1st Grade Math Homework: Addition with Carrying, Subtraction with Borrowing & Multiplication as a Sum (20 Questions)

Answer: B) 42

Explanation. Stack them and add the ones column first.

Ones: 7 + 5 = 12. Write the 2, carry the 1 to the tens column. Tens: 1 + 2 + 1 (carried) = 4. Answer: 42.

Theory — Carrying (regrouping). When the ones digits add up to 10 or more, we cannot fit the answer in one place. So we “trade” ten ones for one ten and move that new ten into the tens column. This is called regrouping, and it works because 10 ones and 1 ten are the same amount.

Answer: B) 64

Explanation.

Ones: 8 + 6 = 14. Write the 4, carry the 1. Tens: 3 + 2 + 1 = 6. Answer: 64.

Theory — Place value. The digit “3” in 38 is worth 30, not 3. Lining numbers up in columns is how we keep ones with ones and tens with tens so we only add pieces that are the same size.

Answer: C) 84

Explanation.

Ones: 6 + 8 = 14. Write the 4, carry the 1. Tens: 4 + 3 + 1 = 8. Answer: 84.

Theory — Why carrying works. 14 ones is the same as 1 ten and 4 ones. So instead of trying to write “14” in the ones column, we keep the 4 ones in place and slide the 1 ten over into the tens column, where it belongs.

Answer: B) 85

Explanation.

Ones: 8 + 7 = 15. Write the 5, carry the 1. Tens: 5 + 2 + 1 = 8. Answer: 85.

Theory — Watch out for choice C. A common mistake is writing “815” — that happens when a child forgets to carry and just writes both digits of the ones-column sum. Regrouping is what stops that.

Answer: B) 35

Explanation. We cannot do 3 − 8 in the ones column, so we borrow.

Take 1 ten from the 6 tens, leaving 5 tens. Add it to the 3 ones to make 13 ones. Now the ones column is 13 − 8 = 5. Tens: 5 − 2 = 3. Answer: 35.

Theory — Borrowing (regrouping in subtraction). Borrowing is the mirror of carrying. When the top digit is too small to subtract, we trade one ten for ten ones. The total amount doesn’t change — we’re just re-writing 63 as “5 tens and 13 ones” instead of “6 tens and 3 ones.”

Answer: B) 25

Explanation.

Ones: 4 − 9 doesn’t work, so borrow. 74 becomes “6 tens + 14 ones.” Now 14 − 9 = 5. Tens: 6 − 4 = 2. Answer: 25.

Theory — Same number, new form. 74 = 6 tens + 14 ones = 5 tens + 24 ones = 4 tens + 34 ones … all equal to 74. Regrouping is legal because the total is preserved.

Answer: B) 36

Explanation.

Ones: 1 − 5 doesn’t work — borrow. 71 becomes “6 tens + 11 ones.” Ones: 11 − 5 = 6. Tens: 6 − 3 = 3. Answer: 36.

Theory — Check with addition. Subtraction and addition are inverses. Check: 36 + 35 should equal 71. Ones: 6 + 5 = 11 (write 1, carry 1). Tens: 3 + 3 + 1 = 7. Total = 71. ✔

Answer: B) 44

Explanation.

Ones: 0 − 6 doesn’t work — borrow. 90 becomes “8 tens + 10 ones.” Ones: 10 − 6 = 4. Tens: 8 − 4 = 4. Answer: 44.

Theory — Borrowing from a “0.” When the top ones digit is 0, we still borrow from the tens. The 0 becomes 10 (because we added a full ten to it), and the tens digit drops by 1.

Answer: B) 8

Explanation. Add the 2s one at a time: 2, 4, 6, 8. There are four 2s, and the total is 8. Another way to say this: 4 × 2 = 8.

Theory — Multiplication is repeated addition. When we add the same number over and over, we can write it as a multiplication. “4 groups of 2” means “2 + 2 + 2 + 2,” which we shorten to “4 × 2.” The × sign is a shortcut for the sum.

Answer: C) 15

Explanation. Skip-count by 5s: 5, 10, 15. Three 5s make 15. Same idea: 3 × 5 = 15.

Theory — Skip-counting. Skip-counting is a first look at multiplication. Counting by 5s (5, 10, 15, 20…) is the same as saying “1×5, 2×5, 3×5, 4×5…” This is why the 5-times table is easier — children already know it from counting nickels or fingers.

Answer: B) 284

Explanation.

Ones: 6 + 8 = 14. Write the 4, carry the 1. Tens: 5 + 2 + 1 = 8. Hundreds: 1 + 1 = 2. Answer: 284.

Theory — The algorithm scales. The same carrying rule that worked for two-digit numbers also works for three-digit, four-digit, and beyond. That’s why we say math “stacks.” Once you have place value + carrying, you can add any two whole numbers.

Answer: C) 422

Explanation.

