Apply Properties of Multiplication - 3rd Grade Math

Which expression correctly uses the distributive property to find $7 \times 6$ , given that $7 \times 5 = 35$ ?
Answer: $7 \times 6 = 7 \times 5 + 7 \times 1 = 35 + 7 = 42$ Breaking 6 into 5+1, 7 x 6 = 7 x 5 + 7 x 1 = 35 + 7 = 42, so Choice A is correct. Choice B subtracts the extra group instead of adding it. Choice C adds only 1 instead of 7 x 1, forgetting to multiply the extra group by 7. Choice D adds 7 and 6 instead of multiplying them, confusing the two operations entirely.
You can think of 8 as 5 + 3. What is $7 \times 8$ ?
Answer: 56 56 is correct because 7 times 8 equals 56. 48 is incorrect because that is the product of 6 and 8, not 7 and 8. 58 is incorrect because it is close to 56 but not the actual product. 54 is incorrect because it does not match 7 times 8.
Use the distributive property: $6 \times 9 = 6 \times (10-1)$ . What is $6 \times 9$ ?
Answer: 54 Six times 10 is 60, and subtracting 6 times 1, or 6, gives 54. Choice A comes from a subtraction error. Choice B comes from an addition error instead of using the distributive property. Choice D stops after computing 6 times 10 without subtracting.
Which regrouping correctly uses the associative property to solve $2 \times 3 \times 7$ ?
Answer: $(2 \times 3) \times 7 = 42$ Grouping 2 times 3 first, then multiplying by 7, gives 6 times 7, or 42. Choice A wrongly adds 2 and 3 instead of multiplying them. Choice B wrongly adds 3 and 7 instead of keeping the multiplication grouping. Choice D wrongly adds 7 instead of multiplying by it.
Use the distributive property to find $8 \times 7$ by breaking apart 7.
Answer: $8 \times 7 = 8 \times (5+2) = 8 \times 5 + 8 \times 2 = 40 + 16 = 56$ Choice A is correct: 8 x 7 = 8 x (5+2) = 8 x 5 + 8 x 2 = 40 + 16 = 56. Choice B distributes to only the second term and drops the 8 x 5 part. Choice C finds both partial products but forgets to add them together. Choice D multiplies the second part by 2 instead of by 8, giving an incorrect partial product.
Use the distributive property to find $6 \times 9$ by thinking $9=10-1$ .
Answer: $6 \times 9 = 6 \times (10-1) = 60-6 = 54$ Breaking 9 into 10 minus 1 gives 6 times 9 equals 6 times 10 minus 6 times 1, which is 60 minus 6, or 54. Choice A restates the numbers in reverse order but reaches an incorrect product without using the distributive breakdown. Choice C adds the two numbers instead of multiplying them. Choice D subtracts 1 directly from the product instead of subtracting 6, which is 6 times 1.
Carlos knows that $9 \times 4 = 36$ and $9 \times 3 = 27$ . He wants to find $9 \times 7$ using the distributive property. Which equation shows the correct way to break apart this problem?
Answer: $9 \times 7 = 9 \times (4 + 3) = 36 + 27 = 63$ D is correct because the distributive property breaks 7 into 4+3, multiplies each part by 9, then adds: 36 + 27 = 63. Choice A incorrectly multiplies 4 and 3 together inside the parentheses instead of adding them. Choice B incorrectly multiplies the two partial products instead of adding them. Choice C replaces multiplication with addition throughout, which isn't how the distributive property works.
Use the associative property to solve $4 \times 2 \times 5$ efficiently.
Answer: $4 \times (2 \times 5) = 4 \times 10 = 40$ This question tests applying properties of operations to multiply and divide (CCSS.3.OA.5), specifically using associative property as a strategy. The associative property of multiplication states the way factors are grouped doesn't change the product. (3×5)×2 = 3×(5×2). Both equal 30. Helpful for three or more factors: regroup to make easier calculations (2×5=10, then 10×3=30 is easier than 2×3=6, then 6×5=30). In this problem, we can solve 4×2×5 by regrouping to make it easier. The associative property helps by grouping 2×5 first to get 10, then multiply by 4. Choice A is correct because it regroups 4×(2×5) as 4×10=40 using associative property. This demonstrates proper use of the associative as a strategy. Choice D is incorrect because it has a calculation error, 4×10=40 not 14. This error occurs when students make computational mistakes within the property application. To help students apply properties: Explicitly teach and name properties with examples. Associative: Model with nested containers (boxes/bags/items). Show both groupings equal same total. Practice as strategies, not just as abstract properties: "How can associative property help you?" Connect to real situations. Watch for students who memorize property names but don't apply them strategically, or who make errors within the property application.
If $7 \times 5 = 35$ , what is $7 \times 6$ ?
