MATH SOLVE

2 months ago

Q:
# 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]

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]