Coimisiún na Scrúduithe Stáit State Examinations Commission LEAVING CERTIFICATE 010 MARKING SCHEME MATHEMATICS (PROJECT MATHS) ORDINARY LEVEL
Contents Page INTRODUCTION... MARKING SCHEME FOR PAPER 1... MODEL SOLUTIONS PAPER... 4 MARKING SCHEME PAPER, SECTION 0 (QUESTION 1)... 59 MARKING SCHEME PAPER, SECTION A AND SECTION B... 65 Structure of the marking scheme... 65 Summary of mark allocations and scales to be applied... 66 Detailed marking notes... 67 MARCANNA BREISE AS UCHT FREAGAIRT TRÍ GHAEILGE... 75 Page 1
Introduction The Ordinary Level Mathematics examination for candidates in the 4 initial schools for Project Maths shared a common Paper 1 and one common question on Paper with the examination for all other candidates. The marking scheme used for these common elements was identical for the two groups. This document contains the complete marking scheme for both papers for the candidates in the 4 schools. Readers should note that, as with all marking schemes used in the state examinations, the detail required in any answer is determined by the context and the manner in which the question is asked, and by the number of marks assigned to the question or part. Requirements and mark allocations may vary from year to year. Page
Marking scheme for Paper 1 GENERAL GUIDELINES FOR EXAMINERS PAPER 1 1. Penalties of three types are applied to candidates work as follows: Blunders - mathematical errors/omissions (-) Slips - numerical errors (-1) Misreadings (provided task is not oversimplified) (-1). Frequently occurring errors to which these penalties must be applied are listed in the scheme. They are labelled: B1, B, B,, S1, S,, M1, M, etc. These lists are not exhaustive.. When awarding attempt marks, e.g. Att(), note that any correct, relevant step in a part of a question merits at least the attempt mark for that part if deductions result in a mark which is lower than the attempt mark, then the attempt mark must be awarded a mark between zero and the attempt mark is never awarded.. Worthless work is awarded zero marks. Some examples of such work are listed in the scheme and they are labelled as W1, W, etc. 4. The phrase hit or miss means that partial marks are not awarded the candidate receives all of the relevant marks or none. 5. The phrase and stops means that no more work is shown by the candidate. 6. Special notes relating to the marking of a particular part of a question are indicated by an asterisk. These notes immediately follow the box containing the relevant solution. 7. The sample solutions for each question are not intended to be exhaustive lists there may be other correct solutions. Any examiner unsure of the validity of the approach adopted by a particular candidate to a particular question should contact his/her advising examiner. 8. Unless otherwise indicated in the scheme, accept the best of two or more attempts even when attempts have been cancelled. 9. The same error in the same section of a question is penalised once only. 10. Particular cases, verifications and answers derived from diagrams (unless requested) qualify for attempt marks at most. 11. A serious blunder, omission or misreading results in the attempt mark at most. 1. Do not penalise the use of a comma for a decimal point, e.g. 5.50 may be written as 5,50. Page
APPLYING THE GUIDELINES Examples (not exhaustive) of the different types of error: Blunders (i.e. mathematical errors) (-) Algebraic errors : 8x + 9x = 17x or 5p 4 p = 0 p or ( ) = 6 Sign error ( 4) = 1 Decimal errors Fraction error (incorrect fraction, inversion etc); apply once. Cross-multiplication error Operation chosen is incorrect, (e.g. multiplication instead of division) Transposition error, e.g. x k + x = + k or x = 6 x = or 4 x = 1 x = 8; each time. Distribution error (once per term, unless directed otherwise) e.g. (x + 4) = 6x + 4 or 1 ( x ) = 5 6 x = 5 Expanding brackets incorrectly, e.g. (x )( x + 4) = 8x 1 Omission, if not oversimplified. Index error, each time unless directed otherwise Factorisation: error in one or both factors of a quadratic: apply once, e.g. x x = (x 1)( x + ) Root errors from candidate s factors: error in one or both roots: apply once. Error in formula Error(s) in transcribing formulae from tables (assuming it generates mathematical acceptable answer(s)) Serious errors or over simplifications will merit Attempt marks at most ( check relevant section of scheme) Central sign error in uv or u/v formulae Omission of v or division not done in u/v formula (apply once) Vice-versa substitution in uv or u/v formulae (apply once) Quadratic formula (acceptable) and its application apply a maximum of two blunders Slips (-1) Numerical slips: 4 + 7 = 10 or 6 = 4, but 5 + = 15 is a blunder. An omitted round-off or incorrect round off to a required degree of accuracy, or an early round off, is penalised as a slip each time. However an early round-off which has the effect of simplifying the work is at least a blunder Omission of units of measurement or giving the incorrect units of measurement in an answer is treated as a slip, once per part (a), (b) and (c) of each question. Only applies where a candidate would otherwise have achieved full marks Misreadings (-1) Writing 46 for 46 will not alter the nature of the question so M(-1) However, writing 5000 for 506 will simplify the work and is penalised as at least a blunder. Page 4
QUESTION 1 Part (a) 10 marks Att Part (b) 5 (15, 5, 5) marks Att (5,, ) Part (c) 15 (10, 5) marks Att (, ) Part (a) 10 marks Att Express 40 metres as a fraction of 1 kilometre. Give your answer in its simplest form. (a) 10 marks Att 1 km = 1000 m [] or 40 1 4 1 [7] = [9] = [10] 0.04 [7] = [9] = [10] 1000 50 5 100 5 * Accept correct answer without work for full marks. Accept 1:5 * Accept without work 0.04, 4%, 5:1 or 1 5 for [7] marks * Accept without work 40, 100 0, 50 4 10, 5 or 0.4 for [4] marks these only Blunders (-) B1 Mathematical error e.g. conversion/decimal error B Fraction error B No simplification Slips (-1) S1 Simplification not completed to simplest form, between 40 1000 and 1 5 Attempts ( marks) A1 1 40 40 or 1 A Some effort at conversion A Mentions 5 without supporting work Worthless (0) W1 Incorrect answer with no work Page 5