Answer: 42 42 is correct because $7 \times 6 = 7 \times 5 + 7 = 35 + 7 = 42$ . 67 is incorrect; it isn't a valid product for $7 \times 6$ . 40 is incorrect because it adds only 5 instead of 7 to 35. 36 is incorrect because it adds just 1 instead of 7 to 35.
If $7 \times 9 = 63$ , what is $9 \times 7$ ?
Answer: 63 63 is correct because multiplying the same two numbers in either order gives the same product, so $9 \times 7 = 7 \times 9 = 63$ . 79 is incorrect; it combines the digits 7 and 9 rather than multiplying them. 72 is incorrect because it doesn't match the given product. 16 is incorrect; it comes from adding 9 and 7 instead of multiplying.
Lisa wants to solve $3 \times 4 \times 2$ by using the associative property. She calculates $4 \times 2 = 8$ first. What should she do next to complete the problem?
Answer: Multiply $3 \times 8 = 24$ When you see multiplication problems with three or more numbers, the associative property lets you group and multiply them in any order to make the calculation easier. This property says that $(3 \times 4) \times 2 = 3 \times (4 \times 2)$ , so you can choose which two numbers to multiply first. Lisa chose to multiply $4 \times 2 = 8$ first, which means she's using the grouping $3 \times (4 \times 2)$ . Now she has simplified the original problem $3 \times 4 \times 2$ into $3 \times 8$ . To finish solving, she needs to multiply $3 \times 8 = 24$ , making choice C correct. Let's see why the other choices don't work. Choice A suggests multiplying $8 \times 2 = 16$ , but Lisa already used the 2 when she calculated $4 \times 2 = 8$ . Using the 2 again would be counting it twice. Choice B says to add $3 + 8 = 11$ , but the original problem uses multiplication throughout, so switching to addition changes the problem entirely. Choice D suggests subtracting $8 - 3 = 5$ , which also changes the operation from multiplication to subtraction. Remember: when using the associative property with multiplication, you're just changing the order of operations, not the operations themselves. Once you multiply two numbers together, you multiply that result by the remaining number. The key is to keep multiplying until you've used each number exactly once.
Solve $3 \times 5 \times 2$ by regrouping using the associative property.
Answer: $(3 \times 5) \times 2 = 15 \times 2 = 30$ This question tests applying properties of operations to multiply and divide (CCSS.3.OA.5), specifically using associative property as a strategy. The associative property of multiplication states that the way factors are grouped doesn't change the product: (a×b)×c = a×(b×c). In this problem, we can solve 3×5×2 by regrouping to make easier calculations. The associative property helps by allowing us to group factors in the most convenient way. Choice A is correct because it regroups (3×5)×2 = 15×2 = 30, calculating 3×5 first, then multiplying by 2. This demonstrates proper use of the associative property as a strategy. Choice D is incorrect because it shows 3×(5×2) = 3×7 = 21, which contains a calculation error (5×2=10, not 7). This error occurs when students apply the property correctly but make computational mistakes within the regrouping. To help students apply properties: Model with nested containers (boxes/bags/items) to show both groupings equal the same total. Practice choosing strategic groupings like pairing factors that make 10.
Use a near fact: $9 \times 8 = 10 \times 8 - 1 \times 8$ . What is $9 \times 8$ ?
Answer: 72 Ten times 8 is 80, and subtracting 1 times 8, or 8, gives 72. Choice B comes from a subtraction error. Choice C stops after computing 10 times 8 without subtracting. Choice D comes from adding instead of using the near-fact strategy correctly.
Jake is solving $6 \times 5 \times 3$ . He decides to group the numbers differently to make it easier. Which two expressions will give Jake the same answer?
Answer: $(6 \times 5) \times 3$ and $6 \times (5 \times 3)$ Choice A shows the associative property of multiplication, where grouping doesn't change the product: (6 × 5) × 3 = 30 × 3 = 90, and 6 × (5 × 3) = 6 × 15 = 90. Choice B uses addition instead of multiplication in the groupings. Choice C changes multiplication to addition, giving different results. Choice D uses subtraction and gives different results: 6 × 2 = 12 vs. 1 × 3 = 3.
Tommy knows that $5 \times 6 = 30$ . He wants to find $5 \times 8$ by thinking of 8 as $6 + 2$ . What should Tommy's next step be to use the distributive property correctly?
