End of Year Examination Paper 2 Instruction to andidates: Marks Obtained 1. Answer all questions. 2. Write your answers and working in the spaces provided. 3. Omission of essential working will result in loss of marks. 50 4. alculators may be used in this paper. 5. If the degree of accuracy is not specified in the question, and if the answer is not exact, give the answer correct to three significant figures. Give answers in degrees correct to one decimal place. Duration: 1h 30 min 1 (a) A worker received a salary increase of %. If his new salary is $1605, find his original salary. (b) The original price of a table is $56. Due to inflation, the price increased the following month to become $610. Find the percentage increase, giving your answer correct to 4 significant figures. Ans: (a) $ [2] (b) % [2] 2 (a) Simplify 4x x + 3 x 2 x x 2 + 2x 3. (b) Express 2 x + 4 as a single fraction in its simplest form. Ans: (a) [3] (b) [2]
3 The histogram below shows the average number of books each student puts in his/her school bag everyday. Number of students 16 14 12 10 8 6 4 2 1 2 3 4 5 6 Number of books (a) What is the fraction of students who carry an average of two books to school daily? (b) Find the angle of the sector on a pie chart representing five books. Ans: (a) [1] (b) º [2] 4 In the figure below, AB is similar to DEA. Given that AB = 3 cm, DE = 10 cm, A = 4 cm and BA = ADE = 90º. Find A 3 B D (a) the length of AD, 4 (b) the area of the trapezium ADE, (c) the length of AE. 10 E Ans: (a) AD = cm [2] (b) cm 2 [1] (c) AE = cm [2]
5 In the diagram, AB is similar to DE. AB = 6 cm, B = cm, A = 5 cm, D = (x + 1) cm and E = 2x cm. Find (a) the value of x, (b) the length DE. B D x + 1 6 A 5 2x E Ans: (a) x = [3] (b) DE = cm [2] 6 The stem-and-leaf diagram shows the weight of 26 gold coins during a quality check. Stem Leaf 25 8 9 26 0 0 0 1 1 3 5 8 8 9 9 2 0 1 1 1 2 2 6 Key: 25 4 represents 25.4 grams Find (a) the ratio of the number of coins weighing less than 26 g to those weighing more than 2 g, (b) the modal weight, (c) the median weight. Ans: (a) : [2] (b) grams [1] (c) grams [1]
1 A map has a scale of 20 000. (a) Express the scale of the map in cm : km. (b) Find the length of a highway on the map, if its actual length is 4 km long. (c) Find the actual area of a field, if its area on the map is 10 cm 2. Ans: (a) : [1] (b) cm 2 [2] (c) km 2 [2] 8 In the diagram, triangle AB is a reduction of triangle PQR. The dimensions given are in cm. (a) Show that AB PQ = 1 3. (b) Hence, form an equation in terms of x and solve it. (c) Find the length of PQ. P A 3x 6 reduction x 2 Q 24 R B x + 3 Ans: (a) on answer space [1] (b) x = [2] (c) PQ = cm [1]
9 James and Pat took an examination and scored x marks and y marks respectively. They scored a total of 145 marks. If Pat scored 15 more marks than James, (a) form a pair of simultaneous equations in x and y, (b) find Pat s score. Ans: (a) [2] (b) [2] 10 The diagram shows a solid cylinder inscribed within a spherical ball of radius 13 cm and centre A. The base radius of the cylinder is 5 cm with centre B. alculate the (a) volume of the sphere, (b) height of the cylinder, (c) volume of empty space within the sphere not occupied by the cylinder. B 5 A 13 Ans: (a) cm 3 [1] (b) cm [2] (c) cm 3 [1]
