Pearson Edexcel International GCSE Mathematics A Paper 1HR Higher Tier May 2024 Study Notes

General Examination Information
  • Total Marks: 100.

  • Time Duration: 2 hours.

  • Paper Reference: 4MA1/1HR (Higher Tier).

  • Date: Thursday 16 May 2024.

  • Required Tools: Ruler graduated in centimeters and millimeters, protractor, pair of compasses, pen, HB pencil, eraser, calculator, and optional tracing paper.

  • Core Instructions:     * Show all stages of working; correct answers without sufficient working may be awarded no marks.     * Calculators are permitted.     * Nothing should be written on the formulae page; such markings will not gain credit.

Mathematical Formulae - Higher Tier
Arithmetic Series
  • Sum to nn terms: Sn=n2[2a+(n1)d]S_n = \frac{n}{2} [2a + (n - 1)d]

Algebra
  • Area of a Trapezium: 12(a+b)h\frac{1}{2}(a+b)h

  • The Quadratic Equation: For ax2+bx+c=0ax^2 + bx + c = 0 where a0a \neq 0, the solutions are: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Trigonometry and Geometry
  • Sine Rule: asin(A)=bsin(B)=csin(C)\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}

  • Cosine Rule: a2=b2+c22bccos(A)a^2 = b^2 + c^2 - 2bc\cos(A)

  • Area of a Triangle: 12absin(C)\frac{1}{2}ab\sin(C)

Mensuration
  • Volume of a Cone: 13πr2h\frac{1}{3}\pi r^2 h

  • Curved Surface Area of a Cone: πrl\pi rl

  • Volume of a Prism: area of cross section×length\text{area of cross section} \times \text{length}

  • Volume of a Cylinder: πr2h\pi r^2 h

  • Curved Surface Area of a Cylinder: 2πrh2\pi rh

  • Volume of a Sphere: 43πr3\frac{4}{3}\pi r^3

  • Surface Area of a Sphere: 4πr24\pi r^2

Number and Arithmetic Properties
  • The Mean Calculation (Problem 1): Five cards have numbers 16,15,3,2,916, 15, 3, 2, 9. To find the sixth card so the mean of all six is 1111, calculate the total sum: 11×6=6611 \times 6 = 66. The sum of the known cards is 16+15+3+2+9=4516 + 15 + 3 + 2 + 9 = 45. The missing card is 6645=2166 - 45 = 21.

  • Mixed Fractions (Problem 4): Requirement to show that 213×514=12142\frac{1}{3} \times 5\frac{1}{4} = 12\frac{1}{4}.

  • Compound Interest (Problem 5): Slavomir invests 5200 euros5200\text{ euros} for 4 years4\text{ years} at 2.5%2.5\% interest per year. The formula used is 5200×(1.025)45200 \times (1.025)^4. The final amount must be rounded to the nearest euro.

  • Highest Common Factors and Prime Factors (Problem 11):     * Given A=25×57×75A = 2^5 \times 5^7 \times 7^5 and B=23×53×74B = 2^3 \times 5^3 \times 7^4.     * Evaluate the HCF of 5A5A and 2B2B.     * Calculate the value of AB2AB^2 as a product of prime factors.

Algebra and Functions
  • Simplification and Expansion (Problem 7):     * g9÷g2=g7g^9 \div g^2 = g^7     * Expand 5k(4k2+3)=20k3+15k5k(4k^2 + 3) = 20k^3 + 15k

  • Quadratic Equations:     * Factorize x22x63x^2 - 2x - 63. Finding factors of 63-63 that sum to 2-2: (x9)(x+7)(x - 9)(x + 7).     * Solve x22x63=0x^2 - 2x - 63 = 0 to find x=9x = 9 or x=7x = -7.

  • Inequalities: Solve 7 - 2y < 3(y - 1).

  • Linear Equations (Problem 9): Determine the equation of Line L from a grid in the form y=mx+cy = mx + c.

  • Simultaneous Equations (Problem 12):     1. 4x+3y=9.64x + 3y = 9.6     2. 6x+5y=16.86x + 5y = 16.8

  • Algebraic Manipulation (Problem 14):     * Expand and simplify (x4)(x+3)(x+1)(x-4)(x+3)(x+1).     * Simplify algebraically: (a3b9)13(a^3 b^9)^{\frac{1}{3}}.

  • Index Laws and Surds (Problem 17):     * Solve k4×(k12)5=knk^4 \times (k^{12})^5 = k^n to find nn.     * Express 723\frac{7}{2 - \sqrt{3}} in the form c+d3c + d\sqrt{3} by rationalizing the denominator.

