Form 2 Mathematics Practice Paper: Algebra, Equations, Percentages And Mensuration

Form 2 Mathematics Practice Paper: Algebra, Equations, Percentages and Mensuration

2026-11-05

Form 2 Mathematics – 12 questions – 100 marks – 75 minutes

Form 2 Maths practice paper: 12 questions, 100 marks, 75 minutes on algebra, equations, inequalities, percentages, ratios, area, volume, statistics and rates.

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Paper details　Form 2 Maths practice paper with 12 questions on algebra, inequalities, percentages, ratios, area, volume, statistics and rates, 100 marks in 75 minutes.

Section A: Algebraic Expressions, Factorisation and Equations

Answer all questions and show your working.

#	Question	Answer and marking notes
1	Simplify the algebraic expression 3a + 5b – a + 2b. (a) Write the simplest form. (b) Find its value when a = 2 and b = -3. (c) In the simplified expression, what is the coefficient of a?	Answer: (a) 3a – a = 2a and 5b + 2b = 7b, so the simplest form is 2a + 7b. (b) 2(2) + 7(-3) = 4 – 21 = -17. (c) The coefficient of a is 2. Marking: 4 marks for collecting like terms in (a), 3 marks for the substitution and answer in (b), 1 mark in (c); the value alone without substitution earns 1 mark.
2	Expand and factorise: (a) expand 3(2x – 5). (b) factorise 8x – 12. (c) expand -2(x – 4).	Answer: (a) 3(2x – 5) = 6x – 15. (b) 8x – 12 = 4(2x – 3), since 4 is the highest common factor of 8 and 12. (c) -2(x – 4) = -2x + 8. Marking: 3 marks in (a), 3 marks in (b) where a factor other than the highest common factor earns 1 mark, 2 marks in (c); a sign error scores zero for that part.
3	Solve the equation 5x – 7 = 3x + 11. (a) Find x step by step. (b) Check your answer by substituting it into the equation and writing the value of each side. (c) The sum of three consecutive integers is 63. Find the three integers.	Answer: (a) 5x – 3x = 11 + 7, so 2x = 18 and x = 9. (b) Left side = 5(9) – 7 = 38 and right side = 3(9) + 11 = 38, so both sides are equal and the answer is correct. (c) Let the smallest integer be n: n + (n + 1) + (n + 2) = 63, so 3n + 3 = 63 and n = 20. The integers are 20, 21 and 22. Marking: 3 marks in (a), 1 mark for each side in (b), 1 mark for the unknown and 2 marks for the equation and answer in (c).
4	Solve these inequalities: (a) 4x + 3 > 19. (b) -2x >= 8. (c) The inequality sign reverses in (b). Explain why.	Answer: (a) 4x > 16, so x > 4. (b) Dividing both sides by -2 reverses the sign, so x <= -4. (c) Dividing by a negative number reverses the order on the number line: 6 > 4 but -6 is less than -4, so the inequality sign must be reversed. Marking: 2 marks in (a), 2 marks in (b) where a wrong direction earns 1 mark, 4 marks in (c) for the example and the explanation.

Section B: Percentages, Ratios and Formulas

Answer all questions and state the units in your answers.

