Two constructors working together finished building a room in 6 days. How long would it take for each constructor to build the room by himself, if it is known that one of them would require 5 more days than the other?
Accepted Solution
A:
If the first constructor works alone for one day he will do (1/x) part of work. If the second constructor works alone for one day he will do (1/(x+5)) part of work. If they work together, they will do for one day
[tex] \frac{1}{x} + \frac{1}{x+5}
part of work. Also, if they work together,
for one day they will do 1/6 part of work.
So,
[/tex] [tex] \frac{1}{x} + \frac{1}{x+5} = \frac{1}{6} , \frac{x+5+x}{x(x+5)} = \frac{1}{6} , \frac{2x+5}{ x^{2} +5x} = \frac{1}{6} ,
6(2x+5)=x^{2} + 5x
12x+30 =x^{2}+5x
x^{2} -7x-30=0
(x-10)(x+3)=0
x_{1}= 10, x_{2}=-3
For our problem we can use only positive value, x=10 days will work
the first worker.
The second worker will work x+5=10+5= 15 days.
First worker alone will work 10 days,
second worker alone will work 15 days.
Check,
(1/10)+(1/15)=(1/6)
(3+2)/30=(1/6)
1/6=1/6 True
[/tex]