WAEC / WASSCE
WAEC Algebra Practice Questions, Solved in WASSCE Style
WAEC algebra theory questions usually test simultaneous linear equations, quadratic equations by factorisation, change of subject, variation, and word problems about prices or ages that turn into equations. This page gives five original questions in that exact WASSCE style, each solved step by step with notes on how the method and accuracy marks are shared.
Algebra is the most predictable part of the WASSCE core maths theory paper, which is exactly why it is worth drilling. Simultaneous equations, quadratics and change of subject appear in some form nearly every year, either directly or hidden inside a word problem about prices, ages or a journey. The five questions below are my own, written in WASSCE style at WASSCE difficulty, not copied from any past paper. Each is solved in full, with a note on where the method marks sit, because in a WAEC theory paper the working earns more than the final answer does.
How algebra shows up in the theory paper
Algebra rarely gets a whole Section B question to itself; instead it is spread through the paper. A Section A question might ask you to solve a pair of simultaneous equations or make a variable the subject of a formula. A Section B question often opens with a word problem, two unknowns and two conditions, that you must translate into equations before solving. Quadratics appear both as solve-this equations and inside area or number problems. Variation, y varies directly as x and inversely as z squared, is a WAEC favourite because it tests whether you can find the constant first. If you can translate words into equations quickly, algebra becomes the most reliable marks on the paper.
Showing working the way examiners want
For simultaneous equations, state your method, elimination or substitution, and show the step where one variable disappears; that line usually carries an M mark. For quadratics, show the factorised form before writing the roots, because the factorisation is where the method mark lives. When changing the subject, write one manipulation per line rather than jumping three steps at once; a skipped step that goes wrong loses both the M and the A mark. Finally, substitute your answers back into an original equation when you have a spare minute. WAEC does not award marks for checking, but it catches the sign slips that otherwise cost accuracy marks silently.
Worked questions, step by step
Solve the simultaneous equations 3x + 2y = 12 and 2x - y = 1.
- From the second equation, y equals 2x minus 1.
- Substitute into the first: 3x plus 2(2x minus 1) equals 12.
- Expand: 3x plus 4x minus 2 equals 12, so 7x equals 14 and x equals 2.
- Then y equals 2(2) minus 1 equals 3.
- Check in the first equation: 3(2) plus 2(3) equals 12. Correct.
Answer: x = 2, y = 3
Where marks slip: M1 for a correct substitution or elimination step, A1 for x, A1 for y. Writing only the answers with no visible method can score just one mark out of three or four.
Try one yourself: Solve 2x + y = 7 and 3x - y = 8. [Answer: x = 3, y = 1]
Solve the equation 2x^2 - 5x - 3 = 0.
- Look for two numbers that multiply to 2 times negative 3, which is negative 6, and add to negative 5: they are negative 6 and 1.
- Rewrite: 2x^2 minus 6x plus x minus 3 equals 0.
- Factor in pairs: 2x(x minus 3) plus 1(x minus 3) equals 0, so (2x plus 1)(x minus 3) equals 0.
- Set each factor to zero: x equals negative 1/2 or x equals 3.
Answer: x = -1/2 or x = 3
Where marks slip: M1 for a correct factorisation or correct use of the quadratic formula, A1 for each root. Unless the question says by factorisation, the formula is equally acceptable, but show the substitution line.
Try one yourself: Solve 3x^2 + x - 2 = 0. [Answer: x = 2/3 or x = -1]
Make x the subject of the formula y = (2x + 3)/(x - 5).
- Multiply both sides by (x minus 5): y(x minus 5) equals 2x plus 3.
- Expand: xy minus 5y equals 2x plus 3.
- Collect x terms on one side: xy minus 2x equals 5y plus 3.
- Factor out x: x(y minus 2) equals 5y plus 3.
- Divide: x equals (5y plus 3)/(y minus 2).
Answer: x = (5y + 3)/(y - 2)
Where marks slip: M1 for clearing the fraction, M1 for collecting the x terms together, A1 for the final form. The mark scheme wants x isolated once, so leaving an x on both sides scores no A mark.
Try one yourself: Make t the subject of s = (3t - 1)/(t + 2). [Answer: t = (2s + 1)/(3 - s)]
y varies directly as x and inversely as the square of z. When x = 4 and z = 1, y = 8. Find (a) the equation connecting x, y and z, (b) the value of y when x = 27 and z = 3.
- Write the relationship with a constant: y equals kx/z^2.
- Substitute the given values: 8 equals k times 4 divided by 1, so k equals 2.
- The equation is y equals 2x/z^2.
- For part (b): y equals 2 times 27 divided by 3^2 equals 54 divided by 9 equals 6.
Answer: (a) y = 2x/z^2 (b) y = 6
Where marks slip: B1 for the correct relationship with k, M1 A1 for finding k, M1 A1 for part (b). Skipping straight to numbers without ever writing y = kx/z^2 forfeits the first mark.
Try one yourself: y varies directly as x^2 and inversely as z. When x = 2 and z = 1, y = 12. Find y when x = 4 and z = 6. [Answer: y = 8]
Four pens and three exercise books cost N1,100. Two pens and five exercise books cost N900. Find the cost of one pen and of one exercise book.
- Let a pen cost p naira and a book cost b naira: 4p plus 3b equals 1,100 and 2p plus 5b equals 900.
- Multiply the second equation by 2: 4p plus 10b equals 1,800.
- Subtract the first equation: 7b equals 700, so b equals 100.
- Substitute back: 4p plus 300 equals 1,100, so 4p equals 800 and p equals 200.
Answer: A pen costs N200 and an exercise book costs N100
Where marks slip: B1 for defining variables and forming both equations, that step alone often carries two marks, then M1 for elimination, A1 for each value. Always answer in words with units at the end.
Try one yourself: Three oranges and two mangoes cost GHc 13, while one orange and four mangoes cost GHc 11. Find the cost of each fruit. [Answer: orange GHc 3, mango GHc 2]
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Solve my question →Quiz me on this topicQuestions students ask
Can I use the quadratic formula instead of factorising?
Yes, unless the question explicitly says by factorisation or by completing the square. Write the formula, show the substitution line, then simplify. The method mark attaches to that substitution, not to the memorised formula.
How many marks is a typical WAEC algebra question worth?
A Section A algebra question is usually around 8 marks; inside Section B, algebra is often one part of a 12 or 13 mark question. Forming the equations from a word problem typically carries two or three marks before you solve anything.
Are these real WAEC past questions on algebra?
No. They are original questions I wrote in the WASSCE format at the same difficulty, because real past papers are WAEC copyright. The topics and mark structure mirror what the theory paper tests year after year.