Elementary School Math Quiz: Explain Addition And Subtraction Strategies
20 questions · exam conditions
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Explain Addition And Subtraction StrategiesQuestion 1 of 20

Complete this make ten strategy: 47+6=47+3+  ‾=50+  ‾47+6=47+3+\underline{\ \ }=50+\underline{\ \ }.

3 and 0
3 and 3
6 and 0
2 and 4
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Elementary School Math Quiz

Elementary School Math Quiz: Explain Addition And Subtraction Strategies

Practice Explain Addition And Subtraction Strategies in Elementary School Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Explain Addition And Subtraction Strategies, giving you a quick way to practice the rules, question types, and explanations that matter most for Elementary School Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Complete this make ten strategy: 47+6=47+3+  ‾=50+  ‾47+6=47+3+\underline{\ \ }=50+\underline{\ \ }.

  1. 3 and 0
  2. 3 and 3 (correct answer)
  3. 6 and 0
  4. 2 and 4
Explanation: This question tests 2nd grade understanding of addition and subtraction strategies, including explaining why strategies work, comparing strategies, and selecting appropriate strategies for different problems (CCSS 2.NBT.B.9: Explain why addition and subtraction strategies work, using place value and the properties of operations). Common 2nd grade strategies include: (1) Counting on—start at larger number, count up by smaller (best for adding small numbers like 47+3); (2) Make ten—decompose to make 10 first (8+5 → 8+2+3 → 10+3=13; best when number close to 10); (3) Break apart—separate tens and ones, add separately (47+28 → 40+20=60, 7+8=15, 60+15=75; best for two-digit numbers); (4) Compensation—adjust to friendly number (47+28 → 50+28-3=78-3=75; good for numbers near decades); (5) Counting back/up for subtraction. Strategies work because they use place value (break apart), number relationships (make ten), or counting principles (counting on/back). Understanding why helps students choose and apply strategies flexibly. In this problem, a partial make ten strategy is shown, and the question asks to complete the blanks. To analyze, examine steps (for 47+6, decompose 6 into 3+3 to make 47+3=50, then 50+3=53). Choice A is correct because the blanks are 3 and 3 (splitting 6 to make next ten at 50), demonstrating understanding of mathematical reasoning. Choice C represents a specific error (incorrect split like 2 and 4 which wouldn't make 50 from 47), which typically happens when students don't understand strategy characteristics. To help students: Explicitly teach and name strategies. Model each strategy with think-aloud: 'I'm using make ten. 47+6: Need 3 to 50, break 6 into 3+3, 50+3=53.' Show same problem solved multiple ways, compare: 'Make ten vs. others.' Discuss when each strategy works best: counting on for small addends, break apart for two-digit, make ten when close to 10/decade. Use number lines to show counting strategies visually. Have students explain their thinking: 'How did you solve this? Why did you choose that strategy?' Teach error checking within strategies (recount, check if makes sense). Compare student work: 'Correct split. Why?' Practice strategy selection: 'For this problem, which strategy makes most sense?' Build flexibility—multiple strategies often work, but efficiency varies. Watch for: naming strategies without understanding why they work, claiming one strategy always best (it depends on problem), weak mathematical explanations ("because it's easier" instead of reasoning about place value or number relationships), confusing strategy names, selecting inefficient strategies for problem type.

Question 2

Which strategy did Amir use for 47+847+8: 47+3=5047+3=50, then 50+5=5550+5=55?

  1. Make ten (correct answer)
  2. Counting on
  3. Counting back
  4. Break apart by tens and ones
Explanation: Amir broke 8 into 3 and 5 so that 47 plus 3 made a friendly ten, 50, then added the remaining 5 to reach 55, which is the make ten strategy. Choice B, counting on, would mean adding one at a time instead of jumping to a ten. Choice C, counting back, involves subtracting, which does not fit an addition problem. Choice D, break apart by tens and ones, would split both numbers by place value rather than building to a ten.

Question 3

Which strategy is most efficient for 68 + 29: break apart, counting on, counting back, or make ten?

  1. Make ten, because every problem starts with 10
  2. Counting on, because you count 29 numbers
  3. Counting back, because addition goes backward
  4. Break apart, because you add tens and ones (correct answer)
Explanation: Breaking apart 68 and 29 into tens and ones (60 and 8, 20 and 9) makes the addition simple and efficient. Choice A is wrong because not every problem starts with making a ten. Choice B describes counting on by 29, which is slow and error-prone for larger numbers. Choice C is incorrect because addition doesn't work by counting backward.

