question 1

JUNIOR CERTIFICATE EXAMINATION 2012

MARKING SCHEME MATHEMATICS HIGHER LEVEL PAPER 1

Page 1

MARKING SCHEME JUNIOR CERTIFICATE EXAMINATION 2012 MATHEMATICS - HIGHER LEVEL - PAPER 1 GENERAL GUIDELINES FOR EXAMINERS 1.

Penalties of three types are applied to candidates’ work as follows:  Blunders - mathematical errors/omissions (-3)  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, B2, B3,…, S1, S2,…, M1, M2,…etc. These lists are not exhaustive.

2.

When awarding attempt marks, e.g. Att(3), 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.

3.

Worthless work is awarded zero marks. Some examples of such work are listed in the scheme and they are labelled as W1, W2,…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.

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.

12.

Do not penalise the use of a comma for a decimal point, e.g. €5·50 may be written as €5,50.

Page 2

BONUS MARKS FOR ANSWERING THROUGH IRISH Bonus marks are applied separately to each paper as follows: If the mark achieved is 225 or less, the bonus is 5% of the mark obtained, rounded down. (e.g. 198 marks  5% = 9.9  bonus = 9 marks.) If the mark awarded is above 225, the following table applies: Bunmharc Marc Bónais Bunmharc Marc Bónais (Marks obtained) (Bonus Mark) (Marks obtained) (Bonus Mark) 226

11

261 – 266

5

227 – 233

10

267 – 273

4

234 – 240

9

274 – 280

3

241 – 246

8

281 – 286

2

247 – 253

7

287 – 293

1

254 – 260

6

294 – 300

0

Page 3

QUESTION 1 Part (a) Part (b) Part (c)

10 marks 20 marks 20 marks

Att (2,2) Att (3,3) Att (2,3,2)

Part (a) (a) (i) (ii)

10(5,5) marks

Att (2,2)

List the divisors of 30. State which of these divisors are prime numbers.

(a) (i) Divisors of 30 *



5 marks 1, 2, 3, 5, 6, 10, 15, 30.

Att 2

Accept correct answer for full marks. No work required, no 

Slips (-1) S1 Each incorrect or missing number to a maximum of -3, must have at least one correct Misreadings (-1) M1 Misreads 30, but continues correctly, provided oversimplification does not occur Attempts (2 marks) A1 Any correct divisor A2 Any relevant step Worthless (0) W1 Incorrect answer, but note A1 W2 Multiples of 30 (a) (ii) 5 marks Prime divisors of 30  2, 3, 5. * Answer may be dependent on part (i)

Att 2

Slips (-1) S1 Each incorrect or missing prime number to a maximum of -3, must have at least one correct S2 Includes 1 as a prime number Attempts (2 marks) A1 Any correct prime divisor A2 Shows some knowledge of a prime number Worthless (0) W1 Incorrect answer with no work of merit

Page 4

Part (b) (i)

(ii)

20 (10,10) marks €900 is invested for two years at 3% per annum compound interest.



Find the value of the investment at the end of the second year

John has a gross weekly wage of €600. After tax his net weekly wage is €554.



Calculate his tax credits if he is taxed at the standard rate of 20%.

(b) (i) I

10 marks P1

=

€900

Att 3 I1=

P2

=

€900 + €27 = €927

A2

= =

€927 + €2781 €95481

€27 = €2781 7m

I2 =

1 €900 1

or €900 1

· 03

€900 1 · 03 A = €900(1·0609) A = €95481

3m 4m 7m 10m

Do not penalise for the omission of € symbol Ignore missing brackets if final answer is not affected Final answer of €954 is 4 marks

Blunders (-3) B1 B2 B3 B4 B5 B6 B7 B8 B9

3m 4m 7m 10m

II

* * *

Att (3,3)

Correct answer, no work shown  3% of an incorrect number Decimal error Incorrect operation Mathematical error Error in squaring Error in formula Precedent error Fails to calculate final step

Slips (-1) S1 Numerical error to a max of -3 Misreadings (-1) M1 Misreads a digit provided it doesn’t oversimplify the question

Page 5

Attempts (3 marks) A1 Relevant correct formula which is not in log tables A2 Finds 3% of some number other than €900 and stops A3 Divides by 100 A4 900 × 3 A5 = 54, oversimplification A6 Identifies P = €900 and/or R =3% or ·03 A7 3% = or ·03 A8 Any correct substitution A9 Any relevant step Worthless (0) W1 Incorrect answer no work shown W2 1 with no correct substitution

