Lesson 22 · Synthesis Story

The Horse Business

A multi-step story problem - every lesson in one adventure.
Skills used in this story: Proportionality (L13) · Area & perimeter (L21) · Writing equations (L01-07) · Rate of change (L19-19c) · Percentages · Break-even analysis

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 €0.50 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?

Daily profit is revenue minus cost Eight horses produce ninety-six euros revenue. Sixteen euros feed cost is subtracted, leaving eighty euros daily profit. REVENUE 8 × €12 = €96 FEED COST 8 × 4 × €0.50 = €16 = PROFIT €80/day
Questions - work through the story step by step
  1. Question 1 · Proportionality (Lesson 13)
    Complete the feed cost table for Mia's first $15$ days. She buys every $3$ days for $2$ horses.
    Purchase daykg of hay boughtTotal cost €
    Day 1
    Day 4
    Day 7
    Day 10
    Day 13
    (a) Calculate kilograms of hay in one purchase.
    (b) Calculate the cost of one purchase at €0.50/kg.
    (c) Fill the five table rows and calculate the total cost.
    Total kg per purchase $= 2 \times 4 \times 3$
    fill in table → sum the costs →
  2. Question 2 · Rate of Change (Lesson 19)
    Cumulative feed cost is €0 on day $0$ and €60 on day $15$. Let $x=$ time in days and $y=$ cumulative feed cost in euros.
    (a) Calculate $\Delta y$.
    (b) Calculate $\Delta x$.
    (c) Calculate the rate of change with units.
    (d) Complete: The cumulative feed cost ______ by ______ euros per day.
    (e) Choose: Is the relationship linear? YES / NO.
    given pointsfind Δyfind Δxdivideinterpret
  3. Question 3 · Area and Perimeter (Lesson 21)
    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 →
  4. Question 4 · Writing an Equation (Lessons 01-05)
    Under the new deal: $8$ horses at €12 per horse per day; feed costs €0.50/kg and each horse eats $4$ kg/day.
    (a) Write an equation for Mia's daily profit $P$ (revenue minus feed cost).
    (b) Calculate $P$ and state whether it is positive.
    revenue per day → feed cost per day → P = revenue − cost →
  5. 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 →
  6. Question 6 · Slope (Lessons 16-17)
    Mia plots cumulative profit after the tack-room cost is recovered. The line passes through $(0,0)$ and $(5,400)$, where $x=$ days and $y=$ profit in euros.
    (a) Calculate $\Delta y$.
    (b) Calculate $\Delta x$.
    (c) Calculate the slope with units.
    (d) Explain what the slope means.
    (e) Calculate her cumulative profit after $30$ days.
    Δy → Δx → divide → units → interpret → extend →
  7. Question 7 · Percentages + Rate of Change
    After week 3, Herr Berger raises Mia's rate by $15\%$ per horse per day.
    (a) Calculate the new rate per horse per day.
    (b) Calculate Mia's new daily profit after the €16 feed cost.
    (c) Compare the new slope with the old slope of €80/day. State the increase in euros per day.
    new rate = 12 × 1.15 → new profit → compare slopes →
  8. Question 8 Open
    Mia has two options for one year.
    Option A: Earn €80/day for $365$ days, then receive $2.5\%$ simple interest on that year's profit.
    Option B: Add one horse at €12/day; its feed costs €2/day, so it adds €10/day profit.
    (a) Convert $2.5\%$ to a decimal rate.
    (b) Calculate Option A's profit before interest.
    (c) Calculate Option A's interest using $I=Prt$ with $t=1$ year.
    (d) Calculate Option A's total amount.
    (e) Calculate Option B's annual profit.
    (f) Choose an option and explain one mathematical reason and one practical reason.
    percent ÷ 100 → principal → interest → total · Option B: daily profit × 365 → compare → justify →
Show answers (Q1-7)
Question 1a
$2\times4\times3=24$ kg per purchase.
Question 1b
$24\times€0.50=€12$ per purchase.
Question 1c
Each row is $24$ kg and €12. Total cost $=5\times€12=€60$.
Question 2a
$\Delta y=60-0=€60$.
Question 2b
$\Delta x=15-0=15$ days.
Question 2c
Rate $=\frac{60}{15}=€4$/day.
Question 2d
The cumulative feed cost increases by €4 per day.
Question 2e
YES. Its rate of change is constant.
Question 3a
$A = 6 \times 4.5 = 27$ m².
Question 3b
$P = 2(6)+2(4.5) = 21$ m.
Question 3c
Tiling cost $=27 \times 12 = €324$.
Question 4a
$P=(8\times12)-(8\times4\times0.50)$.
Question 4b
Revenue $=€96$/day and feed cost $=€16$/day, so $P=€80$/day. It is positive.
Question 5
$80d=200$, so $d=2.5$. She must complete $3$ whole days to recover at least €200.
Question 6a
$\Delta y=400-0=€400$.
Question 6b
$\Delta x=5-0=5$ days.
Question 6c
$m=\frac{400}{5}=€80$/day.
Question 6d
Cumulative profit increases by €80 for each day worked.
Question 6e
$80\times30=€2\,400$.
Question 7a
New rate $=12\times1.15=€13.80$ per horse per day.
Question 7b
New profit $=(8\times€13.80)-€16=€94.40$/day.
Question 7c
The slope increases from €80/day to €94.40/day, a rise of €14.40/day.
Question 8a
$2.5\div100=0.025$, so $r=0.025$.
Question 8b
Option A principal: $P=80\times365=€29\,200$.
Question 8c
$I=Prt=29\,200\times0.025\times1=€730$ interest.
Question 8d
Option A total: $A=P+I=€29\,200+€730=€29\,930$.
Question 8e
Daily profit $=€80+€10=€90$. Annual profit $=90\times365=€32\,850$.
Question 8f
Example: choose B because €32,850 is €2,920 more, but note that another horse adds work and client risk. A well-supported choice of either option is valid.
Coming up next → Lesson 23: Simple Interest The savings account question in Q8 used simple interest. Lesson 23 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.
← All lessons
25:00

Ask your tutor