Part (b) 5 (15, 5, 5) marks Att (5,, ) (i) Calculate the value of 57 6 + 80 44 4 1 10 and write your answer correct to three decimal places. (ii) An importer buys an item for 1 sterling when the rate of exchange is 1 = 0 85 sterling. He sells it at a profit of 14% of the cost price. Calculate, in euro, the price for which he sells the item. (i) 15 marks Att 5 57.6 + 80.44 = 18.04 [5] 57 6 + 80 44 18 04 = = 0 0106 = 0 011 4 1 10 1 000 or 1. 10 = 1000 [5] Both [9] 18 04 1 000 [1] = 0 0106184154 [14] = 0.011 [15] * Accept correct answer without work for 15 marks 0.01...[ 14 marks] without work * Accept without work for 1 marks : 57.606, 94.1756-94.176, 61886.807-61886.81, These only [1] [11] [1] [11] [1] Blunders (-) B1 Mathematical error Slips (-1) S1 Incorrect or no rounding off S Numerical slips which are not mathematical errors Misreading (-1) M1 Must not make work easier see guidelines Page 6
(ii) Exchange 5 marks Att Percentage 5 marks Att 1 1 14 51 94 = = 96 40 0 85 0 85 [] + [] [5] + [] or Sterling to Euro exchange Percentage 1 1 = = 60 [5] 0 85 14% of 60 = 6.40 [4] 60 + 6.4 = 96.40 [5] or 60 1.14 = 96.40 [5] ----------------------------------------------------------------------------------------------- Percentage Sterling to Euro exchange 1 0.14 = 0.94 (14% of 1) [4] 51.94 = 96. 40 0 85 [5] 1 + 0.94 = 51.94 [5] 1 1.14 = 51.94 [5] * Accept correct answer without work for full marks [5] + [5] * No penalty if not included Blunders (-) B1 Error in finding percentage e.g. decimal or inversion B Error in currency conversion e.g. incorrect operation Slips (-1) S1 Fails to add percentage profit Attempts ( marks) A1 Any relevant step, may get both Worthless (0) W1 Incorrect answer without work Page 7
Part (c) 15 (10, 5) marks Att (, ) (i) (ii) What sum of money invested at 5% per annum compound interest will amount to 868 in years? Give your answer correct to the nearest euro. A sum of P was invested at r % per annum compound interest. The interest for the first year was 0 The interest for the second year was 8 80 Calculate r and P. (c) (i) 10 marks Att I t 868 F = P(1 + i) 868 = P(1.05) = 7499.8 P = 7500 1.15765 [4] [7] [9] [10] II F 868 868 P = = [4] = n (1 + i) (1 + 0.05) 1.15765 [7] = 7499.8 = 7500 III 868 at end year P year = 868.57 = 868. 57 = 7874. 8 1.05 1.05 7874.8 P year 1 = = 7499.8 = 7500 1.05 IV P year 1 = 100%; P year = 105% ; P year = 110.5% ; P year 4 = 115.765% 115.765% = 868 [7] 868 100 % = 100 = 7499.8 = 7500 1.15765 * Candidates may offer other correct versions r * Formulae and Tables, page 0, use F for A and i for 100 Blunders (-) B1 Mathematical error e.g. percentages or index Note 868(1.05) = 10 050.50 = 10 051 [7] B Incorrect number of years B Fails to finish method IV Slips (-1) S1 Incorrect or no rounding off Attempts ( marks) A1 No compounding of interest - offers 868 15% ( 780) Work must be shown A Answer found by trial and error A 5% or 15% of 868 or mentions 1.05 or 1.15 A4 7499.8 or 7500 without work Worthless (0) W1 Incorrect answer without work Page 8
(c) (ii) 5 marks Att Finding r I II t F = P( 1+ i) 0(1 + i) = 8.80 (1 + i) = 1.04 r = 4 Interest on 0 = 8.80 0 = 8.80 8.80 100 = 4 0 Finding P P( 0 04) = 0 P = 5500 4% = 0 1% = 55 100% = 5500 * Candidates may offer other correct versions r * Formulae and Tables, page 0, use F for A and i for 100 Blunders (-) B1 Mathematical error B Error in finding % from 1.04, method I Attempts ( marks) A1 Finds 8.80 A Finds by trial and error or r = 4% verified A Correct answer without work Worthless (0) W1 Incorrect answer without work Note Award 5 marks for fully correct with work Award marks for some relevant work Otherwise 0 marks Page 9
QUESTION Part (a) 15 marks Att 5 Part (b) 5 (10, 5, 5, 5 )marks Att (,,, ) Part (c) 10 (5, 5) marks Att (, ) Part (a) 15 marks Att 5 Find the values of x which satisfy ( + 4x ), where x N. (a) 15 marks Att 5 ( + 4x ) 6 + 8x [9] or ( + 4x ) or + 4x 11 [9] 8x 16 x [1] 4x 11 x [1] x {1, } * Correct answer without work, full marks * No penalty for including 0 * Accept marked correctly on a number-line Blunders (-) B1 Mathematical error e.g. distribution error, transposing - once if consistent B x not a natural number, e.g x 1 1 gives negative value B Only identifies one element of the solution set, 1 or B4 Verifies one correct value in the inequality, 1 or B5 Stops at x, x = or x < Attempts (5 marks) A1 Any correct relevant multiplication or division A Tests a non solution in the inequality e.g A 0 on its own verified or not Page 10
Part (b) 5 (10, 5, 5, 5) marks Att (,,, ) Solve for x and y x y = 1 x xy = 6. (b) 5 (10, 5, 5, 5) marks Att (,,, ) x y = 1 y = x 1 Step 1 Isolates x or y [10] x xy = 6 Step Forms quadratic equation [5] x x(x 1) = 6 (Penalise error in simplification at Step ) x x + x + 6 0 Step Roots of quadratic [5] x x 6 = 0 ( x )( x + ) = 0 x = or x = y = 5 or y = 5 Step 4 Values of other coordinate [5] * Error(s) in simplification of quadratic equation apply at the Step * If equation at Step becomes linear award at most Att + Att for Steps and 4 * Apply similar scheme if candidate isolates x at Step 1 * Random value(s) of x award attempt marks at most (Step 4) if no work of merit in previous steps Blunders (-) B1 Mathematical error apply at relevant step see note B Incorrect factors Step B Incorrect roots from factor Step B4 Only finds one value of x Step Note B5 will also apply at Step 4 B5 Only finds one value of y Attempts ( or marks) A1 Some relevant work Note: Don t award multiple Attempts to the same piece of work Page 11
Part (c) 10 (5, 5) marks Att (, ) (i) Show, by division, that x + 1 is a factor of x + 4x 89x 0. (ii) Hence, or otherwise, solve the equation x + 4x 89x 0 = 0. (c) (i) 5 marks Att x + 1 x x + 4x + x 0 89x 0 x + x x x 89x + x 90x 0 90x 0 0 0 Blunders (-) B1 Each error in division f B Shows clearly ( 1 ) = 0 Attempts ( marks) A1 Some correct division and stops A Substitutes 1 A Sets up division correctly into expression or mentions f ( 1 ) Page 1
