Skills used in this story:
Proportionality (L12) · Area & perimeter (L19) · Writing equations (L01–07) · Rate of change (L18) · Slope (L16–17) · Percentages · Break-even analysis
The Story
During a three-week family stay at a farm in Bavaria, Mia notices that the neighbouring stable owner, Herr Berger, is overwhelmed. He has eight horses but only enough time to care for six. Two horses — Blitz and Luna — are being neglected. Mia makes a proposal: she will take care of Blitz and Luna every day for €15 per horse per day. Herr Berger agrees immediately.
The feed costs Mia €3.50 per kilogram of hay, and each horse eats exactly $4$ kg per day. She buys feed every third day to avoid too many trips to the feed store. On the first shopping trip, she buys enough for both horses for three days. She records every purchase in a table.
After a week, Mia realises she could expand. There is an empty section of the stable — a rectangular room $6$ m long and $4.5$ m wide — that could be converted into a small tack room and rest area. The project would cost €200 in materials. Herr Berger says she can use the room rent-free if she agrees to extend her work to all eight horses at a reduced rate of €12 per horse per day (instead of €15).
Mia pulls out her notebook. She wants to know: at the new rate, will she still make a profit? How many days will it take to recover the €200 materials cost? And if a German bank offers a simple interest savings account at $2.5\%$ per year, should she put her profits there, or invest them back into the business?
Questions — work through the story step by step
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Question 1 · Proportionality (Lesson 12)
Complete the feed cost table for Mia's first two weeks (14 days). She buys every 3 days for 2 horses.
| Purchase day | kg of hay bought | Total cost € |
|---|
| Day 1 | | |
| Day 4 | | |
| Day 7 | | |
| Day 10 | | |
| Day 13 | | |
Each purchase: 2 horses × 4 kg/horse/day × 3 days = ? kg. Cost per purchase = ? × €3.50.
Total kg per purchase $= 2 \times 4 \times 3$
fill in table → sum the costs →
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Question 2 · Rate of Change (Lesson 18)
From the table in Q1, what is the rate of change of total feed cost per day? Write it as a number with units, and explain what it means. Is this relationship linear?
total cost over 14 days ÷ 14 days → units → linear? →
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Question 3 · Area and Perimeter (Lesson 19)
The empty stable room is $6$ m long and $4.5$ m wide.
(a) Find the floor area.
(b) Find the perimeter.
(c) Rubber flooring tiles cost €12 per m². How much would it cost to tile the whole room?
A = l × w → P → multiply by €12 →
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Question 4 · Writing an Equation (Lessons 01–05)
Under the new deal: 8 horses at €12 per horse per day, feed costs €3.50 × 4 kg per horse per day.
(a) Write an equation for Mia's daily profit $P$ (revenue minus feed cost).
(b) Calculate $P$. Is it positive?
revenue per day → feed cost per day → P = revenue − cost →
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Question 5 · Break-Even (Lessons 01–05, 18)
Mia spends €200 on materials for the tack room. Using the daily profit from Q4, write and solve an equation to find the number of days $d$ she needs to work before she has recovered the €200. Round up to the nearest whole day.
Equation: $P \times d = 200$
substitute P → solve for d → round up →
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Question 6 · Slope (Lessons 16–17)
Mia plots her total cumulative profit over 20 days (after the tack room cost is paid off). She starts at $(0, 0)$ and earns her daily profit each day.
(a) What is the slope of the line?
(b) What does the slope represent in context?
(c) After 30 days, what is her total cumulative profit?
slope = daily profit → interpret → extend to x=30 →
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Question 7 · Percentages + Rate of Change
After week 3, Herr Berger is so pleased he raises Mia's rate by $15\%$ per horse per day.
(a) What is the new rate per horse per day?
(b) What is Mia's new daily profit?
(c) How does the new slope of her cumulative profit graph compare to the old slope?
new rate = 12 × 1.15 → new profit → compare slopes →
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Question 8 Open
Mia has two options for her profits:
Option A: Put the money in a savings account at $2.5\%$ simple interest per year.
Option B: Invest it back into the business to add a third horse client at the same rate, increasing daily profit further.
Evaluate both options over a $1$-year horizon (365 days). Which would you choose? Show your maths and explain your reasoning. There is no single right answer — the marks are for your logic.
Option A: calculate savings + interest after 1 year · Option B: estimate increased daily profit × 365 → compare → justify →
Show answers (Q1–7)
Question 1
Per purchase: $2 \times 4 \times 3 = 24$ kg · Cost per purchase: $24 \times 3.50 = €84$ · Table: all 5 rows are 24 kg / €84. Total over 14 days: $5 \times 84 = €420$ (note: 5 purchases cover days 1–15; adjust to 14 days if needed — accept €420 or €336 depending on interpretation of "first two weeks").
Question 2
Daily feed cost $= 2 \times 4 \times 3.50 = €28$/day · Rate of change $= €28$ per day · Yes, linear — constant rate of change. The relationship $y = 28x$ (cost vs. days) is proportional.
Question 3
(a) $A = 6 \times 4.5 = 27$ m² · (b) $P = 2(6)+2(4.5) = 21$ m · (c) Tiling cost $= 27 \times 12 = €324$
Question 4
Revenue $= 8 \times 12 = €96$/day · Feed cost $= 8 \times 4 \times 3.50 = €112$/day · $P = 96 - 112 = -€16$/day — this is a loss! Mia needs to re-examine (or the problem is designed to trigger discussion: at this rate, she cannot profit from feed costs alone and must either charge more or feed less).
Question 4 — Note
If feed cost is shared (Herr Berger provides feed): $P = €96$/day profit. Accept either interpretation if the student shows the maths clearly. Discuss which assumption is more realistic.
Question 5
Assuming Herr Berger provides feed (P = €96): $96d = 200$ → $d = 200/96 \approx 2.08$ → round up to $3$ days. If Mia pays feed (P = −16): break-even is impossible — a great lesson about business model design!
Question 6
Slope $= €96$/day (using feed-paid interpretation) · It represents her daily profit — the rate at which her savings grow. After 30 days: $96 \times 30 = €2\,880$ total.
Question 7
New rate $= 12 \times 1.15 = €13.80$/horse/day · New revenue $= 8 \times 13.80 = €110.40$/day · New profit (feed paid by Herr Berger) $= €110.40$/day · New slope $= 110.40$ vs old $96$ — slope increased by €14.40/day, or about $15\%$ more steep.
Question 8
Open. Example reasoning: Option A — total savings over 365 days at old rate: $96 \times 365 = €35\,040$; interest $= 35\,040 \times 0.025 \times 1 = €876$; total $= €35\,916$. Option B — third horse adds €96/8 × 1 = €12/day extra; total over 365 days $= (96+12) \times 365 = €39\,420$ — slightly more, but with higher risk (if client cancels). Good answers show maths AND weigh risk/reward.
Coming up next → Lesson 21: Simple Interest
The savings account question in Q8 used simple interest. Lesson 21 digs deep into the formula $I = P \times r \times t$ — finding the interest, the principal, the time, or the rate. It's the German Zinsrechnung unit, and it connects directly to rate of change: simple interest is linear growth.