Math Enrichment Problems
Welcome to the 4th month of threeringsconnections.orgMonthly Math Enrichment Problems post, Each month I post some Math Enrichment problems for grades 2-3. I hope you will find them useful with your students in class or your kids at home.
Don’t forget to use 1 of your 6 problem solving strategies:
- Draw a picture
- Guess and Check
- Guess and Check
- Use a table or list
- Find a pattern
- Logical reasoning
- Working backwards (try a simpler version first)
Problem Solving – Here we go!
- Abby has 60 cents in pennies, nickels and dimes. She has at least one of each coin. What is the difference between the largest number of coins that she could have and the smallest number of coins that she could have?
- There are 18 children in Room A and 6 children in Room B. If ____ children move from Room A to Room B, there will be twice as many children in Room B as in Room A.
- Amy and Ann were playing tic-tac-toe. Amy won 4 games and Ann won 4 more games than Amy. If there were 9 ties, how many games of tic-tac-toe did they play?
- A parking meter gives 30 minutes of parking for a quarter and 10 minutes for a dime. Mrs. Jaynor feels she will need 2 hours and 15 minutes on the meter. What is the least amount of money she should put in the meter?
- Brooklyn added up all the single-digit odd numbers and all the single-digit even numbers. What was her sum?
- There were 9 coins on a table totaling $1.20, consisting of nickels, dimes and quarters. Al, Ben and Casey each pick up 3 coins. Al has 3 times as much money as Casey. Al has as much money as Ben and Casey together. What 3 coins did Al pick up?
- Find 2 two numbers that multiply to 9 and add to 10?
- (36) Most coins: 1D, 1N, 45P = 47 coins Least: 5D, 1N, 5P = 11 coins 47-11 = 36
- (10) There are 24 total children (18 +6). The result must be 8 children in Room A and 16 in Room B. This can be done by moving 10 children from Room A to Room B.
- (21) Amy won 4 games, Ann won 8 games, (4 more than Amy) and 9 ties. 4 + 8 + 9 = 21.
- ($1.20) Four quarters (2 hours) and two dimes (20 minutes) would be the least expensive way to cover the time period.
- (45) 1 + 3 + 5 + 7 + 9 + 0 + 2 + 4 + 6 + 8 = 45
- (2Q, 1D) Al has 60cents, Ben has 40 cents and Casey has 20 cents. Al’s 3 coins are 2 quarters and 1 dime.
- (9 and 1) 9 X 1 = 9 and 9+1 =10