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Maths for SG Primary Schools

Primary 2 maths · Operations · About 8 minutes

Mentally add or subtract ones, tens and hundreds from three-digit numbers

Name the size of the units being added, then change only that column.

Start the interactive lesson →Open practice activity

What you will learn

Models used: number sentences, stories, bar models.

Before you start

Change one place, keep the others

Start with 247. We want to add 100.

  1. The number at the top of each column is the value of one counter in that column.

    Start with 247. We want to add 100.

    523 + 30 is 553. The hundreds and ones are untouched, because only tens were added.

    100210417
  2. The number at the top of each column is the value of one counter in that column.

    Add one more hundred. The other columns stay just as they were.

    A digit tells us how many units there are. Its position tells us what size those units are; that is why a zero can matter.

    100310417
  3. The number at the top of each column is the value of one counter in that column.

    No exchange was needed. Take that same group away to return to the starting number.

    247 + 100 = 347

    A digit tells us how many units there are. Its position tells us what size those units are; that is why a zero can matter.

    100330010440177
What did the picture help us see?

No exchange was needed. Take that same group away to return to the starting number.

Worked example: A new example, together

What is 523 − 100?

  1. The complete bar is the whole. Each labelled section is one part of it.

    Name the size of the units being added, then change only that column.

    Removing place-value parts one at a time gives the same difference.

  2. The complete bar is the whole. Each labelled section is one part of it.

    Line up the place values. Split the second number into place-value parts.

    What is 523 − 100?

    423

  3. The complete bar is the whole. Each labelled section is one part of it.

    Line up the place values. Split the second number into place-value parts.

  4. The complete bar is the whole. Each labelled section is one part of it.

    523 − 100 = 423.

  5. The complete bar is the whole. Each labelled section is one part of it.

    The result is 423. Check with the inverse operation.

  6. The complete bar is the whole. Each labelled section is one part of it.

    Check the result against the question we started with.

    423 + 100 = 523.

    We found 423. Removing place-value parts one at a time gives the same difference.

Common mistakes to discuss

Check your understanding

Explain your reasoning to an adult, then try a fresh question in the interactive lesson. A result from one attempt is not proof of lasting mastery.

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