(c) (ii) 5 marks Att x + 4x 89x 0 = 0 (x + 1)( x + x 0) = 0 (x + 1)( x 5)( x + 6) = 0 x =, x = 5, x = 6 1 * Accept candidates answer from part (i) provided it does not over simplify question 5 f 6 fully verified for 4 marks * Accept f ( ) and ( ) Blunders (-) B1 Incorrect factors of quadratic B Incorrect or missing roots from factors, but see S1 Slips ( -1) S1 Omits x = 1 as a root, if left out [4] at most Attempts ( marks) A1 States x = 1 is a root and stops in part (ii) A Attempt at factorising quadratic from (i) A Some correct use of b formula [Note: Stating formula does not merit attempt mark] A4 Correct answers without relevant work A5 Sets up using answer from (i) A6 Finds f ( k), k 5, k 6 Worthless (0 marks) W1 Attempts at factorising x + 4x 89x 0 = 0 such as x (x + 4) = 89x + 0 W Differentiation Page 1
QUESTION Part (a) 15 (10, 5) marks Att (, ) Part (b) 0 (10, 10) marks Att (, ) Part (c) 15 (10, 5) marks Att (, ) Part (a) 15 (10, 5) marks Att (, ) Given that (b + a) = t(6 a), calculate the value of a when t = and b = 4. (a) 15 (10, 5) marks Att, I Substitution for t and b: 10 marks Evaluation of a: 5 marks (b + a) = t(6 a) ( 4 + a ) = (6 a) 1 + a = 18 a 6a = 0 a = 5 II b + a = 6t at substitution merits [10] a + at = 6t b a( + t) = 6t b 6t b 6 4 18 + 1 a = = = = 5 rest of work [5] + t + 6 * Accept correct answer without work. * Once a candidate has substituted correctly for t and b he/she is entitled to [10] marks Blunders (-) B1 Mathematical error e.g transposition, distribution, from 5 marks B Substitution error B Substitutes one value only B4 Interchanges t and b Attempts ( marks) A1 Some correct effort at isolating /evaluating a, from 5 marks Worthless (0) W1 Incorrect answer without work Page 14
Part (b) 0 (10, 10) marks Att (, ) Solve for x 5( x + 1) = ( x + 1) + 5. Give your answer correct to two decimal places. (b) 0 (10, 10) marks Att (, ) Step 1, forming quadratic equation: 10 marks Step, solving quadratic equation: 10 marks I 5( x + 1) = ( x + 1) + 5 Let y = x + 1 5y = y + 5 5y y = 5 [10] 5y y 5 = 0 ± 4 4(5)( 5) ± 104 ± 10.198 1.198 8.198 y = [4] = = [7] = or (5) 10 10 10 10 y = 1.198 or y = 0.8198 [9] x = 0. or x = 1.8 [10] II 5( x + 1) = ( x + 1) + 5 5x + 10x + 5 = x + 7 5x + 8x = [10] 5x + 8x = 0 8 ± 64 4(5)( ) 8 ± 104 8 ± 10.198.198 18.198 x = [4] = = [7] = or (5) 10 10 10 10 x = 0. or x = 1.8 [10] * Accept candidate s quadratic equation for second 10 marks if not factorisable * If quadratic equation reduced to a linear attempt marks at most in Step Blunders (-) B1 Mathematical error each time B Error in use of quadratic formula to a maximum of (Step ) Slips ( -1) S1 Fails to round off or rounds off incorrectly once only S Early rounding off that affects answer S Fails to find x from y in method I Attempts ( marks) A1 Some effort at multipling out equation - Step 1 Method 11 A If equation becomes linear, maximum possible mark from Step is Attempt A Solves a factorisable quadratic equation even if they use formula A4 Attempts to factorise the quadratic Page 15
Part (c) 15 (10, 5) marks Att (, ) (i) + is a root of the equation x 4x + c = 0, where c is a real number. Find the value of c and write down the other root. (ii) The equation x + 10x + k = 0 has equal roots. Find the value of the real number k and write down the value of each root. (i) 10 marks Att x 4x + c ( + ) = 0 4( + ) + c = 0 4 + 4 + 8 4 + c = 0 c = 1 [7] Other root: [10] * Accept any valid method Blunders (-) B1 Mathematical error B Using decimals c 1 Attempts ( marks ) A1 Some correct substitution A Some correct substitution into b formula A States nd root is and stops must be in surd form A4 c =1 without work even if second root found (ii) 5 marks Att I Let root = p ( x p)( x p) = 0 x px + p p = 10 p = 5 k = ( 5) = 5 [4] II b 4ac = 0 100 4(1)( k) = 0 k = 5 [4] x = 5, [5] * Accept any valid method Blunders (-) B1 Mathematical error Note: x + 10x + 5 [Att ] Slips (-1) ( x + 5)( x + 5) [Att ] S1 Value of root omitted k = 5 [4] x = 5 [5] Attempts ( marks) A1 Correct answer for k without work A Roots found without work A Correct answer without work Page 16 [4]
QUESTION 4 Part (a) 15 marks Att 5 Part (b) 0 (10, 10) marks Att (, ) Part (c) 15 (5, 5, 5) marks Att (,, ) Part (a) 15 marks Att 5 Given that i = 1, simplify ( 4 + i)( i) and write your answer in the form x + yi, where x, y R. (a) 15 marks Att 5 (4 + i)( i) = 4( i) + i( i) = 1 4i + 6i i = 1 + i + = 14 + i [9] [1] [14] [15] Blunders (-) B1 Mathematical error B Error in multiplication maximum of blunders B i 1, mis-use of i or avoids use of i B1 and B can apply B4 Mixes up real and imaginary terms Slips (-1) S1 Numerical slips Attempts (5 marks) A1 Any correct relevant multiplication Worthless (0) W1 Incorrect answer without work Page 17
Part (b) 0 (10, 10) marks Att (, ) Let u = 4 + i and w = 6 8i (i) Find the value of the real number k such that u = k w. (ii) Express u w in the form x + yi. (b) (i) 10 marks Att u = k w 4 + i = k 6 8i 16 + 9 = k 6 + 64 5 = k 100 1 5 5 k = accept k = = 10 100 Note modulus: One correct 5 or 100 [4] Two correct 5 and 100 [7] * No penalty for using 8 for 8 in formula * Accept distance from (4, ) to (0, 0) or ( 6, 8) to (0,0) Blunders (-) B1 Incorrect formula e.g. omitted B Incorrect substitution e.g. has ( i ) and /or B Mathematical error ( 8i ) in a + b - once only Attempts ( marks) A1 Incorrect formula with some correct substitution A Plots u and/or w A Correct answer without work A4 Correct modulus formula and stops A5 Correct substitution for u and/or v Worthless (0) W1 Incorrect answer without work Page 18
(b) (ii) 10 marks Att w 6 8i 6 8i 4 i = = u 4 + i 4 + i 4 i [] 4 18i i + 4i = 16 + 9 [7] 0 50i = 5 [9] = 0 i or 50i = 0 5 [10] Note: 0 required in answer * Can use multiple of conjugate i.e. n( 4 i), n a real number, n 0 * Calculates numerator or denominator, merits 4 marks * Calculates numerator and denominator, merits 7 marks Blunders (-) B1 i 1 or misuse of i B Mathematical error in multiplying out numerator maximum 1 blunder B Mathematical error in multiplying out denominator maximum 1 blunder B4 Error in formation of u w at final stage e.g. may multiply numerator and denominator Attempts ( marks) A1 Substitutes for u and/or w and stops A Finds conjugate of u and stops A Any correct relevant multiplication Page 19