Question 4

Which strategy is most efficient for 58+358+3: counting on, standard algorithm, break apart, or counting back?

  1. Break apart
  2. Standard algorithm
  3. Counting back
  4. Counting on (correct answer)
Explanation: For 58+358+3, adding on just 3 more from 58 is the quickest way to reach the sum without extra steps, so counting on is the most efficient strategy, matching D. A is incorrect because breaking apart to add tens first is unnecessary when adding a single small number. B is incorrect because the standard algorithm involves more steps than needed for such a small addition. C is incorrect because counting back is used for subtraction, not addition.

Question 5

Which strategy did Jamal use: counting on, make ten, or break apart? 47+8→47+3+5→50+547+8\rightarrow 47+3+5\rightarrow 50+5

  1. Break apart
  2. Counting on
  3. Make ten (correct answer)
  4. Counting back
Explanation: Jamal split 8 into 3 and 5 so that 47 plus 3 made 50, a friendly ten, then added the last 5 to reach 55, which is the make ten strategy. Choice A, break apart, would split both addends by place value instead of building to a ten. Choice B, counting on, means adding one at a time rather than jumping to a ten. Choice D, counting back, involves subtracting, which does not apply here.

Question 6

Sofia solved 55−4755 - 47 by starting at 47 and adding small jumps until she reached 55. Which strategy did she use?

  1. Counting back
  2. Counting up (find the difference) (correct answer)
  3. Make ten
  4. Break apart by tens and ones
Explanation: Sofia used counting up, since she started at the smaller number and added jumps until she reached the larger number, finding the difference between them. Counting back would mean starting at 55 and subtracting down to 47, which is not what she did. Make ten involves adjusting numbers to reach a friendly ten, which is not shown here. Breaking apart by tens and ones means splitting numbers by place value, which is also not what Sofia did.

Question 7

Why does the break-apart strategy work for solving 34+2534+25 by adding 30+2030+20 and 4+54+5 separately?

  1. It adds all the digits together
  2. It uses tens with tens and ones with ones (correct answer)
  3. It always makes a ten first
  4. It counts backward to subtract
Explanation: Break apart works because it adds the tens together (30+20) and the ones together (4+5) separately, then combines the results. Choice A is too vague and does not explain the actual strategy. Choice C describes a different strategy (making a ten). Choice D describes subtraction, not addition.

Question 8

Which way to solve 68+368+3 takes the fewest steps?

  1. Count on from 68: 69, 70, 71 (correct answer)
  2. Break 3 apart into smaller pieces and add each piece separately
  3. Line up the ones and tens in columns and add
  4. Change 68 to 70 first, then adjust the answer
Explanation: Counting on from 68 (69, 70, 71) is a quick way to add a small number like 3, so it fits this problem well. Breaking 3 into smaller pieces adds extra steps that aren't needed for such a small number. Lining up ones and tens in columns works but takes longer for a simple problem like this. Changing 68 to 70 and then adjusting also adds steps that aren't necessary here.

Question 9

Sofia solves 75−2875-28 by first changing it to 77−3077-30. Why does this still give the same answer?

  1. She added 22 to both numbers, so the answer did not change. (correct answer)
  2. She added 22 to both numbers, so the answer got bigger.
  3. She subtracted 22 from both numbers, so the answer did not change.
  4. She added 22 to only one number, so the answer changed.
Explanation: Choice A is correct because adding 2 to both 75 and 28 keeps the gap between them the same, so 77-30 has the same answer as 75-28. Choice B is wrong because adding the same amount to both numbers does not make the answer bigger. Choice C is wrong because Sofia added 2 to both numbers, she did not subtract 2. Choice D is wrong because Sofia changed both numbers, not just one, and that is what keeps the answer the same.

Question 10

Which is more efficient for 47+2847+28: number line jumps or counting on by ones?

