INSIGHT YEAR 12 Trial Exam Paper 2013 FURTHER MATHEMATICS Written examination 2 s This book presents: correct solutions with full working explanatory notes mark allocations This trial examination produced by Insight Publications is NOT an official VCAA paper for the 2013 Further Mathematics 2 written examination. This examination paper is licensed to be printed, photocopied or placed on the school intranet and used only within the confines of the purchasing school for examining their students. No trial examination or part thereof may be issued or passed on to any other party, including other schools, practising or non-practising teachers, tutors, parents, websites or publishing agencies, without the written consent of Insight Publications.
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SECTION A Core: Data Analysis Question 1a. 3 Dorpers Merinos 0 5 10 15 20 25 30 35 40 45 Sheep weights (kg) Question 1b. 1 mark for five-number summary points in the correct position 1 mark for the correct convention used to draw the box plot The distribution of the weights of the Dorpers is approximately symmetrical (or slightly positively skewed) with an outlier at 43, whereas the Merino weights are negatively skewed. The centre of the Dorper weights as described by the median is 21, compared to the median of the Merinos, which is 31. The Dorper weights are more variable with an interquartile range of 10 and a range of 30, compared to an interquartile range of 8 and a range of 26 for the Merinos. Mark allocation: 3 marks 1 mark for comparing the shape of the distributions 1 mark for comparing the centres (medians) of the distributions 1 mark for comparing the spreads of the distributions Note: Values must be used when making comparisons. Core
4 Question 1c. upper fence Q (1.5 IQR) 3 upper fence 27 (1.5 10) upper fence 42 Since 43 is higher than the upper fence, 42, it is an outlier. Tip Values equal to a fence are not considered to be outliers. Question 2a. The graph is non-linear. Question 2b. residual actual value predicted value 103 (49.66 3.7912) 10395.14 7.86 8 Or, alternatively, use the lists and spreadsheet on your calculator to list the residuals. Question 2c. log x or y 2 Core
Question 2d. 1 Sale price 137.48 454.03 age 5 Question 2e. 1 sale price 137.48 454.03 20 sale price $114.779 sale price $115 Core
6 Question 3a. June SI 12 (0.8 0.83 0.85 0.92 0.98 1.06 1.10 1.15 1.15 1.12 1.00) 1.04 Question 3b. 67 July deseasonalised rainfall 1.06 63.21 Question 3c. September 2013 is month number 21. Deseasonalised rainfall 31.72 2.04 (21) 74.56 Rainfall predicted for September 2013 74.561.15 85.74 mm 86 mm 1 mark for obtaining 74.56 for deseasonalised rainfall 1 mark for obtaining 86 mm as the annual rainfall prediction for September 2013 Core
SECTION B Module 1: Number patterns Question 1a. 500 g for the original 5 bantam hens + 3110 g for the 3 Isa Browns 500 3110 830 g Question 1b. 2000 g less 500 g for the bantams leaves 1500 g. If each Isa Brown requires 110 g, Henrietta 7 has enough food for 1500 13.6 Isa Browns, but you cannot have 0.6 of a chicken. Therefore, 110 Henrietta can afford to feed 5 bantams and 13 Isa Browns, which is 18 in total. 1 mark for obtaining 13.6 1 mark for interpreting this as 18 chickens in total. Question 2a. a = 2, b = 50, c = 100 Question 2b. 2 marks for all three values correct OR 1 mark for two out of three values correct t 1 = 100, t 2 = 250, t 3 = 550, t 4 = 1150, t 5 = 2350, 250 100 550 250, therefore not arithmetic 250 550, therefore not geometric 100 250 1 mark for the first 5 terms 1 mark for proof that the sequence is neither arithmetic nor geometric Module 1: Number patterns
Question 2c. n1 ba tn a t1 ( a 1) t t t n n n n1 2 100 n1 ( 1) n1 50(2 1) n1 n1 100 2 50 2 50 n1 150 2 50 (2 1) 8 1 mark for substituting the values of a, b and t 1 into the given equation (line 2) 1 mark for simplifying correctly Question 2d. t n t t 7 7 7 n1 150 2 50 6 150 2 50 15064 50 t 9550 Module 1: Number patterns
Question 3a. t r t 2 t t 3 1 2 300 450 r 200 300 r 1.5 9 Question 3b. Generate the sequence 200, 300, 450, 675, 1012.5, 1518.75, 2278.125, These are the duck numbers at the start of each year. At the start of the 7th year she will have 2278 ducks. Therefore during the 6th year she will first exceed 2000 ducks. Question 3c. S S S n 10 10 n ar ( 1) ( r 1) 10 200(1.5 1) (1.5 1) 22 666 Question 3d. a = 1.5 and b = 0 1 mark for obtaining a = 1.5 1 mark for obtaining b = 0 Module 1: Number patterns
