Math For MATHCOUNTS - The Solutions
03/01/2004

We need to solve the equation 5931 = 0.623x. Dividing both sides by 0.623 gives us that there are approximately 38,100 students registered for MATHCOUNTS this year.

We could use the proportion (550/11) = (5931/x), where x is the number of schools only registering individuals. Then our final answer would be 5931 + x. Alternately, we could solve the proportion (561/550) = (x/5931), where x is the total number of schools registered for MATHCOUNTS. If we multiply both sides by 5931, we see that there are 6050 schools registered for MATHCOUNTS.

Notice that 8.5 × 2 = 17, and we have the 8.5 side corresponding to 1.7 inches. It appears that we have multiplied 8.5 by 2 and then divided by 10. Performing the same operations on the 11-inch side, we get 11 × 2 ÷ 10 = 2.2 inches.

You may or may not have learned about some trigonometry. Since we are working with an isosceles triangle, the altitude from the vertex angle B divides the base AC in half and divides the isosceles triangle into two right triangles. Now using our trig ratios we can find the degree measure of angle A with (cos A) = 15/20, which results in angle A measuring 41 degrees.

Though we don’t know ahead of time how many of the competitors will be males or females, we do know that there will either be an even number of both, or an odd number of both. If there is an even number of both, then we would only need 228 ÷ 2 = 114 rooms. However, if there is an odd number, then you might think we’d need some extra rooms. Notice, however, that an odd number of girls and an odd number of boys will only lead to needing one additional room. Therefore, the greatest number of rooms that would be necessary under these guidelines is 115.