Learning Objectives
- Read and understand word problems carefully to identify unknowns.
- Translate age relationships, quantity/cost relationships, and mixture proportions into simultaneous equations.
- Solve the resulting systems using substitution or elimination.
- Interpret the solution in the context of the problem.
- Check answers by substituting back into the original word problem.
Introduction
Many DSE students find word problems challenging because they involve translating English into algebra.
The key is to identify the unknowns, look for relationships,
and form two equations.
In this article, we will cover three common types:
- Age problems: Relationships between ages now, in the past, or in the future.
- Quantity problems: Two types of items with different prices or values (e.g., coins, tickets, fruits).
- Mixture problems: Combining two or more substances with different concentrations or proportions.
For each type, we will use a step-by-step strategy:
- Define variables (choose \(x\) and \(y\)).
- Set up two equations based on the given information.
- Solve the system.
- Check and interpret the answer.
Part 1: Age Problems
Age problems involve comparing the ages of two or more people at different times. The key is to
add or subtract the same number of years to both ages for past or future scenarios.
Tip for Age Problems
Let the current ages be variables. For "in \(n\) years", add \(n\) to both ages.
For "\(n\) years ago", subtract \(n\) from both ages.
Worked Example 1 – Age Problem
Example 1
A father is three times as old as his son. In 10 years, the father will be twice as old as his son.
Find their current ages.
Solution:
Let the son's current age be \(x\) and the father's current age be \(y\).
Equation 1 (current): \(y = 3x\).
In 10 years: son \(x+10\), father \(y+10\).
Equation 2: \(y + 10 = 2(x + 10)\).
Substitute \(y = 3x\): \(3x + 10 = 2x + 20 \Rightarrow x = 10\).
Then \(y = 3(10) = 30\).
Answer: Son is 10 years old, father is 30 years old.
Part 2: Quantity Problems
Quantity problems involve two types of items with different prices, values, or characteristics.
The two equations are usually:
- Quantity equation: total number of items.
- Value equation: total cost or value.
Tip for Quantity Problems
Let \(x\) be the number of one type and \(y\) be the number of the other.
The value equation is usually: \( \text{price}_1 \times x + \text{price}_2 \times y = \text{total} \).
Worked Example 2 – Quantity Problem
Example 2
A shop sells pens for $8 each and notebooks for $5 each. A customer buys 12 items and pays $84.
How many pens and notebooks did the customer buy?
Solution:
Let \(x\) = number of pens, \(y\) = number of notebooks.
Equation 1 (quantity): \(x + y = 12\).
Equation 2 (cost): \(8x + 5y = 84\).
From Eq 1: \(y = 12 - x\).
Substitute: \(8x + 5(12 - x) = 84 \Rightarrow 8x + 60 - 5x = 84 \Rightarrow 3x = 24 \Rightarrow x = 8\).
Then \(y = 12 - 8 = 4\).
Answer: 8 pens and 4 notebooks.
Part 3: Mixture Problems
Mixture problems involve combining two or more substances to get a mixture with a certain concentration or proportion.
The two equations are:
- Quantity equation: total amount of mixture.
- Content equation: total amount of a specific substance (e.g., acid, salt, alcohol).
Tip for Mixture Problems
The content of a substance = concentration × volume (or amount) of that solution.
For example, 20% acid solution of \(x\) litres contains \(0.2x\) litres of acid.
Worked Example 3 – Mixture Problem
Example 3
A chemist has two acid solutions: 20% and 50% concentration. How many litres of each must be mixed to get 30 litres of a 40% solution?
Solution:
Let \(x\) = litres of 20% solution, \(y\) = litres of 50% solution.
Equation 1 (total volume): \(x + y = 30\).
Equation 2 (acid content): \(0.2x + 0.5y = 0.4(30) = 12\).
From Eq 1: \(y = 30 - x\).
Substitute: \(0.2x + 0.5(30 - x) = 12 \Rightarrow 0.2x + 15 - 0.5x = 12 \Rightarrow -0.3x = -3 \Rightarrow x = 10\).
Then \(y = 30 - 10 = 20\).
Answer: 10 L of 20% and 20 L of 50%.
General Strategy & DSE Exam Techniques
Paper 1 Section A(1) & A(2)
- Define your variables clearly at the start – this gains method marks.
- Write the equations explicitly and label them (1) and (2).
- Choose the method (substitution or elimination) that seems simpler for the system.
- Check your answer by substituting into the original word problem.
