Question

Homer and Mike were replacing the boards on​ Mike's old deck. Mike can do the job alone in 1 hour less time than Homer. They worked together for 7 hours until Homer had to go home. Mike finished the job working by himself in an additional 5 hours. How long would it have taken Homer to fix the deck​ himself?

Answer 1

Answer: 19.65 hours

Step-by-step explanation:

Mike: $\frac{1}{x-1}\ \text{hours}$

Homer: $\frac{1}{x}\ \text{hours}$

The each work 7 hours plus Mike works 5 hours alone:

$7\bigg(\dfrac{1}{x-1}\bigg)+7\bigg(\dfrac{1}{x}\bigg)+5\bigg(\dfrac{1}{x-1}\bigg)=1\ job\\\\\\12\bigg(\dfrac{1}{x-1}\bigg)+7\bigg(\dfrac{1}{x}\bigg)=1\ job\\\\\\\bigg(\dfrac{12}{x-1}\bigg)+\bigg(\dfrac{7}{x}\bigg)=1\\\\\\(x)(x-1)\bigg(\dfrac{12}{x-1}\bigg)+(x)(x-1)\bigg(\dfrac{7}{x}\bigg)=1(x)(x-1)\\\\\\12(x) + 7(x-1)=x^2-x\\\\12x + 7x - 7 = x^2-x\\\\.\qquad \qquad \ \ 0=x^2-20x+7$

Use quadratic formula to solve for x:

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\.\quad=\dfrac{-(-20)\pm \sqrt{(-20)^2-4(1)(7)}}{2(1)}\\\\.\quad=\dfrac{20\pm \sqrt{400-28}}{2}\\\\.\quad=\dfrac{20\pm \sqrt{372}}{2}\\\\.\quad=\dfrac{20\pm 19.3}{2}\\\\.\quad=10\pm9.65\\\\.\quad=19.65\quad or\quad 0.35$

We know that 0.35 cannot be a valid answer since it took at least 12 hours to complete the job.

So, Homer can get 1 job done in 19.65 hours.