Part (c) 15 (5, 5, 5) marks Att (,, ) Let z = a + bi, where a, b R. Find the value of a and the value of b for which z 10i = ( i) z. (c) 15 (5, 5, 5) marks Att (,, ) I z 10i = ( i) z ( a + bi) 10i = ( i)( a + bi) [5] a + bi 10i = a + bi ai bi a + bi 10i = a + bi ai + b [5] II Real parts: a = a + b a = b Imaginary parts: b 10 = b a a + b = 10 a + b = 10 10b = 10 b = 1 a = [5] z 10i = ( i) z z = 10i zi a + bi 10i = i( a + bi) [5] III a + bi = 10i ai bi a + bi = 10 i ai + b [5] Real parts: a = a + b a = b Imaginary parts: b 10 = b a a + b = 10 a + b = 10 10b = 10 b = 1 a = [5] z 10i = ( i) z z = 10i zi z + zi = 10i z( 1+ i) = 10i [5] 10i z = [5] 1+ i z = + i = a + bi a = and b = 1 [5] Blunders (-) B1 Mathematical error - once per step Attempts ( marks) A1 Any relevant work for a given step Page 0
QUESTION 5 Part (a) 10 marks Att Part (b) 0 (10, 5, 5) marks Att (,, ) Part (c) 0 (10, 5, 5) marks Att (,, ) * Do not penalise notation Part (a) 10 marks Att The first term of a geometric sequence is 4 and the common ratio is 0 5. Write down the first five terms of the sequence. (a) 10 marks Att I T 1 = a = 4, T = ar = 4 0.5 = T = ar = 4 0.5 = 1 or [ 0.5] T 4 = ar = 4 0.5 = 0.5 or [1 0.5] 4 4 T 5 = ar = 4 0.5 = 0.5 or [0.5 0.5] II List 4,, 1, 0.5, 0.5 * Accept correct answers with no work * Accept in fractional form Blunders (-) B1 Decimal error once if consistent e.g. 0.5 taken as 5 or r = B Indices error each time B Error in formula see guidelines Misreading (-1) M1 r taken as 0.05 Attempts ( marks) A1 Identifies a as 4 and/or r as 0.5 and stops A States T 1 = 4 Worthless (0) W1 Treats as an arithmetic sequence but see A1 and A W Incorrect answer(s) without work Note: Answers without work 1 term correct marks terms correct 4 marks terms correct 4 marks 4 terms correct 7 marks 5 terms correct 10 marks Page 1
Part (b) 0 (10, 5, 5) marks Att (,, ) In an arithmetic series, the first term is 6 and the fifth term is. (i) Find d, the common difference. (ii) Find T 14, the fourteenth term. (iii) Find S 0, the sum of the first twenty terms. * Answers to parts of questions must be clearly identified (i) 10 marks Att I T 1 = a = 6 [] T 5 = a + 4d = [4] 4d = 6 [7] d = 4 [10] II 6, 10, 14, 18, [7] * Accept correct answer without work * Acceptable formula - see guidelines Blunders (-) B1 Mathematical error Slips ( -1) S1 Numerical slips Attempts ( marks) A1 Correct relevant work A 4 = 16 and stops or d = 16 (ii) 5 marks Att I T 14 = a + 1d = 6 + 1(4) = 6 + 5 = 58 II List: 6 + 10 + 14 + 18 + + 6 + 0 + 4 + 8 + 4 + 46 + 50 + 54 + 58 (Assume final term is answer, otherwise must indicate term 14) * Accept candidates d from (i) * Accept correct answer without work Blunders (-) B1 Mathematical error B Incorrect term from list B Finds S 14 by formula Slips (-1) S1 Numerical slips Attempts ( marks) A1 Identifies a as 6 for this part of question Worthless (0) W1 Treats as a geometric series but may have identified a as 6 as part of this question Page
(iii) 5 marks Att I 0 S 0 = ( a + 19d ) = 10( 1 + 76) = 10( 88) = 880 II List: 6+10+14+18++6+0+4+8+4+46+50+54+58+6+66+70+74+78+8 = 880 * Accept candidate s answers from (i) and (ii) Blunders (-) B1 Finds T0 and stops B Writes complete list but fails to sum B Finds S 14 from (ii) B4 Incorrect number of terms in list Slips (-1) S1 Numerical slips Attempts ( marks) A1 Identifies a and/or d A Correct answer without work. Worthless (0) W1 Treats as a geometric series but identification of a will merit A1 Page
Part (c) 0 (10, 5, 5) marks Att (,, ) In a geometric series, the fourth term is 9 and the seventh term is 4. (i) Find r, the common ratio. (ii) Find a, the first term. (iii) Find S 8, the sum of the first eight terms. (c) (i) 10 marks Att I T 4 = ar = 9 [] Note: ar 4 = 9 for T 4 6 T 7 = ar = 4 [4] ar 7 = 4 for T 7 6 ar = 4 r = 7 ar 9 [7] r = 7 etc. Accept r = [10] II List [ 1, 1,,] 9, 7, 81, 4 [7] r = [10] Blunders (-) B1 Mathematical error B Error in use of formula Attempts ( marks) A1 T 4 or T 7 expressed in algebraic form and stops A Finds 4/9 = 7 and stops A Correct answer without work A4 Partial list (c) (ii) 5 marks Att I II 1 ar = 9 a( ) = 9 7a = 9 a = 1, 1,, 9 a = [] [5] [] [5] * Accept candidate s r from (i) as long as it does not oversimplify work Blunders (-) B1 Mathematical error Attempts ( marks) A1 Any relevant step A Correct answer without work but allow if full list given in (i) Page 4
(c) (iii) 5 marks Att I n a( r 1) 1 (6561 1) 1 (6560) 80 1 S8 = = = = = 109 r 1 1 [] [5] II List: 1 + 1 + + 9 + 7 + 81 + 4 + 79 = 109. [] [5] * Accept candidate s a and r from (i) and (ii) provided they do not over simplify work Blunders (-) B1 Mathematical error B Fails to sum list in method II B Missing or extra terms in list method Slips (-1) S1 Numerical slips Attempts ( marks) A1 Finds T 8 A Identifies a as 1/ in this part A Correct answer without work Worthless (0) W1 Treats as an arithmetic series but identification of a will merit A Page 5
QUESTION 6 Part (a) 10 marks Att Part (b) 0 (10, 10) marks Att (, ) Part (c) 0 (10, 10) marks Att (, ) Part (a) 10 marks Att Let h ( x) = x + 1, where x R. Write down a value of x for which h (x) = 50. Part (a) 10 marks Att h ( x) = 50 x + 1 = 50 x = 49 x = ± 7 [] [7] [10] * Accept correct answer without work. Accept 49 * Only one value for x is required. Blunders (-) B1 Mathematical errors B Evaluates h ( 50) = 501 Attempts ( marks) A1 Unsuccessful trial and error, e.g. h ( 5) = 5 + 1 A Any correct relevant step Worthless (0) W1 50( x + 1) whether continues or not W Incorrect answer with no work W Differentiates Page 6
Part (b) 0 (10, 10) marks Att (, ) Let g ( x) = 1, where x R and x. x (i) Copy and complete the following table: x 0 1 1 5 1 75 5 5 4 g(x) 1 4 (ii) Draw the graph of the function g in the domain0 x 4. (b) (i) 10 marks Att * Values of g( x) = x calculated ( all/some correct ) misreading which oversimplifies, Att * Accept values as fractions; must be 1 or 1 Singleton Single number e.g. for x = 1.5 accept 1 1 but not 0.5 1.5 Blunders (-) 1 1 B1 Treats the function as f ( x) =, even if g ( x) = written. x x The relevant values for f ( x) = 1 are: x (0, undefined), ( 1.5, 4 ), (.5, 14 9), (, 5 ),, 1 B Treats as g ( x) = avoids error with sign x + Slips (-1) S1 Each un-simplified value to a maximum of Attempts ( marks) A1 Copies table and stops A Treats g (x) as x Note: Answers without work 1 value correct marks values correct 4 marks values correct 4 marks 4 values correct 7 marks 5 values correct 10 marks x 0 1 1 5 1 75 5 5 4 g(x) 0 5 1 4 4 1 0 5 ( 4 7 ) 4 Page 7
