20 014 Examination n Report 2014
Further Mathemaatics GA 2: 2 Examination 1
GENERA AL COM MMENTS The majorityy of students were w generallyy well prepared for the 20144 Further Mathhematics exam mination 1.
SPECIFIIC INFOR RMATIO ON The tables b below indicatee the percenttage of studen nts who chosee each option. The correctt answer is sh haded. ng errors resultting in a total less than 100 per cent. The statisticss in this reportt may be subjeect to roundin
Section A Core: Daata analyssis Question
%A
%B
%C
%D
%E
1 2 3 4 5 6 7 8 9 10 11 12 13
3 1 2 1 1 37 7 8 12 6 6 26 7
85 8 9 12 95 30 6 1 58 6 13 9 11
4 10 86 52 2 4 77 24 10 5 4 30 37
1 76 6 2 29 9 1 4 5 39 9 11 6 3 32 2 30 0
7 4 1 5 1 25 5 28 8 77 73 3 14
% No Answer 0 1 0 0 0 0 0 0 1 0 0 0 0
The Core secction was geneerally well ansswered. Studeents performedd well on quesstions that req quired a routinne application of a skill in a familiar circuumstance; how wever, they sttruggled with the graphical location of medians m (Questtion 13). Students perfformed less well w on questioons that requirred deeper connceptual underrstanding to obtain o an answ wer. This was clear in Quesstions 4 and 8, both of whicch required moore than a supperficial underrstanding of thhe differences between categorical aand numerical variables. Question 4 mber of categoorical variablees in a data sett. There were three: t sex, In Question 4 students werre asked to ideentify the num type of car annd postcode (ooption D); how wever, the maajority of studeents did not ch hoose this opttion. Most stuudents decidedd that there weere only two caategorical varriables (option n C), possibly rejecting posttcode becausee its data valuees were numbers. Posstcodes are nuumbers, as aree phone numbeers. However,, in both casess, these numbeers only servee as identifiers. T They have no other o numerical properties. If students aree in doubt abo out classifying g a variable ass categorical or numerical, they should ask, ‘Does it make m sense to o calculate thee mean of this variable?’ If the t answer is ‘No’, the variable is caategorical. Forr postcodes, thhe answer is ‘N No’. Question 6 m a box ploot to a given dot d plot. An in nitial inspectio on of the dot plot p indicated In Question 66, students weere asked to match that there migght be up to th hree outliers. However, H by using the infoormation in thee plots to locaate the upper fence f (Q3 + 1.5 IQR I = 50 + 1.5 20 = 80), it was clear th hat only two of o these pointss could be regarded as outliers, making option A the correct answeer. The majoriity of studentss chose one off the two optioons showing th hree outliers, options that could have been eliminated by making a quick estimaation of the loocation of the outer o fence. Question 8 An understannding of typess of variables was w also requuired to answerr Question 8. The key to annswering this qquestion was to recognise that the type of o variable plaays a role in chhoosing an apppropriate stattistical plot whhen displayingg data. Of the plots listed inn this question n, only a backk-to-back stem m plot was suittable for displaaying the assoociation betweeen a car’s speed (a num merical variablle) and the sexx of the driverr (a categoricaal variable withh two categorries). A paralleel box plot would also have h been apprropriate, but was w not given as an option. Further Maths 1 GA 2 Exam
© VICTORIAN CURRICULUM AND A ASSESSME ENT AUTHORIT TY 2015
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201 14 Examin nation Rep port Question 12 In Question 112 students were asked to determine d the percentage p byy which an acttual sales figuure would channge when the seasonal indeex is 1.25. Few w students ansswered this quuestion correcctly. The majority of studentts chose one of o the two options with a 25% in the answer. This was not a queestion that couuld be answereed by inspection and a calcuulation was required. mula sheet: From the form
seasonal inddex =
actuual figure deseasonnalised figuree
Making the ddeseasonalisedd sales the subbject of the forrmula: deseasonalissed figure
acctual figure seaasonal index
Here, SI = 1..25, so deseasonalissed figure =
acctual figure 0.80 actual figure 1.25
Thus, to obtaain the deseasoonalised saless for summer, the actual salees figure mustt be decreasedd by 20%.
