Year12- MATH-1-3-1 Measurement, Scales, and Data. & Compound Interest comparison problem

Unit 3:

This moves us into Unit 3: Measurement, Scales, and Data. This unit is the bread and butter of the PSMT (Problem-Solving and Modelling Task), which is that big report you have to write in Year 12.

In this topic, you aren’t just doing math; you’re acting like a project manager or a builder.

The Practice Problem: “The Deck Extension”

The Scenario: You are looking at a house plan for a new outdoor deck. On the paper plan, the deck is drawn as a rectangle. You need to calculate the cost of the timber flooring.

The Data:

Scale on Plan: $1:50$
Plan Dimensions: The deck measures $12 \text{ cm}$ long and $8.4 \text{ cm}$ wide on the paper.
Flooring Cost: The timber decking costs $\$85 \text{ per square metre (m}^2\text{)}$ .

Your Tasks:

Calculate the actual real-life length and width of the deck in metres.
Calculate the total real-life area of the deck in $\text{m}^2$ .
Calculate the total cost of the timber required for the deck.

Worked Solution (The “QCAA Way”)

Step 1: Convert Plan to Real Life

A scale of $1:50$ means $1 \text{ unit}$ on the page equals $50 \text{ units}$ in real life.

Actual Length: $12 \text{ cm} \times 50 = 600 \text{ cm}$ Convert to metres: $600 \div 100 = \mathbf{6 \text{ m}}$
Actual Width: $8.4 \text{ cm} \times 50 = 420 \text{ cm}$ Convert to metres: $420 \div 100 = \mathbf{4.2 \text{ m}}$

Step 2: Find the Area

Now that we have the real-world dimensions in metres, we find the area ( $A = L \times W$ ):

$A = 6 \times 4.2$

$A = \mathbf{25.2 \text{ m}^2}$

Step 3: Calculate the Total Cost

$Cost = Area \times \text{Rate per m}^2$

$Cost = 25.2 \times 85$

Total Cost = $\$2,142$

Essential Study Tips for Unit 3

1. The “Square Rule” Trap

One of the most common mistakes in Year 12 exams is trying to convert area by the same scale as length.

Wrong: “The area on paper is $100.8 \text{ cm}^2$ , so I’ll just multiply by $50$ .” (This gives you the wrong answer!)
Right: Always convert the individual side lengths to metres first, then multiply them to get the area. It prevents a world of pain.

2. Units, Units, Units!

In Essential Maths, you will constantly jump between $\text{mm}$ , $\text{cm}$ , $\text{m}$ , and $\text{km}$ .

Check: Did you divide by $100$ to get metres?
Check: Does the answer look right? A $6 \text{ metre}$ deck is reasonable. A $600 \text{ metre}$ deck is a runway for a Boeing 747.

3. The “Wastage” Factor (Complex Unfamiliar)

In a real QCAA exam, they might add a “Complex” twist: “Allow 10% extra for timber wastage.” * To solve this, you would take your area ( $25.2$ ) and multiply by $1.10$ before calculating the cost.

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Unit 4:

Since Unit 4: Loans and Interest is a massive part of the Year 12 Essential Maths course (and a very handy life skill), let’s tackle a Compound Interest comparison problem.

This type of question is a classic “Complex Familiar” task you might see in an exam.

The Practice Problem: “The Car Fund”

>> Problem 1:

The Scenario: Alex has just finished Year 12 and wants to save $5,000 for a second-hand car. They have $4,000 to invest right now and plan to leave it in the bank for 3 years. Alex is comparing two different savings accounts:

Account A: Offers 4.8% p.a. (per annum) Simple Interest.
Account B: Offers 4.5% p.a. Compound Interest, compounded annually.

Your Tasks:

Solve: Calculate the total amount Alex will have in Account A after 3 years.
Solve: Calculate the total amount Alex will have in Account B after 3 years.
Evaluate: Which account should Alex choose to get closer to their $5,000 goal?

