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Form 5 Mathematics Practice Paper: Functions, Sequences, Counting, Probability and Statistics
Form 5 Mathematics – 12 questions – 100 marks – 75 minutes
Form 5 Mathematics paper: 12 questions, 100 marks, 75 minutes on quadratic functions, arithmetic and geometric sequences, counting, probability and statistics, with answers.
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Paper details Form 5 Mathematics practice paper with 12 structured questions on functions, sequences, counting, probability and statistics; full answers and two printable PDFs.
Section A: Functions and Graphs (Q1-Q4)
Answer all questions; show the expression, substitution and unit for every calculation.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A function is defined as f(x) = 2x² − 8x + 5. (a) Express f(x) in completed square form and state the vertex and the axis of symmetry of its graph. (b) Find the minimum value of f(x) and explain whether it has a maximum value. (c) Solve f(x) = 0, giving your answers correct to three significant figures. |
(a) f(x) = 2 times (x² − 4x) + 5 = 2 times (x − 2)² − 8 + 5 = 2 times (x − 2)² − 3. The vertex is at (2, −3) and the axis of symmetry is x = 2. (b) The coefficient 2 is positive so the parabola opens upwards: the minimum value is -3. There is no maximum value because f(x) can grow without bound. (c) 2x² − 8x + 5 = 0 gives discriminant 64 − 40 = 24, so x = (8 plus or minus the square root of 24) divided by 4, that is x = 3.225 or x = 0.775. Marking notes: completed square, vertex and axis (3 marks); minimum value with reasoning (3 marks); quadratic formula and significant figures (3 marks). |
| 2 | The number of bacteria y in a culture after x hours is modelled by y = 3 times 2 to the power x. (a) Find y when x = 4. (b) Find x when y = 96. (c) Describe the transformation that turns the graph of y = 2 to the power x into the graph of y = 3 times 2 to the power x. |
(a) y = 3 times 16 = 48. (b) 3 times 2 to the power x = 96 gives 2 to the power x = 32 = 2 to the power 5, so x = 5. (c) The graph of y = 2 to the power x is stretched vertically by a factor of 3; its y intercept moves from 1 to 3. Marking notes: substitution (3 marks); solving the exponential equation (3 marks); transformation description (2 marks). |
| 3 | The graph of y = x² is translated 3 units to the right and then 4 units upwards to give the graph of y = g(x). (a) Write down the equation of g(x). (b) State the coordinates of the vertex of the graph of g. (c) Find g(0) and explain its connection with the y intercept of the graph. |
(a) A shift 3 units right replaces x by (x − 3) and a shift 4 units up adds 4 outside the function, so g(x) = (x − 3)² + 4. (b) The vertex moves from (0, 0) to (3, 4). (c) g(0) = 9 + 4 = 13, so the graph cuts the y axis at (0, 13); this value is the y intercept. Marking notes: equation (3 marks); vertex (3 marks); evaluation and y intercept explanation (3 marks). |
| 4 | A small bakery sells handmade cakes. When the price is x dollars per cake, it sells (120 − 2x) cakes per day and its daily revenue is R(x) = x(120 − 2x) dollars. (a) Write R(x) in the form a(x − h)² + k. (b) Find the price that maximises revenue and state the maximum revenue. (c) Find the integer range of prices for which the daily revenue is at least 1600 dollars. |
(a) R(x) = −2x² + 120x = −2(x² − 60x) = −2(x − 30)² + 1800. (b) Since −2 is negative, revenue is a maximum when x = 30, and the maximum revenue is 1800 dollars. (c) −2(x − 30)² + 1800 ≥ 1600 gives (x − 30)² ≤ 100, so 20 ≤ x ≤ 40; the integer price range is 20 dollars to 40 dollars. Marking notes: completed square (4 marks); maximum value and price (4 marks). |
Section B: Sequences and Summation (Q5-Q8)
Data: the vertex x coordinate of the parabola y = ax² + bx + c is −b divided by 2a; the sum to infinity of a geometric series is a divided by (1 − r) and exists only when the modulus of r is below 1.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | An arithmetic sequence has first term 7 and common difference −3. (a) Find the 20th term. (b) Find the sum of the first 20 terms. (c) Find an expression for the sum of the first n terms and explain whether that sum can equal 0. |
(a) The 20th term is 7 + 19 times (−3) = 7 − 57 = -50. (b) The sum of the first 20 terms is 20 times (7 − 50) divided by 2 = 10 times (−43) = -430. (c) S(n) = n divided by 2 times [14 + (n − 1)(−3)] = n(17 − 3n) divided by 2. Setting S(n) = 0 gives n = 0 or n = 17 thirds, which is not a whole number, so the sum of the first n terms never equals 0 for a positive integer n. Marking notes: general term (2 marks); sum formula and value (3 marks); expression and conclusion (3 marks). |
| 6 | A geometric series has first term 5 and common ratio 0.6. (a) Find the sum of the first five terms. (b) Find the sum to infinity of the series. (c) Explain why the sum to infinity exists for ratio 0.6 but not for ratio 1.2. |
