A movie ticket costs $\$15$ for adults and $\$10$ for children. A family of 2 adults and 3 children pays a total. How much do they pay? Now, if you only knew the total was $60 and there were 4 people, how could you work backwards?
Word problems are just equations wearing a disguise. The key is to read carefully, define one variable, express everything in terms of that variable, write an equation, solve it, and check that the answer makes sense. Every word problem can be broken down into these steps.
Read → Define → Write → Solve → Check
- Translate word problems into linear equations
- Solve problems involving consecutive numbers
- Set up and solve money and pricing problems
- Solve distance, speed and time problems
- Verify answers by checking against the original problem
Wrong: Using two different variables for related quantities without connecting them. For example, using x and y for two consecutive numbers without stating y = x + 1.
Right: Express all unknowns in terms of a single variable. If the first number is n, the next consecutive number is n + 1.
Wrong: Forgetting to check if the answer makes sense. Getting a negative age or a fractional number of people.
Right: Always ask: "Does this answer make sense in the real world?"
Every word problem can be solved with the same five-step process. Skipping a step invites errors. Rushing to write the equation before understanding the problem is the most common cause of wrong answers.
Read → Define → Write → Solve → Check
Consecutive integers follow each other: n, n+1, n+2. Consecutive even or odd numbers are two apart: n, n+2, n+4. Always express every unknown in terms of one variable. This gives you one equation with one unknown.
Sum = 54: $n + (n+2) = 54$ → $n = 26$.
Money problems involve fixed costs (flag falls, base fees) and variable costs (per item, per kilometre). Write one equation for the total quantity and another for the total cost. Substitute to solve. Always convert all amounts to the same unit before solving.
$22 adults, $14 kids, $152 total, 10 people.
The formula d = rt (distance = rate � time) is the foundation of all motion problems. In catch-up problems, the faster object covers the same distance as the slower one, but starts later. Set the distances equal and solve for time.
Train 90 km/h at 9am. Car 120 km/h at 10am. When does car catch up?
Watch Me Solve It · 3 examples
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1Define the variableLet $n$ = the first even integer.Consecutive even integers are 2 apart, so the next one is $n + 2$.
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2Write the equation$n + (n + 2) = 54$The sum of the two integers equals 54.
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3Solve and state the answer$2n + 2 = 54$ → $2n = 52$ → $n = 26$The integers are 26 and 28.
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4Check the answer$26 + 28 = 54$ ✓ Both are even. ✓The sum matches and both numbers are even. The answer is correct.
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1Define variables and write two equationsLet $a$ = adults, $c$ = children. $a + c = 10$ and $22a + 14c = 152$One equation for total people, one for total cost.
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2Express one variable in terms of the otherFrom $a + c = 10$: $c = 10 - a$Substitute into the cost equation to get one variable.
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3Substitute and solve$22a + 14(10 - a) = 152$ → $22a + 140 - 14a = 152$ → $8a = 12$ → $a = 4$Simplify: 22a - 14a = 8a. Subtract 140: 152 - 140 = 12.
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4Find the other value and check$c = 10 - 4 = 6$. Check: $4 + 6 = 10$ ✓ $22(4) + 14(6) = 88 + 84 = 152$ ✓Both the quantity and the cost match. The answer is correct.
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1Calculate the head startIn 1 hour, the train travels $90 \times 1 = 90$ km.The car starts 1 hour later, so the train has a 90 km head start.
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2Write distance equations and set them equalLet $t$ = hours after 10:00 am. Train distance: $90 + 90t$. Car distance: $120t$.At catch-up, both have travelled the same distance from Sydney.
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3Solve for t$120t = 90 + 90t$ → $30t = 90$ → $t = 3$ hoursSubtract 90t from both sides, then divide by 30.
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4State the catch-up time and verify10:00 am + 3 hours = 1:00 pm. Train: $90 + 90(3) = 360$ km. Car: $120(3) = 360$ km.Both distances match. The car catches up at 1:00 pm.
Brain Trainer · 4 problems
Four word problems covering translation, consecutive numbers, money and distance. Show each step clearly, then reveal the answer to check.
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1 Five more than twice a number is 17. Find the number.
