Q1. Diagram NOT accurately drawn CEAY and BDAX are straight lines. XY, ED and CB are parallel. AE = 5 cm. AX = 9 cm. AD = 4 cm. BC = 4 cm. BD = 2 cm. CE = x cm. XY = y cm. Find the value of x and the value of y . x = ..................................... y = ..................................... (Total 4 marks) Page 1 of 23 Q2. Diagram NOT accurately drawn A cylinder has base radius x cm and height 2x cm. A cone has base radius x cm and height h cm. The volume of the cylinder and the volume of the cone are equal. Find h in terms of x . Give your answer in its simplest form. h = ............................. (Total 3 marks) Page 2 of 23 Q3. Diagram NOT accurately drawn The sketch shows a curve with equation y = kax where k and a are constants, and a > 0 The curve passes through the points (1, 7) and (3, 175). Calculate the value of k and the value of a . k = ................................ a = ................................ (Total 3 marks) Page 3 of 23 Q4. (a) Show that the equation can be rearranged to give 3x 2 + 7x – 13 = 0 (3) (b) Solve 3x 2 + 7x – 13 = 0 Give your solutions correct to 2 decimal places. x = ............................... or x = ................................ (3) (Total 6 marks) Q5. Katy drove for 238 miles, correct to the nearest mile. She used 27.3 litres of petrol, to the nearest tenth of a litre. Petrol consumption = Page 4 of 23 Work out the upper bound for the petrol consumption for Katy’s journey. Give your answer correct to 2 decimal places. ...................................... miles per litre (Total 3 marks) Q6. Simplify fully ................................................ (Total 3 marks) Page 5 of 23 Q7. Solve the simultaneous equations x2 + y2 = 5 y = 3x + 1 x = .................... y = ..................... or x = .................... y = ..................... (Total 6 marks) Page 6 of 23 The diagram shows a sketch of the curve y = sin x ° for 0 ≤ x ≤ 360 Q8. The exact value of sin 60° = (a) Write down the exact value of (i) sin 120°, .................... (ii) sin 240°. .................... (2) (b) On the grid below, sketch the graph of y = 4 sin 2x ° for 0 ≤ x ≤ 360 (2) (Total 4 marks) Page 7 of 23 Q9. Express the recurring decimal as a fraction. .................... (Total 3 marks) Q10. Prove that the difference between the squares of consecutive odd numbers is a multiple of 8 (Total 6 marks) Page 8 of 23 Q11. Diagram NOT accurately drawn OAB is a triangle. =a =b (a) Find the vector in terms of a and b. = ........................................ (1) P is the point on AB such that AP : PB = 3 : 2 (b) Show that (2a + 3b) (3) (Total 4 marks) Page 9 of 23 Prove that (3n + 1)2 – (3n –1)2 is a multiple of 4, for all positive integer values of n . Q12. (Total 3 marks) Q13. Fred and Jim pay Malcolm to do some gardening. Fred has £x . Jim has ten pounds less than Fred. Fred pays one third of his money to Malcolm. Jim pays half of his money to Malcolm. (a) Show that the amount that Malcolm is paid is . (1) Page 10 of 23 Malcolm is paid a total of £170. (b) Use algebra to show how much money Fred has left. ..................................... (4) (Total 5 marks) Q14. The time, T seconds, for a hot sphere to cool is proportional to the square root of the surface area, A m 2, of the sphere. When A = 100, T = 40. Find the value of T when A = 60. Page 11 of 23 Give your answer correct to 3 significant figures. ..................................... seconds (Total 4 marks) Page 12 of 23 Q15. Here is a rectangle. Diagram NOT accurately drawn a = 8.3 cm correct to 1 decimal place. b = 3.6 cm correct to 1 decimal place. (a) Calculate the upper bound of the area of this rectangle. Write down all the figures on your calculator. ..................................... cm2 (2) (b) Find the area of this rectangle correct to an appropriate number of significant figures. ..................................... cm2 (2) (Total 4 marks) Page 13 of 23 M1. Working Answer Mark x = 2.05 4 Additional Guidance M1 a correct expression for x involving ratios of sides, e.g. oe A1 cao y = 11.25 M1 oe OR A1 cao Total for Question: 4 marks M2. Working 2 πx (2x) = Answer Mark 6x 3 Additional Guidance M1 for a correct volume formula in terms of x , e.g. 2 π(x ) h πx2(2x) or π x 2h A1 for π(2x ) = πh or 3Πx2(2x) = πx2h or x2(2x) = x2h (or better) A1 for 6x cao Total for Question: 3 marks Page 14 of 23 M3. Working 7 = ka 1; 175 = ka 3 k = , 175 = , 175 = 7a a2 = 25, so a = 5, k = 1.4 Or 73 = k 3a 3, 175 = ka 3 k2 = 2 Answer Mark k = 1.4 a=5 3 Additional Guidance M1 either a 2 = 25 or 7 = ka (or 7 = ka 1) and 175 = ka 3 A1 k = 1.4 oe A1 a = 5 SC Either a = 5 or k = 1.4 oe gets B2 , k = 1.4, a = 5 Total for Question: 3 marks Page 15 of 23 M4. Working (a) 5(x – 1) = (4 – 3x )(x + 2) Answer Mark Additional Guidance Proof 3 M1 multiply through by (x – 1)(x + 2) and cancel correctly M1 expand 5(x – 1) and (4 – 3x )(x + 2) correctly, need not be simplified A1 rearrange to give required equation (dep on both Ms and fully correct algebra) 1.22 –3.55 3 M1 correct substitution in formula of a = 3, b = 7 and c = ±13 2 5x – 5 = 4x + 8 – 3x – 6x (= 8 – 2x – 3x 2) (3x 2 + 6x + 5x – 4x – 5 – 8 = 0) 3x 2 + 7x – 13 = 0 (b) a = 3, b = 7, c = – 13 x= M1 reduction to A1 1.215 to 1.22 and –3.55 to –3.555 = x = 1.2196… or – 3.55297…. Or M1 Or M1 A1 1.215 to 1.22 and –3.55 to –3.555 SC T&I 1 mark for 1 correct root, 3 marks for both correct roots x = 1.2196… or –3.55297…. Total for Question: 6 marks Page 16 of 23 M5. Working 238 has an UB 238.5, a LB of 237.5 27.3 has an UB of 27.35, a LB of 27.25 Upper: Answer Mark 8.75 3 Additional Guidance B1 for one of 238.5, 237.5, 27.35, 27.25, seen M1 for ‘UB no of miles’÷’LB no of litres’ Where 238 < ‘UB no of miles’≤238.5 and 27.25 ≤ 'LB no of litres’ < 27.3 A1 8.75 or 8.752 or 8.7522 or 8.7523 or better = 8.75229 SC 238.4 ÷ 27.25 which leads to 8.748…B1 M1 A0 Total for Question: 3 marks M6. Working Answer Mark 3 Additional Guidance B1 for (x – 3)(x – 5)or x (x – 5)-3(x – 5) M1 for (2x ± 3)(x ± 5) or 2x (x + 5) ± 3(x + 5) or 2x (x – 5) ± 3(x – 5) A1 for cao as final answer Total for Question: 3 marks Page 17 of 23 M7. Working y2 = (3x + 1)2 x2 + 9x2 + 6x + 1 = 5 10x 2 + 6x + 1 = 5 10x 2 + 6x – 4 = 0 2(5 x 2 + 3x – 2) = 0 2(5x – 2)(x + 1) = 0 Answer Mark Additional Guidance x = 0.4 y = 2.2 6 M1 for (3x + 1)2 seen or implied by sight of 9x 2 + 1 A1 for x 2 + 9x 2 + 6x + 1 = 5 or equivalent expanded form M1 (dep) for correct attempt to solve a 3-term quadratic equation (condone omission of = 0) A1 for x = 0.4, x = –1 M1 (dep on previous Ms) for sub one