5th Grade Continental Math: Problems and Solutions
Many problems posed at Continental Math League competitions require advanced reasoning skills and quick thinking. Help your child develop these skills by practicing at home using the sample problems provided here, or develop your own using these examples as a guide.
What Is Continental Math?
The Continental Math League is an extracurricular program for students who love math. Schools form teams of students and compete in 30minute math meets at the regional and national level by solving six math problems. In 5th grade, Continental Math League teams can participate in either Euclidean or Pythagorean division. Euclidean division challenges your child to use logic, while Pythagorean division problems require advanced levels of reasoning and analysis.
Many 5th graders who participate in Continental Math are advanced and benefit from creative problems. In 5th grade, students learn about fractions; Pythagorean division problems require your child to take fractions to the next level by using them in proportions.
When you write your own problems for athome practice, avoid writing simple drills, like 30 ÷ 13. Instead, write out word problems: find the number that results from the product of 15 and 2 divided by the sum of 6 and 7. Such problems will challenge your child to use higherorder thinking skills, rather than rote memorization.
Sample Problems and Solutions
Euclidean
 1) If three A's are equal to one B and four B's are equal to one C, then how many A's are equal to 15 C's?
The solution is 36 A's are equal to 15 C's. To begin, write out the known equations: 3 A's = 1 B and 4 B's = 5 C's. Then, multiply the second equation by three to find out how many B's are equal to 15 C's. The equation should look like this: 12 B's = 15 C's.
Finish up by multiplying the first equation by 12 to determine how many A's equal 12 B's. You'll find that 36 A's = 12 B's and 12 B's = 15 C's. Therefore, 36 A's are equal to 15 C's.
 2) Find the missing number, A, if A always represents the same number: 61A  5A2 = 36.
The solution to this problem is A = 8. The first clue is when we subtract in the ones' column, A  2 = 6. After solving this equation, you can hypothesize that A may equal 8. Next, test out your answer by comparing the tens' and hundreds' place values. Does 61  58 = 3? Because this equation is accurate, you can conclude that A equals 8.
Pythagorean
 1) Solve for A: 500,000 ÷ 1,000 = 5,000 ÷ A
This problem can be solved using cross multiplication; however, because the Continental Math contests are limited to 30 minutes, it's important to figure it out quickly. Begin by solving the first equation, 500,000 ÷ 1,000, which equals 500. This can be easily solved in your head. Without writing anything down, calculate what 5,000 would be divided by to equal 500. The answer is 10.
 2) A person can run one mile in 30 minutes. How many miles will the same person run in one hour and 15 minutes?
Solving this problem requires understanding proportions. Begin by converting one hour and 15 minutes into one unit of measure. There are 60 minutes in one hour, so the total amount of minutes for one hour and 15 minutes is 75 minutes. Then, set up the proportion: 1 mile / 30 minutes = x miles / 75 minutes. Continue by crossmultiplying: (30)(x) = (1)(73). Isolate x by dividing both sides by 30, so that x = 75/30, or 2.5 miles.
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