Suggested Solutions TAK SUN SECONDARY SCHOOL MOCK EXAMINATION ONE MATHEMATICS COMPULSORY PART PAPER 1 FORM 5 THIS PAPER MUST BE ANSWERED IN ENGLISH INSTRUCTIONS 1 Write your Name, Class and Class Number in the spaces provided on this cover. This paper consists of THREE sections, A(1), A() and B. Each section carries 35 marks. 3 Answer ALL questions in this paper. Write your answers in the spaces provided in this Question-Answer Book. Do not write in the margins. Answers written in the margins will not be marked. 4 Graph paper and supplementary answer sheets will be supplied on request. Write your Name, Class and Class Number on each sheet and put them inside this book. 5 Unless otherwise specified, all working must be clearly shown. 6 Unless otherwise specified, numerical answers should be either exact or correct to 3 significant figures. 7 The diagrams in this paper are not necessarily drawn to scale. Question No. 1 3 4 5 6 7 8 9 10 11 1 13 14 15 16 17 18 19 0 1 Total Marks
Section A(1) (35 marks) x 1. Simplify 3 xy 5 4 and express your answer with positive indices. (3 marks) x 5 4 xy 3 5 x 3 1 x y 5 3 x 1 y x. 1 y. Make x the subject of the formula x y 3 x xy 3 x. (3 marks) y 6 xy 3x x 3. Factorize (a) x 11x 4 y 6 xy x 3 xy y y 3 x. y (b) y 11y y 4 y (3 marks) (a) 11x 4 x 3x 8 x (b) y y 11y y 4 y 3 y y y 8 y y 3 y y 8 1 y 3 y y 4 y TSSS F5 MATH-CP1 MOCK EXAM 1 015-016 [ 1 ]
4. The marked price of a smartphone is $4000. It is then sold at a discount of 10% on its marked price and with another discount of $300. If the percentage profit is 10%, find the cost of the smartphone. (4 marks) The selling price is $ 4000 1 10% 300 $ 3300 + Let x be the cost of the shoes. Hence, 1 10% $ 3300 x x $3000. The cost of the shoes is $3000. x 1 5. Find the range of the values of x which satisfies both 5 x 3 x 1 5 x 3 x 1 3 5 x x 1 15 6x 8x 16 and x 5 x. x. x 5 x x 5 5 x. 5 Hence, x. (3 marks) 6. In a polar coordinate system, with pole O, the polar coordinates of the points A and B are o o 6, 66 and, 8 x, where 0 x 360, respectively. (a) If AOB is a right-angled triangle, find the possible values of x? (b) Write down the length of AB and the area of AOB. (3 marks) (a) o o o x 66 90 156 or o o o x 66 70 336. (b) Since AOB is a right-angled triangle, we have AB 6 8 10 units 8 6 Area 4 sq. units TSSS F5 MATH-CP1 MOCK EXAM 1 015-016 [ ]
7. The table shows the waiting times for bank service in a certain bank. Waiting time (min) 5 6 9 10 13 14 17 Frequency 36 5 14 5 Find the range, the mean and the standard deviation of the waiting times. (4 marks) Range = 17.5 min - 1.5 min = 16 min Mean = 6.9 min Standard Deviation = 36 3.5 6.9 57.5 6.9 14(11.5 6.9) 80 5(15.5 6.9) = 3.69 min (3 sig. fig.) 8. In a rectangular plane, P is a moving point such that P is always equidistant from the point A 1, 3 and the straight line y 4. (a) Find the equation of the locus of the point P. (b) Describe the locus of the point P. (4 marks) (a) The required equation is 1 y 3 y 4 x x x + x 1 y 3 y 4 x 1 y x y 6 0 1 6y 9 y 8y 16 y x x 6 (b) Parabola 9. The coordinates of the point A is, 3. B is the reflection image of A with respect to the x axis. C is the image of A after a clockwise rotation by o 90 about the origin O. (a) Write down the coordinates of B and C. TSSS F5 MATH-CP1 MOCK EXAM 1 015-016 [ 3 ]
