GCSE Maths Daily Practice (Higher)
Mark Scheme — 42 marks total
Higher TierMarking Instructions
- • Accept equivalent answers unless the mark scheme states otherwise.
- • M marks are for method — award even if the final answer is wrong.
- • A marks are for accuracy — only award if the method mark has been earned.
- • B marks are independent and can be awarded without other marks.
- • Where a range of answers is acceptable, this is indicated in the mark scheme.
Simplify fully (2x³y²)³ ÷ (4x²y⁵)
Acceptable Answer
(2x³y²)³ = 8x⁹y⁶. Then 8x⁹y⁶ ÷ 4x²y⁵ = 2x⁷y
Solve the simultaneous equations: 3x + 2y = 1 x² + y = 5
Acceptable Answer
From eq1: y = (1 - 3x)/2. Substitute: x² + (1 - 3x)/2 = 5 → 2x² + 1 - 3x = 10 → 2x² - 3x - 9 = 0 → (2x + 3)(x - 3) = 0. x = 3, y = -4 or x = -3/2, y = 11/4
After a 20% reduction, a laptop costs £680. (a) Work out the original price. (b) The shop then increases the sale price by 20%. Show that this does not give the original price and explain why.
Acceptable Answer
(a) £680 = 80% of original → original = 680 ÷ 0.8 = £850 (b) New price = 680 × 1.20 = £816 ≠ £850. The 20% decrease is calculated on £850 (= £170 off), but the 20% increase is calculated on the smaller amount £680 (= £136 on). The base amounts are different so the changes are not equal.
The first four terms of a quadratic sequence are: 5, 12, 23, 38, ... Find the nth term.
Acceptable Answer
1st differences: 7, 11, 15. 2nd differences: 4, 4. So coefficient of n² = 4/2 = 2. Sequence 2n²: 2, 8, 18, 32. Original − 2n²: 3, 4, 5, 6 → linear: n + 2. nth term = 2n² + n + 2
A frustum is formed by removing a small cone of height 6 cm from a large cone of height 18 cm. The radius of the base of the large cone is 12 cm. (a) Show that the radius of the top of the frustum is 4 cm. (b) Calculate the volume of the frustum in terms of π. [Volume of cone = ⅓πr²h]
Acceptable Answer
(a) By similar triangles: r/12 = 6/18 = 1/3, so r = 4 cm ∎ (b) Large cone: V = ⅓π(12²)(18) = ⅓π(144)(18) = 864π. Small cone: V = ⅓π(4²)(6) = ⅓π(16)(6) = 32π. Frustum = 864π − 32π = 832π cm³
Prove algebraically that (n + 3)² − (n − 2)² is always an odd multiple of 5 for all positive integers n.
Acceptable Answer
(n+3)² − (n−2)² = (n² + 6n + 9) − (n² − 4n + 4) = 10n + 5 = 5(2n + 1). Since (2n + 1) is always odd for integer n, the expression is always 5 × an odd number, i.e., an odd multiple of 5. ∎
A bag contains n counters. 4 are red and the rest are blue. Tom takes two counters at random without replacement. The probability that both are red is 2/15. (a) Show that n² − n − 90 = 0. (b) Find the number of blue counters.
Acceptable Answer
(a) P(both red) = 4/n × 3/(n−1) = 12/(n(n−1)) = 2/15. Cross multiply: 12 × 15 = 2 × n(n−1) → 180 = 2n² − 2n → n² − n − 90 = 0 ∎ (b) (n−10)(n+9) = 0. n = 10 (reject −9). Blue = 10 − 4 = 6 counters.
The line L passes through (−1, 5) and is perpendicular to y = ²⁄₃x + 1. (a) Find the equation of L. (b) L intersects the x-axis at point P. Find the coordinates of P. (c) Calculate the area of the triangle formed by L, the line y = ²⁄₃x + 1, and the x-axis.
Acceptable Answer
(a) Perpendicular gradient = −3/2. y − 5 = −3/2(x + 1) → y = −3/2 x + 7/2 (b) At y = 0: 3/2 x = 7/2 → x = 7/3. P = (7/3, 0) (c) Intersection of lines: −3/2 x + 7/2 = 2/3 x + 1 → −9x + 21 = 4x + 6 → 13x = 15 → x = 15/13, y = 23/13. x-intercept of y = 2/3 x + 1: x = −3/2. Base = 7/3 − (−3/2) = 23/6. Height = 23/13. Area = ½ × 23/6 × 23/13 = 529/156 ≈ 3.39 square units
The graph of y = f(x) passes through the point (4, 7). Write down the coordinates of the corresponding point on the graph of: (a) y = f(x) + 3 (b) y = f(x − 2) (c) y = 2f(x) (d) y = f(−x)
Acceptable Answer
(a) (4, 10) (b) (6, 7) (c) (4, 14) (d) (−4, 7)