Answer: Calculate $5 \times 2 = 10$ , then add $30 + 10 = 40$ When you see a problem asking you to use the distributive property, you're breaking apart one number to make multiplication easier. The distributive property says that when you multiply a number by a sum, you can multiply by each part separately and then add the results. Tommy wants to find $5 \times 8$ by thinking of 8 as $6 + 2$ . This means $5 \times 8 = 5 \times (6 + 2)$ . Using the distributive property, this becomes $5 \times 6 + 5 \times 2$ . Since Tommy already knows $5 \times 6 = 30$ , he needs to calculate $5 \times 2 = 10$ , then add these results: $30 + 10 = 40$ . This makes choice D correct. Let's see why the other answers don't work. Choice A calculates $5 + 6 = 11$ , but the distributive property doesn't involve adding the factors together before multiplying. Choice B adds $5 + 2 = 7$ and then multiplies by 30, which completely misapplies the property and gives an unreasonably large answer. Choice C calculates $6 \times 2 = 12$ , but this multiplies the parts of 8 together instead of multiplying each part by 5. Remember this pattern: when using the distributive property to break apart multiplication, multiply the outside number by each part separately, then add those products together. So $a \times (b + c) = a \times b + a \times c$ . This strategy helps make difficult multiplication problems much easier!
Maria knows that $4 \times 9 = 36$ . She wants to find $4 \times 11$ using what she already knows. Which strategy shows the correct use of the distributive property?
Answer: $4 \times 11 = 4 \times (9 + 2) = (4 \times 9) + (4 \times 2) = 36 + 8 = 44$ Choice A correctly uses the distributive property by breaking 11 into (9 + 2), then distributing the 4: 4 × (9 + 2) = (4 × 9) + (4 × 2) = 36 + 8 = 44. Choice B incorrectly adds 36 + 11 instead of using properties. Choice C mistakes 11 as 9 × 2 and uses an incorrect operation. Choice D incorrectly adds 4 + 9 and creates an unrelated expression.
Use the distributive property: $7 \times 6 = 7 \times 5 + 7 \times 1$ . What is $7 \times 6$ ?
Answer: 42 This question tests applying properties of operations to multiply and divide (CCSS.3.OA.5), specifically using distributive property as a strategy. The distributive property of multiplication states that you can break apart one factor and distribute the multiplication. 7×6 = 7×(5+1) = (7×5)+(7×1) = 35+7 = 42. Helpful when you know part of the fact (7×5=35) and can add on the rest (7×1=7). In this problem, we can solve 7×6 by breaking 6 into 5+1. The distributive property helps by allowing us to compute easier multiplications and add them. Choice C is correct because it breaks 7×6 into 7×5 + 7×1 = 35+7 = 42 using distributive property. This demonstrates proper use of the distributive as a strategy. Choice B is incorrect because 41 might come from a miscalculation like 35+6 instead of 35+7. This error occurs when students make computational mistakes within the property application. To help students apply properties: Explicitly teach and name properties with examples. For distributive: Use area models (split rectangle) or break-apart strategies. Teach: "Use facts you know to figure out facts you don't." Practice as strategies, not just as abstract properties: "How can distributive property help you?" Connect to real situations. Watch for students who memorize property names but don't apply them strategically, or who make errors within the property application.
Which strategy correctly breaks apart the numbers to solve $7 \times 9$ ?
Answer: $7 \times 9 = 7 \times (10-1) = 70 - 7$ Breaking 9 into 10 minus 1 is correct because 7 times 9 equals 7 times 10 minus 7, which is 63. Reversing the order of the factors is incorrect because it does not break either number apart. Adding 7 and 9 is incorrect because it replaces multiplication with addition. Adding the numbers first and multiplying by 1 is incorrect because it also replaces multiplication with addition.
Ben knows that $8 \times 3 = 24$ and $8 \times 4 = 32$ . He needs to find $8 \times 7$ . If Ben uses $7 = 3 + 4$ , which calculation shows the correct use of the distributive property?
Answer: $8 \times 7 = 8 \times (3 + 4) = (8 \times 3) + (8 \times 4) = 24 + 32 = 56$ C is correct because the distributive property splits 7 into 3+4, then multiplies each part by 8 before adding: (8x3)+(8x4) = 24+32 = 56. Choice A incorrectly multiplies 3 and 4 together instead of adding them inside the parentheses. Choice B incorrectly multiplies the two partial products instead of adding them. Choice D replaces multiplication with addition throughout, which doesn't represent the distributive property.
Ana is solving $7 \times 5 \times 4$ . She wants to make the calculation easier by changing how the numbers are grouped. Which grouping will give Ana the same answer as the original problem?
Answer: $(7 \times 5) \times 4 = 35 \times 4 = 140$ The correct regrouping is B, since the associative property lets Ana regroup the factors as (7x5)x4 = 35x4 = 140, the same value as the original 7x5x4. Choice A also multiplies throughout but contains an arithmetic error (7x4=28, not 35), giving a wrong product. Choices C and D change one of the operations to addition, which changes the value of the expression entirely.

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