(b) (ii) 10 marks Att 4 1-1 1 4 * Consider graph as having features LHS/branch, asymptote (actual or implied) and RHS/branch. * Asymptote x = need not be drawn; an implied vertical asymptote (or visible gap ) will suffice * Has graph of x : oversimplified Att * Accept candidate s values from (i) if not over simplified * Ignore any graph errors outside the given range e.g. graph cutting the horizontal asymptote * Points plotted and not joined and not showing asymptote [4] marks * Only one branch without a vertical asymptote [4] marks at most Blunders (-) B1 Left and right branches joined B Points joined incorrectly B LHS or RHS branch missing or asymptote missing or not implied B4 Serious incorrect scaling of axes e.g. equal distance on x-axis for given values Slips (-1) S1 Each point clearly incorrectly plotted or each point clearly omitted to a maximum of per side Attempts ( marks) A1 Draws axes and stops A One point correctly plotted A Any mention of asymptotes A4 Table from (i) does not give rise to two branches - - -4 Note: If B1 or B applied at (i) graph at (ii) will merit attempt mark at most Page 8
Part (c) 0 (10, 10) marks Att (, ) Let f ( x) = x 5, where x R and x 0. x (i) Find f (x), the derivative of f (x). (ii) Find the co-ordinates of the two points at which the tangent to the curve y = f (x) is parallel to the line y = 6x. (c) (i) 10 marks Att I f ( x) = x 5 1 5 = x 5x f ( x) = 1+ 5x = 1+ x x [4] [10] II f ( x) = x 5 x 5 = x x [4] u = x 5 v = x du dv = x = 1 dx dx x(x) ( x 5)(1) f ( x) = [10] x (Maximum of blunders in differentiation simplification not necessary, penalise errors in part (ii)) * Candidates may offer other correct versions e.g. may treat x 5 as a v u. * Apply differentiation penalties as per guidelines * Answer need not be simplified, penalise in (ii) if necessary but see B 5 0 5 * f ( x) = 1 or 1 merits 4 marks i.e. not treated as a quotient 1 1 x Blunders (-) B1 Differentiation error once per term B Indices error B Simplification error at start of method II Attempts ( marks) A1 Any correct step at simplification and stops Page 9
(c) (ii) 10 marks Att 5 f ( x) = 6 1+ = 6 5 = 5x x = 1 x = x [] [7] ± 5 f ( 1) = 1 = 1 5 = 4. 1 Point ( 1, 4) 5 f ( 1) = 1 = 1+ 5 = 4. 1 Point ( 1, 4) [10] 1 * Accept candidates answer from (i) unless it is oversimplified * Penalise simplification of f (x) errors in this part if necessary Blunders (-) B1 Mathematical errors B f ( x) 6 B Only one solution found for x, B4 will also apply B4 Only one value of f (x) /y found Slips (-1) S1 Numerical slips Attempts ( marks) A1 Mentions slope of y = 6x is 6 A Answer from (i) = 6 and stops A Mentions connection of slope and derivative and stops Page 0
QUESTION 7 Part (a) 15 marks Att 5 Part (b) 0 (5, 15) marks Att (, 5) Part (c) 15 (5, 5, 5) marks Att (,, ) Part (a) 15 marks Att 5 Differentiate x 6x + 1 with respect to x. (a) 15 marks Att 5 dy = x 6 dx * Correct answer without work or notation: full marks. * If done from first principles, ignore errors in procedure just mark the answer. * Only one non zero term correct, award 1 marks Blunders (-) B1 Differentiation error once per term Attempts (5 marks) A1 A correct step in differentiation from 1 st principles A A correct coefficient or a correct index of x in one of the term(s) A Mentions f (x ) dx Worthless (0) W1 No differentiation Page 1
Part (b) 0 (5, 15) marks Att (, 5) (i) Differentiate 5 x with respect to x from first principles. (ii) Given that ( dy y = x 4)(x 1), find the value of when x =. dx (b) 5 marks Att f ( x) = 5 x I f ( x + h) = 5 ( x + h) = 5 x h II f ( x + h) f ( x) = 5 x h (5 x) = h f ( x + h) f ( x) h III = = h h f ( x + h) f ( x) lim = h 0 h y = 5 x I y + y = 5 ( x + x) = 5 x x y = 5 x II y = x y = x y III lim = x 0 x * Accept use of ( x h) Blunders (-) B1 Any error once per step I, II or III Note: Must have correct LHS and RHS Attempts ( marks) A1 f ( x ± h) on LHS or some substitution of x ± h for x on RHS, or equivalent; these only A Mentions x or y or similar Worthless (0) W1 Answer without work; no attempt at first principles Page
(b) (ii) 15 marks Att 5 I or II y = ( x 4)(x 1) y = ( x 4)(x 1) = x u 4 v = x 1 y = x x 1x + 4 [9] du dv = x = [9] dx dx dy dy = ( x 1)( x) + ( x 4)( ) [1] = 9x x 1 [1] dx dx At x = At x = dy dy = ( 6 1)( 4) + ( 4 4)( ) = 0 [15] = 9 (4) () 1 = 6 4 1 = 0 [15] dx dx u du dv * Uses merits 9 marks at most allow for u = =... and v = =... better than A5 v dx dx Blunders (-) B1 Differentiation error B Errors in expanding brackets once only unless over simplifies. B Error in substitution, once only Slips (-1) S1 Numerical slips Attempts (5 marks) A1 u and/or v correctly identified and stops (I) A Any correct differentiation A At least one term multiplied correctly A4 Uses x + 4 even if completed correctly A5 dy = ( x)() dx Worthless (0) W1 Substitutes x = into y and stops W uv formula written and stops Page
Part (c) 15 (5, 5, 5) marks Att (,, ) The speed, v, of an object at time t is given by v = 96 + 40t 4t where t is in seconds and v is in metres per second. (i) At what times will the speed of the object be 96 metres per second? (ii) What will the acceleration of the object be at t = 5 seconds? (iii) At what value of t will the acceleration become negative? * Units: Penalise as per guidelines. * No retrospective marking. * No penalty for incorrect notation. * If parts of (c) are unlabelled, and the context doesn t identify which part is which, assume the questions were answered in sequence from (c) (i) to (c) (iii). (c) (i) 5 marks Att 96 = 96 + 40t 4t 40t = 0 t( t 10) = 0 t = 0, t = 10 s 4t * One or both answers correct without work, Att Blunders (-) B1 Equation 96 B Incorrect factors B Incorrect roots from factors but see S Slips (-1) S1 No units or incorrect units S t = 0 not included Attempts ( marks) A1 Attempt at factorising A Trial and error on 96 + 40t 4t even if correct Worthless (0) W1 Differentiation Page 4
