Explain Addition and Subtraction Strategies - 2nd Grade Math

Why does the break-apart strategy work for solving $34+25$ by adding $30+20$ and $4+5$ separately?
Answer: It uses tens with tens and ones with ones 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.
To solve $75 - 28$ , Chen rewrote it as $77 - 30 = 47$ . Which strategy did Chen use?
Answer: Adjusted both numbers to make friendlier numbers before subtracting Chen added 2 to both 75 and 28 to get the friendlier problem 77 minus 30, which is compensation: adjusting both numbers so the subtraction is easier while keeping the difference the same. Splitting each number into tens and ones would look like working with 70 and 5, and 20 and 8, separately, which isn't what Chen did. Making a ten first would mean adjusting only one of the numbers, not both. Counting up from 28 to 75 would involve counting the gap in small jumps, which also isn't shown here.
Which strategy is best for solving $74 - 6$ , and why?
Answer: Counting back, because you subtract a small number Counting back works well here because 6 is a small number, so subtracting it by counting back a few steps is efficient. Choice A is wrong because breaking apart tens is not needed for such a small subtraction. Choice C is wrong because 6 does not combine with 74 to make a new ten. Choice D is wrong because subtraction does not always require counting up.
Which way to solve $68+3$ takes the fewest steps?
Answer: Count on from 68: 69, 70, 71 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.
Which strategy did Amir use for $47+8$ : $47+3=50$ , then $50+5=55$ ?
Answer: Make ten 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.
Which strategy is most efficient for $48+29$ : break apart, counting on, counting back, or make ten?
Answer: Break apart, because you can add tens and ones Breaking 48 and 29 apart into tens and ones ( $40+20$ and $8+9$ ) 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.
Maya solved $46+27$ by calculating $46+30-3$ . Which strategy did she use?
Answer: Compensation 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.
Maya tried to solve $8+7$ by breaking 7 apart into 2 and 5, so that $8+2=10$ and then she could add the leftover 5. But she wrote: $8+7=8+2+7=10+7=17$ . What mistake did Maya make?
Answer: She kept the whole 7 instead of using the leftover 5 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.
Which strategy did Marcus use for $23+25$ : $20+20$ and $3+5$ ?
Answer: Break apart Marcus split 23 and 25 into tens and ones, 20 plus 20 and 3 plus 5, which is the break apart strategy. Choice A, counting on, would mean adding one at a time instead of splitting by place value. Choice C, counting back, involves subtracting, which does not fit an addition problem. Choice D, make ten, would involve adjusting one number to reach a friendly ten, which is not what Marcus did here.
Sofia used counting on to solve $35+6$ . She wrote: $35, 36, 37, 38, 39, 40$ . Which statement best describes her work?
Answer: She counted 5 numbers, but needed 6 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.
Which way to solve $92 - 89$ takes the fewest steps?
Answer: Count up from 89 to 92: 90, 91, 92 Since 92 and 89 are very close together, counting up from 89 to 92 takes just three quick steps (90, 91, 92), making it the fastest method here. Breaking 89 apart into a friendly ten adds extra steps that aren't needed for numbers this close. Lining up the ones and tens in columns works but takes longer for such a small difference. Splitting both numbers into tens and ones separately also adds steps that a close comparison like this doesn't require.
Sofia solved $55 - 47$ by starting at 47 and adding small jumps until she reached 55. Which strategy did she use?
Answer: Counting up (find the difference) 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.
7 is split into 2 and another number so that $8+2$ makes ten. Complete the equation: $8+7=8+2+\underline{\ \ }$
Answer: 5 7 is split into 2 and 5 so that 8 plus 2 makes 10, and then adding the remaining 5 gives $8+7=15$ . Ten is the sum after adding 8 and 2, not the amount still needed to complete the equation. Two is the piece already added to make ten, not the piece still missing. Seven is the original number being split apart, so using it again would double-count part of it.
Which strategy is best to solve $62-59$ ?
Answer: Counting up, because the numbers are close Counting up is best here since 62 and 59 are close together, making it easy to count up just 3 steps. Breaking apart into tens and ones adds unnecessary steps for numbers this close. Making a ten is not needed since neither number requires regrouping to reach a friendly ten first. Counting back could work too, but the reasoning given is wrong, since the numbers are actually close together, not far apart.
Which is more efficient for $47+28$ : number line jumps or counting on by ones?
Answer: Number line jumps, because you can jump tens then ones 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.
Which strategy did Chen use: compensation, break apart, counting on, or counting back? $49 + 20 \rightarrow 50 + 20 - 1$
Answer: Compensation Chen changed 49 to 50 (adding 1), then subtracted that same 1 at the end to keep the total correct. This adjust-then-correct approach is called compensation. Choice A (counting on) usually means counting up from a number one at a time. Choice B (break apart) means splitting numbers into place values before adding. Choice C (counting back) is a subtraction strategy, not what Chen used here.
Which strategy did Jamal use: counting on, make ten, or break apart? $47+8\rightarrow 47+3+5\rightarrow 50+5$
Answer: Make ten 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.
Which strategy is most efficient for 68 + 29: break apart, counting on, counting back, or make ten?
Answer: Break apart, because you add tens and ones 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.
Which strategy did Jamal use: counting on, make ten, or break apart? $47+8\rightarrow 47+3+5\rightarrow 50+5$
Answer: Make ten 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.
Which strategy is most efficient for $58+3$ : counting on, standard algorithm, break apart, or counting back?
Answer: Counting on For $58+3$ , 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.

Start this quiz