NCEA Level 1AS 91945Mathematical methodsNCEA Level 1 Maths

Mathematical methods: 7 problems solved step by step

What this lesson covers

  • A real problem has to be turned into mathematical questions before any method can be chosen.
  • Methods from different strands of mathematics can work on the same problem.
  • In relational thinking the result of one process becomes the input to the next.
  • A connected chain is only as good as its weakest step, so every link must be correct.
  • Every model rests on assumptions, and changing one can change the answer.
  • Extending a problem means asking what happens in general, or what the model cannot tell you.
Key idea

7 problems, one hāngī

  • A kapa haka fundraiser at the marae.
  • Each problem: a different method.
  • Write the assumption and its limit.
  • Excellence: connect two methods.
Your turn

Question 1 · Achieved

  • Last year, 80 meals used 30 kg of meat.
  • How much meat for 120 meals?
  • State your assumption.
Worked example

Q1: meat for 120 meals

  • Method: a rate.
  • One meal first.
  1. 30 ÷ 80 = 0.375 kg per mealrate
  2. 0.375 × 120 = 45 kgscale up
  3. 120 ÷ 80 = 1.5, 30 × 1.5 = 45
  4. Assume: same portion per meal
  5. Limit: appetites and waste vary
  6. Order about 45 kg of meat
Your turn

Question 2 · Achieved

  • Meat costs $12 per kg.
  • Orders of 40 kg or more: 20% off.
  • Find the cost of 45 kg, and
  • the meat cost per meal (120 meals).
Worked example

Q2: the cost of the meat

  • Method: a percentage.
  • 45 kg ≥ 40 kg: discount.
  1. 45 × $12 = $540full price
  2. 0.2 × $540 = $10820% of $540
  3. $540 − $108 = $432take it off
  4. 0.8 × $540 = $432pay 80%
  5. $432 ÷ 120 = $3.60 a meal
  6. Assume: price holds till the day
  7. Limit: prices can change
  8. Meat $432, or $3.60 a meal
Your turn

Question 3 · Merit

  • Other food: $2.40 a meal.
  • Fixed costs: $200. Meals sell at $15.
  • Write the profit P for n meals.
  • Find P for 120 meals.
Worked example

Q3: a profit model

  • Method: a linear model.
  • Profit = income − costs.
  1. food = 3.60 + 2.40 = $6 a meal
  2. income = 15n
  3. costs = 6n + 200
  4. P = 15n − (6n + 200)income − costs
  5. P = 9n − 200collect like terms
  6. 9 × 120 − 200 = $880n = 120
  7. Assume: every meal is sold
  8. Limit: unsold meals still cost
  9. Profit $880 from 120 meals
Your turn

Question 4 · Merit

  • The trip needs $1000 of profit.
  • Using P = 9n − 200, how many
  • meals must they sell?
  • Careful with the rounding.
Worked example

Q4: meals for $1000

  • Method: an equation.
  • Round up, then check.
  1. 9n − 200 = 1000P = 1000
  2. 9n = 1200+200 both sides
  3. n = 1200 ÷ 9 = 133.3…÷ 9
  4. 9 × 133 − 200 = 997133 meals: $3 short
  5. 9 × 134 − 200 = 1006134 meals: target met
  6. Assume: every meal is sold
  7. Limit: 134 buyers, not 120
  8. Sell at least 134 meals
Your turn

Question 5 · Merit, in context

  • The pit is a cuboid: 1.5 m long,
  • 1.2 m wide and 0.8 m deep.
  • A wheelbarrow holds 0.09 m³.
  • How many barrow loads of soil?
Worked example

Q5: digging the pit

  • Method: volume.
  • Then divide by a load.
  1. V = l × w × dcuboid
  2. V = 1.5 × 1.2 × 0.8
  3. = 1.44 m³
  4. 1.44 ÷ 0.09 = 16 loadsbarrow holds 0.09 m³
  5. 16 × 0.09 = 1.44
  6. Assume: straight sides, flat floor
  7. Limit: dug soil swells
  8. About 16 barrow loads of soil
Your turn

Question 6 · Excellence

  • Dug soil takes up 25% more space.
  • How many barrow loads now?
  • Evaluate your answer to question 5.
Worked example

Q6: soil that swells

  • Volume, then a percentage.
  • Then judge question 5.
  1. 1.44 × 1.25 = 1.8 m³+25%
  2. 1.8 ÷ 0.09 = 20 loads
  3. 16 × 1.25 = 2025% more loads
  4. 16 loadsignored the swelling
  5. Assume: 25% suits this soil
  6. Limit: sloped sides hold less
  7. Plan for about 20 barrow loads
Your turn

Question 7 · Excellence

  • They cook 140 meals, but only 85% sell.
  • Do they still make $1000 profit?
  • Evaluate the model P = 9n − 200.
Worked example

Q7: cooked is not sold

  • Percentage, then the model.
  • Costs: cooked. Income: sold.
  1. sold = 0.85 × 140 = 119
  2. income = 119 × $15 = $1785
  3. costs = 6 × 140 + 200 = $1040
  4. profit = $1785 − $1040 = $745
  5. $1000 − $745 = $255 short
  6. P = 9n − 200treats cooked as sold
  7. Assume: exactly 85% sell
  8. Better: pre-sell, cook to order
  9. No: $745 profit, $255 short
Summary

The methods that solved all seven

  • Choose, connect, state the assumption, then evaluate.
StepHow
Ratefind one meal, then scale up
Percent20% off means pay 80%
Modelprofit = income − costs
Equationsolve, round up, then check
Volumel × w × d, then divide
Connectone answer feeds the next
Evaluateassumption, limit, better plan

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