Algebra Word Problems Worksheets - 15 Worksheets Library - Free Printable

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 $ C = 5r $

✔ Answer: $ \boxed{C = 5r} $

> Note: This is not the standard formula ($ C = 2\pi r $), but according to the problem, it's given as 5 times the radius.

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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 $ k $ kilometers.

Solution:
- For $ k \leq 5 $: cost is $ 2.50 $
- For $ k > 5 $: base cost + extra for each km beyond 5

Extra distance: $ k - 5 $ km
Extra cost: $ 0.20(k - 5) $

Total cost:
$$
c = 2.50 + 0.20(k - 5)
$$

Simplify:
$$
c = 2.50 + 0.20k - 1.00 = 0.20k + 1.50
$$

But this only applies when $ k > 5 $. So we need a piecewise function or assume $ k > 5 $.

However, since the problem asks for a general equation, we use:

✔ Answer: $ \boxed{c = 2.50 + 0.20(k - 5)} $ for $ k > 5 $

Or simplified: $ \boxed{c = 0.20k + 1.50} $, valid for $ k > 5 $

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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:
Base 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:
Distance = Speed × Time
So:
$$
d = 60t
$$

✔ Answer: $ \boxed{d = 60t} $

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10. Perth Sunset Time

Problem: The time, $ t $ minutes, after 6:00 PM that 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?

Solution:
7:00 PM is 60 minutes after 6:00 PM → $ m = 60 $

But wait — let's read carefully.

It says:
$t = 4m + 120$, where $ m $ is the number of minutes past 6:00 PM.

Wait — this seems inconsistent. Let's interpret:

Actually, if $ m $ is minutes past 6:00 PM, then $ t $ is the sunset time after 6:00 PM.

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

That's straightforward:
7:00 PM is 60 minutes after 6:00 PM.

So $ m = 60 $

But maybe the equation is meant to be used to find $ t $, but here we're being asked to find $ m $, given sunset at 7:00 PM.

But the equation is $ t = 4m + 120 $. That would mean $ t $ depends on $ m $, which doesn't make sense unless $ m $ is something else.

Wait — perhaps there's confusion in variable naming.

Let me re-read:

> "The time, $ t $ minutes, after 6:00 PM that the sun sets... is given by $ t = 4m + 120 $, where $ m $ is the number of minutes past 6:00 PM."

That can't be right — if $ m $ is minutes past 6:00 PM, and $ t $ is also minutes after 6:00 PM, then $ t = m $? But the equation says $ t = 4m + 120 $, which would imply $ t $ is much larger than $ m $, which is illogical.

So likely, the variables are mislabeled.

Alternative interpretation:

Maybe $ m $ is not minutes past 6:00 PM, but some other variable (like month or day). But the problem says:

> "...where $ m $ is the number of minutes past 6:00 PM"

So if $ m $ is minutes past 6:00 PM, and $ t $ is the sunset time after 6:00 PM, then $ t = m $, so $ m = t $.

But the equation says $ t = 4m + 120 $. Then:

Set $ t = m $, so:
$$
m = 4m + 120 \\
m - 4m = 120 \\
-3m = 120 \\
m = -40
$$

Negative? Doesn't make sense.

So either the equation is wrong, or the interpretation is off.

Wait — perhaps the equation is meant to model sunset time based on month, not minutes.

But the problem says: "where $ m $ is the number of minutes past 6:00 PM"

This seems like a typo.

Alternatively, maybe the equation is:

> The sunset time $ t $ (in minutes after 6:00 PM) is given by $ t = 4m + 120 $, where $ m $ is the month (e.g., January = 1, etc.)

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

That’s 60 minutes.

So regardless of the equation, the answer is simply:

7:00 PM = 60 minutes after 6:00 PM.

So $ \boxed{60} $

But the equation might be a red herring, or misprinted.

Let’s assume the question is asking:
"If the sun sets at 7:00 PM, how many minutes past 6:00 PM is that?"

Then the answer is:

✔ Answer: $ \boxed{60} $

> Note: The equation $ t = 4m + 120 $ may be intended for another part, but based on the wording, this is a direct conversion.

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✔ Final Answers Summary:

1. $ \boxed{b = \frac{x}{3}} $
2. $ \boxed{2y + 300,000} $
3. $ \boxed{500(0.90)^t} $
4. $ \boxed{y = 134} $
5. $ \boxed{405} $ meters
6. $ \boxed{C = 5r} $
7. $ \boxed{c = 2.50 + 0.20(k - 5)} $
8. $ \boxed{c = 30 + 5r} $
9. $ \boxed{d = 60t} $
10. $ \boxed{60} $ minutes

Let me know if you'd like these explained further!