Ones: 3 + 9 = 12. Write the 2, carry the 1. Tens: 4 + 7 + 1 = 12. Write the 2, carry the 1. Hundreds: 2 + 1 + 1 = 4. Answer: 422.

Theory — Carrying more than once. A single problem can carry in multiple columns. Each carry moves one ten of that column’s worth into the next column to the left.

Answer: B) 184

Explanation. Borrow twice.

Ones: 2 − 8 doesn’t work — borrow from tens. 4 tens becomes 3 tens; the 2 ones becomes 12 ones. Ones: 12 − 8 = 4. Tens: 3 − 5 doesn’t work — borrow from hundreds. 3 hundreds becomes 2; the 3 tens becomes 13. Tens: 13 − 5 = 8. Hundreds: 2 − 1 = 1. Answer: 184.

Theory — Chained borrowing. Some subtraction problems need borrowing in more than one column. Do them one column at a time, from right to left, and check by adding the answer back to 158 — you should get 342.

Answer: C) 233

Explanation. Borrowing across zeros is the trickiest kind. 500 has no tens and no ones to borrow from directly, so we regroup all the way across.

Think of 500 as “4 hundreds + 9 tens + 10 ones.” (Same total: 400 + 90 + 10 = 500.) Now subtract normally. Ones: 10 − 7 = 3. Tens: 9 − 6 = 3. Hundreds: 4 − 2 = 2. Answer: 233.

Theory — Borrowing across zeros. When the middle digit is a 0, we borrow “through” it. The hundreds drop by 1, the middle 0 becomes 9 (borrowed from), and the ones digit gains 10. This is the same rule applied twice in one move.

Answer: C) 18

Explanation. Skip-count by 3s: 3, 6, 9, 12, 15, 18. That’s six 3s, and the total is 18. In multiplication language: 6 × 3 = 18.

Theory — Rows and columns. Picture 6 rows of 3 dots each (or 3 rows of 6 dots). Either way you get an array with 18 dots total. Arrays show why 6 × 3 and 3 × 6 give the same answer — the multiplication table is symmetric.

Answer: C) 756

Explanation.

Ones: 7 + 9 = 16. Write the 6, carry the 1. Tens: 6 + 8 + 1 = 15. Write the 5, carry the 1. Hundreds: 4 + 2 + 1 = 7. Answer: 756.

Theory — Carry-chains. When both ones and tens overflow, you carry twice in a single problem. Careful children write the small carried “1” above the next column so they don’t forget to add it.

Answer: B) 257

Explanation. Borrowing across the 0 in the tens place.

Think of 605 as “5 hundreds + 9 tens + 15 ones.” (5×100 + 9×10 + 15 = 605.) Ones: 15 − 8 = 7. Tens: 9 − 4 = 5. Hundreds: 5 − 3 = 2. Answer: 257.

Theory — Two-step borrow. To take 1 ten from 605, we first have to break up the 6 hundreds. One hundred becomes 10 tens, then one of those tens becomes 10 ones. It looks like two moves, but it’s the same idea — regroup a bigger unit into smaller units of the same total value.

Answer: B) 179

Explanation. Two steps.

Step 1 — add what she received.

Ones: 8 + 6 = 14 → write 4, carry 1. Tens: 2 + 9 + 1 = 12 → write 2, carry 1. Hundreds: 1 + 0 + 1 = 2. Total so far: 224.

Step 2 — subtract what she gave away.

Ones: 4 − 5 doesn’t work, so borrow: 2 tens becomes 1 ten, 4 ones becomes 14. Ones: 14 − 5 = 9. Tens: 1 − 4 doesn’t work, borrow again: 2 hundreds becomes 1, 1 ten becomes 11. Tens: 11 − 4 = 7. Hundreds: 1. Answer: 179.

Theory — Word problems as a sequence of steps. The hardest part of a word problem isn’t the arithmetic — it’s deciding which operation to use and in what order. “Received” = add. “Gave away” = subtract. Write both steps down so you don’t lose track.

Answer: C) 32

Explanation. “8 boxes of 4 cookies” means 8 groups of 4. Write it as a sum:

4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 = 32

Skip-count by 4s to check: 4, 8, 12, 16, 20, 24, 28, 32. In multiplication language: 8 × 4 = 32.

Theory — Multiplication is the “how many total?” question. Whenever a word problem says “X groups of Y,” the total is X × Y — which is the same as adding Y to itself X times. First graders should be able to write the sum out; the times-table symbol is a shortcut they’ll fully learn in 2nd–3rd grade.

Answer: C) 30

Explanation. Two steps.

Step 1 — marbles from the bags. “3 bags of 6” = 6 + 6 + 6 = 18. (Or 3 × 6 = 18.)

Step 2 — add the gift.

Ones: 8 + 2 = 10 → write 0, carry 1. Tens: 1 + 1 + 1 = 3. Total: 30 marbles.

Theory — Combining operations. Real problems don’t come with just one operation. This one uses multiplication (repeated addition) and addition with carrying in the same problem. Do them in the order the story tells you.