- For mixture problems, be careful with percentages – convert to decimals (e.g., 20% = 0.2).
Paper 2 (MC)
- Quickly form the equations and test the options.
- Watch for tricky wording – "more than" and "less than" can be reversed.
- For age problems, remember to add/subtract the same number of years to both ages.
DSE-Style Practice Questions
Form and solve simultaneous equations for each word problem.
Practice Set 1 – Age
- A mother is 4 times as old as her daughter. In 8 years, the mother will be 3 times as old as her daughter. Find their current ages.
- John is 5 years older than Mary. In 3 years, the sum of their ages will be 39. Find their current ages.
- A man is twice as old as his son. 10 years ago, the man was 4 times as old as his son. Find their current ages.
Practice Set 2 – Quantity
- A vending machine accepts $2 and $5 coins. There are 30 coins in total, with a total value of $102. How many of each coin are there?
- A shop sells apples for $3 each and oranges for $2 each. A customer buys 15 fruits and pays $38. How many apples were bought?
- A theatre has 200 seats, with adult tickets at $50 and child tickets at $30. If the total revenue is $8400, how many adult tickets were sold?
Practice Set 3 – Mixture
- A solution contains 30% acid. How much pure acid must be added to 20 litres to make a 50% solution?
- A chemist mixes 10% and 40% salt solutions to get 50 litres of a 25% solution. How many litres of each are needed?
- A coffee shop blends two types of coffee: Type A at $60/kg and Type B at $80/kg. How many kg of each are needed to make 100 kg of a blend worth $70/kg?
Solutions to Practice Sets
Set 1 Solutions
- Let daughter = \(x\), mother = \(4x\). In 8 years: \(4x+8 = 3(x+8) \Rightarrow 4x+8=3x+24 \Rightarrow x=16\). Daughter 16, mother 64.
- Let Mary = \(x\), John = \(x+5\). In 3 years: \((x+3)+(x+5+3)=39 \Rightarrow 2x+11=39 \Rightarrow x=14\). Mary 14, John 19.
- Let son = \(x\), man = \(2x\). 10 years ago: \(2x-10 = 4(x-10) \Rightarrow 2x-10=4x-40 \Rightarrow 2x=30 \Rightarrow x=15\). Son 15, man 30.
Set 2 Solutions
- Let $2 coins = \(x\), $5 coins = \(y\). \(x+y=30\), \(2x+5y=102\). Solve: \(y=30-x\), \(2x+5(30-x)=102 \Rightarrow 2x+150-5x=102 \Rightarrow -3x=-48 \Rightarrow x=16\), \(y=14\).
- Let apples = \(a\), oranges = \(15-a\). \(3a+2(15-a)=38 \Rightarrow 3a+30-2a=38 \Rightarrow a=8\).
- Let adult = \(a\), child = \(200-a\). \(50a+30(200-a)=8400 \Rightarrow 50a+6000-30a=8400 \Rightarrow 20a=2400 \Rightarrow a=120\).
Set 3 Solutions
- Let \(x\) = litres of pure acid added. Initial acid = \(0.3 \times 20 = 6\) L. New total acid = \(6+x\), new volume = \(20+x\). \(\frac{6+x}{20+x} = 0.5 \Rightarrow 6+x = 10 + 0.5x \Rightarrow 0.5x = 4 \Rightarrow x = 8\).
- Let \(x\) = litres of 10%, \(y\) = litres of 40%. \(x+y=50\), \(0.1x+0.4y=12.5\). Solve: \(y=50-x\), \(0.1x+0.4(50-x)=12.5 \Rightarrow 0.1x+20-0.4x=12.5 \Rightarrow -0.3x=-7.5 \Rightarrow x=25\), \(y=25\).
- Let \(x\) = kg of Type A, \(y\) = kg of Type B. \(x+y=100\), \(60x+80y=7000\). Solve: \(y=100-x\), \(60x+80(100-x)=7000 \Rightarrow 60x+8000-80x=7000 \Rightarrow -20x=-1000 \Rightarrow x=50\), \(y=50\).
9. Exercise
Click the following link to have
An Exercise on Simultaneous Equations Word Problems: Age, Quantity & Mixture Problems
Summary Checklist for Revision
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Define variables: I identify the unknowns and assign variables.
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Form equations: I translate age, quantity, and mixture relationships into equations.
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Solve: I use substitution or elimination to solve the system.
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Check: I verify my answer in the context of the problem.
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Exam ready: I can tackle age, quantity, and mixture problems confidently.