$36 \cdot 56$ A factory has a clock that is running slow, so that the minute hand meets the hour hand at the usual clock face positions (such as 12 o'clock, etc.) only once every 69 minutes. A worker is paid $4.00 per hour, and $6.00 per hour for any time worked over the specified hours. After the worker completes the specified 8-hour work according to the slow clock, the additional amount that should be paid to the worker is
(A) $2.30.
(B) $2.60.
(C) $2.80.
(D) $3.00.
(E) $3.30.
(22nd American High School Mathematics Examination, 1971)
[Solution] The slow clock's hour and minute hands meet 11 times, that is,
12 hours $=69 \cdot 11$ minutes $=12$ hours + 39 minutes.
$\therefore$ The slow clock runs 8 hours $=8$ hours + 26 minutes, exceeding the specified time by 26 minutes. At a rate of 6 dollars per hour, or 10 cents per minute, the total cost is 2.60 dollars. Therefore, the answer is $(B)$.