  1. Number line jumps, because you can jump tens then ones (correct answer)
  2. They are always the same speed
  3. Counting on by ones, because it is always fastest
  4. Counting back, because subtraction is easier
Explanation: This question tests 2nd grade understanding of addition and subtraction strategies, including explaining why strategies work, comparing strategies, and selecting appropriate strategies for different problems (CCSS 2.NBT.B.9: Explain why addition and subtraction strategies work, using place value and the properties of operations). Common 2nd grade strategies include: (1) Counting on—start at larger number, count up by smaller (best for adding small numbers like 47+3); (2) Make ten—decompose to make 10 first (8+5 → 8+2+3 → 10+3=13; best when number close to 10); (3) Break apart—separate tens and ones, add separately (47+28 → 40+20=60, 7+8=15, 60+15=75; best for two-digit numbers); (4) Compensation—adjust to friendly number (47+28 → 50+28-3=78-3=75; good for numbers near decades); (5) Counting back/up for subtraction. Each strategy has strengths—counting on quick for small numbers, break apart good for larger, make ten leverages 10 as friendly number. Strategies work because they use place value (break apart), number relationships (make ten), or counting principles (counting on/back). Understanding why helps students choose and apply strategies flexibly. In this problem, two strategies compared for efficiency. To analyze, compare (counting on efficient for 47+3 because only 3 counts; break apart better for 47+28 because too many counts otherwise). Choice B is correct because number line jumps is more efficient than counting on for 47+28 because you can jump tens then ones (jump +20, then +8—quick). This demonstrates understanding of efficiency comparison. Choice A represents wrong efficiency judgment (said counting on best for 47+28 when this would require 28 counts—inefficient compared to break apart or number line jumps). This error typically happens when students misjudge efficiency. To help students: Explicitly teach and name strategies. Model each strategy with think-aloud: 'I'm using make ten. I have 8, I need 2 more to make 10. So I break 5 into 2 and 3. Now I have 10 + 3 = 13.' Show same problem solved multiple ways, compare: 'For 47+3, I could count on (48, 49, 50—quick!), or use standard algorithm (write vertically, add—more work). Which is faster? Counting on!' Discuss when each strategy works best: counting on for small addends, break apart for two-digit, make ten when close to 10/decade. Use number lines to show counting strategies visually. Have students explain their thinking: 'How did you solve this? Why did you choose that strategy?' Teach error checking within strategies (recount, check if makes sense). Compare student work: 'Emma used make ten, Jamal used break apart. Both got 13. Why did both work?' Practice strategy selection: 'For this problem, which strategy makes most sense?' Build flexibility—multiple strategies often work, but efficiency varies. Watch for: naming strategies without understanding why they work, claiming one strategy always best (it depends on problem), weak mathematical explanations ("because it's easier" instead of reasoning about place value or number relationships), confusing strategy names, selecting inefficient strategies for problem type.

Question 11

Sofia solves 9+69+6 by thinking 9+1=109+1=10, then 10+5=1510+5=15. What strategy did she use?

  1. Doubling the number
  2. Counting all from one
  3. Make a ten (correct answer)
  4. Counting back
Explanation: Sofia turned 9 into 10 by adding 1, then added the leftover 5 to get 15. This is called the make a ten strategy, so Choice C is correct. Choice A is wrong because doubling means adding a number to itself, which is not what Sofia did. Choice B is wrong because counting all from one means starting over at 1, not breaking apart numbers. Choice D is wrong because counting back is used for subtraction, not for adding on to reach ten.

Question 12

Which strategy did Jamal use: counting on, make ten, or break apart? 47+8→47+3+5→50+547+8\rightarrow 47+3+5\rightarrow 50+5​

