A and B working together can do a work in 6 days. If A takes 5 days less than B to finish the work, in how many days B alone can do the work.

Respuesta :

frika

Answer:

15 days

Step-by-step explanation:

Let x be the number of days needed for B to complete the job. Then x-5 is the number of days needed for A to complete the job.

In 1 day,

  • A completes [tex]\dfrac{1}{x-5}[/tex] of all work;
  • B completes [tex]\dfrac{1}{x}[/tex] of all work.

Hence, in 1 day both A and B complete [tex]\dfrac{1}{x-5}+\dfrac{1}{x}[/tex] of all work. A and B working together can do a work in 6 days. Then

[tex]6\cdot \left(\dfrac{1}{x-5}+\dfrac{1}{x}\right)=1.[/tex]

Solve this equation:

[tex]\dfrac{6x+6x-30}{x(x-5)}=1,\\ \\12x-30=x^2-5x,\\ \\x^2-17x+30=0,\\ \\D=(-17)^2-4\cdot 30=289-120=169=13^2,\\ \\x_{1,2}=\dfrac{17\pm 13}{2}=2,\ 15.[/tex]

If [tex]x=2,[/tex] then [tex]x-5=-3[/tex] that is impossible. So, B needs 15 days to complete the work alone.

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