Dividing Multi-Digit Whole Numbers: practice solutions, Grade 4 – download the PDF

1 Sharing stickers ★★★

Divide the total by the number of friends: \(36 \div 4 = 9\), because \(4 \times 9 = 36\).

Each friend gets 9 stickers.

2 Fact family ★★★

A multiplication fact gives a fact family: \(9 \times 7 = 63\), \(63 \div 7 = 9\) and \(63 \div 9 = 7\).

3 Tens and hundreds ★★★

1. \(18 \div 3 = 6\), so 18 tens \(\div 3\) = 6 tens: \(180 \div 3 = 60\).
2. \(56 \div 7 = 8\), so \(560 \div 7 = 80\).
3. \(24 \div 8 = 3\), so 24 hundreds \(\div 8\) = 3 hundreds: \(2{,}400 \div 8 = 300\).
4. \(35 \div 5 = 7\), so \(3{,}500 \div 5 = 700\).
5. \(48 \div 6 = 8\), so \(4{,}800 \div 6 = 800\).

Check: \(3 \times 60 = 180\), \(7 \times 80 = 560\), \(8 \times 300 = 2{,}400\).

4 Fill the area model ★★★

First part: \(160 \div 4 = 40\). Second part: \(8 \div 4 = 2\).

Whole length: \(40 + 2 = 42\) inches. So \(168 \div 4 = 42\). Check: \(4 \times 42 = 168\).

5 Two partial quotients ★★★

\(3 \times 30 = 90\) and \(96 - 90 = 6\). Then \(3 \times 2 = 6\) and nothing is left.

Add the partial quotients: \(30 + 2 = 32\). So \(96 \div 3 = 32\). Check: \(3 \times 32 = 96\).

6 First long divisions ★★★

1. \(4 \div 2 = 2\), \(6 \div 2 = 3\), \(8 \div 2 = 4\): the quotient is 234.
2. \(6 \div 3 = 2\), \(3 \div 3 = 1\), \(9 \div 3 = 3\): the quotient is 213.
3. \(9 \div 5 = 1\) with 4 left; bring down the 0 to make 40, \(40 \div 5 = 8\); bring down the 5, \(5 \div 5 = 1\): the quotient is 181.

Checks: \(2 \times 234 = 468\), \(3 \times 213 = 639\), \(5 \times 181 = 905\).

7 True or false? ★★★

1. True. \(5 \times 600 = 3{,}000\).
2. False. \(9 \times 5 = 45\), so \(450 \div 9 = 50\).
3. True. Multiplication and division are inverse operations.
4. True. \(4 \times 6 + 1 = 25\) and \(1 < 4\).
5. False. A remainder must be smaller than the divisor. The correct answer is \(6 \text{ R } 1\), because \(5 \times 6 + 1 = 31\).

8 Estimate, then divide ★★★

1. 623 is close to 630, and \(630 \div 7 = 90\). About 90 apples per basket.
2. \(62 \div 7 = 8\) with 6 left (\(7 \times 8 = 56\)). Bring down the 3 to make 63, and \(63 \div 7 = 9\). The quotient is 89.

Check: \(7 \times 89 = 623\). Each basket gets 89 apples, close to the estimate of 90.

9 Quotients with remainders ★★★

1. \(6 \times 7 = 42\) and \(47 - 42 = 5\): 7 R 5.
2. \(8 \times 11 = 88\) and \(92 - 88 = 4\): 11 R 4.
3. \(9 \times 27 = 243\) and \(245 - 243 = 2\): 27 R 2.
4. \(7 \times 142 = 994\) and \(1{,}000 - 994 = 6\): 142 R 6.

In each case the remainder is smaller than the divisor.

10 Spot the mistake ★★★

His multiplication check works (\(528 + 10 = 538\)), but a remainder of 10 is bigger than the divisor 6, so there are still whole groups of 6 left to take out.

\(10 = 6 + 4\), so one more group fits: \(88 + 1 = 89\) with 4 left. The correct answer is 89 R 4. Check: \(6 \times 89 + 4 = 534 + 4 = 538\).

