Every introduction to algebra question from 12 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 2.1. 13 questions covering Core and Extended tiers, ordered from Core to Extended. Each question is labelled with its question paper code (for example 0580/11/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Algebra and Graphs past paper questions.
0580 Introduction to Algebra past paper questions: Core (Papers 1 and 3)
May/June 2026 · Paper 1 · 0580/11/M/J/26 · Q2
(a) Bella is b years old. Alan is 5 years older than Bella. Write an expression, in terms of b, for how old Alan is.
(b) Write an expression for the total cost, in cents, of x bags of sweets costing 65 cents each.
Show solution
- “5 years older” means add 5 to Bella’s age.
- Alan’s age = b + 5
- Each bag costs 65 cents, so x bags cost 65 × x.
- Total cost = 65x cents
- Tip: An expression has no equals sign. Write only b + 5 or 65x.
May/June 2026 · Paper 1 · 0580/13/M/J/26 · Q4
A car hire company uses this formula to work out the hire cost, in dollars: cost = fixed charge + rate per day × number of days. Find the cost of hiring a car for 7 days when the fixed charge is $140 and the rate per day is $30.
Show solution
- Substitute the values into the formula: cost = 140 + 30 × 7.
- Multiply before adding: 30 × 7 = 210.
- Cost = 140 + 210 = $350
May/June 2026 · Paper 1 · 0580/13/M/J/26 · Q10
K = 2x + √y. Find the value of K when x = 5 and y = 169.
Show solution
- Substitute x = 5 and y = 169 into K = 2x + √y.
- 2 × 5 = 10 and √169 = 13 (because 13 × 13 = 169).
- K = 10 + 13 = 23
Feb/Mar 2026 · Paper 1 · 0580/12/F/M/26 · Q7
Krish has 6x + 4 marbles. Jo has 9 − x marbles.
(a) Write down an expression for the total number of marbles that Krish and Jo have. Give your answer in its simplest form.
(b) Krish has more marbles than Jo. Write down an expression for how many more marbles Krish has than Jo. Give your answer in its simplest form.
Show solution
- Total = Krish + Jo = (6x + 4) + (9 − x).
- Collect the x terms: 6x − x = 5x. Collect the numbers: 4 + 9 = 13.
- Total = 5x + 13
- Difference = Krish − Jo = (6x + 4) − (9 − x).
- Remove the bracket. The minus sign changes every sign inside it: 6x + 4 − 9 + x.
- Collect like terms: 6x + x = 7x and 4 − 9 = −5.
- Difference = 7x − 5
Oct/Nov 2025 · Paper 3 · 0580/31/O/N/25 · Q3
Total cost of a journey = number of litres of fuel used × cost of fuel per litre. A journey uses 128 litres of fuel. The cost of fuel is $1.52 per litre. Calculate the total cost of this journey.
Show solution
- Use the formula: total cost = litres × cost per litre.
- Total cost = 128 × 1.52 = $194.56
Oct/Nov 2025 · Paper 3 · 0580/32/O/N/25 · Q11
Y = 3t² + 2m. Find the value of Y when t = 6 and m = −3.
Show solution
- Replace t with 6 and m with −3: Y = 3 × 6² + 2 × (−3).
- Square first: 6² = 36, so 3 × 36 = 108.
- 2 × (−3) = −6.
- Y = 108 − 6 = 102
Oct/Nov 2025 · Paper 3 · 0580/33/O/N/25 · Q12
(a) Simplify 5b − 8c + 2b − 3c. (b) q = 3r + 5t. Find the value of t when q = 37 and r = 4.
Show solution
- Collect the b terms: 5b + 2b = 7b.
- Collect the c terms: −8c − 3c = −11c.
- 7b − 11c
- Substitute q = 37 and r = 4: 37 = 3 × 4 + 5t, so 37 = 12 + 5t.
- Subtract 12 from both sides: 25 = 5t.
- Divide by 5: t = 5
May/June 2025 · Paper 1 · 0580/11/M/J/25 · Q7
(a) A ticket costs $18. Write down an expression, in dollars, for the cost of t tickets.
(b) A bag contains n red balls and 16 green balls. Write down an expression for the total number of balls in the bag.
Show solution
- One ticket costs $18, so t tickets cost 18 × t.
- Cost = 18t
- Total balls = red balls + green balls.
- Total = n + 16
0580 Introduction to Algebra past paper questions: Extended (Papers 2 and 4)
May/June 2026 · Paper 2 · 0580/21/M/J/26 · Q3
Find the value of 4x² + 3y when x = 5 and y = 10.
Show solution
- Substitute x = 5 and y = 10 into 4x² + 3y.
- Square first: x² = 5² = 25, so 4x² = 4 × 25 = 100.
- 3y = 3 × 10 = 30.
- 4x² + 3y = 100 + 30 = 130
May/June 2026 · Paper 2 · 0580/22/M/J/26 · Q6(a)
E = 5c² − d². Find the value of E when c = √18 and d = 2.
Show solution
- Substitute c = √18 and d = 2 into E = 5c² − d².
- (√18)² = 18, because squaring undoes the square root. And 2² = 4.
- E = 5 × 18 − 4 = 90 − 4 = 86
May/June 2026 · Paper 2 · 0580/23/M/J/26 · Q17(a)
c = 4a² + 2b. Find the value of c when a = −2 and b = 3.
Show solution
- Substitute a = −2 and b = 3 into c = 4a² + 2b.
- Square first: (−2)² = 4 (a negative number squared is positive).
- c = 4 × 4 + 2 × 3 = 16 + 6 = 22
Oct/Nov 2025 · Paper 2 · 0580/22/O/N/25 · Q8
Kemi buys p pens that each cost 40 cents. She pays with $20. Write an expression, in terms of p, for the change, in cents, Kemi receives from the $20.
Show solution
- Change $20 to cents so both amounts use the same unit: $20 = 2000 cents.
- p pens at 40 cents each cost 40p cents.
- Change = 2000 − 40p
Oct/Nov 2025 · Paper 4 · 0580/43/O/N/25 · Q3
G = 4⁄5m²n. Find the value of G when m = 6 and n = 15.
Show solution
- Substitute m = 6 and n = 15: G = 4⁄5 × 6² × 15.
- Square first: 6² = 36, and 36 × 15 = 540.
- 4⁄5 × 540 = 540 ÷ 5 × 4 = 108 × 4 = 432
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