Difficult Addition Problems: Solutions and Strategies

Does your child understand how to add, but more difficult problems cause him or her to struggle? Review the strategies below, which can help your child master even the most difficult addition problems. Sample problems and solutions are also included for additional practice.

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What Are Difficult Addition Problems?

What makes an addition problem difficult is subjective; however, there are definitely some factors that make some problems more challenging than others. At the elementary level, difficult addition problems are those that involve regrouping, variables and decimals. For many students, word problems make addition seem more difficult. Since difficult problems involving these factors are present throughout middle school and high school, it is important for your child to learn strategies to use when difficult problems arise.

Counting on or counting back is a strategy that can be used for solving equations with variables. For instance, the problem 1 + ? = 5 could be solved by counting from one to five, which will yield the answer four. Another strategy includes using the knowledge of basic facts, which means that your child will look for simple math facts within harder problems. For instance, 504 + 306 can seem easier if children concentrate on 4 + 6 (4 + 6 = 10 and 504 + 306 = 810). Finally, physically modeling the problem is a strategy that can allow students to visualize equations.

Below, you will find some difficult addition problems for your child to solve. Each problem has the correct answer included, as well as an appropriate strategy your child can implement.

Lower Elementary (K-2)

1. Joan had four stickers in her sticker book. As a surprise, her mom bought her a pack of stickers to add to her book. After putting the new stickers in her book, Joan had ten stickers total. How many stickers were in the pack her mom bought?

To solve this problem, your child would need to solve for the missing number in the equation 4 + x = 10. Your child may choose to use the counting on or counting back strategy to determine that the variable 'x' stands for six. The equation would look like this: 4 + 6 = 10. Physically modeling or drawing the problem would be another appropriate strategy for this problem.

2. Jonah and his friends collect baseball cards. The friends decided to put all of their cards together to make one collection. Jonah has 38 cards, Adam has 24 cards, Abe has 18 cards and Noah has 11 cards. How many cards do the friends have altogether?

This addition problem will require your child to add several numbers and regroup. Your child should add: 38 + 24 + 18 + 11 = 91. To correctly solve this word problem, your child will need to apply his or her knowledge of basic addition facts.

Upper Elementary (3-5)

1. Eddie and his brothers cleaned out their piggy banks and put their money together in hopes of having enough money to buy the newest video game. Eddie had $5.08, Shawn had $12.92, Rob had $28.16 and Ben had $0.97. How much did the brothers have altogether? If Eddie's sister has enough money in her piggy bank for them to have $50.00, how much money must she have?

To solve this addition word problem, your child first needs to add the monetary amounts like this: $5.08 + $12.92 + $28.16 + $0.97 = $47.13. To determine how much money the boys' sister has, your child would need to set up an equation like this: $47.13 + x = $50.00. Your child could use a counting on or counting back strategy to solve for x. In this equation, x = $2.87, which would show the amount of money Eddie's sister had.

2. First Street High School has 982 students in attendance. The ninth grade class has 184 students present, tenth grade has 208 and eleventh grade reports 368 students present. How many students are present in twelfth grade?

To determine how many twelfth grade students are present, your child needs to set up an equation and solve for the unknown. The equation would look like this: 184 + 208 + 368 + x = 982. After your child isolates x, the equation would look like this: x = 982-760. Then, your child should subtract the two amounts, with x = 222. This shows that there are 222 students present in the twelfth grade. In solving this problem, your child could use his or her knowledge of basic facts to obtain the answer.
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