Question

Herbert and Dranreb can thoroughly clean a house in 3 hours. If each work, one of them would do the job in 1 hour less time than the other. How long would it take each person to do the cleaning?

Answer 1

First of all let's write collaboration time formula:

$t_C=\dfrac{V}{\upsilon_1+\upsilon_2}$ , where t_C - collaboration time, V - volume of the work, $\upsilon_1$ - labor productivity of the first man, $\upsilon_2$ - labor productivity of the second man.

Now let's write labor productivity formulas for both men:

$\upsilon_1=\dfrac{V}{t_1}$ and $\upsilon_2=\dfrac{V}{t_2}$ , where V - volume of the work, t₁ - first man-hours used, t₂ - second man-hours used.

Let's put these formulas in collaboration time formula:

$t_C=\dfrac{V}{\dfrac{V}{t_1}+\dfrac{V}{t_2}}=\dfrac{V}{V(\dfrac{1}{t_1}+\dfrac{1}{t_2})}=$

$=\dfrac{1}{\dfrac{1}{t_1}+\dfrac{1}{t_2}}=\dfrac{1}{\dfrac{t_2+t_1}{t_1*t_2}}=\dfrac{t_1*t_2}{t_1+t_2}$

Let's mark t₁ as x, then t₂ will be x - 1. Collaboration time is equal to 3 hours. Now we can write equation:

$\dfrac{x*(x-1)}{x+(x-1)}=3$

Now let's simplify and solve it:

$\dfrac{x^2-x}{2x-1}=3$

$x^2-x=3(2x-1)$

$x^2-x=6x-3$

$x^2-7x+3=0$

a = 1, b = -7, c = 3

$x_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a}=$

=$\dfrac{-(-7)-\sqrt{(-7)^2-4*1*3}}{2*1}=\dfrac{7-\sqrt{49-12}}{2}=$

$=\dfrac{7-\sqrt{37}}{2}\approx0.5$ hours does not satisfy the conditions of our task because man alone can't clean house faster than both men together (0.5 < 3)

$x_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}=$

=$\dfrac{-(-7)+\sqrt{(-7)^2-4*1*3}}{2*1}=\dfrac{7+\sqrt{49-12}}{2}=$

$=\dfrac{7+\sqrt{37}}{2}\approx6.5$ hours is man-hours of the first guy.

6.5 - 1 = 5.5 hours is man-hours of the second guy.

Answer: 6.5 hours and 5.5 hours or 6 hours 30 mins and 5 hours 30 mins.