(c) (ii) 5 marks Att dv a = = 40 8t [4] dt - At t =.5 a = 40 8( 5) = 0 m s [5] * Acceleration as second derivative of v i.e correct d v / dt merits 4 Blunders (-) B1 Differentiation error Slips (-1) S1 No units or incorrect units S Substitution error Attempts ( marks) A1 Mentions dv / dt or similar Worthless (0) W1 Substitutes t =.5 into v (c) (iii) 5 marks Att I dv < 0 40 8t < 0 8t < 40 t > 5 dt or II Acceleration negative (deceleration) after velocity reaches it maximum or similar dv = 0 40 8t = 0 t = 5 dt Acceleration negative after t = 5 * Correct answer without work, Att. Blunders (-) B1 Error solving inequality (I) or equation (II) Slips (-1) S1 t 5 Attempts ( marks) A1 Any correct value offered A Has acceleration d v / dt = 8, therefore acceleration is always negative Worthless (0) W1 t = 8 from d v / dt = 8 W Attempts to solve 96 + 40t 4t < 0 Page 5
QUESTION 8 Part (i) 15 marks Att 5 Part (ii) 10 marks Att Part (iii) 10 marks Att Part (iv) 10 marks Att Part (v) 5 marks Att * Assume answering in order (i)...(v) No retrospective marking Part (i) 15 marks Att 5 Let f ( x) = x x + 1, where x R. (i) Find f ( ), f ( ), f (0), f () and f (). (i) 15 marks Att 5 f ( x) = x x + 1 f ( ) = ( ) ( ) + 1 = 7 + 9 + 1 = 17 [5] f ( ) = ( ) ( ) + 1 = 8 + 6 + 1 = 1 [6] f (0) = (0) (0) + 1 = 0 + 0 + 1 = 1 [9] f () = () () + 1 = 8 6 + 1 = [1] f () = () () + 1 = 7 9 + 1 = 19 [15] * Correct answers without work, full marks. * Don t penalise extra values e.g f (1) and/or f ( 1). Blunders (-) B1 Mathematical errors, each time if different B Use x for x Slips (-1) S1 Arithmetic slips to maximum of Attempts (5 marks) A1 Only finds one value and stops A Some correct substitution into f (x) A f (x) with some correct substitution Worthless (0) W1 All incorrect answers without work Note: Answers without work 1 point/value correct 5 marks points/values correct 6 marks points/values correct 9 marks 4 points/values correct 1 marks 5 points/values correct 15 marks Page 6
Part (ii) 10 marks Att Find f (x), the derivative of f (x). (ii) 10 marks Att f ( x) = x * Correct answer without work or notation, full marks. * If done from first principles, ignore errors in procedure just mark the answer. * Only one non zero term correct, award 7 marks Blunders (-) B1 Differentiation error once per term. Attempts ( marks) A1 A correct step in differentiation from 1 st principles A A correct coefficient or a correct index of x. Page 7
Part (iii) 10 marks Att Find the co-ordinates of the local maximum point and of the local minimum point of the curve y = f (x). (iii) 10 marks Att f ( x) = x =0 [] x 1 = 0 ( x + 1)( x 1) = 0 x = 1 or x = 1. [7] f ( x) = x x + 1 f ( 1) = ( 1) ( 1) + 1 = 1+ + 1 = f (1) = (1) (1) + 1 = 1 + 1 = 1 Local maximum ( 1, ), local minimum ( 1, 1). [10] * Accept candidate s f (x) from (ii) but see A1 * Accept implied = 0 if subsequent work supports it. * Accept distinguishing max from min by comparing y-ordinates. Second derivative not required. * Correct answers without calculus, Att at most. May be from graph. Blunders (-) B1 f ( x) 0 (but see nd asterisk) B Error finding roots B Only finds one root ( B4 will also apply) B4 Error finding f (x) value e.g. fails to find f (x) value or only finds one value or does not use f (x) Slips (-1) S1 Numerical slips S Does not distinguish between maximum and minimum, or indentifies incorrectly Attempts ( marks) A1 f (x) linear and continues A f (x ) Worthless (0) W1 f (x) = 0, whether continues or not Page 8
Part (iv) 10 marks Att Draw the graph of the function f in the domain x. (iv) 10 marks Att 15 10 5 x from previous parts unless oversimplified. * If candidate recalculates points, apply slips and blunders as per guidelines. * Seven (7) points required Only uses 5 points from (i) [8] * Accept candidate s values of (, f ( x) ) - - -1 1-5 -10-15 Blunders (-) B1 Scale error, serious B Points not joined or joined incorrectly or joined with a series of straight lines B Axes not in standard form Slips (-1) S1 Each point incorrectly plotted or omitted Attempts ( marks) A1 Plots f (x) or graph of a non-cubic function A Answers from part (iii) transferred to this part, carries forward max and min values A Effort at calculation of a point with some substitution e.g. f (0) A4 Scaled and labelled axes and stops Page 9
Part (v) 5 marks Att Find the range of values of k for which the equation x x + 1 = k has three real solutions (roots). (v) 5 marks Att 1 k * Accept answer consistent with candidate s graph if cubic * Accept any valid solution * Accept answer clearly indicated on graph * Accept answer using words rather than symbols and [, 1] or [ 1, ] * Accept 1 < k < Blunders (-) B1 Inequalities not as stated Attempts ( marks) A1 One correct end-point identified A Solves f (x) = 0 or finds one correct value of k A Mentions local maximum or local minimum or max. and min. Page 40
Coimisiún na Scrúduithe Stáit State Examinations Commission LEAVING CERTIFICATE 010 MARKING SCHEME MATHEMATICS (PROJECT MATHS) PAPER ORDINARY LEVEL
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010. M18 Coimisiún na Scrúduithe Stáit State Examinations Commission Leaving Certificate Examination 010 Mathematics (Project Maths) Paper Ordinary Level Monday 14 June Morning 9:0 1:00 00 marks Model Solutions Paper Note that the model solutions for each question are not intended to be exhaustive there may be other correct solutions. Any examiner unsure of the validity of the approach adopted by a particular candidate to a particular question should contact his/her advising examiner. Page 4
Section 0 Area and Volume (old syllabus) 50 marks Answer Question 1 from this section. Question 1 (a) A circle is inscribed in a square as shown. The radius of the circle is 9 cm. (50 marks) (i) Find the perimeter of the square. 9 cm l = 9 8 = 7 cm or l = 18 4 = 7cm [5 marks] (ii) Calculate the area of the square. A = (18) = 4 cm [5 marks] (b) The diagram shows a sketch of a field ABCD that has one uneven edge. At equal intervals of 5 m along [BC], perpendicular measurements are made to the uneven edge, as shown on the sketch. D A 7 5 m 6 m 5 m 8 5 m 7 m 4 5 m 9 5 m B 5 m C (i) Use Simpson s rule to estimate the area of the field. Area h ( F + L + ΣO + 4ΣE) = 5 ( 7.5 9.5 + (5 + 7) + 4(6 + 8.5 + 4.5) ) = 5 ( + (1) + 4(19) = 5 ( 17 + 4 + 76) = 5 ( 117) + [10 marks] 17 = 195 m. [5 marks] Page 44
