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LIU PO SHAN MEMORIAL COLLEGE (2010 – 2011) SECOND TERM TEST SECONDARY THREE MATHEMATICS MARKING SCHEME Date of Exam. : 29 – 3 – 2011 Time allowed : 1 hour Name : ________________ Class: ______ Class no.:_____ Marks: / 105 Instructions: 1. 2. 3. 4. 5. Answer ALL questions in this paper. Write your name, class and class number in the spaces provided on this cover. This paper must be answered in English. For Section A, the answers should be written in the boxes provided. For Section B and Section C, the solutions should be shown clearly in the spaces provided below the questions. 6. Calculators in the approved list can be used. 7. This paper consists of 13 pages. 1 Section A: Multiple Choice Questions (30 marks) Choose the correct answer and write the letter in the boxes provided. 1. 2. (4) 3 A. B. –64. 64. C. D. 1 . 64 C Which of the following must have x + y as a factor? I. x2 – y2 II. x2 + y2 III. x(x + y) – x – y A. B. C. D. 3. 1 . 64 I only II only I and III only II and III only C In a purse, there are three $2 coins and x $5 coins. The total value of the coins is not more than $26. Which of the following inequalities can be used to find the range of values of x? 4. A. 6 + 5x < 26 B. C. 6 + 5x 26 6x + 5 < 26 D. 6x + 5 26 B A coin is tossed 400 times and the results are as follows: Outcome Head Tail Frequency 228 172 The coin is tossed once again, find the probability of getting a tail. A. B. C. D. 0.64 0.43 0.34 0.27 B 2 5. 6. 7. 8. If the mode of 2, 2, 2, 4, 6 and x is x, find the mean of this set of data. A. 2 B. 8 3 C. D. 3 Cannot be found A sum of $14 000 is deposited at 4% p.a. for 5 years, compounded yearly. correct to the nearest dollar. A. B. C. $2 378 $2 800 $3 033 D. $3 034 Find the interest C In the figure, AB = AC, BF = CD and BD = CE. Which of the following must be true? A. B. C. AFE ~ DEF BDF CED DF // CE D. ABC B is equilateral In △ABC, AB BC and BD AC. Which of the following must be true? A. B. AB CD = BC BD AB BC = AD BD C. BC 2 AB CD AB AC BC BD D. 9. C A The orthocentre of a triangle is the point of intersection of the three A. B. C. D. perpendicular bisectors. medians. angle bisectors. altitudes. D 3 10. Which net can be folded into the 3-D object below? A. B. C. D. A 11. In the figure, ABCD is a parallelogram. A. B. C. D. If x1 = x2 , find the value of y. 100° 110° 120° 130° x1 x2 12. In the figure, ABCD is a square and ABE is an equilateral triangle. Find∠BEG. A. B. C. D. 40° 45° 55° 65° D DEG is a straight line. B 13. Referring to the figure, find y. A. B. C. D. 3 4 6 8 D 4 14. Referring to the front, top and side views of the given object, choose the corresponding objects on the given information. A. B. C. D. A 15. The angle between the planes BCD and ABC is A. B. C. D. DMA. DAM. ABD. ACD. A 5 Section B : Conventional Questions (70 marks) Solutions should be shown clearly in the spaces provided below the questions. 1. Factorize the following expressions. (a) 4h2 – 81k2 (b) 6x2 + 7xy – 5y2 (4 marks) 1M (a) 4h2 – 81k2 = (2h)2 – (9k)2 1A = (2h + 9k)(2h – 9k) (b) 6x2 + 7xy – 5y2 = (3x + 5y)(2x – y) 2. Simplify ( x 3 y 3 ) 1 ( x 3 y 3 ) 1 y 2 y2 = 1A and express the answer with positive indices. x 3 y 3 y 1A 1A 2 for (ab) a b 1A n n n (3 marks) or am a n a m n or a n = x 3 y 5 = y5 x 3. 1A 3 Express the following numbers in scientific notation. (a) 486 000 000 (b) 0.000 265 (2 marks) 4. (a) 486 000 000 = 4.86 108 1A (b) 0.000 265 = 2.65 10–4 1A In a class of 40 students, there are 22 boys. There are 10 boys and 8 girls wearing glasses. If one student is selected at random, find the probability that (a) the student is a girl, (b) the student does not wear glasses. (4 marks) (a) P(girl) = 40 22 40 18 9 = 20 40 1A for numerator 1A (b) P(does not wear glasses) = = 40 10 8 40 1A 11 20 1A 6 for numerator 1 an 5. The cumulative frequency polygon below shows the numbers of pages of the books on a bookcase. (a) How many books are there on the bookcase? (b) Find the median of number of pages of the books. (3 marks) 6. (a) There are 38 books 1A (b) median = 165 1M 1A (a) Solve the inequality 2x + 13 5x 1. (b) Find the greatest integer that satisfies the inequality in (a). (3 marks) (a) 2x + 13 5x 1 2x – 5x 1 – 13 –3x 14 x (b) 14 3 1M 1A The greatest integer that satisfies the inequality in (a) is 4. 