Math worksheet with Australian-themed word problems for students.

Let's solve each problem one by one from the worksheet titled "Australia". Each question is a real-world application of algebra, and we'll write an appropriate algebraic expression or equation for each.

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1. Kangaroo Jump

Problem: A kangaroo can jump 3 times the length of its own body. If a kangaroo can jump *x* meters, how can you express its body length?

Solution:
Let the body length be $ b $ meters.

Given:
Jump distance = 3 × body length
So:
$ x = 3b $

To find body length in terms of $ x $:
$ b = \frac{x}{3} $

✔ Answer: $ \boxed{b = \frac{x}{3}} $

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2. Sydney Opera House Attendance

Problem: The number of visitors to the Sydney Opera House in a month is 300,000 more than twice the number of visitors in the previous month. If the number of visitors in the previous month is $ y $, write an algebraic expression for the number of visitors in the current month.

Solution:
Twice the previous month’s visitors: $ 2y $
Add 300,000: $ 2y + 300,000 $

✔ Answer: $ \boxed{2y + 300,000} $

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3. Koala Population

Problem: The number of koalas in a national park is decreasing by 10% each year. If there are initially 500 koalas, write an algebraic expression for the number of koalas after $ t $ years.

Solution:
This is exponential decay:
- Initial amount: 500
- Decay rate: 10% per year → multiplier = $ 1 - 0.10 = 0.90 $
- After $ t $ years: $ 500 \times (0.90)^t $

✔ Answer: $ \boxed{500(0.90)^t} $

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4. Sydney Harbour Bridge Height

Problem: The height of the Sydney Harbour Bridge is 134 meters. Write an equation in the form $ y = mx + b $, where $ y $ is the height of the bridge above the water level and $ x $ is the distance from the center of the bridge.

Solution:
The height is constant at 134 meters regardless of distance from the center.

So, $ y = 134 $, which means:
- $ m = 0 $ (no slope)
- $ b = 134 $

Thus:
$ y = 0x + 134 $

✔ Answer: $ \boxed{y = 134} $

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5. Great Barrier Reef Diving Depth

Problem: A diver is exploring the Great Barrier Reef. Depth $ d $ meters, at a distance $ x $ meters from the shore, is given by the equation $ d = 10x + 5 $. If the diver is 40 meters from the shore, what is the diving depth?

Solution:
Substitute $ x = 40 $ into the equation:
$$
d = 10(40) + 5 = 400 + 5 = 405
$$

✔ Answer: $ \boxed{405} $ meters

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6. Ayers Rock Circumference

Problem: The circumference of Ayers Rock is 5 times its radius. If the radius of Ayers Rock is $ r $ meters, write an algebraic expression for its circumference.

Solution:
Circumference of a circle is usually $ C = 2\pi r $, but here it says it's 5 times the radius, so:
$$
C = 5r
$$

✔ Answer: $ \boxed{5r} $

> Note: This is not physically accurate (since $ 2\pi \approx 6.28 $), but based on the problem statement, we follow the given condition.

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7. Melbourne Tram Fare

Problem: The cost of a tram ticket in Melbourne is $2.50 for the first 5 kilometers and an additional $0.20 for each kilometer beyond that. Write an equation to represent the cost, $ c $ dollars, for a tram ride of $ x $ kilometers.

Solution:
- For $ x \leq 5 $: $ c = 2.50 $
- For $ x > 5 $: cost = $ 2.50 + 0.20(x - 5) $

We can write this as a piecewise function:

$$
c =
\begin{cases}
2.50 & \text{if } x \leq 5 \\
2.50 + 0.20(x - 5) & \text{if } x > 5
\end{cases}
$$

But if we assume $ x > 5 $, then simplify:
$$
c = 2.50 + 0.20x - 1.00 = 0.20x + 1.50
$$

✔ Answer: $ \boxed{c = 0.20x + 1.50} $ for $ x > 5 $, or piecewise as above.

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8. Gold Coast Theme Park Tickets

Problem: At a theme park on the Gold Coast, the cost of a ticket is $30 plus an additional $5 for each ride. Write an equation to represent the total cost, $ c $ dollars, for $ r $ rides.

Solution:
Fixed cost: $30
Each ride: $5 → $5r

Total cost:
$$
c = 30 + 5r
$$

✔ Answer: $ \boxed{c = 30 + 5r} $

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9. Perth Bus Travel Time

Problem: A bus in Perth travels at a speed of 60 kilometers per hour. If the distance is $ d $ kilometers and the time is $ t $ hours, write an algebraic equation relating the distance and time.

Solution:
Use the formula:
$$
\text{Distance} = \text{Speed} \times \text{Time}
$$
$$
d = 60t
$$

✔ Answer: $ \boxed{d = 60t} $

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10. Perth Sunset Time

Problem: The time, $ t $ minutes, after 6:00 PM when the sun sets at Uluru is given by the equation $ t = 4m + 120 $, where $ m $ is the number of minutes past 6:00 PM. If the sun sets at 7:00 PM, how many minutes past 6:00 PM is that?

Wait — let's re-read carefully.

It says:
"The time, $ t $ minutes, after 6:00 PM when the sun sets at Uluru is given by $ t = 4m + 120 $" — but this seems confusing.

Actually, likely a typo or misstatement.

Let’s interpret:
- $ m $ = number of minutes past 6:00 PM
- $ t $ = time (in minutes) after 6:00 PM when the sun sets

But if the sun sets at 7:00 PM, that’s exactly 60 minutes after 6:00 PM.

So $ t = 60 $

But the equation is: $ t = 4m + 120 $

Wait — this doesn’t make sense unless $ m $ is something else.

Alternatively, perhaps the equation is meant to be used to find $ m $ such that $ t = 60 $?

But if $ t = 4m + 120 $, and $ t = 60 $, then:
$$
60 = 4m + 120 \\
4m = -60 \\
m = -15
$$

Negative time? That doesn't make sense.

Wait — maybe the equation is wrong, or misinterpreted.

Alternatively, perhaps the equation is meant to model sunset time, and $ m $ is a variable like day of year or something.

But the question says: "If the sun sets at 7:00 PM, how many minutes past 6:00 PM is that?"

That’s directly asking:
How many minutes from 6:00 PM to 7:00 PM?

Answer: $ \boxed{60} $ minutes.

The equation $ t = 4m + 120 $ might be a red herring or incorrectly stated.

But let’s suppose the equation is correct and we're told the sun sets at 7:00 PM, so $ t = 60 $, and we need to solve for $ m $?

Then:
$$
60 = 4m + 120 \\
4m = -60 \\
m = -15
$$

That would mean 15 minutes before 6:00 PM, which contradicts the premise.

So likely, the equation is either:
- Incorrect,
- Or $ m $ is not minutes past 6:00 PM.

But the question says: “where $ m $ is the number of minutes past 6:00 PM”

And asks: “If the sun sets at 7:00 PM, how many minutes past 6:00 PM is that?”

That’s a simple question.

✔ Answer: $ \boxed{60} $ minutes

> The equation $ t = 4m + 120 $ appears to be irrelevant or incorrectly written for this question.

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✔ Final Answers Summary:

1. $ b = \frac{x}{3} $
2. $ 2y + 300,000 $
3. $ 500(0.90)^t $
4. $ y = 134 $
5. $ 405 $ meters
6. $ 5r $
7. $ c = 0.20x + 1.50 $ (for $ x > 5 $)
8. $ c = 30 + 5r $
9. $ d = 60t $
10. $ \boxed{60} $ minutes

Let me know if you'd like these formatted as a printable answer sheet!