Additive Thinking with Decimals

Overview

Students use place value to add and subtract decimals with different numbers of decimal places. They estimate first, select an efficient strategy, and check whether their answers are reasonable in a practical long-jump context.

Learning intentions

Students will:

• use place value to partition, add and subtract decimals;
• align decimal places and use zero placeholders;
• estimate before calculating and check the reasonableness of answers;
• explain their strategy using mathematical language.

Success criteria

• I can partition decimals in more than one way.
• I can align place values when adding or subtracting decimals.
• I can estimate an answer before calculating.
• I can explain whether my answer is reasonable.

Curriculum links

• Number — addition and subtraction of decimals using place value.
• Number — estimation and rounding to check the reasonableness of answers.
• Mathematical problem-solving — applying calculation strategies in a practical measurement context.

Lesson structure (60 minutes)

1. 0–10 min · Warm-up and launch. Open with the decimal place-value introduction slides and display 3.45. Students use mini-whiteboards to show three partitions, such as 3 + 0.4 + 0.05, 2 + 1.45 and 345 hundredths. Invite students to share different representations and ask, “What stays the same when we partition the number?” Emphasise that whole-number strategies, including partitioning and regrouping, work across the decimal point.

2. 10–25 min · Explicit teaching and guided practice. Use the worked-example slides to model 4.2 − 1.85 on an empty number line. Students estimate first: approximately 2.3, or about 2. Model jumping back 1 whole to 3.2, then 0.2 to 3.0, then 0.65 to 2.35. Show that 4.2 can be written as 4.20 so place values align in a vertical calculation. Ask: “Which place value are we subtracting first?”, “Why is the zero useful?” and “How does the estimate help us?” Guide students through 3.6 + 1.27 and 6.05 − 2.8, accepting partitioning, number-line and vertical methods.

3. 25–35 min · Check for understanding. Display the error-analysis prompt in the error-analysis slide: “Tom solved 5.3 + 0.47 and got 0.500. Is he correct? Explain why and fix his work.” Students independently write a response on mini-whiteboards, then compare with a partner. Select responses that show the decimal points have been misaligned and establish 5.30 + 0.47 = 5.77. Students estimate 5 + 0.5 ≈ 5.5 to confirm that 5.77 is reasonable, whereas 0.500 is not.

4. 35–55 min · Open-ended task: The Decathlon Gap. Distribute The Decathlon Gap investigation sheet. Present the challenge using the open-task instruction slides: “An athlete is trying to beat a local long-jump record of 5.5 metres. Design three different sets of two or three jumps whose combined total is within 0.25 metres of 5.5 m.” Students record each jump, calculate the combined total, show addition or subtraction working, and write an estimate for each set. They must explain whether the result is below or above the record and by how much. Confer with pairs, asking, “How did you choose your numbers?”, “What estimate did you make first?” and “Can you prove your set is within 0.25 metres?” Extension within the task: find a combination with a difference from 5.5 metres of less than 0.05 metres. Invite two contrasting solutions to be shared.

5. 55–60 min · Exit ticket and review. Display the two exit-ticket questions on the review and exit-ticket slide. Students complete independently:

• Calculate 7.4 − 2.86. Show how you aligned the place values and give an estimate.
• Calculate 3.25 + 0.8. Explain why 4.05 is or is not reasonable. Collect responses to identify students requiring further place-value support.

Resources

• the complete decimal addition and subtraction slide deck
• The Decathlon Gap investigation sheet
• Mini-whiteboards, markers and erasers
• Empty number-line models
• Place-value chart showing ones, tenths and hundredths
• Board or interactive display
• Calculators for teacher checking only, if required
• Prepared examples and exit-ticket questions

Key vocabulary: place value, decimal point, tenths, hundredths, thousandths, partition, align, estimate, difference, reasonable, total.

Teacher preparation: prepare the slide deck and worksheet, display the place-value chart, arrange students in pairs, and have mini-whiteboards ready. Ensure the long-jump values are written in metres.

Assessment

• During the launch, check whether students can represent 3.45 in equivalent ways.
• During guided practice and error analysis, scan whiteboards for alignment, zero placeholders and sensible estimates; question students who rely only on a procedure.
• Use the investigation sheet and exit ticket to assess accurate calculation, estimation and explanation. Expected exit answers: 7.4 − 2.86 = 4.54, estimated about 4.5; 3.25 + 0.8 = 4.05, which is reasonable because 3.25 + 0.80 is just over 4.

Differentiation

• Support: provide a labelled place-value chart, pre-drawn empty number lines, the sentence stem “I estimate ___ because ___”, and allow students to write decimals with trailing zeroes before calculating.
• Core: require at least one partition or number-line representation and one aligned written calculation for the investigation.
• Extension: require a combination within 0.05 metres of 5.5, or ask students to find two different combinations with the same difference from the record.
• EAL and students needing additional support: explicitly rehearse “total”, “difference”, “above” and “below”; use paired talk, worked examples and teacher modelling before independent recording.

Answer guide

• 4.2 − 1.85: write 4.20 − 1.85 = 2.35. Number-line jumps may be 1, then 0.2, then 0.65.
• 5.3 + 0.47: Tom is incorrect. Align as 5.30 + 0.47 = 5.77; an estimate of about 5.5 supports the answer.
• 3.6 + 1.27 = 4.87.
• 6.05 − 2.8 = 3.25.
• Valid open-task examples include:
• 2.10 + 1.75 + 1.60 = 5.45; difference 0.05 m.
• 3.20 + 2.15 = 5.35; difference 0.15 m.
• 1.85 + 1.90 + 1.70 = 5.45; difference 0.05 m.
• Extension: 2.75 + 2.74 = 5.49; difference 0.01 m. Any accurately calculated sets within 0.25 m are valid.