($27) which is $9.
He then loses a third of his remaining money ($9) which is $3.
He then loses a third of his remaining money ($3) which is $1 left.
You are a mathematical failure.

1/3 of $27 is $9.  He LOSES that 1/3, or $9.  He now has $18.  (27-9)
1/3 of $18 is $6.  He LOSES that 1/3, or $6.  He now has $12.  (18-6)
1/3 of $12 is $4.  He LOSES that 1/3, or $4.  He now has $8.  (12-4)


And so on and so forth.

You are assuming that he has the third remaining at the end of each bet, but in reality he simply lost that third, leaving him two-thirds of the original bet.

Last edited by BlackKoala (2007-11-29 16:02:09)

Surgeons
U shud proabbly f off u fat prik
+3,097|6927|Gogledd Cymru

BlackKoala wrote:

Mr.Surgeons wrote:

A guy has $82 he loses a third of it so he's left with $27.
He then loses a third of his remaining money ($27) which is $9.
He then loses a third of his remaining money ($9) which is $3.
He then loses a third of his remaining money ($3) which is $1 left.
You are a mathematical failure.

1/3 of $27 is $9.  He LOSES that 1/3, or $9.  He now has $18.  (27-9)
1/3 of $18 is $6.  He LOSES that 1/3, or $6.  He now has $12.  (18-6)
1/3 of $12 is $4.  He LOSES that 1/3, or $4.  He now has $8.  (12-4)


And so on and so forth.

You are assuming that he has the third remaining at the end of each bet, but in reality he simply lost that third, leaving him two-thirds of the original bet.
Its 00:04 you can forgive me
OrangeHound
Busy doing highfalutin adminy stuff ...
+1,335|7087|Washington DC

Mr.Surgeons wrote:

No guys, its not $162
Apparently, asking for math homework answers on BF2s is a sure way to get a failing grade ... I will say, however, that the correct answer is in this thread, but you will be better off in the long run if you do your own homework.
williedyna
Member
+7|7083
holy shit... some of you really suck at math... and reading comprehension....

enjoy pushing that broom at Mc'D's

Kid do your own work
Andoura
Got loooollllll ?
+853|7076|Montreal, Qc, Canada

SenorToenails wrote:

0seb--morin0 wrote:

prob 2-
A rectangle has a perimeter of 500 meters and the area is 14400 m2 .... i have to find the lenght of the sides. I need to put this in an equation like f(x)= ......

2x+2y=500
x*y=14400    right? well  how come i get stuck at   14400=250-x
Easy:

2x + 2y = 500

x * y = 14400

so, x = (500 - 2y) / 2

put that into the area equation, and solve for y.  Then use the first equation to get the numerical value of x.  Check to make sure that it works.

prob 3-
we can cover a floor with 2000 tile of a certain size but if they where longer and larger of 1 cm, we would only need 1280. Find the dimension of the tile.
longer and larger?  do you mean longer and wider?  as in the usual tile is x by y, the new tile is x+1 by y+1? (see post further down for answer with assumptions)
For the 3rd question yes its longer and larger.... so yes its x+1 by y+1

Its my friend i just talked with him on the phone, he cant go online at the moment!

And he tells you thx for the help
seb--morin
Im high
+152|7077|Montréal, Québec
Okay guys thks a lot for the help!

I had the answer for the 3 problems, that wasnt what i wanted to know its more how to get to it and the function related to the problem.

1. 162 $

2. x90   y 160

3. 4 and 5

anyone who were right got karma !|
Flecco
iPod is broken.
+1,048|7103|NT, like Mick Dundee

This is cheating.

Dude, get the fuck out of here with your homework. We aren't here to help you through school you lazy sod. Learn some frigging discipline.

Don't give us that 'I'm too tired' bullshit either, you're either too lazy or too stupid to do it. I suggest you take a reality check and think very hard about what you want to do in life.

Then start applying the appropriate amount of effort to your school-work.
Whoa... Can't believe these forums are still kicking.
SenorToenails
Veritas et Scientia
+444|6568|North Tonawanda, NY

Flecco wrote:

This is cheating.

Dude, get the fuck out of here with your homework. We aren't here to help you through school you lazy sod. Learn some frigging discipline.

Don't give us that 'I'm too tired' bullshit either, you're either too lazy or too stupid to do it. I suggest you take a reality check and think very hard about what you want to do in life.

Then start applying the appropriate amount of effort to your school-work.
Woah man, relax.  I can understand if this guy needed some help, as in, see a final solution to reverse engineer the process.  I have done this many times, and I don't think its cheating at all.  I assumed that is what he wanted, and hope that he learned how to do these problems.

When I first learned how to do these word problems, I remember that some people in my class also had a real hard time wrapping their heads around them.  Sometimes you need examples, with the steps laid out there.

Besides, if he cheats on his homework, it'll show when it is time to take the test.
Jenspm
penis
+1,716|7170|St. Andrews / Oslo

Out of curiosity, what year are you in / how old are you?

Last edited by Jenspm (2007-11-30 06:05:29)

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