  • Completing the Square (Problem 20):     * Express 2x211x+92x^2 - 11x + 9 in the form a(xb)2+ca(x - b)^2 + c.     * Find the minimum point PP of the curve y=(2x211x+9)2y = (2x^2 - 11x + 9)^2.

Geometry and Trigonometry
  • Cylinder Mensuration (Problem 6):     * Volume V=1208cm3V = 1208\,cm^3, Radius r=8cmr = 8\,cm. Calculate height hh.     * Given density 1.25g/cm31.25\,g/cm^3, calculate the mass of the cylinder in kilograms (kgkg).

  • Trapezium Perimeter (Problem 8):     * Trapezium ABCDABCD with right angles at AA and DD. AD=15cmAD = 15\,cm, DC=14cmDC = 14\,cm. Area is 360cm2360\,cm^2. Work out the perimeter by first finding the length of ABAB and then the slanted side BCBC.

  • Circle Theorems (Problem 13): A, B, C, and D are points on a circle with centre O. Given angle BCD=128BCD = 128^{\circ}, determine angle OBDOBD. Reason: Opposite angles in a cyclic quadrilateral sum to 180180^{\circ}; angle at the center is twice the angle at the circumference.

  • Isosceles Triangle Area (Problem 15): Triangle EFGEFG where EF=GFEF = GF and EFG=130\angle EFG = 130^{\circ} . Given Area = 74cm274\,cm^2, find length EFEF using Area=12absin(C)\text{Area} = \frac{1}{2} ab \sin(C).

  • Similarity (Problem 18): Vases A and B have heights of 30cm30\,cm and 12cm12\,cm respectively. Area scale factor is the square of the linear scale factor. Given the difference in surface area is 178.5cm2178.5\,cm^2, calculate the surface area of Vase A.

  • 3D Geometry (Problem 22): Cuboid with dimensions 8×12×20cm8 \times 12 \times 20\,cm. Calculate angle BMPBMP where MM is the midpoint of base ADEHADEH and PP is the midpoint of edge CFCF.

Statistics and Probability
  • Biased Spinner (Problem 2):     * Probabilities for numbers 1 to 5: 2x,0.27,0.04,x,0.122x, 0.27, 0.04, x, 0.12.     * Total probability sum: 2x+0.27+0.04+x+0.12=12x + 0.27 + 0.04 + x + 0.12 = 1.     * Calculate xx and then use it to estimate landing on an odd number (1, 3, or 5) over 400400 spins.

  • Ratio and Profit (Problem 3):     * Total loaves: 200200. White to Brown ratio = 3:23:2.     * Profit on White: 40%40\% of £1.50\pounds 1.50. Profit on Brown: 60%60\% of £1.75\pounds 1.75.     * Calculate total profit for all loaves.

  • Interquartile Range (Problem 10): Data set of goals: 0,1,2,2,3,4,4,6,7,9,110, 1, 2, 2, 3, 4, 4, 6, 7, 9, 11. Identify the median, lower quartile (Q1Q_1), and upper quartile (Q3Q_3) to find IQR=Q3Q1IQR = Q_3 - Q_1.

  • Histogram Construction (Problem 16):     * Trees of heights hh metered. Frequency density is calculated as Frequency÷Class Width\text{Frequency} \div \text{Class Width}.     * Intervals: 020-2, 252-5, 5105-10, 102010-20, 203520-35.

  • Probability and Algebra (Problem 21):     * Bag contains 2525 counters: 66 blue, xx orange (x > 9), and remainder are pink (256x=19x25 - 6 - x = 19 - x).     * Probability of picking one orange and one pink counter is 2275\frac{22}{75}.     * Solve algebraically for xx and determine the probability of picking two pink counters.

Advanced Algebra and Calculus
  • Stationary Points (Problem 19):     * Curve CC: y=x33x2212x+5y = \frac{x^3}{3} - \frac{x^2}{2} - 12x + 5.     * Find the stationary points by finding dydx\frac{dy}{dx} and setting it to zero (x2x12=0x^2 - x - 12 = 0).     * Solve for xx coordinates of points A and B. Identify A where x_A > x_B.

  • Arithmetic Sequences (Problem 23):     * Sequence terms: (4x1)(4x - 1), (x+14)(x + 14), (7x+9)(7x + 9). (Note: Transcript suggests terms related to xx).     * Calculate the value of xx based on common difference dd, then find the sum of the first 4040 terms using SnS_n.