Centre Number 71 Candidate Number General Certificate of Secondary Education 2006 Mathematics (Non-calculator) Higher Tier GMM41 Module M4 Paper 1 [GMM41] MONDAY 5 JUNE 1.30 pm – 2.30 pm TIME 1 hour. INSTRUCTIONS TO CANDIDATES Write your Centre Number and Candidate Number in the spaces provided at the top of this page. Write your answers in the spaces provided in this question paper. Answer all ten questions. Any working should be clearly shown in the spaces provided since marks may be awarded for partially correct solutions. You must not use a calculator for this paper For Examiner’s use only Question Number 1 2 3 4 INFORMATION FOR CANDIDATES 11 –– 6.1.06BP 6.1.06BP The total mark for this paper is 49. Figures in brackets printed down the right-hand side of pages indicate the marks awarded to each question or part question. You should have a ruler, compasses, set-square and protractor. The Formula Sheet is on page 2. 5 6 7 8 9 10 Total Marks GMM41S6 2344 Marks Formula Sheet Area of trapezium = 1–2 (a + b)h a h b Volume of prism = area of cross section × length Cross section length In any triangle ABC A Area of triangle = 1–2 ab sin C Sine rule : c b a b c = = sin A sin B sin C B a Cosine rule: a2 = b2 + c2 – 2bc cos A C r Volume of sphere = 4 – πr 3 3 Surface area of sphere = 4πr 2 Volume of cone = 1–3 πr 2h l h Curved surface area of cone = πrl r Quadratic equation: 1 – 6.1.06BP The solutions of ax2 + bx + c = 0, where a ≠ 0, are given by – b ± b 2 – 4 ac x= 2a GMM41S6 2344 2 [Turn over B 1 Examiner Only Marks Remark y C O Diagram not drawn accurately z 100° D x A E O is the centre of the circle. Find the sizes of the angles (a) x Answer _____________° [1] (b) y Answer _____________° [1] (c) z 1 – 6.1.06BP Answer _____________° [1] GMM41S6 2344 3 [Turn over 2 In a closing down sale, all prices were reduced by 70%. The sale price of a camera was £48. What was the original price before the sale? Examiner Only Marks Remark Answer £ _____________ [3] 3 Evaluate (a) 10–3 Answer _____________ [1] (b) 60 1 – 6.1.06BP Answer _____________ [1] GMM41S6 2344 4 [Turn over (a) Solve x2 + 14x – 15 = 0 4 Examiner Only Marks Remark Answer x = _____________ [3] (b) On the grid below, illustrate the inequalities y ⬍ 8, y ⬍ 3x + 4, y ⬎ x + 1, x⬎0 Mark with R the region satisfying all four inequalities. [4] y 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 x 1 – 6.1.06BP 0 GMM41S6 2344 5 [Turn over 5 Evaluate Examiner Only Marks (a) Remark 1– 36 2 Answer _________ [1] (b) 320.2 Answer _________ [1] 3– (c) 164 Answer _________ [2] – 2– 3 (d) 125 1 – 6.1.06BP Answer _________ [2] GMM41S6 2344 6 [Turn over 6 The two triangles shown are similar. Examiner Only Marks Remark Diagram not drawn accurately 7.5 cm xcm 9cm 12cm (a) Calculate the length of the side marked x. Answer _____________cm [2] (b) Given that the area of the small triangle is 22.5 cm2, calculate the shaded area between the two triangles. 1 – 6.1.06BP Answer _____________cm2 [3] GMM41S6 2344 7 [Turn over 7 The table gives information about the heights of plants in a crop test. Examiner Only Marks Height (h cm) Frequency Remark 0 ⬍ h ⭐ 6 6 ⬍ h ⭐ 8 8 ⬍ h ⭐ 10 10 ⬍ h ⭐ 15 15 ⬍ h ⭐ 20 120 320 260 50 [3] 1 – 6.1.06BP (a) Show this information on a histogram. 150 GMM41S6 2344 8 [Turn over (b) The plants are arranged in order of height and every tenth one is selected to form a sample. Explain why this is not a suitable way to select a sample. Examiner Only Marks Remark [2] (c) In a stratified sample of all plants over 9 cm in height, 20 up to 10 cm were selected and 8 over 15 cm were selected. How many plants in the original crop were over 9 cm in height? 1 – 6.1.06BP Answer _____________ [3] GMM41S6 2344 9 [Turn over 8 Factorise fully Examiner Only Marks Remark (a) 6cd – 9c2 Answer _________________ [2] (b) 3a2 – 27d 2 Answer _________________ [3] 9 Solve the equation 4 9 ——— + ——— = 3 (x – 3) (2x – 1) 1 – 6.1.06BP Show your working. Answer _____________ [7] GMM41S6 2344 10 [Turn over 10 Find the values of c and d for which Examiner Only Marks Remark y2 – 12y + c ≡ (y + d)2 1 – 6.1.06BP Answer c = ___________ d = ___________ [3] GMM41S6 2344 11 [Turn over 1 – 6.1.06BP S – 1/06 – 3200 – 302507(133) [Turn over Centre Number 71 Candidate Number General Certificate of Secondary Education 2006 Mathematics (With calculator) Higher Tier GMM42 Module M4 Paper 2 [GMM42] MONDAY 5 JUNE 2.45 pm – 3.45 pm TIME 1 