  1. Counting on
  2. Counting back
  3. Make ten (correct answer)
  4. Break apart
Explanation: This question tests 2nd grade understanding of addition and subtraction strategies, including explaining why strategies work, comparing strategies, and selecting appropriate strategies for different problems (CCSS 2.NBT.B.9: Explain why addition and subtraction strategies work, using place value and the properties of operations). Common 2nd grade strategies include: (1) Counting on—start at larger number, count up by smaller (best for adding small numbers like 47+3); (2) Make ten—decompose to make 10 first (8+5 → 8+2+3 → 10+3=13; best when number close to 10); (3) Break apart—separate tens and ones, add separately (47+28 → 40+20=60, 7+8=15, 60+15=75; best for two-digit numbers); (4) Compensation—adjust to friendly number (47+28 → 50+28-3=78-3=75; good for numbers near decades); (5) Counting back/up for subtraction. Each strategy has strengths—counting on quick for small numbers, break apart good for larger, make ten leverages 10 as friendly number. Strategies work because they use place value (break apart), number relationships (make ten), or counting principles (counting on/back). Understanding why helps students choose and apply strategies flexibly. In this problem, student work is shown and the question asks which strategy was used. To analyze, examine steps in work (decomposing 8 into 3+5 to make 47+3=50 first, then +5). Choice C is correct because the work shown uses make ten strategy (decomposed 8 into 3+5 to make 47+3=50 first), demonstrating understanding of strategy characteristics. Choice A represents wrong strategy identified (said counting on when work clearly shows make ten—no simple counting sequence). This error typically happens when students don't recognize strategy from work shown. To help students: Explicitly teach and name strategies. Model each strategy with think-aloud: 'I'm using make ten. I have 8, I need 2 more to make 10. So I break 5 into 2 and 3. Now I have 10 + 3 = 13.' Show same problem solved multiple ways, compare: 'For 47+3, I could count on (48, 49, 50—quick!), or use standard algorithm (write vertically, add—more work). Which is faster? Counting on!' Discuss when each strategy works best: counting on for small addends, break apart for two-digit, make ten when close to 10/decade. Use number lines to show counting strategies visually. Have students explain their thinking: 'How did you solve this? Why did you choose that strategy?' Teach error checking within strategies (recount, check if makes sense). Compare student work: 'Emma used make ten, Jamal used break apart. Both got 13. Why did both work?' Practice strategy selection: 'For this problem, which strategy makes most sense?' Build flexibility—multiple strategies often work, but efficiency varies. Watch for: naming strategies without understanding why they work, claiming one strategy always best (it depends on problem), weak mathematical explanations ("because it's easier" instead of reasoning about place value or number relationships), confusing strategy names, selecting inefficient strategies for problem type.

Question 13

Which strategy is best for 62−5962-59: counting up or counting back?

  1. Standard algorithm, because it is the only correct way
  2. Break apart, because you must separate tens and ones
  3. Counting up, because the numbers are close together (correct answer)
  4. Counting back, because you always subtract by tens
Explanation: This question tests 2nd grade understanding of addition and subtraction strategies, including explaining why strategies work, comparing strategies, and selecting appropriate strategies for different problems (CCSS 2.NBT.B.9: Explain why addition and subtraction strategies work, using place value and the properties of operations). Common 2nd grade strategies include: (1) Counting on—start at larger number, count up by smaller (best for adding small numbers like 47+3); (2) Make ten—decompose to make 10 first (8+5 → 8+2+3 → 10+3=13; best when number close to 10); (3) Break apart—separate tens and ones, add separately (47+28 → 40+20=60, 7+8=15, 60+15=75; best for two-digit numbers); (4) Compensation—adjust to friendly number (47+28 → 50+28-3=78-3=75; good for numbers near decades); (5) Counting back/up for subtraction. Strategies work because they use place value (break apart), number relationships (make ten), or counting principles (counting on/back). Understanding why helps students choose and apply strategies flexibly. In this problem, two strategies are compared for a subtraction problem, and the question asks which is best and why. To analyze, compare (counting up efficient for 62-59 because numbers close: from 59 up 60,61,62=3; counting back also works but up good for small differences). Choice B is correct because counting up is best (numbers are close together, quick counts from 59 to 62), demonstrating understanding of efficiency comparison. Choice A represents a specific error (wrong strategy with incorrect reason, said counting back always by tens which isn't true here), which typically happens when students misjudge efficiency. To help students: Explicitly teach and name strategies. Model each strategy with think-aloud: 'I'm using counting up. 62-59: From 59, 60,61,62=3.' Show same problem solved multiple ways, compare: 'Counting up vs. back—which is faster?' Discuss when each strategy works best: counting on for small addends, break apart for two-digit, make ten when close to 10/decade. Use number lines to show counting strategies visually. Have students explain their thinking: 'How did you solve this? Why did you choose that strategy?' Teach error checking within strategies (recount, check if makes sense). Compare student work: 'Both worked. Why?' Practice strategy selection: 'For this problem, which strategy makes most sense?' Build flexibility—multiple strategies often work, but efficiency varies. Watch for: naming strategies without understanding why they work, claiming one strategy always best (it depends on problem), weak mathematical explanations ("because it's easier" instead of reasoning about place value or number relationships), confusing strategy names, selecting inefficient strategies for problem type.

Question 14

Maya solved 46+2746+27 by calculating 46+30−346+30-3. Which strategy did she use?