Page 6

(b) (ii) I

10 marks Gross income Net income Tax paid

= = = = Tax @20% = = Tax credits = = Steps are interchangeable: Tax @20% = = Tax paid = = Tax credits = = II Tax @20% = = = =

Att 3

€600 €554 ................ Given in question €600 − €554 3m €46 4m 20% of €600 €120 7m €120 – €46 €74 10m 20% of €600 €120 €600 − €554 €46 €120 – €46 €74

3m 4m 4m 7m 7m 10m

20% of €600 €120 €600 − €120 €480 €554 − €480 €74

3m 4m 7m 10m

Blunders (-3) B1 B2 B3 B4 B5 B6 B7

Correct answer, no work shown  Decimal error Incorrect operation Inversion Mathematical error Finds 20% of an incorrect figure Fails to calculate final step

Slips (-1) S1 Numerical error to a max of -3 Misreadings (-1) M1 Misreads a digit provided it doesn’t oversimplify the question Attempts (3 marks) A1 Finds 20% correctly of a figure other than €600 and stops A2 20% = or equivalent A3 Divides by 100 A4 Shows some knowledge of tax credits e.g. writes “Tax payable = total tax – tax credits” A5 Any relevant step Worthless (0) W1 Incorrect answer no work shown W2 600 × 554 W3 No work of merit Page 7

Part (c) (i)

20 (5,10,5) marks



Att (2,3,2)

By rounding to the nearest whole number, estimate the value of 3  89 × 7  24  8  94 8  52  3  65

(ii)



Evaluate

(iii)



Simplify

3  89 × 7  24  8  94 , correct to two decimal places. 8  52  3  65 5





2 5 

8





2  5 without the use of a calculator.

Express your answer in the form a  b c , where a, b, c  ℕ.

(c) (i)

=

5 marks

47  9 94 28  3 5

2m 2m

=

2m

= 5 Blunders (-3)

5m

B1 B2 B3 B4 B5 B6 B7 B8 B9 B10

Correct answer, no work shown  Rounds incorrectly, once if consistent Incorrect operation Inversion Mathematical error Precedent error Square root error Invalid cancellation Sign error Fails to calculate final step, stops at

Slips (-1) S1 Numerical errors to a max of -3 Misreadings (-1) M1 Misreads a digit provided it doesn’t oversimplify the question Attempts (2 marks) A1 Some correct rounding A2 5·17 with work Worthless (0) W1 Incorrect answer, no work shown e.g. 5·17 without work W2 No work of merit

Page 8

Att 2

(c) (ii)

10 marks

Att 3

3  89 × 7  24  8  94 8  52  3  65

= (28·1636 − 2·989983278) ÷ 4· 87 = 25·17361672 ÷ 4· 87 = 5·16912 = 517 Blunders (-3)

B1 B2 B3 B4 B5 B6 B7 B8 B9

Correct answer, no work shown  Decimal error Incorrect operation Inversion Mathematical error Precedent error Square root error Sign error Stops at 25·17361672 ÷ 4·87

Slips (-1) S1 Numerical errors to a max of -3 S2 Stops at 5169 or 5·1691 or 5·16912 or similar S3 Early rounding if it affects final answer, but note A2 Misreadings (-1) M1 Misreads a digit provided it doesn’t oversimplify the question Attempts (3 marks) A1 Some correct calculation A2 Rounds to whole numbers and continues A3 Any relevant step Worthless (0) W1 Incorrect answer, no work shown W2 No work of merit

Page 9

4m 7m 9m 10m

(c) (iii) 5





2 5 

8



2 5



=

10  25  16  40

=

10  5  4  2 10

= 1  3 10 Blunders (-3) B1 B2 B3 B4 B5 B6

5 marks

2m 2m 5m

Correct answer, no work shown  Distribution error Sign error Error in surds, once if consistent Mathematical error Fails to finish

Slips (-1) S1 Numerical error to a max of -3 Misreadings (-1) M1 Misreads a digit provided it doesn’t oversimplify the question Attempts (2 marks) A1 10·486832298, no surds used, with work shown A2 Any relevant attempt at handling surds A3 √ = power of ½ A4 Any relevant work Worthless (0) W1 Incorrect answer no work shown W2 √5 = 2·236067977 and/or √8 = 2·828427125 and/or √2 = 1·414213562 and stops W3 No work of merit

Page 10

Att 2

QUESTION 2 Part (a) 10 marks Part (b) 20 marks Part (c) 20 marks Part (a) 10 marks Fuel consumption in a car is measured in litres per 100 km. Alan’s car travels 1250 km on a tank of 68 litres.



