A and B can do a job together in 80 days, B and C in 20 days.

Question

A does the job for 5 days, then B comes over and does it for 15 days, then C does it for 18 days and the job is finished. How much time does C need to do the job alone?

Attempt:

Let their job effeciencies be a, b and c jobs per day respectively.

Then

$a+b=\frac{1}{80}$

$b+c=\frac{1}{20}$

$5a+15b+18c=1$

From equation 2,

$b=\frac{1}{20}-c$

From equation 1,

$a=\frac{1}{80}-b=\frac{1}{80}-(\frac{1}{20}-c)=\frac{1}{80}-\frac{1}{20}+c$

Substituting $a$ and $b$ in equation 3 and solving for $c$,

$5(\frac{1}{80}-\frac{1}{20}+c)+15(\frac{1}{20}-c)+18c=1$

I solved for c and then took its reciprocal $\frac{1}{c}$ and got 18 days as answer.

However, that makes no sense because B and C do it together in 20 days. There's no way C can do it alone in 18 days. Where am I wrong?

Answer 1

If we also solve for $a$ and $b$ we get $\frac{11}{640}$ and $-\frac{3}{640}$ respectively. This means that either there is a mistake in the question or person B's incompetence hinders production! :)