hour. INSTRUCTIONS TO CANDIDAT ES Write your Centre Number and Candidate Number in the spaces provided at the top of this page. Write your answers in the spaces provided in this question paper. Answer all eleven questions. Any working should be clearly shown in the spaces provided since marks may be awarded for partially correct solutions. For Examiner’s use only Question Number 1 2 3 4 5 INFORMATION FOR CANDIDATES The total mark for this paper is 49. Figures in brackets printed down the right-hand side of pages indicate the marks awarded to each question or part question. You should have a calculator, ruler, compasses, set-square and protractor. The Formula Sheet is on page 2. 6 7 8 9 10 11 Total Marks GMM42S6 2345 Marks Formula Sheet Area of trapezium = 1–2 (a + b)h a h b Volume of prism = area of cross section × length Cross section length In any triangle ABC A Area of triangle = 1–2 ab sin C Sine rule : c b a b c = = sin A sin B sin C B a Cosine rule: a2 = b2 + c2 – 2bc cos A C r Volume of sphere = 4 – πr 3 3 Surface area of sphere = 4πr 2 Volume of cone = 1–3 πr 2h l h Curved surface area of cone = πrl r Quadratic equation: The solutions of ax2 + bx + c = 0, where a ≠ 0, are given by x= – b ± b 2 – 4 ac 2a GMM42S6 2345 2 [Turn over 1 The cumulative frequency graph gives information about the percentage marks obtained by 300 candidates in an examination. Examiner Only Marks Remark Use the graph to estimate (a) the median percentage mark, Answer _____________ [1] (b) the interquartile range, Answer _____________ [2] (c) the percentage mark separating the top 20 candidates from the others. Answer _____________ [2] 300 250 Cumulative frequency 200 150 100 50 0 10 20 30 40 50 60 70 80 % marks (less than) GMM42S6 2345 3 [Turn over 90 2 (a) A is (0, 4). B is (3, 13) Find the equation of the line AB. Examiner Only Marks Remark Answer _________________ [3] (b) Find the equation of the line parallel to the line AB and passing through (0, 8). Answer _________________ [2] (c) Find the equation of the line perpendicular to the line AB and passing through (9, 0). Answer _________________ [2] GMM42S6 2345 4 [Turn over Q 3 Examiner Only Marks Diagram not drawn accurately 8cm P Remark R 11cm In the triangle PQR, angle PQR is 90°, PR = 11 cm and PQ = 8 cm. Calculate the size of the angle QPR. Answer _________________° [3] 4 Calculate the volume of a sphere of radius 12 cm. Answer _____________ [3] 5 Solve the simultaneous equations 6x + 7y = 1 2x – y = 7 Show your working. A solution by trial and improvement will not be accepted. Answer x = ___________ y = ___________ [3] GMM42S6 2345 5 [Turn over D 6 Examiner Only G Marks Remark 7 cm E F C B 8 cm O 10 cm A (a) Calculate the length of the space-diagonal OG in the cuboid OABCDEFG. Answer _____________ cm [2] (b) Calculate the size of the angle GOA between the lines OA and OG. Answer _____________° [2] 7 Prove that opposite angles of a cyclic quadrilateral add up to 180°. GMM42S6 2345 6 [3] [Turn over 8 The area of a rectangle is 6 cm2 and the perimeter is 20 cm. Examiner Only Marks Remark (a) If x is the length of one side of the rectangle, show that x2 – 10x + 6 = 0 [3] (b) Find the lengths of the sides of the rectangle. You may give your answers correct to two decimal places. Answer ______________________ cm [3] GMM42S6 2345 7 [Turn over 9 y 1 0 90 180 270 360 x –1 y = cosx Use the graph to find, for 0° ⬍ x ⬍ 360°, the solutions of Examiner Only Marks Remark (a) cos x = – 0.45 Answer __________________________ [2] (b) 8 cos x = 5 Answer __________________________ [2] GMM42S6 2345 8 [Turn over Start 10 5km Bill 3 km Examiner Only Marks Remark Diagram not drawn accurately 3k m Ben Bill and Ben are hiking. They hike together 5 km due East. They then split up. Bill continues 3 km due East but Ben travels 3 km in a straight line in a different direction. When they have both stopped Ben is 1 km nearer the starting point than Bill. Calculate Ben’s bearing from the starting point. A solution by scale drawing will not be accepted. Answer _____________ [4] 11 Solve simultaneously the equations x2 + 4y2 = 10 2y + 3x = 10 A solution by trial and improvement will not be accepted. Show your working. Answer x = ____________ y = __________ [7] GMM42S6 2345 9 [Turn over THIS IS THE END OF THE QUESTION PAPER GMM42S6 2345 10 [Turn over S – 1/06 – 3200 – 302507(134) [Turn over
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