Module 2: Geometry and trigonometry Question 1a. 10 The height of B above the ground is the opposite side of the triangle (with respect to 20 ) and the sloping edge is the hypotenuse. opposite hypotenusesin height 3sin 20 height 30.3420 height 1.026 height 1.0 m Question 1b. The horizontal distance is the adjacent side of the triangle (with respect to 20 ) and the sloping edge is the hypotenuse. adjacent hypotenusecos horizontal 3cos horizontal 30.9397 horizontal 2.8191 horizontal 2.8 m Module 2: Geometry and trigonometry
Question 2a. The three angles of a triangle add up to 180. angle at B = 180º 30º 47º angle at B = 103 11 Question 2b. To find this length, use the sine rule. a c sin A sin C BC 5 sin 30 sin 47 5 BC sin 30 sin 47 BC 3.418 m BC 3.42 m or 342 cm 1 mark for substituting the values correctly (line 2) 1 mark for obtaining BC = 3.42 m Question 2c. 1 area of triangle absin C 2 1 area of triangle 58.45sin 30 2 area of triangle 10.5625 m area of triangle 7.04 1.5 Therefore, Bill can fit a maximum of 7 steers. 1 mark for finding the correct area of the triangle 1 mark for obtaining a maximum of 7 steers Module 2: Geometry and trigonometry
Question 3a. ABC is 35 + 60 = 95 Using the cosine rule to find a side length. 2 2 2 b a c 2accosB b 120 100 2120100 cos 95 2 2 2 CA 162.76 m 12 1 mark for correct substitution into the cosine rule 1 mark for obtaining CA = 162.76 m Question 3b. Find ACB using the sine rule. sin C sin B c b sin 95 sin ACB 100 162.76 sin ACB 0.612 1 sin 0.612 37.7389 ACB 38 Therefore the bearing of A from C is 270 8, which is 262 T. 1 mark for correct substitution into sine rule 1 mark for the correct bearing Module 2: Geometry and trigonometry
Question 4a. front 2 30 62 360 end2 1062 120 roof slant 2 3062 360 13 roof end 2 ss ( a)( sb)( sc) 2 roof end 2 11(11 10)(11 6)(11 6) 2 roof end 2 33.17 total surface area 873.17 m 2 abc Given that s 2 Question 4b. volume area of end 30 volume (60 11(11 10)(11 6)(11 6) 30 volume 76.5830 volume 2297.4 m Question 4c. 30 metres : 0.3 metres 100 :1 3 Question 4d. Area ratio is length ratio squared. Area ratio is 10 000 :1. 2 873.17 m area of the model 10 000 2 2 area of the model 0.087317 m or 873.17 cm Alternatively, calculate the surface area of the model (as in the solution to part a) but using the model dimensions instead of the actual dimensions; e.g. length of 30 m becomes 30 cm, etc. Module 2: Geometry and trigonometry
Module 3: Graphs and relations Question 1a. R = 900n 14 Question 1b. C = 1500 + 500n Question 1c. P = 900n 1500 500n P = 400n 1500 Question 1d. 400n 1500 0 400n 1500 15 n 4 n 3.75 Therefore, Stacey will make a profit after she makes four or more surfboards. Module 3: Graphs and relations
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16 Question 2 80x50y 8000 40x60y 6000 1 mark for each correct constraint Question 3a. If Jake can hire out Malibus no more than 100 times per month (20 days), he must have only five Malibus available for hire. Question 3b. y 200 100 50 100 x Module 3: Graphs and relations
Question 3c. y 17 200 100 50 100 x Question 3d. P = 50x + 70y Question 3e. P = 50x + 70y Test vertices: (0, 0) P = $0; (0, 200) P = $14 000; (50, 200) P = $16 500; (100, 100) P = $12 000; (100, 0) P = $5000 Therefore, the maximum profit Jake can earn in a month is $16 500. Module 3: Graphs and relations
Question 3f. 18 50 Malibus and 200 short boards (working shown in part e). Question 3g. New constraint will be y 300. Any higher than 300 and it will not be a boundary of the feasible region. This means Jake can hire out any extra short boards 100 times per month; divide this by 20 days and it means he should purchase an extra 5 short boards. 1 mark for obtaining y 300 1 mark for an extra 5 short boards Question 3h. P = 50x + 70y P = 50(0) + 70(300) P = $21 000 Module 3: Graphs and relations
19 Module 4: Business-related mathematics Question 1a. 21 125 100 2.5% 845 000 Question 1b. Using the TiNspire calculator page: Tony must pay $85 500 in Capital Gains Tax. Module 4: Business-related mathematics
