2018 Year 10/10A Mathematics v1 & v2 exam structure

Transcription

1 018 Year 10/10A Mathematics v1 & v eam structure Section A Multiple choice questions Section B Short answer questions Section C Etended response Mathematics 10 0 questions (0 marks) 10 questions (50 marks) 3 questions (30 marks) Mathematics 10A etra questions 1 questions (1 marks) 7 questions (8 marks) 3 questions (30 marks) Total 100 marks 70 marks Teachers please note: our 10 & 10A eams cover the entire Year 10/10A content all eams are ed in pdf format some schools asked about the two versions of our eams, so we would like to clarify: version 1 and version eams consist of completely different questions. if you purchased a single version for $100 (say version 1), you will receive two of the following eams: 10 eam version 1 10A eam version 1 if you purchased both versions of eams for $00, you receive all of the following eams: 10 eam version 1 10A eam version 1 10 eam version 10A eam version please feel free to modify the time allocated to 10/10A eams if necessary

2 MATHEMATICS 10 Written eamination Reading time Writing time : 15 minutes : 3 hours 018 version 1 QUESTION BOOK structure of book Section Number of questions Number of questions to be answered Number of marks A B C Total 100

3 ELITE MATHS 018 v1 MATHEMATICS 10 EXAM SECTION A Question 1 Instructions for Section A Answer all questions. Choose the response that is correct for the question. A correct answer scores 1, an incorrect answer scores 0. Marks will not be deducted for incorrect answers. No marks will be given if more than one answer is completed for any question. The solution to the equation + 1 = is A. 1 B. 3 C. D. 3 E. 5 3 Question Simplifying the epression A B. + 1 C. + 5 D E results in Question 3 The formula 1 u + 1 v = 1 f is used in optics. Which one of the following is the correct result of making u the subject? A. u = 1 f v B. u = fv v f C. 1 u = 1 f v D. u = v f fv E. u = f v

4 ELITE MATHS v1 MATHEMATICS 10 EXAM Question 4 The smallest number from the following is: A. 0.4 B. 1 4 C. 50% D. 3 1 E. 6 7 Question 5 Which one of the following number lines represents the set { R : ( + ) < 3( + )}? A B C D E Question 6 Consider the right angled triangle below. 15 cm 9 cm The length of is A. 3 cm B. 1 cm C. 6 cm D. ±1 cm E. 144 cm

5 ELITE MATHS v1 MATHEMATICS 10 EXAM Question 16 If triangles ABC and DEF are simliar, what is the value of? D A B C E F A. 5 B. 4 C. 60 D E. 15 Question 17 Which one of the following states the correct value for y and its justification? 50 y A. y = 50, since the angles are vertically opposite B. y = 40, since the angles are complementary C. y = 130, since the angles are on a line D. y = 130, since the angles are co-interior angles E. y = 40, since the angles are corresponding angles Question 18 A workplace runs a raffle in which 100 tickets are sold. There is one winning ticket, and the tickets are drawn at random one after the other without replacement. If an employee buys tickets, the probability that he has a winning ticket is A B C D. 0.0 E. 0.04

6 ELITE MATHS v1 MATHEMATICS 10 EXAM SECTION B Answer all questions. Instruction In all questions where a numerical answer is required, an eact value must be given unless otherwise specified. In questions where more than one mark is available, appropriate working must be show Unless otherwise indicated, the diagrams in this book are not drawn to scale. Question 1 (5 marks) Simplify the following epressions. a mark b y y. 1 mark c. ( + 3) ( + )( ). 1 mark d marks + 6

7 ELITE MATHS v1 MATHEMATICS 10 EXAM Question 3 (6 marks) Sketch the inequalities y and y on the set of aes below. Indicate the point of intersection and shade the region of intersection. y

8 ELITE MATHS v1 MATHEMATICS 10 EXAM Question 5 (5 marks) There are two cylinders of the same height h cm. One cylinder has radius 1 m, and the other cylinder has radius 0.5 m. Show that the volume of one cylinder is 4 times the volume of the other. h cm h cm 1 m 0.5 m

9 ELITE MATHS v1 MATHEMATICS 10 EXAM Question 7 (4 marks) Find the value of an eterior angle of a regular pentagon.

10 ELITE MATHS v1 MATHEMATICS 10A EXAM Question 8 (6 marks) Match each equation (1-6) with its correct graph (A - I), if it is present. 1. y = 1. + y = 5 3. y = 3 4. y = 5. y 1 = 0 6. y = 3 + A. y B. y C. y O O O D. y E. y F. y O O O G. y H. y I. y O O O