Module 11: Numbeer pattern ns Question
%A
%B
%C
%D
%E
1 2 3 4 5 6 7 8 9
99 9 5 8 6 2 6 23 10
1 2 19 2 63 2 11 40 17
0 85 5 9 10 15 42 11 17
0 2 677 6 166 6 199 11 444
0 1 5 75 4 74 21 13 12
% Noo Answ wer 0 0 0 0 1 0 0 0 1
The questionns in Module 1: 1 Number pattterns were veery well answeered. Studentss performed well w on questioons that required a routine applicattion of arithmeetic and geom metric sequences in familiar circumstancees. Students deemonstrated a d equuations and theeir applicationns. general comppetence with difference Question 7 The questionn stated that thhe first two terrms in a Fibonnacci-related sequence s weree p and q. Thee question requuired studentss to find the diifference in vaalue between tthe fourth andd fifth terms inn terms of p annd q. A straightforrward way to answer a this quuestion was too form a table and use the baasic property of a Fibonaccci-related sequence to ggenerate the first fi five termss of this sequeence. For exam mple: term numberr
1
2
3
4
5
value
p
q
p+q
p + 2q q
2p + 3q
ms is 2p + 3q – (p + 2q) = p + q, which coorresponded too option C. Thus, the diffference betweeen the fourth and fifth term
Further Maths 1 GA 3 Exam
Published: 266 May 2015
2
201 14 Examin nation Rep port Module 22: Geomeetry and trigonome t etry Question
%A
%B
%C
%D
%E
1 2 3 4 5 6 7 8 9
5 2 6 15 10 16 57 8 9
83 2 4 17 11 14 8 69 25
8 90 5 49 6 14 4 8 26
1 4 822 155 677 122 144 122 333
2 1 2 3 4 43 17 1 6
% Noo Answeer 0 0 0 1 1 1 1 1 1
The questionns in Module 2: 2 Geometry and a trigonomeetry were veryy well answereed. Students were w generallyy correct in answering quuestions that required r a routtine applicatioon of geometriic and trigono ometric techniques in a rangge of contexts including thee use of bearinngs. Question 9 Many studennts struggled too correctly annswer Questionn 9, which invvolved the scaaling of surfacce areas. The task in Question 9 was w to determiine the surfacee area of the middle m sectionn of a cone, givven that the suurface area off the top section was 1180 cm2. One solutionn strategy was to use scalingg to first find the t total surfacce area of the middle and toop sections coombined. The surface area oof the top secttion could theen be subtracteed to find the shaded area. SA middle + top = 180 (15/9)2 = 500 cm2 SA middle = SA A middle + top– SA A middle = (5500 –180) cm2 = 320 cm2 (ooption D)
Module 33: Graphs and relaations Question
%A
%B
%C
%D
%E
1 2 3 4 5 6 7 8 9
4 19 3 9 65 2 3 11 17
81 5 14 68 8 4 64 55 12
12 18 7 7 11 7 15 13 45
2 7 74 4 7 11 75 5 11 13 3 7
1 50 1 9 5 11 6 7 19
% Noo Answer 0 0 0 0 0 1 0 1 1
The majorityy of questions in Module 3 were w very welll answered.
