Certain work is completed in 12 days when $A, B$, and $C$ work together. With the same work, $A, B$, and $D$ complete it in 15 days, $A, C$, and $D$ in 18 days, and $B, C$, and $D$ in 20 days. How long will it take to complete the work if all four workers work together, but
a) $A$ is absent on the first day?
b) $C$ is absent on the last day?
I. solution: Let's consider the whole work as a unit, and denote the daily performance of $A, B, C, D$ by $x, y, z, u$ respectively, then according to the problem

\[
\begin{aligned}
& x+y+z=\frac{1}{12} \\
& x+y+u=\frac{1}{15} \\
& x+z+u=\frac{1}{18} \\
& y+z+u=\frac{1}{20}
\end{aligned}
\]

By adding these four equations:

\[
3(x+y+z+u)=\frac{1}{12}+\frac{1}{15}+\frac{1}{18}+\frac{1}{20}=\frac{138}{540}=\frac{23}{90}
\]

from which

\[
x+y+z+u=\frac{23}{270}
\]

Thus, the four workers working together would complete the entire work in $270: 23=11 \frac{17}{23}$ days.

a) If we subtract (4) from (5), we get the daily performance of $A$

\[
x=\frac{23}{270}-\frac{1}{20}=\frac{19}{540}
\]

This work, which was missed on the first day, needs to be compensated for at the end, after $11 \frac{17}{23}$ days, by the joint work. Since the four of them together complete $\frac{46}{540}$ in one day, it takes them $\frac{19}{46}$ days to complete $\frac{19}{540}$, so if $A$ is absent on the first day, the work will be completed in

\[
11 \frac{34}{46}+\frac{19}{46}=12 \frac{7}{46}
\]

days.
b) In 11 workdays, they complete together $11 \cdot \frac{23}{270}=\frac{253}{270}$, so $\frac{17}{270}$ remains for the last day. On this last day, only $A, B, D$ work. The daily performance of $A, B$ and $D$ is $\frac{1}{15}$, so they complete $\frac{17}{270}$ in $\frac{17}{270}: \frac{1}{15}=\frac{17 \cdot 15}{270}=\frac{17}{18}$ days.

Thus, in case b), the work is completed in $11 \frac{17}{18}$ days.

András Magos (Bp. II., Rákóczi g. II. o. t.)

II. solution: a) Without $A$ on the first day, as (4) shows, $\frac{1}{20}$ is completed, so $\frac{19}{20}$ remains. Since (5) indicates that they complete $\frac{23}{270}$ together in one day, it takes $\frac{19}{20}: \frac{23}{270}=11 \frac{7}{46}$ days to complete the remaining $\frac{19}{20}$, so the entire work is completed in

\[
1+11 \frac{7}{46}=12 \frac{7}{46}
\]

days.

b) Unchanged.

Miklós Ortutay (Hajdunánás, Kőrösi Csoma g. I. o. t.)

III. solution: a) If we subtract (4), (3), (2), and (1) from (5) respectively, we get that in one day

\[
\begin{array}{llrl}
A & \text { performance } & \frac{23}{270}-\frac{1}{20}=\frac{19}{540} \\
B & , & \frac{23}{270}-\frac{1}{15}=\frac{16}{540} \\
C & , & \frac{23}{270}-\frac{1}{18}=\frac{10}{540} \\
D & , & \frac{23}{270}-\frac{1}{12}=\frac{1}{540}
\end{array}
\]

In $u$ days, if $A$ does not work for one full day, the total performance is

\[
(u-1) \frac{19}{540}+u \frac{16}{540}+u \frac{10}{540}+u \frac{1}{540}=1
\]

from which

\[
46 u=540+19, \quad \text { or } \quad u=\frac{559}{46}=12 \frac{7}{46} \text { days. }
\]

b) Unchanged. The above method cannot be applied here, because the absence of $C$ on the last workday is not a full day.

Péter Tatai (Bp. XIV., I. István g. II. o. t.)

Note: Here, in the third solution, we see that the performance of $D$ is disproportionately low compared to the other three, and thus it is understandable that while $A, B, C$ together would complete the work in 12 days, the four of them together would complete it in only slightly less time $\left(11 \frac{17}{23}\right.$ days $)$.