Goal: apply I=Pin with a yearly rate and labelled time, then test a changed term. 25 = 2 + 5 + 6 + 7 + 5.
- 0–2: Name P, annual i and years n in Card F. Say these are fictional assumptions, not a real account or return.
- 2–7: Model
I=2400×0.035×2=$168; final amount$2,568. Check 3.5% of 2400 is $84 for one year, so two simple-interest years give $168. - 7–13: Learners recompute one year: interest $84/final $2,484. Explain why multiplying the first year's new amount by 3.5% would be compounding, which is not this stated model.
- 13–20: Routes: two equal $84 annual strips attached to original P only; fill the interest timeline; use symbolic
P×i×nthen verbalise units and no-compounding limit. All routes compare one and two years. - 20–25: Exit “Which amount is the base in year 2 of this simple model?” Key: original $2,400. Move: if learner compounds, draw both $84 strips from the same P.
Alternative domain: Fictional invoice late-charge maths without a real contract. Optional/home: find I for invented P=$1,000, i=2%, n=2 years; no financial recommendation.