value of x into either equation A1 for y = 2.2, y = –2 (correctly paired with x values) [SC: B1 for one correct pair of solutions if M0 scored] x = –1 y = –2 Total for Question: 6 marks M8. Answer (a)(i) Mark 2 (ii) (b) Additional Guidance B1 cao B1 cao 2 B2 cao [B1 for sine curve,starting from the origin with amplitude 4, OR B1 cuts x axis at 90, 180, 270, 360 and starts from 0] Total for Question: 4 marks Page 18 of 23 M9. Working Answer x = 0.213131313… Mark Additional Guidance 3 M1 for 0.2131313…. or 0.2 + 0.0131313…. (dots MUST be included) M1 for two correct recurring decimals that, when subtracted, leave a terminating decimal 10x = 2.13131313… 1000x = 213.131313… 990x = 211 A1 for Total for Question: 3 marks M10. Working QWC (2n +1)2 – (2n –1)2 = ii, iii 4n2 + 4n + 1 – (4n2 – 4n + 1) = 8n Answer Mark Fully algebraic argument, set out in a logical and coherent manner 6 Additional Guidance B2 the nth term for consecutive odd numbers is 2n – 1 oe (B1 2n + k, k ≠ –1 or n = 2n – 1 or 2x – 1 B1 use of 2n + 1 and 2n – 1 oe M1 (2n +1)2 – (2n –1)2 M1 4n2 + 4n + 1 – (4n2 – 4n + 1) C1 conclusion based on correct algebra QWC: Conclusion should be stated, with correct supporting algebra. OR (2n +1)2 – (2n –1)2 = ((2n +1) – (2n –1))(2n +1 + 2n – 1) =2 × 4n = 8n OR B1 use of 2n + 1 and 2n –1 oe M1 (2n +1)2 – (2n –1)2 M1 ((2n +1) – (2n –1))(2n +1 + 2n –1) C1 conclusion based on correct algebra QWC: Conclusion should be stated, with correct supporting algebra. Total for Question: 6 marks Page 19 of 23 M11. Working Answer Mark (a) b–a 1 (b) proof 3 Additional Guidance B1 for b – a or – a + b oe M1 for oe M1 for × “(b – a)” oe or × “(a – b)” oe A1 for a + × (b – a) oe or b + × (a – b) oe leading to given answer with correct expansion of brackets seen Total for Question: 4 marks M12. Working (9n 2 + 6n + 1) – (9n 2 – 6n + 1) = 12n Answer 12n correct comment Mark 3 Additional Guidance M1 for (3n )2 + 3n + 3n + 1 or (3n )2 – 3n – 3n + 1 or ((3n + 1) – (3n – 1))((3n + 1) + (3n – 1)) A1 for 12n from correct expansion of both brackets A1 for 12n is a multiple of 4 or 12n = 3 × 4n or 12n = 4 × 3n or = 3n or = 4n NB: Trials using different values for n score no marks. Total for Question: 3 marks Page 20 of 23 M13. Working (a) Fred pays and Jim pays Answer Mark Additional Guidance Clear and coherent explanation 1 C1 a clear and coherent explanation £140 4 M1 multiply through by 6 and cancels fractions Malcolm gets £170 for Fred and Jim, so Malcolm gets + = 170 (b) Fred has using left, so solving for x M1 (dep)expand 3(x − 10) + = 170 2x + 3(x – 10) = 170 × 6 5x = 1050 M1 (dep)collect terms on each side correctly A1 cao x = 210 OR OR M1 collects terms over 6 M1(dep) expand 3(x − 10) M1(dep) multiply through by 6 and collect terms 5x = 1050 A1 cao x = 210 Total for Question: 5 marks Page 21 of 23 M14. Working Answer Mark 31.0 4 k=4 Additional Guidance M1 M1 A1 A1 for 30.98... or 31(.0) OR M2 for oe M1 for oe A1 for 30.98... or 31(.0) Total for Question: 4 marks M15. (a) Working Answer Mark UB 8.35 × 3.65 = 30.4775 30.4775 2 Additional Guidance M1 sight of 8.35 or 3.65 A1 30.7445 (b) LB 8.25 × 3.55 = 29.2875 30 2 M1 8.25 × 3.55 A1 30 (dep on 8.25 × 3.55) Total for Question: 4 marks Page 22 of 23 Page 23 of 23
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