11 ELITE MATHS v1 MATHEMATICS 10 EXAM SECTION C Instructions for Section C Answer all questions in the spaces provided. In all questions where a numerical answer is required, an eact value must be given unless otherwise specified. In questions where more than one mark is available, appropriate working must be shown. Unless otherwise indicated, the diagrams in this book are not drawn to scale. Question 1 (10 marks) Consider the points A(, 5), B(4, 3) and C(5, 4), as shown below. Point M is the midpoint of line segment AB. The line segment CM is perpendicular to the line segment AB. A(, 5) y C(5, 4) 1 M O B(4, 3) 5 6 a. Find the length of the line segment AB. marks b. Show that the coordinates of M is (1, 1). 1 mark

12 ELITE MATHS 018 v1 MATHEMATICS 10 EXAM Question (10 marks) Consider a rectangular prism with a square base of side length cm and height h cm. h cm cm cm a. Show that the surface area of the prism is A = + 4h. marks b. If the side length of the base is 3 cm and the height of the prism is 4 cm, find the surface area. 1 mark

13 ELITE MATHS v1 MATHEMATICS 10 EXAM Question 3 (10 marks) Consider the data set A with values, 3, 3, 4, 4, 5, 5, 6, 6, 7 and 8. a. Complete the following five-figure summary: marks Minimum value Lower quartile (Q1) Median Upper quartile (Q3) Maimum value b. Find the range and the interquartile range of data A. marks c. Show that 8 is not an outlier. 3 marks

14 ELITE MATHS 6 SOLUTIONS SOLUTIONS SECTION A Question Answer 1 A D 3 B 4 A 5 A 6 B 7 D 8 C 9 A 10 C 11 B 1 D 13 C 14 A 15 B 16 E 17 D 18 D 19 B 0 E

15 ELITE MATHS 7 SOLUTIONS Question = = = 3 = 1 Answer is A. Question Answer is D. = 3( + 3) 15 5( ) (5 10) = = 15 = Question 3 1 u + 1 v = 1 f 1 u = 1 f 1 v 1 u = v f fv u = fv v f Answer is B. Question 4 50% = = = = = Therefore, 0.4 is the smallest number. Answer is A.

16 ELITE MATHS 9 SOLUTIONS Question 8 Let the two numbers be and y. + y = 11 (1) y = 105 () Substituting () into (1) + y = 11 + y = = = = 0 ( 7)( 15) = 0 = 7, 15 When = 7, y = 15 (or = 15, y = 7 which are the same numbers). Answer is C. Question 9 The equation + y = 10 represents a circle. When we substitute = 6 then solve for y, this means finding where the circle intersects with the line = 6. Answer is A. Question 10 Since the point ( 1, 6) is on the parabola, a( 1) + b( 1) + = 6 a b + = 6 a b = 4 (1) Since the point (1, 0) is also on the parabola, a(1) + b(1) + = 0 a + b = () Adding the equations (1) and () gives: a b + a + b = 4 a = a = 1. It follows that b = 3. Answer is C.

17 ELITE MATHS 31 SOLUTIONS Question 16 Since the scale factor is 1 4 = 3: = 3 5 = 15 Answer is E. Question 17 Answer is D. Question 18 Either one ticket or the other ticket can be a winning ticket = 0.0 Answer is D. Question 19 Since A B = {0, 1} P r(a B) = n(a B) n(ɛ) Answer is B. = 7 Question 0 P r(b A) = = P r(a B) P r(a) = 5 Answer is E.

18 ELITE MATHS 34 SOLUTIONS Question 3 (6 marks) We first find the corresponding line equations, and their aes intercepts. y + 6 = 0 y = 6 (A0.5) -intercept is 3 and y-intecept is 6. (A0.5) y = (A0.5) -intercept is 1 and y-intecept is. (A0.5) The point of intersection of the two lines is: 6 = + = 6 4 = 4 = 1 (1, 4). (A1) y 1 O 3 6 (1, 4) correct straight lines with solid lines (A1) correct aes intercepts and the point of intersection indicated (A1) correct intersecting region shaded (A1)

