F3 MATH OPENER TERM 3 2026 MS

FORM 3 TERM 3 OPENER EXAM 2026

MATHEMATICS

MARKING SCHEME

SECTION I — 50 MARKS

1. 3 1/2 − 1 3/4 ÷ 7/8 = 7/2 − 7/4 × 8/7 = 7/2 − 2 = 3/2. Award M1 for correct conversion/division, A1 for the subtraction, A1 for 3/2. [3 marks]

2. Profit = 12.5/100 × 4,800 = 600. Selling price = 4,800 + 600 = KSh 5,400. Award M1, A1, A1. [3 marks]

3. (3x²−12x)/(x²−16)=3x(x−4)/[(x−4)(x+4)]=3x/(x+4), x≠4. Award M1 factorisation, A1 cancellation, A1 final form. [3 marks]

4. 2x−5≤3x+7 gives x≥−12. Also 3x+7<5x+1 gives x>3. Therefore x>3. Award M1, A1, A1. [3 marks]

5. m=(−5−3)/(4−(−2))=−8/6=−4/3. Using A: y−3=−4/3(x+2), hence y=−4x/3+1/3. Award M1 gradient, A1 gradient, A1 substitution, A1 equation. [4 marks]

6. 0.0000725=7.25×10⁻⁵. √(7.25×10⁻⁵)=√7.25×10⁻²·⁵≈8.515×10⁻³. Award marks for correct standard form and correct square-root process. [3 marks]

7. P=kQ². 72=36k, so k=2. When Q=10, P=2(10²)=200. Award M1, A1, A1. [3 marks]

8. From 4x−y=13, y=4x−13. Substitute: 2x+3(4x−13)=17 →14x=56 →x=4. Then y=4(4)−13=3. Award M1 substitution, A1 x, A1 y. [3 marks]

9. Area=90/360×π×14²=49π cm². Award B1 formula/substitution, A1 simplification, A1 answer. [3 marks]

10. AB=6 cm, AC=8 cm. BC²=6²+8²=100; BC=10 cm. Award M1 Pythagoras, A1 substitution, B1 answer. [3 marks]

11. Total=5+3+2=10. Neither red nor green means blue: P=3/10. Award B1 total, A1 favourable outcome, A1 probability. [3 marks]

12. (2x−3)(x+4)=2x²+5x−12; (x−2)²=x²−4x+4. Difference =2x²+5x−12−x²+4x−4=x²+9x−16. Award M1, A1, A1. [3 marks]