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Problem Solving - Developing mental strategies for calculation

Grade 3IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Partitioning: Breaking numbers into hundreds, tens, and units (e.g., 34+2534 + 25 as 30+2030+20 and 4+54+5).

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Compensation: Adding or subtracting a near multiple of 10 and then adjusting (e.g., to add 19, add 20 then subtract 1).

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Bridging through 10 or 100: Adding or subtracting to the nearest 10 or 100 first, then adding the remainder.

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Near Doubles: Using known doubles to solve calculations (e.g., 15+1615 + 16 is double 15 plus 1).

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Inverse Operations: Understanding that addition is the opposite of subtraction and can be used to check answers.

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Counting On: For subtraction, starting at the smaller number and counting up to the larger number to find the difference.

📐Formulae

a+b=b+aa + b = b + a (Commutative Law)

a−b=c  ⟹  c+b=aa - b = c \implies c + b = a (Inverse Relationship)

n+9=(n+10)−1n + 9 = (n + 10) - 1 (Compensation Strategy)

n+n=2×nn + n = 2 \times n (Doubling)

💡Examples

Problem 1:

Calculate 45+3745 + 37 mentally using partitioning.

Solution:

8282

Explanation:

Partition the numbers into tens and ones: (40+30)=70(40 + 30) = 70 and (5+7)=12(5 + 7) = 12. Then add the results together: 70+12=8270 + 12 = 82.

Problem 2:

Find the result of 156−19156 - 19 using the compensation strategy.

Solution:

137137

Explanation:

Instead of subtracting 19, subtract 20 (the nearest ten): 156−20=136156 - 20 = 136. Since you subtracted 1 too many, add 1 back: 136+1=137136 + 1 = 137.

Problem 3:

Solve 82−7582 - 75 by counting on.

Solution:

77

Explanation:

Start at 75. Count on to the next ten: 75+5=8075 + 5 = 80. Then count from 80 to 82: 80+2=8280 + 2 = 82. The total difference is 5+2=75 + 2 = 7.

Problem 4:

Use near doubles to solve 25+2625 + 26.

Solution:

5151

Explanation:

Recognize that 25+2625 + 26 is (25+25)+1(25 + 25) + 1. Since double 25 is 50, adding 1 gives 51.