19 ELITE MATHS 36 SOLUTIONS Question 6 (5 marks) a. Since the base ( 1) and the height (4 8) both represent length, they must be greater than 0. (A1) marks They simplify to > 1 and >, which results in the final inequality >. (A1) b. 3 marks A = 1 base height 1 = 1 ( 1) (4 8) (A1) 1 = ( 1) ( 4) 1 = = = = = ( 4)( + 1) (A1) Therefore, = 4 cm is the only solution since > from part a. (A1) penalise if = 1 was not ecluded Question 7 (4 marks) The sum of the interior angles of the pentagon is S = 180 (5 ) = = 540 (A1) Since the pentagon is regular, then the interior angle is: = 108 (A1) Thus, the angle of an eterior angle is: = 5 (A1) since the angle is at a point. (A1) Question 8 (6 marks) 1. not present (A1). F (A1) 3. E (A1) 4. not present (A1) 5. D (A1) 6. C (A1)

20 ELITE MATHS 37 SOLUTIONS Question 9 (5 marks) a. 1 mark y all given points are indicated on the aes (A1) b. There is a postive correlation between and y, since as increases y increases. (A1) 1 mark c. There is a clear upwards trend in the scatter plot, indicating that the positive correlation is strong. (A1) 1 mark d. The line of best fit sketched must be well balanced between the given points. (A0.5) 1 mark y = (A0.5) allow small variations in the values for m and c e. 1 mark y = = 17. (A1)

21 ELITE MATHS 39 SOLUTIONS SECTION C Question 1 (10 marks) a. marks AB = ( 4) + (5 + 3) (A1) = ( 6) + (8) = = 100 = 10 (A1) b. 1 mark ( + 4 M =, 5 3 ) = (1, 1) (A1) c. marks m AB = (A1) = 8 6 = 4 3 (A1) d. marks m AB m CM = 1 (A1) 4 3 m CM = 1 m CM = 3 4 (A1) e. 3 marks CM = (5 1) + (4 1) = = = 5 = 5 (A1) A = 1 AB CM (A1) = = 5 unit (A1)

22 ELITE MATHS 40 SOLUTIONS Question (10 marks) a. marks A = + 4 h (A1) = + 4h (A1) b. 1 mark A = (3) = = 66 cm (A1) c. marks (5) + 4(5)h = 90 (A1) h = 90 0h = 40 h = cm (A1) d. i. marks A O sketch is part of part of a parabola (A1) graph with restriction > 0 (A1) ii. 3 marks A = 10 (A1) + 8 = = 0 ( + 5)( 1) = 0 (A1) = 1 cm (A1)

23 MATHEMATICS 10A

24 MATHEMATICS 10A Written eamination Reading time Writing time : 15 minutes : 3 hours 018 version 1 QUESTION BOOK structure of book Section Number of questions Number of questions to be answered Number of marks A B C Total 170

25 ELITE MATHS v1 MATHEMATICS 10A EXAM Question 3 What are the solutions to the quadratic equation 3 1 = 0? A. there is no solution B. = 1 3 and = 1 C. = 1 3 and = 1 D. = 1 3 and = 1 E. = 1 3 and = 1 Question 4 Which one of the following is a polynomial? A B C D. + 1 E Question 5 What is the remainder when is divided by? A. 0 B. 5 C. 11 D. 1 E. 14 Question 6 If the total volume of identical spheres with radius R cm is 480πR 3 cm 3, then is R cm A. 3 B. 36 C. 160 D. 360 E. 3600

26 ELITE MATHS v1 MATHEMATICS 10A EXAM Question 30 The solution of the equation + 1 = 4 is A. 3 B. log (3) C. log (3) D. 1 log 3() E. 1 log (3) Question 31 Find the value of y in the circle below. 45 y A. 90 B. 10 C. 130 D. 180 E. 360 Question 3 The standard deviation of the dataset 1.5,, 5.5, 6, 7.5,,.5, 3, 4 and 8.5 is closest to A..48 B. 4.5 C D. 10 E. 4.5

27 ELITE MATHS 018 v1 MATHEMATICS 10A EXAM Question 10 (5 marks) A computer program is designed to randomly select a letter from the first 10 letters of the alphabet: A, B, C, D, E, F, G, H, I and J. a. Find the probability of obtaining a vowel. 1 mark b. Find the probability of obtaining a consonant. 1 mark c. If the event of obtaining a vowel and the event of obtaining a consonant independent? Justify your answer. marks Suppose the computer program eecutes twice, selecting two letters, one after the other, without replacement. d. Find the probability of obtaining a consonant and a vowel. 1 mark