  1. Compensation (correct answer)
  2. Doubles
  3. Counting back
  4. Counting on
Explanation: Compensation is correct because Maya changed 27 into the friendlier number 30, then subtracted 3 to adjust for the extra amount she added. Doubles means adding two equal numbers, which is not what happened here. Counting back is a subtraction strategy, not an addition one. Counting on usually means adding small jumps one at a time, not adjusting to a friendlier number first.

Question 15

Maya tried to solve 8+78+7 by breaking 7 apart into 2 and 5, so that 8+2=108+2=10 and then she could add the leftover 5. But she wrote: 8+7=8+2+7=10+7=178+7=8+2+7=10+7=17. What mistake did Maya make?

  1. She kept the whole 7 instead of using the leftover 5 (correct answer)
  2. She added tens before ones, so it changed the sum
  3. She should have counted back instead of adding
  4. She forgot to subtract after adding to 10
Explanation: Maya meant to break 7 into 2 and 5, using the 2 to make a ten and adding the leftover 5. Instead, she added the 2 but then added the whole 7 again, which is why her answer came out to 17 instead of 15. Choice B doesn't describe what actually happened in her work. Choice C suggests a completely different strategy than the one Maya was using. Choice D isn't what her written steps show.

Question 16

Which strategy is most efficient for 48+2948+29: break apart, counting on, counting back, or make ten?

  1. Break apart, because you can add tens and ones (correct answer)
  2. Counting on, because counting 29 ones is quickest
  3. Counting back, because addition goes backward
  4. Make ten, because you must always start with 10
Explanation: Breaking 48 and 29 apart into tens and ones (40+2040+20 and 8+98+9) makes the addition simpler and more efficient. Choice B is slower since it means counting up by ones 29 times. Choice C (counting back) is a subtraction strategy, not addition. Choice D's reasoning that you must always start with 10 is not how make ten actually works.

Question 17

Sofia used counting on to solve 35+635+6. She wrote: 35,36,37,38,39,4035, 36, 37, 38, 39, 40. Which statement best describes her work?

  1. She counted 5 numbers, but needed 6 (correct answer)
  2. She should have added tens to tens first
  3. There is no mistake; this shows 6 counts
  4. She should have counted backward, not forward
Explanation: Sofia only counted 5 numbers forward from 35 (36, 37, 38, 39, 40), but adding 6 requires 6 counts, so her work has a mistake. Choice B is wrong because this problem is being solved by counting on, not by adding tens. Choice C is wrong because only 5 counts are shown, not 6. Choice D is wrong because counting forward is the correct direction for this addition problem.

Question 18

Which strategy did Chen use: break apart, counting back, compensation, or counting on? 49+20→50+20−149+20\rightarrow 50+20-1

  1. Break apart
  2. Counting back
  3. Compensation (correct answer)
  4. Counting on
Explanation: Chen rounded 49 up to 50, added 20, then subtracted 1 to adjust, which is the compensation strategy. Choice A, break apart, would split the numbers into parts rather than round one up. Choice B, counting back, is a subtraction strategy, not what Chen used here. Choice D, counting on, would mean counting up one at a time rather than rounding and adjusting.

Question 19

Which strategy is more efficient for solving 58+358+3?

  1. Counting on (correct answer)
  2. Standard algorithm
  3. Counting back
  4. Break apart
Explanation: Counting on works best because 58 + 3 only needs a few small steps forward, so it is quick and simple. Standard algorithm involves lining up columns and regrouping, which takes longer than needed for such a small addition. Counting back moves in the wrong direction since this problem is addition, not subtraction. Break apart is helpful for larger numbers but adds extra steps that are not needed here.

Question 20

Jamal solved 47 + 8 by first finding 47 + 3 = 50, then 50 + 5 = 55. Which strategy did he use?

  1. Counting on
  2. Make ten (make a new ten) (correct answer)
  3. Counting back
  4. Break apart by tens and ones
Explanation: Jamal broke 8 into 3 and 5 so that 47 plus 3 made a new ten, 50, then added the remaining 5 to reach 55, which is the make ten strategy. Choice A, counting on, would mean adding one at a time instead of jumping to a ten. Choice C, counting back, involves subtracting, which does not fit an addition problem. Choice D, break apart by tens and ones, would split both numbers by place value rather than building to a ten.