March 11, 2013 Problems & Solutions

```MATHCOUNTS&reg; Problem of the Week Archive
More State Problems – March 11, 2013
Problems &amp; Solutions
This month, students from 56 states and territories are competing for a chance to go to Washington, DC
in May and participate in the 2013 Raytheon MATHCOUNTS National Competition. During the
Countdown Round, students are asked to solve problems like these from the 2012 State Competition in
45 seconds or less.
A set S contains 19 different integers with a mean of 101. The mean of the ten smallest integers in S is
50, and the mean of the ten largest integers in S is 150. What is the median of all the integers in S?
(Countdown #30)
The mean of the integers in S is 101, so the sum of the integers in S is 101 &times; 19 = 1919. The mean of the 10
smallest integers in S is 50, so the sum of these 10 integers is 50 &times; 10 = 500. The mean of the 10 largest
integers in S is 150, so the sum of these 10 integers is 150 &times; 10 = 1500. Each of these sums includes the tenth
integer in S, which is also the median. Therefore, the median of S is 500 + 1500 – 1919 = 2000 – 1919 = 81.
Eight swimmers, numbered from 1 to 8, stand in ascending order clockwise around the edge of a circular
swimming pool. Going around clockwise, every second swimmer still standing will jump into the pool.
Swimmer #2 is the first to jump in the pool. What is the number of the swimmer who is the seventh to
jump into the pool? (Countdown #31)
Based on the information provided, we see that the swimmers will jump into the pool in the
following order: swimmer #2, swimmer #4, swimmer #6, swimmer #8, swimmer #3, swimmer #7,
swimmer #5, at which time only swimmer 1 remains. Therefore, the seventh swimmer to jump in
the pool was swimmer #5.
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Square ABCD has sides of length 22 cm. What is the area, in square centimeters, of the