(a) Sum of the first five terms = 5 times (1 − 0.6 to the power 5) divided by 0.4 = 5 times (1 − 0.07776) divided by 0.4, about 11.528. (b) Sum to infinity = 5 divided by (1 − 0.6) = 5 divided by 0.4 = 12.5. (c) The ratio 0.6 has absolute value below 1, so the terms tend to 0 and the partial sums converge. The ratio 1.2 has absolute value above 1, so the terms grow and the partial sums diverge, hence no sum to infinity. Marking notes: finite sum (3 marks); sum to infinity (3 marks); convergence explanation (2 marks). |
| 7 | Consider all the multiples of 3 between 100 and 500. (a) State the first and the last term of this sequence. (b) Find how many such numbers there are. (c) Find their total and state the condition for the formula you use. |
(a) The first term is the smallest multiple of 3 above 100, namely 102; the last term is the largest multiple of 3 below 500, namely 498. (b) Number of terms n = (498 − 102) divided by 3 + 1 = 133. (c) Total = n(first plus last) divided by 2 = 133 times 600 divided by 2 = 39900. The formula requires an arithmetic sequence with the correct number of terms. Marking notes: first and last term (3 marks); number of terms (3 marks); total with condition (3 marks). |
| 8 | A town has a population of 5000 and grows at 3% per year, compounded yearly like compound interest. (a) Find the population after 5 years, correct to the nearest integer. (b) Find the least number of years for the population to become more than double. (c) If the growth rate were 5% per year, would that number of years rise or fall? Explain briefly. |
(a) Population = 5000 times 1.03 to the power 5, about 5796.37, that is about 5796 people. (b) We need 1.03 to the power n above 2. Since 1.03 to the power 23 is about 1.974 and 1.03 to the power 24 is about 2.033, the least number of years is 24. (c) A higher growth rate means a larger yearly factor (1.05 instead of 1.03), so the number of years falls, to about 15 years. Marking notes: population after five years (4 marks); doubling time (4 marks). |
Section C: Counting, Probability and Statistics (Q9-Q12)
Give probabilities as fractions or decimals; decimals must be given to three significant figures, with units stated where needed.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | A class has 5 boys and 4 girls. Three students are chosen to form a committee. (a) Find the total number of possible committees. (b) Find the number of committees that contain at least one girl. (c) Find the number of committees with exactly one girl and check it against part (b). |
(a) Number of committees = C(9, 3) = 84. (b) All-boy committees number C(5, 3) = 10, so committees with at least one girl = 84 − 10 = 74. (c) Exactly one girl gives C(4, 1) times C(5, 2) = 4 times 10 = 40. Two girls give 30 and three girls give 4, and 40 + 30 + 4 = 74, matching part (b). Marking notes: total (3 marks); at least one girl (3 marks); exactly one girl and the check (3 marks). |
| 10 | Seven students stand in a row for a photograph, and two particular students must stand next to each other. (a) Find the total number of arrangements. (b) Find the number of arrangements in which the two students are adjacent. (c) Hence find the probability that they are adjacent in a random arrangement, to three significant figures. |
(a) All arrangements of seven people = 7! = 5040. (b) Treat the pair as one unit, giving 6 units to arrange, and multiply by the 2 internal orders: 2 times 6! = 2 times 720 = 1440. (c) Probability = 1440 divided by 5040 = 2 sevenths, about 0.286. Marking notes: total arrangements (3 marks); adjacent arrangements (3 marks); probability (2 marks). |
| 11 | A bag contains 4 red balls and 6 blue balls. Two balls are drawn at random without replacement. (a) Find the probability that both balls are red. (b) Find the probability that at least one ball is red. (c) Compare your answer to part (a) with the case of drawing with replacement. |
(a) Probability both are red = C(4, 2) divided by C(10, 2) = 6 divided by 45 = 2 fifteenths, about 0.133. (b) Probability of at least one red = 1 − C(6, 2) divided by C(10, 2) = 1 − 15 divided by 45 = 2 thirds, about 0.667. (c) With replacement the probability of two reds is (4 divided by 10) squared = 0.16, which is larger than 0.133, because removing a red ball without replacement lowers the chance of a red ball on the second draw. Marking notes: combination calculation (4 marks); at least one red (4 marks). |
| 12 | A set of data is 4, 7, 7, 9, 13. (a) Find the mean and the standard deviation, using the population standard deviation and three significant figures. (b) Find the new mean and standard deviation if 5 is added to every value. (c) Explain how the mean and the standard deviation change if every value is multiplied by 2. |
(a) Mean = 35 divided by 5 = 8; the sum of squared deviations is 16 + 1 + 1 + 1 + 25 = 44.0, so the variance is 44.0 divided by 5 = 8.8 and the standard deviation is about 2.966. (b) Adding 5 to every value gives a mean of 13 and leaves the standard deviation unchanged at about 2.966. (c) Multiplying every value by 2 doubles the mean to 16 and doubles the standard deviation to about 5.93, since a constant multiplier scales both statistics by the same factor. Marking notes: mean and standard deviation (4 marks); effect of adding a constant (4 marks). |
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