# Two diggers

Two diggers should dig a ditch. If each of them worked just one-third of the time that the other digger needs, they'd dig up a 13/18 ditch together. Find the ratio of the performance of this two diggers.

Result

r =  3:2

#### Solution:

$r = t_{ 2 }:t_{ 1 } \ \\ \ \\ \dfrac{ t_{ 2 } }{ 3 } /t_{ 1 } + \dfrac{ t_{ 1 } }{ 3 } /t_{ 2 } = 13/18 \ \\ \dfrac{ t_{ 2 } }{ t_{ 1 } } + \dfrac{ t_{ 1 } }{ t_{ 2 } } = 13/6 \ \\ \ \\ r + 1/r = 13/6 \ \\ \ \\ r^2 + 1 = 13/6 \ r \ \\ r^2 -2.167r +1 = 0 \ \\ \ \\ a = 1; b = -2.167; c = 1 \ \\ D = b^2 - 4ac = 2.167^2 - 4\cdot 1 \cdot 1 = 0.6944444444 \ \\ D>0 \ \\ \ \\ r_{1,2} = \dfrac{ -b \pm \sqrt{ D } }{ 2a } = \dfrac{ 2.17 \pm \sqrt{ 0.69 } }{ 2 } \ \\ r_{1,2} = 1.08333333 \pm 0.41666666666667 \ \\ r_{1} = 1.5 \ \\ r_{2} = 0.66666666666667 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (r -1.5) (r -0.66666666666667) = 0 \ \\ r = r_{ 1 } = 1.5 = \dfrac{ 3 }{ 2 } = 3:2$

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