11 (a) Using a pair of compass, construct a quadrilateral ABD in which AB = 8 cm, B = cm, D = 6 cm, AD = 5 cm and BAD = 80º. (b) onstruct the (i) perpendicular bisector of AB, l 1, (ii) angle bisector of AB, l 2. (c) Find, by measurement, the area of the triangle formed by AB, l 1 and l 2. Ans: (a) on answer space [3] (b) on answer space [2] (c) cm 2 [2]
Solutions to: End of Year Examination Paper 2 1. (a) Original salary = $1605 10 100 = $1500 610 56 (b) Percentage increase = 56 100% 5.9028 = 5.903% (4 s.f.) 2. (a) 4x x + 3 x2 x x 2 + 2x 3 = 4x x + 3 + 2x 3 x2 x 2 x = 4x + 3)(x 1) x + 3 (x x(x 1) = 4 (b) 2_ x + 4 2 (x + 4) = = 2 x 4 = x 2 or x + 2 3. (a) Total number of students = 4 + 10 + 14 + + 8 + 2 = 45 Required fraction = 10 45 = 2_ 9 (b) Angle of sector = 8 45 360º = 64º 4. (a) Since AB is similar to DEA, A AD = AB DE corr. sides are proportional 4 AD = 3 10 3AD = 40 AD = 13 1_ 3 cm (b) Note that A // DE. Area of trapezium ADE = 1_ 2 (A + DE) AD AD = height of trapezium = 1_ 2 (4 + 10) 13 1_ 3 = 93 1_ 3 cm2 (c) Using the Pythagoras Theorem on ADE, AE 2 = AD 2 + DE 2 = ( 13 1_ 3 ) 2 + 10 2 = 2500 9 S AE = 2500 9 = 50 3 = 16 2_ 3 cm 5. (a) Since AB is similar to DE, B E = A D corr. sides are proportional 2x = 5 x + 1 (x + 1) = 5(2x) x + = 10x x 10x = 3x = (b) x = _ 3 = 2 1_ 3 AB DE = A D 6 DE = 5 2 1_ 3 5DE = 6 ( 2 1_ 3 + 1 ) 5DE = 20 DE = 4 cm 6. (a) Required ratio = 6 : 8 = 3 : 4 (b) Modal weight = 25. grams corr. sides are proportional + 1 D = x + 1 (c) Median = ( 26 + 1 2 ) th number cross-multiplication = 13.5 th number mean of 13 th and 14 th values 26.5 + 26.8 = 2 = 26.65 grams. (a) Scale = 1 cm : 20 000 cm R.F. = 1 20 000 = 1 cm : 0.2 km 1 km = 100 000 cm (b) Map length of highway = 4 0.2 cm = 20 cm (c) Area scale = 1 2 cm 2 : 0.2 2 km 2 = 1 cm 2 : 0.04 km 2 Actual area of field = 10 0.04 km 2 = 0.4 km 2
8. (a) AB PQ = x 2 3x 6 = x 2 3(x 2) = 1_ 3 (shown) (b) 1_ 3 = x + 3 24 B QR = 1_ 3 24 = 3(x + 3) cross-multiplication 24 = 3x + 9 3x = 15 x = 5 (c) PQ = 3x 6 = 3(5) 6 subst. x = 5 = 9 cm 9. (a) x + y = 145 y = x + 15 (b) Substitute into, x + (x + 15) = 145 2x + 15 = 145 2x = 130 x = 65 Substitute x = 65 into, y = 65 + 15 = 80 Pat scored 80 marks. 10. (a) Volume of sphere = 4_ 3 ϖ(133 ) = 2929 1 3 ϖ 9202. = 9200 cm 3 (3 s.f.) (b) Since AB is a right-angled triangle, by the Pythagoras Theorem, AB 2 = 13 2 5 2 = 144 AB = ± 144 = 12 cm or 12 cm (rej.) Height of cylinder = 2 12 cm by symmetry = 24 cm (c) Volume of empty space = 2929 1 3 ϖ ϖ(52 )(24) 31.816 = 320 cm 3 (3 s.f.) Step 5: Draw an arc, a 2, of radius 6 cm from point D. Step 6: Draw an arc, a 3, of radius cm from point B. Step : Label the intersection a 2 and a 3 as. Step 8: onnect A, B, and D to form the required quadrilateral. a 1 5 cm D r 1 6 cm P 80º M A 8 cm B cm (b) See diagram for the construction of l 1 and l 2. (c) Label the intersection of l 1 and l 2 as P. l 1 The triangle formed by AB, l 1 and l 2 is PMB. By measurement, PM = 3.2 cm (± 0.1 cm) and MB = 4 cm (± 0.1 cm) Area of PMB = 1_ 2 PM MB PMB = 90º a 2 = 1_ 2 3.2 4 = 6.4 cm l 2 a 3 11. (a) onstruction Steps Step 1: Draw line segment AB, 8 cm long. Step 2: Measure BAD = 80º and draw a ray, r 1, through D (dotted line). Step 3: Draw an arc, a 1, of radius 5 cm from point A. Step 4: Label the intersection of a 1 and r 1 as D. S