(b) Let P be a moving point in the rectangular coordinate plane such that P is equidistant from the midpoint of BC and A. Find the equation of the locus of P. (a) The coordinates of B is (b) Mid-point of BC is So, y 3 5 5,, 3. The coordinates of C is, 5 5 x y. Let the coordinates of P be x, y. (4 marks) 3. + x x 4x 4 y 6y 9 x 5x 5 4 y 5y 1 x 11y 0 or x 1 0. 10. In Figure 1, O is the centre of the circle. Find the values of p and q. (4 marks) 5 4 Figure 1 o o p 180 60 ( at centre twice at ce ) o p 40. o o o o o 360 p 70 180 60 360 q o q 50. TSSS F5 MATH-CP1 MOCK EXAM 1 015-016 [ 4 ]
Section A() (35 marks) 11. The stem-and-leaf diagram shows the ages of all the twenty-six members of a reading club. Stem (tens) Leaf (units) 1 5 8 8 3 5 5 7 8 9 4 1 1 5 5 6 6 7 7 7 8 9 5 0 3 3 4 (a) Find the mode and the median of the ages of the twenty-six members. ( marks) (b) Four new members join the club. It is given that the mean age of the four new members is 4. Suppose that the ages of two of these four new members are 38 and 39. (i) Write down the mean ages of the thirty members. (ii) Is it possible that the median of the ages of the thirty members is the same as the median found in (a)? Explain your answer. (4 marks) (a) Mode == 47 Median =45 (b) (i) Mean = 4 (ii) Let a and b be the ages of the other two members. Note that a b 38 39 4. 4 Therefore, we have a + b = 91. If the two medians are the same, then we have a 45 and b 45. Hence, we have a + b 90. It is possible that the two medians are the same. TSSS F5 MATH-CP1 MOCK EXAM 1 015-016 [ 5 ]
1. Let $ C be the cost of printing of a book with n pages. It is given that C is partly constant and partly proportional to the square root of n. When n 100, C 80000 n 144, C 86000., and when (a) Find C in terms of n. (4 marks) (b) If the cost of printing is $95000, find the number of pages of that book. ( marks) (a) We have C 1 k k n, where k 1 and k are constants When n 100, C 80000. 80000 k1 k 100 (1) When n 144, C 86000. 86000 k1 k 144 () () (1): 6000 1 10 From (1): 80000 1 300010 Therefore, k k 3000. k k 50000. 1 C 50000 3000 n. (b) When C 95000, we have 95000 50000 3000 n 15 n n 5. Therefore, there are 5 pages in the book. TSSS F5 MATH-CP1 MOCK EXAM 1 015-016 [ 6 ]
13. Consider the equation of circle C : x y 6x 4y 1 0. (a) Find the coordinates of the centre O and the radius of the circle C. (3 marks) (b) The circle cuts the line L : x y 4 0 at two points A and B. Find the coordinates of A and B. (3 marks) (c) Hence, find the area of OAB. ( marks) (a) Centre of the circle C is 6 4, 3,. 6 3 Radius of the circle C is 1 9 4 1 5 units. + (b) L : x y 4 0 y x 4 Substitute L into C. We have x 6x 4x 41 0 x x 4 x x x 1 0 Hence, from L, we have y 7 or y. 8x 16 6x 4x 16 1 0 x 3 or x (or y) Therefore, the coordinates of A and B are 3, 7 and,. (c) 3, 7 is vertically below the centre of the circle with a distance of 5 units. Also,, is horizontally on the right of the centre with a distance of 5 units So AOB is a right angle. Area of OAB 5 5 5 sq. units TSSS F5 MATH-CP1 MOCK EXAM 1 015-016 [ 7 ]
