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Computation and Operations - Multiplication of 2-digit by 2-digit numbers

Grade 4IB

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

🔑Concepts

Place Value Understanding: In 2-digit by 2-digit multiplication, numbers are broken down into tens and ones. For example, 24×3524 \times 35 is thought of as (20+4)×(30+5)(20 + 4) \times (30 + 5). Visualizing this as a large rectangle partitioned into four smaller sections helps track the different parts of the calculation.

The Area Model: This is a visual strategy where a rectangle represents the product. The width is split into tens and ones (e.g., 2020 and 44) and the height is split into tens and ones (e.g., 3030 and 55). The area of the four smaller boxes (20×3020 \times 30, 20×520 \times 5, 4×304 \times 30, and 4×54 \times 5) are calculated and then added together.

Partial Products: When multiplying, we calculate four separate products and find their sum. For 42×1342 \times 13, the partial products are 3×23 \times 2, 3×403 \times 40, 10×210 \times 2, and 10×4010 \times 40. In a vertical column layout, these appear as separate lines of calculation before the final addition.

The Placeholder Zero: When using the standard algorithm, we place a 00 in the ones column when multiplying by the digit in the tens place. This is because we are actually multiplying by a multiple of 1010. Visually, this shifts the entire second row of digits one place to the left.

The Distributive Property: This mathematical rule allows us to multiply a sum by a number. For 2-digit multiplication, it is expressed as a×(b+c)=(a×b)+(a×c)a \times (b + c) = (a \times b) + (a \times c). We use this twice to break down complex problems into four simpler multiplication steps.

Estimation for Reasonableness: Before solving, round both numbers to the nearest ten to estimate the product. If calculating 28×3128 \times 31, estimate 30×30=90030 \times 30 = 900. If your final answer is significantly different from 900900, you should re-check your steps.

Regrouping (Carrying): When a partial product in a specific column exceeds 99, we carry the extra value to the next higher place value column. Visually, this is often noted as a small digit written above the tens or hundreds column to be added later.

📐Formulae

Product=Factor 1×Factor 2\text{Product} = \text{Factor 1} \times \text{Factor 2}

(a+b)×(c+d)=(a×c)+(a×d)+(b×c)+(b×d)(a + b) \times (c + d) = (a \times c) + (a \times d) + (b \times c) + (b \times d)

Total Product=Partial Product 1+Partial Product 2+Partial Product 3+Partial Product 4\text{Total Product} = \text{Partial Product 1} + \text{Partial Product 2} + \text{Partial Product 3} + \text{Partial Product 4}

💡Examples

Problem 1:

Calculate 23×1423 \times 14 using the Area Model.

Solution:

Step 1: Decompose the numbers: 23=20+323 = 20 + 3 and 14=10+414 = 10 + 4. \nStep 2: Multiply the four parts: \n20×10=20020 \times 10 = 200 \n20×4=8020 \times 4 = 80 \n3×10=303 \times 10 = 30 \n3×4=123 \times 4 = 12 \nStep 3: Add the partial products together: \n200+80+30+12=322200 + 80 + 30 + 12 = 322. \nFinal Answer: 23×14=32223 \times 14 = 322.

Explanation:

We use the distributive property to break the multiplication into four simpler calculations based on place value, then sum the results.

Problem 2:

Solve 45×2645 \times 26 using the standard vertical algorithm.

Solution:

Step 1: Multiply 4545 by 66 (the ones digit of 2626). \n6×5=306 \times 5 = 30 (write 00, carry 33) \n6×4=246 \times 4 = 24, plus the carried 3=273 = 27. First line: 270270. \nStep 2: Multiply 4545 by 2020 (the tens digit of 2626). \nPlace a 00 in the ones place. \n2×5=102 \times 5 = 10 (write 00, carry 11) \n2×4=82 \times 4 = 8, plus the carried 1=91 = 9. Second line: 900900. \nStep 3: Add the two lines together: \n270+900=1170270 + 900 = 1170. \nFinal Answer: 45×26=117045 \times 26 = 1170.

Explanation:

This method uses two rows for partial products. The first row calculates 45×645 \times 6 and the second row calculates 45×2045 \times 20 (using a placeholder zero), which are then summed.