Further Maths 1 GA 3 Exam
Published: 266 May 2015
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201 14 Examin nation Rep port Module 44: Busineess-related d mathem matics Question
%A
%B
%C
%D
%E
1 2 3 4 5 6 7 8 9
1 20 8 3 8 23 10 6 5
9 10 16 2 9 18 55 13 13
4 59 57 6 76 17 20 12 10
85 5 4 13 3 7 5 36 6 5 57 7 53 3
1 7 5 83 2 5 9 10 18
% Noo Answer 0 0 1 0 1 1 1 1 1
Student perfoormance on Module M 4: Busiiness-related mathematics m w similar too the performaance of studennts in other was modules. Stuudents generallly performed well on questtions that requuired the routinne applicationn of percentagge change, the principles off simple and coompound inteerest, and straight line/flat raate and reduciing balance deepreciation. Questions Q that required the rroutine appliccation of a finaancial solver for f solution were w also well answered. Students struuggled with thhe calculation of effective innterest and shoowed a poor understanding u of how intereest is calculated at each stage in a reducing baalance loan. Question 6 d not correcttly answer thiss question on determining an a effective in nterest rate. The majorityy of students did Calculating aan effective innterest problem m is a two-stepp process. Step 1 involvved calculating g the flat rate of interest. interest paid = total amoun nt repaid – am mount owed = 61 80 – 10000 = $1080 = $80 intereest paid 80 100 flat rate of innterest ( rf ) = 100 16% 10000 0.5 amount owedd time in yearss
Step 2 involvved converting g this flat rate of interest intto an effectivee rate of intereest. For n paymennts,
2n 26 effective interest = rf 16 27.422 ...% (option D) n 1 6 1 Question 9 ment for a reduucing balancee loan. Question 9 asked students to calculate thhe amount of the final paym was a two-stepp problem. Again, this w Step 1 involvved calculating g the future vaalue of the loaan after 47 payyments. Using a finanncial solver, th his amount is found to be $802.39... Step 2 requirred adding a month’s m interest to this amouunt to find thee final paymennt.
4.75 final paymennt = 802.3911.... + 802.3911... E = $805.57 to the nearest cent (option E) 0 12 100 While most sstudents couldd correctly usee their financiaal solver to finnd the amountt still owed affter the secondd last payment hadd been made ($$802.39), theyy apparently failed fa to realisee that this amoount would atttract interest during d the last month off the loan.
Further Maths 1 GA 3 Exam
Published: 266 May 2015
4
201 14 Examin nation Rep port Module 55: Networrks and decision d mathemati m ics Question
%A
%B
%C
%D
%E
1 2 3 4 5 6 7 8 9
1 11 2 59 84 3 58 12 4
10 0 2 3 5 69 10 12 39
1 0 75 22 3 15 15 17 9
87 7 88 19 9 10 0 3 10 0 10 0 18 23 3
1 0 2 6 5 4 7 40 24
% Noo Answer 0 0 0 0 0 0 0 0 0
Questions in the Networkss and decisionn mathematics module weree generally verry well answeered, with the notable n exceptions off Questions 7 and 9. Students gennerally perform med well on quuestions that required r know wledge of the general g properrties of graphss, and eulerian and hamiltonian paths p and circuuits. They werre also generaally able to sollve problems involving i the application of Prim’s alggorithm and deetermining thee shortest pathh by inspectionn. Students struuggled with recognising a pllanar graph when w drawn in non-planar foorm (Question n 7) and correcctly identifying thhe minimum cut c in a practiccal flow probllem with moree than one sou urce (Question n 9). Question 7 In this questiion, students were w given fouur graphs and asked how many m were plannar; however, few students answered correctly (opption E, 4). Th he majority of students chosse option A, 0.. This suggestts that most stuudents were unaware u that intersecting edges e in a grap ph do not autoomatically preeclude the graaph from beingg planar. Question 9 This questionn assessed thee implicit assuumption made about the prooperties of min nimum cut whhen applying thhe minimum cut/maximum m flow theorem m, but many students s did not n answer corrrectly. A flow netwoork with two sources s (two ccar parks) and d a single sinkk (the exit) wass provided. While W Cut C haad the minimum cappacity, neitherr it nor Cut A separated botth sources (carr parks) from the sink. Thuus, neither of thhese cuts could be usedd to determinee the minimum m flow. Of thee three cuts thhat did separatte both sources from the sinnk, Cut D had the minimum m capacity and d hence determ mined the maxximum flow.
Module 66: Matricces Question
%A
%B
%C
%D
%E
1 2 3 4 5 6 7 8 9
92 2 11 10 10 16 15 11 14
1 80 68 2 5 17 6 9 17
0 12 7 77 69 7 3 10 40
1 5 8 2 8 16 6 64 4 8 19 9
5 1 5 9 8 42 12 60 11
% Noo Answer 0 0 1 0 0 2 1 1 1
The majorityy of questions in Module 6: Matrices werre generally veery well answered.
Further Maths 1 GA 3 Exam
Published: 266 May 2015
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