20 Question 2a. interest paid total repayments amount borrowed interest paid 12024 2400 interest paid $480 Question 2b. (interest 100) R f = (principal time) (480100) R f = (2400 2) R =10% f Question 2c. R R e e Rf 2 n, where n is equal to the total number of repayments. ( n 1) 48 10 25 R e 19.2% Question 2d. Interest is charged at the flat interest rate (10%) on the initial principal ($2400) for the duration of the loan (24 months), even though this principal is being reduced during the loan. The effective interest rate (19.2%) takes into account the fact that the principal is being reduced while the interest charges remain unchanged and, therefore, the effective rate is higher. Module 4: Business-related mathematics
Question 3a. investment principal interest (55 0004.755) investment 55 000 100 investment 55 000 13 062.50 investment $68 062.50 21 Question 3b. 1 4.25 A 55000 12100 67 996.60 512 Or use TVM solver: Question 3c. Use TVM solver: Monthly deposit required is $152.26. Module 4: Business-related mathematics
Question 4a. Using TVM solver: 22 Quarterly instalment is $13 822.79. Question 4b. total repayments 13 822.7980 $1 105 823.20 interest paid $1 105 823.20 $520 000 $585 823.20 1 mark for obtaining correct value for total repayments 1 mark for obtaining the correct value for interest paid Module 4: Business-related mathematics
23 Question 5a. yearly depreciation 0.7 40 000 $28 000 Question 5b. book value new value depreciation book value 125 000 30.24125 000 book value $35 000 Question 5c. Therefore, after 6 years Fred s truck will be valued at less than $40 000. Module 4: Business-related mathematics
24 Module 5: Networks and decision mathematics Question 1a. ABEFG = 35 + 11 + 11 + 9 = 66 m Question 1b. A minimum spanning tree. Question 1c. B E A C F H Question 1d. D G Question 1e. F E H G D A B C F length = 219 m 1 mark for a correct Hamilton circuit 1 mark for correct length Module 5: Networks and decision mathematics
25 Question 2a. Yes. Start at classroom A and finish at E, or vice versa. Question 2b. Between A and E. Euler circuit. 1 mark for A and E 1 mark for Euler circuit Module 5: Networks and decision mathematics
26 Question 3a. 4 weeks because the earliest start time for H is 8, which means we must be waiting 4 weeks for F to be completed. Question 3b. Task EST LST A 0 1 B 0 3 C 0 0 D 3 4 E 3 5 F 4 4 G 8 9 H 8 8 I 12 12 J 10 11 K 4 11 Question 3c. float = LST EST, so 1 week Question 3d. 14 weeks. Critical path is CFHXI. Module 5: Networks and decision mathematics
Question 3e. 27 New minimum completion time is 13 weeks. Minimum extra cost is $10 000. The new critical path is ADI. Path CFHXI, if not reduced, takes 14 weeks, so take 1 week off C to make it another critical path. Any further reduction of C or any reduction of B has no reducing effect on the shortest completion time, and, so, would be a waste of money. 1 mark for 13 weeks 1 mark for $10 000 Module 5: Networks and decision mathematics
Module 6: Matrices Question 1a. 4 1 (4 rows and 1 column) 28 Queston 1b. The matrix product NP gives the total sales. N is a 1 4 matrix and P is a 4 1 matrix resulting in a 1 1 matrix, the element of which represents total sales. PN results in a 4 4 matrix for which, in the context of this question, the elements have no meaning. Question 1c. 550 85 N (203) (567) (45) (32) 180 120 N (171 785) Total sales at market = $171 785 Module 6: Matrices
Question 2a. 29 27 250 Question 2b. Question 2c. 1 b 50 20 15 1140 g 25 30 25 985 p 45 0 25 925 b 15 g 12 p 10 Beef is $15, goat $12 and pork $10. 1 mark for matrix solution (line 2) 1 mark for stating the prices of each meat Module 6: Matrices
Question 3a. 250 + 375 + 120 = 745 30 Question 3b. Since the columns of the transition matrix add up to 1, the total number of shoppers each week remains constant at 745; i.e. the total number of shoppers stays the same each week. Question 3c. Accept 0.1 or 10%. Question 3d. S TS 1 0 0.7 0.0 0.1 250 S1 0.1 0.9 0.3 375 0.2 0.1 0.6 120 187 S1 398.5 159.5 Question 3e. S 1 is the second week since S 0 is the first week. Question 3f. 0.1 375 = 37.5, therefore 38 people switched. Module 6: Matrices
Question 3g. S 5 will be the 6th week of the study. S 5 = T 5 S 0 31 485 people will shop at Myway in the 6th week. Question 3h. Find the steady state matrix. 53 at Safeless, 532 at Myway and 160 at Coldstore. END OF SOLUTIONS BOOK Module 6: Matrices