#	Question	Answer and marking notes
5	A jacket marked at $480 is sold at a 20% discount. (a) Find the selling price. (b) Find the discount. (c) The cost of the jacket is $320. Find the percentage profit when it is sold at the selling price.	Answer: (a) 480 x 0.8 = $384. (b) 480 – 384 = $96. (c) Profit = 384 – 320 = $64, so percentage profit = 64 / 320 x 100% = 20%. Marking: 3 marks in (a), 2 marks in (b), 2 marks for the profit and 2 marks for the percentage in (c); using the marked price as the denominator earns only 1 mark.
6	The population of an estate rises from 25 000 to 27 500. (a) Find the percentage increase. (b) If the population then falls by 10% from 27 500, find the population a year later. (c) Does the 10% fall bring the population back to 25 000? Explain.	Answer: (a) (27 500 – 25 000) / 25 000 x 100% = 10%. (b) 27 500 x 0.9 = 24 750. (c) No. The two changes use different bases: the rise is measured from 25 000 while the fall is measured from 27 500, so the two 10% amounts differ and do not cancel. Marking: 4 marks in (a), 3 marks in (b), 2 marks in (c) for identifying the different bases.
7	Concentrated juice and water are mixed in the ratio 2 : 7. (a) To make 900 mL of juice drink, how much concentrate and how much water are needed? (b) If 300 mL of concentrate is used, how much water is needed and what is the total volume?	Answer: (a) Total parts = 2 + 7 = 9, so 900 / 9 = 100 mL per part; concentrate = 2 x 100 = 200 mL and water = 7 x 100 = 700 mL. (b) Water = 300 / 2 x 7 = 1050 mL and the total is 300 + 1050 = 1350 mL. Marking: 2 marks for the value of one part and 2 marks for each quantity in (a), 3 marks in (b).
8	The perimeter of a rectangle is P = 2(l + w), where l is the length and w is the width. (a) Given P = 36 cm and l = 11 cm, find w. (b) Make w the subject of the formula. (c) Use the formula in (b) to find w when P = 50 cm and l = 16 cm.	Answer: (a) 36 = 2(11 + w), so 11 + w = 18 and w = 7 cm. (b) P = 2(l + w), so P / 2 = l + w and w = P/2 – l. (c) w = 50 / 2 – 16 = 9 cm. Marking: 3 marks in (a), 3 marks in (b) for changing the subject, 3 marks in (c); deduct 1 mark for a missing unit.

Section C: Mensuration, Statistics and Rates

Answer all questions and write down the working and the answer for each problem.

#	Question	Answer and marking notes
9	Find the area of each figure, taking pi as 22/7: (a) a trapezium with parallel sides 8 cm and 14 cm and height 6 cm. (b) a circle of radius 7 cm. (c) a semicircle of radius 7 cm.	Answer: (a) (8 + 14) / 2 x 6 = 66 cm squared. (b) 22/7 x 7 x 7 = 154 cm squared. (c) 154 / 2 = 77 cm squared. Marking: 3 marks in (a) for the formula and answer, 3 marks in (b), 2 marks in (c); deduct 1 mark if the unit is written as cm.
10	A rectangular container is 12 cm long, 8 cm wide and 5 cm high. (a) Find its volume. (b) Find its total surface area. (c) If 360 mL of water is poured in, find the depth of the water (1 mL = 1 cm cubed).	Answer: (a) 12 x 8 x 5 = 480 cm cubed. (b) 2 x (12×8 + 12×5 + 8×5) = 2 x (96 + 60 + 40) = 392 cm squared. (c) Base area = 12 x 8 = 96 cm squared, so the depth = 360 / 96 = 3.75 cm. Marking: 2 marks in (a), 3 marks in (b), 3 marks in (c); a correct answer without finding the base area earns 2 marks.
11	The test scores of six pupils are 12, 15, 15, 18, 20 and 25. (a) Find the mean. (b) Find the median. (c) Find the mode. (d) If a pupil scoring 27 joins the group, find the new mean to one decimal place.	Answer: (a) (12 + 15 + 15 + 18 + 20 + 25) / 6 = 105 / 6 = 17.5 marks. (b) The middle two values are 15 and 18, so the median = (15 + 18) / 2 = 16.5 marks. (c) The mode is 15 marks, which appears twice. (d) Total = 105 + 27 = 132, so the new mean = 132 / 7 = 18.857…, that is 18.9 marks. Marking: 2 marks in (a), 2 marks in (b), 1 mark in (c), 3 marks in (d); deduct 1 mark if not given to one decimal place.
12	A car travels 180 km in 2.5 hours. (a) Find its average speed in km/h. (b) At the same speed, how long does it take to travel 270 km? Give the answer in hours and minutes. (c) Convert the speed to m/s.	Answer: (a) 180 / 2.5 = 72 km/h. (b) 270 / 72 = 3.75 hours, that is 3 hours 45 minutes. (c) 72 x 1000 / 3600 = 20 m/s. Marking: 3 marks in (a), 2 marks for the time and 1 mark for the conversion in (b), 2 marks in (c); 3.75 hours alone without conversion earns 1 mark less.

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