28 ELITE MATHS v1 MATHEMATICS 10A EXAM Question 11 (5 marks) a. Epand and simplify ( + 5). 1 mark b. Epand ( 1)( 1). marks c. Evaluate log 10 ( ). 1 mark d. Evaluate log 1 (7). 1 mark 3

29 ELITE MATHS v1 MATHEMATICS 10A EXAM Question 14 (4 marks) Consider the circles below. One circle has two intersecting chords and the other circle has an inscribed quadrilateral. Find the sum of a, b and c c a b

30 ELITE MATHS v1 MATHEMATICS 10A EXAM Question 17 (4 marks) Show that the equation y 8y 110 = 0 represents a circle. States the coordinates of its centre and the length of its radius.

31 ELITE MATHS v1 MATHEMATICS 10A EXAM Question 4 (10 marks) A researcher has proposed a mathematical model for a population of mosquitos in a particular area during Summer: = e t 15 where is the number of mosquitos, and t is the time in days. a. Find the initial population of the mosquitos. marks b. Find the popluation of the mosquitos after 5 days, rounded to the nearest whole number. marks

32 ELITE MATHS v1 MATHEMATICS 10A EXAM Question 5 (10 marks) The square-based pyramid below has base side length m and vertical height h m. h m m m a. Draw the net of the pyramid. 3 marks b. Find the height of the triangular face in terms of and h. 1 mark

33 ELITE MATHS v1 MATHEMATICS 10A EXAM Complete the flow chart below so that it can represent an algorithm for solving all linear equations in the form a 1 = b, where a and b are constants.

34 ELITE MATHS v1 MATHEMATICS 10A EXAM b. i. Completely factorise the polynomial p() = marks ii. Sketch the graph of y = below, indicating the coordinates of any aes intercepts. y 3 marks

35 ELITE MATHS 43 SOLUTIONS SOLUTIONS SECTION A Question Answer 1 A D 3 B 4 A 5 A 6 B 7 D 8 C 9 A 10 C 11 B 1 D 13 C 14 A 15 B 16 E 17 D 18 D 19 B 0 E 1 C C 3 E 4 A 5 C 6 D 7 B 8 C 9 D 30 E 31 A 3 B

36 ELITE MATHS 60 SOLUTIONS Question 13 (4 marks) a. marks = (A0.5) 1 ( 1 ) ( 1) = ( 1) (A0.5) = ( 1) = 0 = 0 = ( ) (A0.5) 0, = (0, 0) and (, ) (A0.5) b. marks y y = 1 O (, ) = 1 correct straight line (A0.5) correct hyperbola (A0.5) correct points of intersection (A0.5) correct equations for the asymptotes (A0.5)

37 ELITE MATHS 66 SOLUTIONS Question 4 (10 marks) a. marks (0) = e 0 (A1) = = 300 (A1) b. marks (5) = e 5 15 = e 1 3 = (A1) 640 mosquitos. (A1) c. 3 marks (t) = 1000 (A1) e t 15 = e t 15 = 500 e t 15 = 5 1 t 15 = log e ( 5 1 t = 15 log e ( 5 1 ) (A1) ) ) 1 ( 5 = 15 log e 1 ( 1 ) = 15 log e (A1) 5 d. 3 marks = t correct eponential graph (A1) horizontal asymptote = 1500 (A1) y-intercept at (0, 300) (A1)

38 ELITE MATHS 67 SOLUTIONS Question 5 (10 marks) a. 3 marks m m square base (A1) triangles adjacent to each of the side of the square (A1) square side length labelled as m (A1) b. ( ) + h m (A1) 1 mark c. i. marks A = ( ) + h (A1) = ( 7 ) = = = + 4 = + 5 = + 5 = (1 + 5 ) (A1)

39 ELITE MATHS 68 SOLUTIONS ii. A = 49 (1 + 5 ) = = = (A1) = = 49(1 5 ) 1 (5 ) = 49(1 5 ) 1 50 = 49(1 5 ) 49 = (1 5 ) = = m (A1) marks iii. Let V 1 be the volume of the original pyramid and V be the volume of the enlarged pyramid. marks V 1 = = (A0.5) V = 1 3 (3) 7 (3) = = (A0.5) Therefore V 1 : V = : = : = 1 : 7. (A1)