14. In the figure, D is a point on the side BC of the acute-angled triangle ABC. It is given that the area of ABD is 770.4 cm. (a) Find ABD. ( marks) (b) Find the value of x and the area of ACD. (5 marks) (a) (b) Area of ABD= 770.4 = sin ABD = 1 AB BD sin ABD 1 45 36 sin ABD [] 770.4 810 ABD = 7.010 1 = 7.0 (correct to 3 sig. fig.) [] In ABC, by the cosine formula, x = 45 + 70 (45)(70) cos 7.010 1 x = 695 6 300cos 7.0101 [] = 70.6 (correct to 3 sig. fig.) [] Area of ABC = 1 AB BC sin ABC = 1 770. 45 70 8104 cm [] = 1 498 cm [] Area of ACD = Area of ABC Area of ABD = (1 498 770.4) cm = 77.6 cm [] TSSS F5 MATH-CP1 MOCK EXAM 1 015-016 [ 8 ]
15. In Figure, A, B, C and D are four points on the circle. PAQ is the tangent to the circleat A. It is given that CDA = 80, BAP = x and CD : DA = 7 : 13. Figure (a) Find DAQ. (3 marks) (b) (a) Given x = 40, show that (i) ABC is an isosceles triangle, ( marks) (ii) DA: AB = 13 : 8. ( marks) Join AC. CD : DA = 7 : 13 (given) CAD : ACD = 7 : 13 (arcs prop. to s) [] CAD = 13 7 ACD... (1) In ACD, 180 = ACD + CAD + ADC ( sum of ) 180 = ACD + 13 7 ACD + 80 (from (1)) ACD = 65 DAQ = 65 ( in alt. segment) [ + ] (b) (i) ABC = 180 80= 100 (opp. s, cyclic quad.) (ii) ACB = PAB = 40 ( in alt. segment) BAC = 180 100 40 = 40( sum of ) = ACB ABC is an isosceles triangle. [] Join BD. ADB= ACB = 40 (s in the same segment) ABD= ACD = 65 (s in the same segment) DA : AB = ABD : ADB = 65 : 40 = 13 : 8 [1] TSSS F5 MATH-CP1 MOCK EXAM 1 015-016 [ 9 ] []
Section B (35 marks) 16. A student estimates that his chances of passing Chinese, English, Mathematics and Liberal 4 9 Studies are,, 5 3 10 3 and 4 respectively. (a) What is the probability that he failed all subjects? ( marks) (b) What is the probability that he can pass at least three subjects? ( marks) (c) Given that he has passed at least three subjects, what is the probability that he passed all subjects? 4 9 3 P 5 3 10 4 (a) failed all subjects 1 1 1 1 (b) P Passed at least 3 subjects 4 1 5 3 1 600 9 10 (c) PPassed all Passed at least 3 subjects 4 9 1 4 1 3 5 3 10 4 5 3 10 4 3 1 9 3 4 9 3 4 5 3 10 4 5 3 10 4 ( marks) 79 100 P Passed all & at least 3 subjects PPassed at least 3 subjects 4 9 3 5 3 10 4 79 100 36. + 79 17. 8 boys and 6 girls pose for photo-taking in rows of 7. (a) Find the number of ways of posing if boys and girls pose alternatively. ( marks) (b) If 3 of the girls must be in the first row while the other 3 must be in the second row and also all girls cannot stand at the end, find the number of ways of posing. ( marks) (a) Number of ways P 9030400. + 6 8 6 P8 (b) Number of ways C 7 7 6 8 3 C3 P6 P8 903040000. + TSSS F5 MATH-CP1 MOCK EXAM 1 015-016 [ 10 ]
18. The figure shows a cube with side 6 cm. EG and FH meet at P. (a) Find the length of AH and AP. (b) Hence find the angle between the plane AFH and the base EFGH.. (a) Consider ADH, AH AD DH (Pyth. theorem) 6 6 cm 6 cm Since EG AH and P is the mid-point of EG, we have EP 1 6 cm AP 3 cm AE EP (Pyth. theorem) ( marks) (3 marks) 6 (3 ) cm 3 6 cm (b) APE is the angle between the plane AFH and the base EFGH. In APE, AE tan APE EP tan APE 3 6 tan APE APE 54.7 (cor. to 3 sig. fig.) The angle between the plane AFH and the base EFGH is 54.7. TSSS F5 MATH-CP1 MOCK EXAM 1 015-016 [ 11 ]
