Answer to Question #118029 in Algebra for Angelica

Question

Herbert and Dandreb 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

let's x-Herbert's performance, y-Dandreb's, P-square of work

together they work (x+y)*3=P - > (x+y)/P=1/3 -> x/P+y/P=1/3 -> y/P= 1/3-x/P

and each: P/x+1=P/y -> y/P=1/(P/x+1)

we need to find P/x-time Herbert, P/y- Dandreb's time

1/3-x/P=1/(P/x+1) -> (1/3-x/P)*(P/x+1)=1 ->

P/(3x)-1+1/3-x/P=1 -> P/(3x)-5/3-x/P=0 lets *(3P/x)

(P/x)^2-5P/x-3=0

Find D=5^2-4*1*(-3)=25+12=37

P/x=(5+sqrt(37))/2

P/x=(5-sqrt(37))/2, but sqrt(37)>5, so P/x will be less 0, but time can't be< 0, so we have one solution:

P/x=(5+sqrt(37))/2

P/y=(5+sqrt(37))/2 + 1=(7+sqrt(37))/2

one of them need (5+sqrt(37))/2 hour it's about 5.54

and another need (7+sqrt(37))/2 hour it's about 6.54