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Year 4 · Multiplication and division · 18 min

Larger calculations

Break larger multiplication and division calculations into known facts and place-value parts.

01 / LEARN

The idea, step by step

Partition larger numbers

A known fact can be scaled to tens or hundreds. Since 6 × 2 = 12, 6 × 20 = 120. To find 6 × 23, split 23 into 20 + 3. Multiply each part, then add the two results.

Division can also use useful parts. For 156 ÷ 3, split 156 into 150 + 6. Both parts divide easily by 3. The two quotients are 50 and 2, making 52 altogether. Choose parts that fit the divisor.

A remainder must fit the situation

Some amounts do not split into the requested groups exactly. For 86 ÷ 4, twenty-one groups use 84, leaving 2. Write 21 remainder 2. The leftover must be smaller than the divisor: another full group would otherwise be possible.

Check by rebuilding the total: divisor × quotient + remainder. In a story, a remainder can mean items left over, or a need for an extra container. Say which meaning fits rather than simply writing the remainder without a unit.

A common trap

For 24 × 3, multiplying only the 4 ones and answering 12.

Instead: Multiply both place-value parts: 20 × 3 = 60 and 4 × 3 = 12, so 24 × 3 = 72.

02 / WORKED EXAMPLES

See the thinking

Example 1

Multiply in parts

Find 23 × 6.

  1. Split 23 into 20 + 3.
  2. 20 × 6 = 120 and 3 × 6 = 18.
  3. 120 + 18 = 138.
138.
Example 2

Divide and interpret

86 counters are shared equally among 4 pupils. How many does each get, and how many remain?

  1. 4 × 20 = 80, leaving 6.
  2. One more counter each uses 4 of those 6, leaving 2.
  3. Check: 4 × 21 + 2 = 86.
21 counters each, with 2 counters left over.

Check your understanding

Find 152 ÷ 4 using 120 + 32.

Show answer

38.

120 ÷ 4 = 30 and 32 ÷ 4 = 8, so the quotient is 38.

03 / WORKSHEET

Your turn

6 questions · 8 marks · Work at your own pace.

Read the questions online
  1. Find 34 × 3.
    1 mark
  2. Find 46 × 5.
    1 mark
  3. Find 168 ÷ 4.
    1 mark
  4. Find 246 ÷ 3.
    1 mark
  5. Put 95 apples into full bags of 6. How many full bags can be made, and how many apples remain?
    2 marks
  6. A coach must seat 47 pupils in minibuses with 8 pupil seats each. How many minibuses are needed? Explain the remainder.
    2 marks
Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

  1. Find 34 × 3.
    1 mark

    30 × 3 = 90; 4 × 3 = 12.

    90 + 12 = 102.

    102.
  2. Find 46 × 5.
    1 mark

    40 × 5 = 200; 6 × 5 = 30.

    200 + 30 = 230.

    230.
  3. Find 168 ÷ 4.
    1 mark

    160 ÷ 4 = 40; 8 ÷ 4 = 2.

    40 + 2 = 42.

    42.
  4. Find 246 ÷ 3.
    1 mark

    240 ÷ 3 = 80; 6 ÷ 3 = 2.

    80 + 2 = 82.

    82.
  5. Put 95 apples into full bags of 6. How many full bags can be made, and how many apples remain?
    2 marks

    6 × 15 = 90.

    95 − 90 = 5; the remainder is less than 6.

    15 full bags and 5 apples left.
  6. A coach must seat 47 pupils in minibuses with 8 pupil seats each. How many minibuses are needed? Explain the remainder.
    2 marks

    47 ÷ 8 = 5 remainder 7.

    Five minibuses seat only 40 pupils. The other 7 need one more minibus.

    6 minibuses.
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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum Mathematics

Facts to 10×10; related division; larger calculations; multiples/powers of ten; efficient strategies.

Official curriculum source ↗
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