# 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.

Correct result:

r =  3:2

#### Solution:

$r=t_{2}:t_{1} \ \\ \ \\ \dfrac{ t_{2} }{ 3 } /t_{1} + \dfrac{ t_{1} }{ 3 } /t_{2}=13/18 \ \\ \dfrac{ t_{2} }{ t_{1} } + \dfrac{ t_{1} }{ t_{2} }=13/6 \ \\ \ \\ r + 1/r=13/6 \ \\ r^2 + 1=13/6 r \ \\ \ \\ r^2 + 1=13/6 \ r \ \\ r^2 -2.167r +1=0 \ \\ \ \\ a=1; b=-2.167; c=1 \ \\ D=b^2 - 4ac=2.167^2 - 4\cdot 1 \cdot 1=0.6944444444 \ \\ D>0 \ \\ \ \\ r_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ 2.17 \pm \sqrt{ 0.69 } }{ 2 } \ \\ r_{1,2}=1.08333333 \pm 0.416666666667 \ \\ r_{1}=1.5 \ \\ r_{2}=0.666666666667 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (r -1.5) (r -0.666666666667)=0 \ \\ \ \\ r=r_{1}=1.5=\dfrac{ 3 }{ 2 }=3:2$

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