Math worksheet featuring Australian-themed algebra problems, including a cartoon kangaroo holding a flag.

Here are the solutions to the problems on the worksheet, worked out step by step.

1. Kangaroo Jump
* Goal: Find an expression for the jump length in body lengths.
* Given: Jump distance = $x$ meters. Body length is $\frac{1}{3}$ of the jump distance.
* Step 1: Calculate the body length. Since the jump is 3 times the body length, the body length is the jump divided by 3.
$$\text{Body Length} = \frac{x}{3}$$
* Step 2: Divide the total jump distance ($x$) by the body length ($\frac{x}{3}$) to find how many body lengths fit into the jump.
$$\frac{x}{\frac{x}{3}} = x \cdot \frac{3}{x} = 3$$
* Result: The kangaroo jumps 3 body lengths.

2. Sydney Opera House Attendance
* Goal: Write an expression for visitors in the current month ($y$).
* Given: Previous month = $x$. Current month is 200,000 more than twice the previous month.
* Step 1: "Twice the number of visitors in the previous month" means $2x$.
* Step 2: "200,000 more than" that amount means we add 200,000.
* Result: $y = 2x + 200,000$

3. Koala Population
* Goal: Write an expression for koalas after $t$ years.
* Given: Start = 500. Decrease by 10% each year.
* Step 1: If the population decreases by 10%, then 90% of the population remains each year.
* Step 2: Convert 90% to a decimal: $0.90$ or just $0.9$.
* Step 3: To find the population after $t$ years, we multiply the starting amount by $0.9$ repeatedly $t$ times (exponential decay).
* Result: $500(0.9)^t$

4. Sydney Harbour Bridge Height
* Goal: Write equation in form $y = mx + b$.
* Given: Total height = 134 m. $y$ = height above water. $x$ = distance from center.
* Logic: This problem describes a bridge arch. Usually, the highest point is at the center ($x=0$). At the center, the height $y$ is the full height, 134. So, $b = 134$. As you move away from the center ($x$ increases), the bridge goes down towards the water. This means the slope $m$ must be negative. Without specific data points for the curve, we use variables for the rate of descent.
* Result: $y = -mx + 134$ (Note: In many school contexts involving simple linear models for slopes, if specific coordinates aren't given, this general form representing a downward slope from the peak is standard).

5. Great Barrier Reef Diving Depth
* Goal: Find depth when diver is 40 meters from shore.
* Given: Equation $d = 10x + 5$. $x = 40$.
* Step 1: Substitute 40 for $x$ in the equation.
$$d = 10(40) + 5$$
* Step 2: Multiply.
$$10 \cdot 40 = 400$$
* Step 3: Add 5.
$$400 + 5 = 405$$
* Result: 405 meters

6. Ayers Rock Circumference
* Goal: Write expression for cost $c$.
* Given: Radius = $r$. Circumference is 5 times radius ($5r$). Cost is $k$ dollars per meter.
* Step 1: Determine the total length around the rock (circumference).
$$\text{Length} = 5r$$
* Step 2: Multiply the length by the cost per meter ($k$) to get total cost.
$$\text{Cost} = k \cdot (5r)$$
* Result: $c = 5kr$

7. Melbourne Tram Fare
* Goal: Write equation for cost $c$ for ride of $k$ km.
* Given: First 5 km cost $\$2.50$. Beyond 5 km, cost is $\$0.20$ per km.
* Step 1: Identify the fixed cost. The first 5 km are a flat fee of $\$2.50$.
* Step 2: Identify the variable cost. The distance *beyond* 5 km is $(k - 5)$. The rate is $\$0.20$. So, extra cost is $0.20(k - 5)$.
* Step 3: Combine them.
$$c = 2.50 + 0.20(k - 5)$$
* Step 4: Simplify (optional but good practice).
$$c = 2.50 + 0.20k - 1.00$$
$$c = 0.20k + 1.50$$
* Result: $c = 0.20(k - 5) + 2.50$ OR $c = 0.20k + 1.50$

8. Gold Coast Theme Park Tickets
* Goal: Write equation for total cost $c$ for $n$ rides.
* Given: Entry = $\$30$. Each ride = $\$5$.
* Step 1: The entry fee is paid once, regardless of rides. This is the constant: 30.
* Step 2: The ride cost depends on $n$. Cost = $5 \cdot n$.
* Step 3: Add them together.
* Result: $c = 5n + 30$

9. Perth Bus Travel Time
* Goal: Equation relating distance $d$ and time $t$.
* Given: Speed = 60 km/h.
* Formula: $\text{Distance} = \text{Speed} \times \text{Time}$.
* Step 1: Plug in the values.
$$d = 60 \cdot t$$
* Result: $d = 60t$

10. Uluru Sunset Time
* Goal: Find minutes past 6:00 PM when sun sets at 7:00 PM.
* Given: Equation $t = 4m + 120$. $t$ = minutes past 6:00 PM. $m$ = minutes past 6:00 PM.
*(Note: The problem defines both $t$ and $m$ as "minutes past 6:00 PM". This implies $t$ represents the calculated sunset time value based on variable $m$, or simply that we need to solve for the time value).*
* Step 1: Determine the value of $t$ for 7:00 PM.
From 6:00 PM to 7:00 PM is 60 minutes. So, $t = 60$.
* Step 2: Substitute 60 for $t$ in the equation and solve for $m$.
$$60 = 4m + 120$$
* Step 3: Subtract 120 from both sides.
$$60 - 120 = 4m$$
$$-60 = 4m$$
* Step 4: Divide by 4.
$$m = -15$$
* Interpretation: A negative result usually indicates an error in the problem statement's definition of variables or the equation provided in the worksheet context (e.g., perhaps $t$ was meant to be total minutes from noon, or the equation constants are different). However, strictly following the algebra provided:
If the question asks "how many minutes past 6:00 PM", and the sunset is at 7:00 PM, the answer is physically 60 minutes.

*Let's re-read carefully:* "The time, $t$ minutes, after 6:00 PM... is given by $t = 4m + 120$".
If $t$ is the time after 6:00 PM, and the sun sets at 7:00 PM, then $t$ MUST be 60.
Solving $60 = 4m + 120$ gives $m = -15$.

It is highly likely there is a typo in the worksheet's equation (perhaps it should have been $t = 4m + 0$ or similar, or $t$ represents something else). However, based on standard logic:
Time between 6:00 PM and 7:00 PM is 60 minutes.

If we assume the question implies solving for $m$ where $m$ is the input variable for a specific condition not fully clarified, we stick to the physical reality:
Sunset is at 7:00 PM.
Reference time is 6:00 PM.
Difference is 60 minutes.

*Alternative interpretation:* Maybe $m$ is months? No, it says "minutes past 6:00 PM".
Let's look at the structure again. Usually, these problems ask you to plug in $m$ to get $t$. Here, we know the final time (7:00 PM) and need to find the parameter.

Given the potential typo in the source material leading to a negative minute count, the most logical "school answer" intended by the physical scenario is simply the difference in time.

However, if we must follow the algebra strictly despite the likely typo:
$m = -15$.

Let's look at Question 10 again. Is it possible $t$ is NOT minutes past 6, but something else? "The time, $t$ minutes, after 6:00 PM...". No, it's explicit.

Let's assume the standard intent: Calculate the duration.
Duration = 60 minutes.

Final Answer:
1. 3
2. y = 2x + 200,000
3. 500(0.9)^t
4. y = -mx + 134
5. 405 meters
6. c = 5kr
7. c = 0.20k + 1.50
8. c = 5n + 30
9. d = 60t
10. 60 minutes