Math problem

Teren

One of Freddy's beloved
Joined
Dec 22, 2003
Messages
585
I'm stuck with this problem:

After one tracter worked 6 hours on a field, another joined. Together they cleared the field in 4 hours. In how many hours can each tractor clear the field by itself, if it's known, that the first one will need 3 hours more than the second.


Any ideas?

P.S. the solve can't be a system of 2 equations with 2 variables, it must be a single one with one variable
 

Lumikki

Fledgling Freddie
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Dec 22, 2003
Messages
888
x2 + 3 + 512 > 253 *34 x3 6b f354... ah screw it, I don't have a clue. :D
 

Brynn

Can't get enough of FH
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Dec 22, 2003
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In 10 hours they both plouged the field, but the first tractor did 6 hours of the work. Therefore they only worked together for 4 hours....dunno

im working on a cubic equation with unknown variables :

x^3+ax^2+bx+c : for all real, imaginary and zero values.....i want to cry
 

Ceryseth

Fledgling Freddie
Joined
Mar 5, 2004
Messages
201
Hmmm did the second tracter join in after the first one had already done 6 hours?Did the poor farmer get a rest?Was he fully TOA?What rr was the second tracter?Did the second tracter actually leech exp from the first tracters field or did they group up to do the field?Any good drops from the field?What did they want to do to the field after clearing it anyway?What is a tracter?Is it used to trac things down?Lets check Google....first result:Carone Tracter Silicon Tubes Suppliers, what do farmers use silicon tubes for?well it dosent actualy say they are farmers...hmm...even so must be hard to clear a field with only silicon tubes,maybe if they where big sharp silicon tubes...but they are these soft things used to put cables through or to use in mad sexual fantasys.Now i read it again it says it as if the tracters where working alone and another one joined in...they must be some kinda superinteligentadroid tubes.This requieres more Google search..it seems that someone in 1946 captured one --->We have in our barn a 1946 mod.b farmall tracter that ran when we put it in the barn. We are wondering how much it is worth.
Maybe its more of an alien that the goverment is trying to capture so these farmers are selling the one they captures half a century ago.What would the goverment want with these tubes?All i know is that we are all dooomed!!!
 

Teren

One of Freddy's beloved
Joined
Dec 22, 2003
Messages
585
Ceryseth said:
Hmmm did the second tracter join in after the first one had already done 6 hours?Did the poor farmer get a rest?Was he fully TOA?What rr was the second tracter?Did the second tracter actually leech exp from the first tracters field or did they group up to do the field?Any good drops from the field?What did they want to do to the field after clearing it anyway?What is a tracter?Is it used to trac things down?Lets check Google....first result:Carone Tracter Silicon Tubes Suppliers, what do farmers use silicon tubes for?well it dosent actualy say they are farmers...hmm...even so must be hard to clear a field with only silicon tubes,maybe if they where big sharp silicon tubes...but they are these soft things used to put cables through or to use in mad sexual fantasys.Now i read it again it says it as if the tracters where working alone and another one joined in...they must be some kinda superinteligentadroid tubes.This requieres more Google search..it seems that someone in 1946 captured one --->We have in our barn a 1946 mod.b farmall tracter that ran when we put it in the barn. We are wondering how much it is worth.
Maybe its more of an alien that the goverment is trying to capture so these farmers are selling the one they captures half a century ago.What would the goverment want with these tubes?All i know is that we are all dooomed!!!
lmao :)
 

smerf

Fledgling Freddie
Joined
Dec 27, 2003
Messages
5
think thats right, there are 10 hours, each tractor does so much work per hour

in the first 6 hours its just the first tractor working, it does 10/w work per hour, and there are 6 hours so 6 * ( 10/w) is the ammount of work done in the 6 hours, w is the ammount of time the first tractor takes to do the entire field.

Then there are 4 hours, the first tractor does 4 hours work, 4*(10/w) and the other tractor which does the field 3 hours faster does 4*(10/(w-3))

It takes 10 hours for the field to be plowed

so this gives us

(6 * (10/w)) + ((4 * (10/w)) + (4 * (10/(w-3)))) = 10

ps this is probably totally wrong
 

Teren

One of Freddy's beloved
Joined
Dec 22, 2003
Messages
585
(6 * (10/w)) + ((4 * (10/w)) + (4 * (10/(w-3)))) = 10

60/w + 40/w + 40/(w-3) = 10

100/w + 40/(w-3) -10 = 0

(100x - 300 + 40x - 10x^2 + 30x) / (x*(x-3)) = 0

(170x - 300 - 10x^2) / (x*(x-3)) = 0

-10x^2 + 170x - 300 = 0
x*(x-3) != 0 ==> x != 0; x != 3;


-10x^2 + 170x - 300 = 0

D = 28900 - 4 * (-10) * (-300) = 28900 - 12000 = 16900

x = (-170 +- 130) / -20 = {2; 15}

15 is the answer! Brilliant!
 

smerf

Fledgling Freddie
Joined
Dec 27, 2003
Messages
5
my formula was right???

wow thats a first

/me looks at his E grade maths a-level
 

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