Multi-Digit Multiplication: practice solutions, Grade 5 – download the PDF

1 Reading exponents ★★★

a) There are three factors of 10, so \(10 \times 10 \times 10 = 10^3\).

b) \(10^5 = 100{,}000\) (a 1 followed by 5 zeros).

c) \(10^2 = 100\) has 2 zeros, the same as the exponent.

2 Shifting digits ★★★

a) \(530\) b) \(5{,}300\) c) \(53{,}000\)

d) \(10^4 = 10{,}000\), so \(8 \times 10{,}000 = 80{,}000\).

3 Multiples of 10 ★★★

a) \(6 \times 4 = 24\), one zero: \(240\).

b) \(7 \times 5 = 35\), two zeros: \(3{,}500\).

c) \(3 \times 8 = 24\), two zeros: \(2{,}400\).

d) \(2 \times 6 = 12\), three zeros: \(12{,}000\).

4 Quick estimate ★★★

\(48 \approx 50\) and \(31 \approx 30\), so \(50 \times 30 = 1{,}500\).

The value 1,488 is very close to 1,500, so it is reasonable. (In fact \(48 \times 31 = 1{,}488\).)

5 Fill in an area model ★★★

\(10 \times 10 = 100\), \(2 \times 10 = 20\), \(10 \times 3 = 30\), \(2 \times 3 = 6\).

Sum: \(100 + 20 + 30 + 6 = 156\). So \(12 \times 13 = 156\).

6 Two partial products ★★★

\(23 \times 4 = 92\) and \(23 \times 10 = 230\).

\(92 + 230 = 322\), so \(23 \times 14 = 322\).

7 Pencil boxes ★★★

Equal groups, so multiply: \(12 \times 24\).

\(24 \times 2 = 48\) and \(24 \times 10 = 240\), so \(240 + 48 = 288\).

There are 288 pencils.

8 Standard algorithm practice ★★★

Estimate: \(40 \times 30 = 1{,}200\) (a bit high, because both numbers were rounded up).

Ones row: \(36 \times 7 = 252\). Tens row: \(36 \times 20 = 720\).

Total: \(252 + 720 = 972\). It is a little under the estimate, as expected.

9 Regrouping twice ★★★

Ones row: \(58 \times 3 = 174\). Tens row (placeholder zero): \(58 \times 40 = 2{,}320\).

Add: \(174 + 2{,}320 = 2{,}494\).

10 Estimate, then calculate ★★★

a) \(70 \times 30 = 2{,}100\).

b) \(73 \times 8 = 584\) and \(73 \times 20 = 1{,}460\), so \(584 + 1{,}460 = 2{,}044\).

c) Yes: 2,044 is only 56 away from 2,100.

11 Find the mistake ★★★

The tens digit 3 stands for 30, so the second row must be \(45 \times 30 = 1{,}350\), not 135. Maya forgot the placeholder zero.

Correct work: \(90 + 1{,}350 = 1{,}440\). So \(45 \times 32 = 1{,}440\).

An estimate (\(50 \times 30 = 1{,}500\)) would have shown that 225 is far too small.

12 Area model with three parts ★★★

Top row (×30): \(200 \times 30 = 6{,}000\), \(10 \times 30 = 300\), \(4 \times 30 = 120\).

Bottom row (×2): \(200 \times 2 = 400\), \(10 \times 2 = 20\), \(4 \times 2 = 8\).

Sum: \(6{,}000 + 300 + 120 + 400 + 20 + 8 = 6{,}848\). So \(214 \times 32 = 6{,}848\).

13 Theater seats ★★★

\(35 \times 8 = 280\) and \(35 \times 20 = 700\), so \(280 + 700 = 980\).

The theater has 980 seats. (Estimate: \(30 \times 30 = 900\), close enough.)

14 True or false? ★★★

a) False. \(10^4 = 10 \times 10 \times 10 \times 10 = 10{,}000\).

b) True. \(6 \times 5 = 30\) and there are three zeros to add: \(30{,}000\).

c) True. \(25 \times 4 = 100\), then one more zero gives \(1{,}000\).

15 Big standard algorithm ★★★

Estimate: \(500 \times 40 = 20{,}000\).

Ones row: \(472 \times 6 = 2{,}832\). Tens row: \(472 \times 30 = 14{,}160\).

Total: \(2{,}832 + 14{,}160 = 16{,}992\). It is below the estimate because both numbers were rounded up.

16 Zero in the middle ★★★

Ones row: \(805 \times 7 = 5{,}635\). Tens row: \(805 \times 40 = 32{,}200\).

Total: \(5{,}635 + 32{,}200 = 37{,}835\).

In \(805 \times 7\), the 0 tens digit gives \(0 \times 7 = 0\), and you add the 3 regrouped from \(5 \times 7 = 35\). That column must not be skipped, or the 8 hundreds would land in the wrong place.

17 Missing factors ★★★

a) \(4 \times 7 = 28\), and \(28{,}000\) has three zeros, one from 40, so the missing number has two zeros: \(700\).

b) \(48 \div 6 = 8\), and \(4{,}800\) has two zeros, one from 60, so \(80\).

c) \(63 \div 9 = 7\), and \(63{,}000\) has three zeros: \(7{,}000\).

18 Juice orders ★★★

a) \(36 \times 24 = 864\) (\(36 \times 4 = 144\), \(36 \times 20 = 720\)).

b) \(18 \times 24 = 432\).

c) \(864 + 432 = 1{,}296\) bottles.

d) \(54 \times 24 = 1{,}296\) (\(216 + 1{,}080\)). Both methods agree.

19 Which is greater? ★★★

\(38 \times 52\): \(38 \times 2 = 76\) and \(38 \times 50 = 1{,}900\), so \(1{,}976\).

\(41 \times 49\): \(41 \times 9 = 369\) and \(41 \times 40 = 1{,}640\), so \(2{,}009\).

Since \(2{,}009 > 1{,}976\), \(41 \times 49\) is greater.

20 Powers of 10 in a product ★★★

\(3{,}600 = 36 \times 10^2\) and \(200 = 2 \times 10^2\).

So \(3{,}600 \times 200 = 36 \times 2 \times 10^2 \times 10^2 = 72 \times 10^4 = 720{,}000\).

21 Factory jars ★★★

Estimate: \(100 \times 20 \times 20 = 40{,}000\) (a rough value, a bit lower than the exact answer).

Jars per day: \(125 \times 16 = 2{,}000\). Jars in 24 days: \(2{,}000 \times 24 = 48{,}000\).

The factory fills 48,000 jars.