(ii) The actual area of the field is 00 m. Find the percentage error in the estimate. 5 Percentage error: 100 =.5%. [5 marks] 00 (c) (i) The diameter of a solid metal sphere is 9 cm. Find the volume of the sphere in terms of π. V = r 4 4 π = ) π (4.5 = 11.5π cm [10 marks] (ii) The sphere is melted down. All of the metal is used to make a solid shape which consists of a cone on top of a cylinder, as shown in the diagram. The cone and the cylinder both have height 8 cm. The cylinder and the base of the cone both have radius r cm. r cm 8 cm Calculate r, correct to one decimal place. 8 cm V = 1 π r h + π r h = 11.5π 4 r (8) = 11.5 11.5 r = = 11.9 4 8 r =.7 r.4 cm [10 marks] Page 45
Section A Concepts and Skills 15 marks Answer all five questions from this section. Question (a) A line crosses the x-axis at x = and the y-axis at y =. Find the equation of the line. Slope = 4 1 (5 marks) OR Points (,0) and (0,) 0 Slope = = 0 [15 marks] - -1 1 4 5-1 - Equation: y = x + OR y 0 = ( x ) y = x + 6 x + y = 6 [5 marks] (b) The equations of two lines l 1 and l are: l : x+ y = 8 1 l : 6x y = 15. Determine whether these lines are perpendicular. Justify your answer clearly. Slope l 1 = 1 Slope l = 1 = 1 l1 l [5 marks] m 1 m = ( ) Page 46
Question (a) A circle has centre (0, 0) and passes through the point (, 4). (5 marks) (i) Find the equation of the circle. x () + y 5 = r + (4) = r = r Equation : x + y = 5 [10 marks] (ii) Find the co-ordinates of the two points at which the circle crosses the y-axis. Let x = 0 y = 5 y = ±5 Points are (0, 5) and (0, -5) [5 marks] (b) A circle has centre (, 4) and touches the y-axis. Find the equation of the circle. Radius = Equation: ( x ) + ( y 4) = () ( x ) + ( y 4) = 4 [10 marks] 6 5 4 1 1 4 Page 47
Question 4 (5 marks) (a) Using a calculator, or otherwise, find the mean and standard deviation of the data in the following frequency table. x 0 0 40 50 f 16 8 6 0 Mean = 5 Standard deviation = 9.84885780 (using n) = 9.89847458 (using n - 1) [5 marks] (b) Below is a stem-and-leaf plot of the heights of a group of students, in centimetres. 1 1 5 6 14 0 0 1 14 6 6 7 8 15 0 1 15 5 5 6 7 Key: 1 means 1 cm. (i) How many students are in the group? 0 students [10 marks] (ii) What is the range of heights in the group? 157 cm 1 cm = 4 cm [5 marks] (iii) What percentage of the students are between 145 cm and 154 cm in height? 10 = 50% [5 marks] 0 Page 48
Question 5 (a) (5 marks) Helen has enough credit to download three songs from the internet. There are seven songs that she wants. (i) How many different possible selections of three songs can she make? 7 C = 5 (ii) If there is one particular song that she definitely wants, how many different selections can she now make? 6 C = 15 [5 marks (parts (i) and (ii)] (b) (i) Two fair coins are tossed. What is the probability of getting two heads? 1 1 1 = [5 marks] 4 (ii) Two fair coins are tossed 1000 times. How often would you expect to get two heads? 1000 4 1 = 50 times [5 marks] (c) Síle hands Pádraig a fair coin and tells him to toss it ten times. She says that if he gets ten heads then she will give him a prize. The first nine tosses are all heads. How likely is it that the last toss will also be a head? Tick the correct answer, and give a reason. Reason: Extremely unlikely Fairly unlikely 50-50 chance Fairly likely Almost certain e.g.: Independent Trials Fair coin Not influenced by previous tosses [10 marks] Page 49
Question 6 (5 marks) The diagram shows a triangle ABC in which AB = 6 cm, CB = 10 cm, and ABC = 50. A B C (a) Calculate the area of triangle ABC, correct to the nearest cm. 1 Area = (6)(10) sin 50 = 0 sin50⁰ =.981 cm [15 marks] (b) Calculate the length of [AC], correct to one decimal place. AC = (10) + (6) = 16 10cos50 = 58.86548684 (10)(6) cos 50 AC = 7. 678 7.7 cm [5 marks] (c) The triangle A BC is the image of triangle ABC under the enlargement with centre B and scale factor. Find the area of A BC, correct to the nearest cm. Image Area = () [] = 9[] = 07 cm OR BA ' = 18 cm, BC' = 0 cm, 1 Area = (18)(0) sin 50 07 cm [5 marks] Page 50
Section B Contexts and Applications 15 marks Answer Question 7, Question 8, and either Question 9A or Question 9B. Question 7 Probability and Statistics (40 marks) The table below gives motor insurance information for fully licensed, 17 to 0-year-old drivers in Ireland in 007. All drivers who had their own insurance policy are included. Number of drivers Number of claims Average cost per claim Male 964 977 6108 Female 674 581 6051 Questions (a) to (e) below refer to drivers in the table above only. (Source: adapted from: Financial Regulator. Private Motor Insurance Statistics 007.) (a) What is the probability that a randomly selected male driver made a claim during the year? Give your answer correct to three decimal places. 977 964 0.101 [10 marks] (b) What is the probability that a randomly selected female driver made a claim during the year? Give your answer correct to three decimal places. 581 674 0.086 [10 marks] (c) What is the expected value of the cost of claims on a male driver s policy? 0.101 6108 = 616.91 [5 marks] (d) What is the expected value of the cost of claims on a female driver s policy? 0.086 6051 = 50.9 [5 marks] Page 51
(e) The male drivers were paying an average of 1688 for insurance in 007 and the female drivers were paying an average of 104. Calculate the average surplus for each group, and comment on your answer. (Note: the surplus is the amount paid for the policy minus the expected cost of claims.) Male Female 1688-616.91 = 1071.09 104-50.9 = 50.61 Comment: e.g. Male drivers are generating a much higher surplus. Insurance companies are making far more money from male drivers. [5 marks] (f) A 40-year-old female driver with a full license has a probability of 0 07 of making a claim during the year. The average cost of such claims is 900. How much should a company charge such drivers for insurance in order to show a surplus of 175 per policy? Expected claims value = 900 0.07 = 7 Charge: 7 + 175 = 448 [5 marks] Page 5
Question 8 Geometry and Trigonometry (40 marks) Windows are sometimes in the shape of a pointed arch, like the one shown in the picture. A person is designing such an arched window. The outline is shown in the diagram below the picture. The centre for the arc AB is C and the centre for the arc AC is B. BD = 4 metres and DE = 1 8 metres. (a) Show that ABC = 60. AB = BC (Radius of arc AC) AC = BC (Radius of arc AB) ABC is an equilateral triangle A ABC = 60 [10 marks] B C (b) Find the length of the arc AB. Give your answer in metres, correct to three decimal places. 60 Length = π (1.8) 60 = 6 1 11.09755 =1.884955 1.885 m [5 marks] D E (c) Find the length of the perimeter of the window. Give your answer in metres, correct to two decimal places. Perimeter = (.4) + 1.8 + (1.885) = 4.8 + 1.8 +.77 = 10.7 m [10 marks] Page 5