Let $n$ be the number. Equation: $2n + 5 = 17$. Subtract 5: $2n = 12$. Divide by 2: $n = 6$. Check: $2(6) + 5 = 17$. -
2 Three consecutive integers sum to 48. Find the largest integer.
Let $n$ be the first. Equation: $n + (n+1) + (n+2) = 48$. $3n + 3 = 48$. $3n = 45$. $n = 15$. Largest: $17$. Check: $15 + 16 + 17 = 48$. -
3 A taxi charges a $\$5$ flag fall plus $\$2.50$ per kilometre. A fare costs $32.50. How many kilometres were travelled?
Let $d$ = distance. Equation: $5 + 2.5d = 32.5$. Subtract 5: $2.5d = 27.5$. Divide by 2.5: $d = 11$ km. Check: $5 + 2.5(11) = 32.5$. -
4 A number is doubled and then 7 is subtracted. The result is 15. What is the original number?
Let $n$ be the number. Equation: $2n - 7 = 15$. Add 7: $2n = 22$. Divide by 2: $n = 11$. Check: $2(11) - 7 = 15$.
Quick Check · 5 questions
Show Your Working · 3 questions
Q6. The sum of three consecutive odd integers is 81. Find the three integers. Define your variable, show all working, and check your answer.
Q7. A train leaves Sydney at 9:00 am travelling at 90 km/h. A car leaves Sydney at 10:00 am travelling at 120 km/h on the same route.
(a) How far has the train travelled by 10:00 am? (1 mark)
(b) Let $t$ be the number of hours after 10:00 am. Write an equation for when the car catches up to the train. (1 mark)
(c) At what time does the car catch up to the train? (1 mark)
Q8. A school fundraiser sells chocolate bars for $\$3$ each and cookies for $\$2$ each. They sell 150 items in total and raise $380.
(a) Define variables and write two equations. (1 mark)
(b) Solve to find how many chocolate bars and how many cookies were sold. (1 mark)
(c) Verify your answer by checking both the total number of items and the total money raised. (1 mark)
Quick Check
1. B"Twice a number" = $2n$, "five more than" = $+ 5$, "is 17" = $= 17$.
2. B$n + (n+1) + (n+2) = 48$ → $3n + 3 = 48$ → $n = 15$. Largest = 17.
3. C$d = rt = 80 \times 2.5 = 200$ km.
4. B$5 + 2.5d = 32.5$ → $2.5d = 27.5$ → $d = 11$ km.
5. C$2n - 7 = 15$ → $2n = 22$ → $n = 11$.
Show Your Working Model Answers
Q6 (3 marks): Let $n$ = first odd integer [0.5]. $n + (n+2) + (n+4) = 81$ [0.5]. $3n + 6 = 81$ [0.5]. $3n = 75$ → $n = 25$ [0.5]. The integers are 25, 27, and 29 [0.5]. Check: $25 + 27 + 29 = 81$ ✓ [0.5].
Q7 (3 marks): (a) By 10:00 am, the train has travelled $90 \times 1 = 90$ km [1]. (b) Distance of train after $t$ hours: $90 + 90t$. Distance of car: $120t$. Equation: $120t = 90 + 90t$ [1]. (c) $30t = 90$ → $t = 3$ hours. The car catches up at 1:00 pm [1].
Q8 (3 marks): (a) Let $c$ = chocolate bars, $k$ = cookies. $c + k = 150$ and $3c + 2k = 380$ [1]. (b) From first equation: $k = 150 - c$. Substitute: $3c + 2(150 - c) = 380$ → $c = 80$, $k = 70$ [1]. (c) Check items: $80 + 70 = 150$ ✓. Check money: $3(80) + 2(70) = 240 + 140 = 380$ ✓ [1].
The Pool Puzzle
A swimming pool has two pipes. Pipe A fills the pool in 6 hours. Pipe B empties the pool in 8 hours. The pool is initially empty. If both pipes are turned on at the same time, how long will it take to fill the pool? Hint: think about the fraction of the pool each pipe handles in one hour.
Reveal solution
Pipe A fills $\frac{1}{6}$ of the pool per hour.
Pipe B empties $\frac{1}{8}$ of the pool per hour.
Combined rate: $\frac{1}{6} - \frac{1}{8} = \frac{4}{24} - \frac{3}{24} = \frac{1}{24}$ of the pool per hour.
Time to fill: $\frac{1}{\frac{1}{24}} = 24$ hours.
It takes 24 hours to fill the pool with both pipes running.
Read Twice
Understand before you calculate
Define Clearly
Write "Let n = ..." in full sentences
One Variable
Express all unknowns in terms of a single letter
Two Equations
Quantity and cost for money problems
Distance = Rate x Time
The foundation of all motion problems
Check with Words
Plug back into the original sentence
Interactive: Problem Solver
Practise translating word problems into equations with scaffolded hints and instant feedback.
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