11 Vans for the field trip ★★★

\(83 \div 9 = 9 \text{ R } 2\), because \(9 \times 9 = 81\) and \(83 - 81 = 2\).

Nine vans carry only 81 students, and 2 students still need a ride. The school needs 10 vans.

12 Ribbon pieces ★★★

\(275 \div 8 = 34 \text{ R } 3\), because \(8 \times 34 = 272\) and \(275 - 272 = 3\).

Mia can cut 34 full pieces, and 3 inches are left over.

13 Three-part area model ★★★

\(700 \div 7 = 100\), \(140 \div 7 = 20\) and \(21 \div 7 = 3\).

Length: \(100 + 20 + 3 = 123\) feet. So \(861 \div 7 = 123\). Check: \(7 \times 123 = 861\).

14 Jumps on a number line ★★★

The jumps are 5, 10, 15, 20, 25, 30, 35, 40, 45: that is 9 jumps, and \(5 \times 9 = 45\). From 45 to 47 there are 2 left.

So \(47 \div 5 = 9 \text{ R } 2\). Check: \(5 \times 9 + 2 = 47\).

15 A zero in the quotient ★★★

1. \(32 \div 8 = 4\) and nothing is left. Bring down the 2: \(2 \div 8 = 0\), so write 0. Bring down the 4: \(24 \div 8 = 3\). The quotient is 403. Check: \(8 \times 403 = 3{,}224\).
2. \(21 \div 7 = 3\), nothing left. Bring down the 0: \(0 \div 7 = 0\). Bring down the 5: \(5 \div 7 = 0\) with 5 left. The quotient is 300 R 5. Check: \(7 \times 300 + 5 = 2{,}105\).

16 Apples, crates and trucks ★★★

Step 1, number of crates: \(1{,}248 \div 8\). Partial quotients: \(8 \times 100 = 800\), left 448; \(8 \times 50 = 400\), left 48; \(8 \times 6 = 48\), left 0. So \(100 + 50 + 6 = 156\) crates.

Step 2, crates per truck: \(156 \div 4 = 39\), because \(4 \times 39 = 156\).

Each truck carries 39 crates.

17 Which quotient is larger? ★★★

\(27 \div 9 = 3\), so \(2{,}700 \div 9 = 300\). \(35 \div 7 = 5\), so \(3{,}500 \div 7 = 500\).

The second quotient is larger, by \(500 - 300 = 200\).

18 Find the dividend ★★★

Use dividend = divisor \(\times\) quotient + remainder.

1. \(6 \times 47 = 282\), and \(282 + 5 = 287\). The number is 287.
2. \(7 \times 52 = 364\), and \(364 + 6 = 370\). The number is 370.

19 Error analysis ★★★

Estimate: \(1{,}200 \div 4 = 300\). Nina’s 39 is far too small, so a digit is missing.

\(12 \div 4 = 3\), nothing left. Bring down the 3: \(3 \div 4 = 0\), so write 0. Bring down the 6: \(36 \div 4 = 9\). The correct quotient is 309. Nina forgot the 0. Check: \(4 \times 309 = 1{,}236\).

20 Bags of beads ★★★

1. \(1{,}000 \div 6 = 166 \text{ R } 4\), because \(6 \times 166 = 996\) and \(1{,}000 - 996 = 4\). The store can make 166 full bags with 4 beads left over.
2. The 4 extra beads need one more bag, so the store needs \(166 + 1 = \) 167 bags.

21 Two ways to solve ★★★

Sample problem: a hiking club has 245 trail bars to put into 5 equal backpacks. How many bars go in each backpack?

Partial quotients: \(5 \times 40 = 200\), left 45; \(5 \times 9 = 45\), left 0. So \(40 + 9 = 49\).

Area model: width 5, parts \(5 \times 40 = 200\) and \(5 \times 9 = 45\), total length \(40 + 9 = 49\).

Each backpack gets 49 trail bars. Check: \(5 \times 49 = 245\).