Worked Solution (The “QCAA Way”)

Part 1: Account A (Simple Interest)

From your QCAA formula sheet, the formula for Simple Interest is:

$I = Prt$

Where $P = 4000$ , $r = 0.048$ (the decimal of 4.8%), and $t = 3$ .

$I = 4000 \times 0.048 \times 3$
$I = 576$
Total Amount $= P + I = 4000 + 576 = \mathbf{\$4,576}$

Part 2: Account B (Compound Interest)

The formula for the total amount ( $A$ ) in Compound Interest is:

$A = P(1 + r)^n$

Where $P = 4000$ , $r = 0.045$ , and $n = 3$ .

$A = 4000(1 + 0.045)^3$
$A = 4000(1.045)^3$
$A = 4000 \times 1.141166…$
Total Amount $= \mathbf{\$4,564.66}$ (rounded to 2 decimal places)

Part 3: Evaluation & Reasonableness

Comparison: Account A yields $4,576, while Account B yields $4,564.66.
Conclusion: Even though Account B’s interest “compounds,” the higher interest rate of Account A (4.8% vs 4.5%) makes it the better choice over this short 3-year period.
Reasonableness Check: Alex still hasn’t reached the $5,000 goal. They are about $424 short. Alex might need to find an account with a higher rate or save for another year.

Study Tip for this Topic

In Essential Maths, they love to change the compounding period. If Account B compounded monthly instead of annually, you would have to:

Divide the interest rate by 12 ( $r \div 12$ ).
Multiply the number of years by 12 ( $n \times 12$ ).

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>> Problem 2:

That is a great idea. In the Queensland Essential Mathematics syllabus, being able to adjust for different compounding periods is a key skill for Unit 4, Topic 3: Loans and Investments.

Here is the modified challenge and the step-by-step “QCAA style” solution.

The Challenge: Monthly Compounding

The Scenario: Alex still has $4,000 to invest for 3 years.

Account A (Reminder): 4.8% p.a. Simple Interest (Total = $4,576)
Account B (Modified): 4.5% p.a. Compound Interest, but this time it is compounded monthly.

The Question: Does compounding monthly make enough of a difference for the 4.5% account to beat the 4.8% simple interest account?

Worked Solution

In an exam, you are taught to “adjust the variables” before plugging them into the formula $A = P(1 + r)^n$ .

1. Adjust the Rate ( $r$ ):

The annual rate is 4.5%. Since it’s monthly, we divide by 12.

$r = 0.045 \div 12 = 0.00375$

2. Adjust the Periods ( $n$ ):

The time is 3 years. Since it’s monthly, we multiply by 12.

$n = 3 \times 12 = 36 \text{ periods}$

3. Apply the Formula:

$A = 4000(1 + 0.00375)^{36}$

$A = 4000(1.00375)^{36}$

$A = 4000 \times 1.14424…$

$A = \$4,576.99$ (rounded to 2 decimal places)

Comparison Table: Which is better?

Investment Type	Interest Rate	Compounding	Total after 3 years
Simple Interest	4.8% p.a.	N/A	$4,576.00
Compound Interest	4.5% p.a.	Annually	$4,564.66
Compound Interest	4.5% p.a.	Monthly	$4,576.99

The Verdict

By compounding monthly, the 4.5% account actually beats the 4.8% simple interest account—but only by $0.99! Why this matters for your Year 12 studies:

Precision: In Essential Maths, rounding too early can lose you marks. Always keep the full decimal in your calculator until the very final step.
Comparison: QCAA often asks you to “Compare and Contrast.” A good answer would state: “While Account B has a lower nominal interest rate, the frequency of compounding (monthly) allows the interest to earn interest more often, eventually yielding a higher return than Account A.”
The Formula Sheet: Remember, the QCAA formula sheet usually provides $A = P(1+r)^n$ . It is up to you to remember to divide the rate and multiply the time if it’s not annual.

Would you like to move on to a practice problem for Unit 3 (like reading a Scale Drawing or House Plan), or would you like to see how credit card interest works?

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