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

Explanation:

Mike:
`(1)/(x-1)\ \text{hours}`

Homer:
`(1)/(x)\ \text{hours}`

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

`7\bigg((1)/(x-1)\bigg)+7\bigg((1)/(x)\bigg)+5\bigg((1)/(x-1)\bigg)=1\ job\\\\\\12\bigg((1)/(x-1)\bigg)+7\bigg((1)/(x)\bigg)=1\ job\\\\\\\bigg((12)/(x-1)\bigg)+\bigg((7)/(x)\bigg)=1\\\\\\(x)(x-1)\bigg((12)/(x-1)\bigg)+(x)(x-1)\bigg((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=(-b\pm √(b^2-4ac))/(2a)\\\\.\quad=(-(-20)\pm √((-20)^2-4(1)(7)))/(2(1))\\\\.\quad=(20\pm √(400-28))/(2)\\\\.\quad=(20\pm √(372))/(2)\\\\.\quad=(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.