Copiers

The new copier copying a folder of papers 5 min. faster than the old. The operator used new, but out of toner and exchange took 5 min. In that time copied on the old. The whole work has been done for 9 min. How long would the work done only by old copier?

Result

t =  11.9 min

Solution:

$\dfrac{9-5}{ t-5} +\dfrac{5}{t}=1 \ \\ 4t+5(t-5)=t(t-5) \ \\ \ \\ t^2 -14t +25 =0 \ \\ \ \\ a=1; b=-14; c=25 \ \\ D = b^2 - 4ac = 14^2 - 4\cdot 1 \cdot 25 = 96 \ \\ D>0 \ \\ \ \\ t_{1,2} = \dfrac{ -b \pm \sqrt{ D } }{ 2a } = \dfrac{ 14 \pm \sqrt{ 96 } }{ 2 } = \dfrac{ 14 \pm 4 \sqrt{ 6 } }{ 2 } \ \\ t_{1,2} = 7 \pm 4.89897948557 \ \\ t_{1} = 11.8989794856 \ \\ t_{2} = 2.10102051443 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (t -11.8989794856) (t -2.10102051443) = 0 \ \\ \ \\ t>5 \ \\ t = 11.9 \ \text{min}$

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