(d) Find the height of the window. Give your answer in metres, correct to two decimal places. A x + ( 0.9) = (1.8) x + 0.81 =.4 x =.4 x = 1.5558 1.56 m x F 0.9 1.8 C Height of Window = 1.56 +.4 =.96 m [5 marks] (e) Make an accurate scaled drawing below of the outline of the window, using the scale 1:0. That is, 1 cm on your diagram should represent 0 cm in reality. A B C [10 marks] D E Page 54
Question 9A Probability and Statistics (45 marks) Students in two schools one in County Kerry and the other in County Offaly were arguing about which county had the nicest weather in the summer. They agreed to record the highest temperature at each school on ten randomly selected days during the summer of 009. The results were as follows: Temperature at Kerry school (/ C) Temperature at Offaly school (/ C) 18 5 17 17 8 1 18 0 19 1 17 6 17 5 17 17 18 4 18 6 17 1 16 9 16 9 19 8 19 0 17 6 17 1 17 0 (a) Construct a back-to-back stem-and-leaf plot of the above data. 9 9 16 8 6 5 1 1 17 0 6 5 18 0 4 6 19 0 1 8 0 1 1 key: 17 0 means 17 0 [15 marks] (b) State two differences between the two distributions. Examples Kerry temperatures are generally lower Greater range of temperatures in Offaly Offaly has an outlier Temperatures in Kerry are largely in the 17⁰C bracket [5 marks] (c) Perform a Tukey Quick Test on the data, stating clearly what can be concluded. Lower Tail = Upper Tail = 5 Tail Count = 7 Conclusion: In general, in the summer of 009, temperatures in Offaly were higher than in Kerry. [5 marks] Page 55
(d) The students in Offaly looked also at the amount of sunshine. They recorded the number of hours of sunshine each day in July 009. The data are summarised in the table below. Hours of sunshine 4 6 8 10 1 Number of days 11 1 0 9 0 1 Draw a cumulative frequency curve to represent this data, using the scale indicated. 0 Number of days (cumulative) 0 10 4 6 8 10 1 Hours of sunshine [10 marks] (e) Use your cumulative frequency curve to estimate: (i) the median number of hours of sunshine 5.5 (ii) the number of days with more than 7 hours of sunshine. 4 [5 marks] (f) The mean amount of sunshine per day in Offaly in July generally is 4 4 hours. A day is chosen at random from the days in July 009, as described in part (d) above. What is the probability that the amount of sunshine on that day was less than the mean? 1 Probability 1 [5 marks] (Data in this question adapted from Monthly Weather Bulletin, July 009, at www.met.ie.) Page 56
Question 9B Geometry and Trigonometry (45 marks) (a) The photograph shows the Dockland building in Hamburg, Germany. The diagram below is a side view of the building. It is a parallelogram. The parallelogram is 9 metres high. The top and bottom edges are 88 metres long. Photo by NatiSythen. Wikipedia Commons. License: CC-SA A B D C. (i) Find the area of this side of the building. Area = 9 88 = 55 m [15 marks] (ii) If BD = AD, find BC. Let BC = x B x x = (9) = 777 + (44) 9 x x = BC 777 = 5.697485 E 44 C BC 5.697 m [5 marks] Page 57
(iii) The lines BC and AD are parallel. Find the distance between these parallel lines. Let [BC] = base and let y = perpendicular height (distance between BC and AD) BC y =55 5.697 y = 55 y = 48.47 m [5 marks] (b) There is a theorem on your geometry course that can be used to construct the tangent to a circle at a given point on the circle. State this theorem and use it to construct the tangent to the circle shown at the point P. Theorem: Each tangent is perpendicular to the radius that goes to the point of contact. [5 marks] P [5 marks] (c) In the diagram, the line l is a tangent to the circle. Find the values of x, y and z. l y z x = 90 x y = 48 4 z = 4 [10 marks] Page 58
Marking scheme Paper, Section 0 (Question 1) N.B. This page applies only to Question 1. The scheme for this question is identical to that used for candidates who are not involved in Project Maths. GENERAL GUIDELINES FOR EXAMINERS 1. Penalties of three types are applied to candidates work as follows: Blunders - mathematical errors/omissions (-) Slips - numerical errors (-1) Misreadings (provided task is not oversimplified) (-1). Frequently occurring errors to which these penalties must be applied are listed in the scheme. They are labelled: B1, B, B,, S1, S,, M1, M, etc. These lists are not exhaustive.. When awarding attempt marks, e.g. Att(), note that any correct, relevant step in a part of a question merits at least the attempt mark for that part if deductions result in a mark which is lower than the attempt mark, then the attempt mark must be awarded a mark between zero and the attempt mark is never awarded.. Worthless work is awarded zero marks. Some examples of such work are listed in the scheme and they are labelled as W1, W, etc. 4. The phrase hit or miss means that partial marks are not awarded the candidate receives all of the relevant marks or none. 5. The phrase and stops means that no more work is shown by the candidate. 6. Special notes relating to the marking of a particular part of a question are indicated by an asterisk. These notes immediately follow the box containing the relevant solution. 7. The sample solutions for each question are not intended to be exhaustive lists there may be other correct solutions. Any examiner unsure of the validity of the approach adopted by a particular candidate to a particular question should contact his/her advising examiner. 8. Unless otherwise indicated in the scheme, accept the best of two or more attempts even when attempts have been cancelled. 9. The same error in the same section of a question is penalised once only. 10. Particular cases, verifications and answers derived from diagrams (unless requested) qualify for attempt marks at most. 11. A serious blunder, omission or misreading results in the attempt mark at most. 1. Do not penalise the use of a comma for a decimal point, e.g. 5.50 may be written as 5,50. Page 59