There are seven children in the Arthur family, including one set of twins. The youngest child is 6 years old, and the oldest is
16 years old. The mean of their ages is 11 years, the median is 10 years, and the mode is 15 years. What are the ages of the Arthur children? How old are the twins?
One is 6 years old, one is 16 as per the rules. Now we have 5 unknowns, of which either the twins can be both a part of, or one can be either 6 or 16 as well.
Since the mode is the most frequently occurring number, let's assume that 15 years old appears twice to make it so.
Now we have one 6 y.o, two 15 y.os, and one 16 y.o.
Out of 7 kids, the middle child has to be 10 years old to secure the median.
Now we just have two missing, and we need to figure out the mean. 6+15+15+10+16 is 62, dividing by 5 is 12.4 The mean is currently 12.4 years old. For the mean to be 11, the combined total has to be 77 (since 77/7 kids is 11 mean) 77-62 is 15. But to make sure the median stays 10, both of those kids have to be less than 10 but more than 6. Therefore, their ages must be 7 and 8.
Their ages are 6, 7, 8, 10, 15, 15, 16
Let's double check to make sure we have all the qualifications. 7 kids, check. Youngest 6, check. Oldest 16, check. Twins, check. Median of 10, check. Mode of 15, check. Average of 11, well 6+7+8+10+15+15+16 is 77, 77/7 is 11, so check. Phew.
The sum is 10s^2t-5st^2. With two separate sections, it is a binomial, and since the sum of the powers of the variables is 2+1=3, it has a degree of 3.
