I believe the intention of the problem is as follows:

Working together, printers A and B can do a certain printing job in 12 days. Working together, printers B and C can do the job in 16 days. Over the past 13 days, printer A worked for 5 days, printer B worked for 7 days, and printer C worked for 13 days, with the result that the job was completed at the end of the 13 days. How many days would it take printer C to do the job on its own?

Let the job = 48 pages.
Since A and B working together can do the job in 12 days, the combined rate for A and B = w/t = 48/12 = 4 pages per day.
Since B and C working together can do the job in 16 days, the combined rate for B and C = w/t = 48/16 = 3 pages per day.

Printer A worked for 5 days and printer B worked for 7 days:
Since A and B each worked for at least 5 days, the amount of work produced by A and B together over these 5 days = r*t = 4*5 = 20 pages.
Since B worked for 2 additional days, the amount of work produced by B and C together over these 2 days = r*t = 3*2 = 6 pages.
Remaining work = 48 - 20 - 6 = 22 pages.

Printer C worked for 13 days:
The work produced by C over 2 of these 13 days has already been counted, leaving 11 days for C to work alone.
Since C prints the remaining 22 pages over these 11 days, C's rate alone = w/t = 22/11 = 2 pages per day.

Thus:
Time for C to do the entire job on its own = w/r = 48/2 = 24 days.

Yes thanks guys this was a tricky problem but u need to use algebrac manipulation.
Since it stated worker A worked for 5 days and worker B worked for 7 days and worker C worked for 13 days,

5A + 7B +13C= 1 work

Since A+ B together took 20 days their combined rates= (1/20)
Since B+C together took sixteen days their combined rates = (1/16)
From the earlier equation we can use distributive property
5(A+B)+ 2(B+C) +11C = 1 work
subsitute the rates into the equation R(A+B)=1/20 and R(B+C)=(1/16)
5(1/20) + 2(1/16)+11C= 1 work
1/4+1/8+11C=1 work
3/8+11C=1 work
11C=5/8
C=5/88
#of days C work= 1 work/rate of C=1/(5/88)=88/5=17.5 days=18days