7 1A 7. Draw the front, top and right views of the following object. 2A 2A Front view 8. (6 marks) 2A Top view Right view Refer to the following triangular prism ABCDEF. ADE is a right-angled triangle. (a) Name the angle between the planes CDEF and ABCD. (b) Name the angle between the planes ADE and ABFE. (c) Find the projection of the line segment DF on the plane ABCD. (d) Find the projection of the line segment DF on the plane ABFE. (4 marks) (a) ADE or BCF 1A (b) BAD 1A (c) BD 1A (d) AF 1A 8 9. The figure below shows a net which can be folded into a cube. If we can see that the numbers on the front and the top of the cube are 4 and 3 respectively, what would be the number shown on each of the following faces? right (a) The right face. (b) The left face. (c) The back face. (d) The bottom face. (4 marks) (a) 5 1A (b) 2 1A (c) 6 1A (d) 1 1A 10. The figure below shows the rhombus ABCD. Find the values of x and y. DA = DC D (property of rhombus) 4 cm 4=9–x y° A (property of rhombus) C 30° 1A DA = DC DAC = DCA = 30 y + DAC + DCA = 180 (base s, isos. ) ( sum of ) y = 180 – 2 30 y = 120 (9 – x ) cm 1A x=5 DAC = BAC = 30 (5 marks) 1A 9 1M 1A B 11. Referring to the figure below, find the values of p and q. (3 marks) C AE = EC (given) DE // BC (given) AD = DB p cm 3 cm 3 cm (intercept theorem) 1A q=2 B 1 DE = BC 2 2 cm D q cm 1A 1A =6 12. (a) Prove that PQRS is a parallelogram. (b) Find the values of x and y. PS = QR (3 marks) (3 marks) 6 6 48o PQRS is a parallelogram. Q (opp. sides equal) PSR = PQR = 48 S x y 1A (given) 10 P 1A (given) PQ = SR (b) 1.5 Acm (mid-point theorem) p=23 (a) E 1.5 cm 10 1A (property of // gram) 1A x = 48 QPS + PQR = 180 (int. s, PS // QR) 1M y + 48 = 180 1A y = 132 10 R 13. In the figure, DC // AB and AD = CD. (a) Find CAB. (b) Prove that ABC is isosceles. (a) DA = DC DAC = DCA (4 marks) (3 marks) (given) (base s, isos. ) DAC + DCA + 130 = 180 DCA = ( sum of ) 180 130 = 25 2 1M 1A CAB = DCA = 25 (alt. s, DC // AB) CBA = 180 – ACB – CAB (b) 1A 1A ( sum of ) 1A = 180 – 25 – 130 = 25 1A CBA = CAB CA = CB (sides opp., equal s) 1A ABC is isosceles. 14. In BDF as shown, AB = 18 cm, BC = 12 cm, CD = 15 cm, DE = 10 cm, AC = 15 cm and CE = 12.5 cm. Prove that (a) ABC ~ CDE, (b) FB = FD. (4 marks) (3 marks) (a) In ABC and CDE, AB 18 6 CD 15 5 BC 12 6 DE 10 5 1A 1A any two AC 15 6 CE 12.5 5 AB BC AC CD DE CE ABC ~ CDE (3 sides prop.) (b) ABC ~ CDE (proved) ABC = CDE (corr. s, ~s) 1 (sides opp., equal s) 1 FB = FD 1 1 11 1 15. In the figure, the diagonals of the quadrilateral and BO = OD, prove that ABCD intersect at O. (a) ABO CDO, (b) AB // DC, (c) ABCD is a parallelogram. (a) (4 marks) (2 marks) (3 marks) In ABO and CDO, OB = OD OAB = OCD = 90 1A (given) 1A (given) AOB = COD (vert. opp. s) 1A ABO CDO (AAS) 1 (b) OAB = OCD = 90 AB // DC (c) 1A (given) 1 (alt. s equal) AB // DC (proved) ABO CDO (proved) AB = CD If AB AC, DC CA 1A (corr. s, s) ABCD is a parallelogram. 1A (2 sides equal and //) 12 1 Section C : Bonus Questions (5 marks) Solutions should be shown clearly in the spaces provided below the questions. 1. Convert the decimal number 210 + 25 + 21 to a binary number. 1A 210 + 25 + 21 = 10000100010 2 2. Calculate (1 mark) 1116 112 and express the answer as a decimal number. 1116 112 = (116 + 1) – (12 + 1) 1A (2 marks) any one = 17 – 3 = 14 1A 3. Factorize x 3 ( x 2)3 . (2 marks) x 3 ( x 2)3 = [ x ( x 2)][ x 2 x( x 2) ( x 2) 2 ] 1M = 2( x 2 x 2 2 x x 2 4 x 4) = 2(3x 2 6 x 4) 1A - END OF PAPER - 13