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Horses
#5
RE: Horses
Forget Hygenically challenged Uncle Jimbo and his pseudo-math, this can be solved by simple algebraic substitution.

The eldest son will be person A, and he gets a 1/2 share of the horses. A = (1/2)
The middle son will be person B, and he gets a 1/3 share of the horses. B = (1/3)
The youngest son will be person C, and he gets a 1/9 share of the horses. C = (1/9)

A = (3/2)B = (9/2)C
B = (2/3)A = 3C
C = (2/9)A = (1/3)B
A + B + C = 17

All of these are true according to the problem.

A + B + C = 17 → The horses are 17, so the sum the all the sons shares must be 17.

A + (2/3)A + (2/9)A = 17 → First subtitute B & C for terms of A so that the equation has one variable.

(17/9)A = 17 → Simplify.

(9/17)*(17/9)A = 17*(9/17) → Multiply both sides by (9/17) (the inverse of (17/9)) to solve for A.

A = 9 → Solution.


After doing this for B and C also you find that A=9, B=6 and C=2.
Has anyone really been far even as decided to use even go want to do look more like?

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Messages In This Thread
Horses - by Darwinian - March 3, 2010 at 4:30 pm
RE: Horses - by Tiberius - March 3, 2010 at 4:38 pm
RE: Horses - by LukeMC - March 3, 2010 at 4:55 pm
RE: Horses - by Welsh cake - March 3, 2010 at 6:52 pm
RE: Horses - by theblindferrengi - March 12, 2010 at 12:26 am
RE: Horses - by Violet - March 15, 2010 at 7:46 am



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