19. Find the range of the values of k such that the circle C : x y 4x 10y 4 0 and the straight line L : 4x 3y k 0 intersect. (5 marks) 4x k L : 4x 3y k 0 y. 3 4x k 4x k Substitute L into C, we have x 4x 10 4 0 3 3 x 16x 8xk k 9 40x 10k 4x 4 0 3 9x 16x 8xk k 36x 10x 30k 36 0 5x 8kx 156x k 30k 36 0 For intersection, we have 8 156 45 30 36 0 k k k 64k 496k 4336 100k 3000k 3600 0 36k 504k 0736 0 k 14k 576 0 18 k 3 0. The High Class airline with two types of airplanes, P-1 and P-, has contracted with a tour group to provide accommodation for at least 000 first-class, 1500 business-class and 400 economy-class passengers. Airplane P-1 costs $1000 per km to operate and can accommodate 40 first-class, 40 business-class and 10 economy-class passengers, whereas airplane P- costs $10000 per km to operate and can accommodate 80 first-class, 30 business-class and 40 economy-class passengers. Let x be the number of P-1 to be used and y be the number of P- used. (a) Write down all the constraints on x and y. ( marks) (b) Draw and shade the region that satisfies all the constraints. ( marks) (c) Find the number of each type of airplane that minimizes the operating cost. ( marks) TSSS F5 MATH-CP1 MOCK EXAM 1 015-016 [ 1 ]
甲 The information is summarized as follows: Accommodation P-1 (x) P- (y) Number of passengers (at least) First-class 40 80 000 Business-class 40 30 1500 Economy-class 10 40 400 The constraints are: (b) 40x 80y 000 40x 30y 1500 10x 40y 400 x, y are non negative integers x y 50 4x 3y 150 3x y 60 x, y are non negative integers any correct all correct any 1 lines correct all lines with region A A (c) C 1000x 10000y. From the graph, we minimum value of C occurs at 30, 10. Therefore, 30 P-1 and 10 P- are used to minimize the operating cost. TSSS F5 MATH-CP1 MOCK EXAM 1 015-016 [ 13 ]
1. In the figure, ABTC is a circle and ABT = 90. AT is the angle bisector of CAB. AC is produced to D so that DT is the tangent to the circle at T. It is given that AB = 8 and CD =. C D T A B (a) (i) Prove that ATC ATB. (ii) Find the length of TB. (6 marks) (b) A rectangular coordinate system is introduced in the figure so that the coordinates of A and B are (0, 0) and (8, 0) respectively. Find the equation of the tangent DT. (3 marks) (a) (i) ACT + ABT = 180 (opp. s, cyclic quad.) ACT + 90 = 180 ACT = 90 In ATC and ATB, CAT = BAT (angle bisector) ACT = ABT (proved) AT = AT (common side) ATC ATB (AAS) (a)(ii) ATC ATC Marking Scheme: Case 1 Any correct proof with correct reasons. 3 Case Any correct proof without reasons. Case 3 Incomplete proof with any one correct step and one correct reason. ATB TC = TB In CDT, CD tan CTD = = TC TB ATB TSSS F5 MATH-CP1 MOCK EXAM 1 015-016 [ 14 ] 1
In ATB, tan BAT = TB TB = AB 8 DT is a tangent, CTD = CAT i.e. CTD = BAT tan CTD = tan BAT TB = TB 8 TB = 16 TB = 4 (b) Coordinates of T= (8, 4) 4 0 1 Slope of AT = = 8 0 ABT = 90 AT is a diameter of the circle. AT DT Slope of DT = 1 = 1 The equation of DT is y 4 = (x 8) y 4 = x + 16 x + y 0 = 0 END OF PAPER TSSS F5 MATH-CP1 MOCK EXAM 1 015-016 [ 15 ]