Att 3 Att (3,2,2) Att (3,2,2) Att 3

Calculate his car’s fuel consumption in litres per 100 km.

(a)

10 marks I 1250 ÷ 100 = 1250 68 litres for 1250 ‘100km’ 

Att 3

3m 4m 68 12  50

7m

= 544 litres / 100km. 10m II 1250 km = 68 litres ........given in question 1 km = 68 ÷ 1250 3m = ·0544 4m 100km = ·0544 × 100 7m = 544 litres / 100km 10m *

× 100 and stops is worth 4m; ·1838 with work is 7m; 18·38 with work is 3m

Blunders (-3) B1 B2 B3 B4 B5 B6 B7 B8

Correct answer, no work shown  Decimal error Incorrect multiplier (check method) Incorrect division Mathematical error Inversion Incorrect operation Fails to complete last step

Slips (-1) S1 Numerical errors to a max of -3 Misreadings (-1) M1 Misreads a digit provided it doesn’t oversimplify the question Attempts (3 marks) A1 Multiplies or divides by 100 A2 1250 ÷ 68 A3 Some knowledge of relationship between fuel consumption and distance indicated A4 Any relevant step Worthless (0) W1 Incorrect answer no work shown W2 No work of merit Page 11

Part (b)

20 (10,5,5) marks

Att (3,2,2)

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} is the universal set. P = {3, 5, 6, 8, 10}, Q = {2, 4, 6, 8, 10, 12} and R = {2, 5, 6, 7, 9, 12} are three subsets of U. (i) Represent the above information on a Venn diagram. Hence list the elements of: (ii) (P  Q  R)΄ (iii) (P  Q) \ R.

(b) (i) Venn diagram:

10 marks U

Att 3

P

Q 8 4 10 6 12 2

3 5 7 R

* *

1 9

11

Ignore notation Sets P, Q, R may be positioned differently from above

Slips (-1) S1 Each incorrect or missing or misplaced element in Venn diagram each time but note A2 S2 Universal box not drawn on diagram Misreadings (-1) M1 Misreads a digit provided it doesn’t oversimplify the question Attempts (3 marks) A1 Draws a Venn diagram with three intersecting circles and stops A2 Any correct entry A3 Universal box with two intersecting circles A4 Three intersecting circles A5 Universal box with a correct entry Worthless (0) W1 Rectangle only W2 Circle with no correct entry W3 Two intersecting circles with no correct entry

Page 12

(b) (ii) 5 marks (P  Q  R)΄ = {1, 11} * Answer may be dependent on candidate’s answer to (b)(i) * Answer may be indicated on diagram * Ignore notation

Att 2

Blunders (-3) B1 Elements of (P  Q  R) given as answer i.e. {2, 3, 4, 5, 6, 7, 8, 9, 10, 12} Slips (−1) S1 Each incorrect or missing or misplaced element to a maximum of -3, must have at least one element correct; note B1 Attempts (2 marks) A1 One correct element A2 Any relevant step Worthless (0) W1 Incorrect answer with no work of merit, note B1 W2 { } but note *1 W3 Draws diagram again, with no further work of merit

(b) (iii) 5 marks (P  Q) \ R = {8, 10} *Answer may be dependent on candidate’s answer to (b)(i) *Answer may be indicated on diagram *Ignore notation

Att 2

Slips (-1) S1 Each incorrect or missing or misplaced element to a maximum of -3, must have at least one element correct Attempts (2 marks) A1 Shades/indicates the correct region on Venn diagram, but elements not clearly identified A2 (R  Q) \ P = {2,12} or candidate’s equivalent or (P  R) \ Q = {5} or candidate’s equivalent A3 Any relevant step Worthless (0) W1 { } but note *1 W2 Draws diagram again, with no further work of merit W3 Incorrect answer with no work of merit, note A2

Page 13

Part (c) 20 (10,5,5) marks Att (3,2,2) An electronics company imports tablet computers from China at a cost of 696 Yuan (元) per tablet. (i)  Find the cost of each tablet, in euro, if €1 = 87 元. The company must also pay a shipping cost on each tablet imported. By selling a tablet at €10540, the company can make a profit of 24%. (ii)  Find the shipping cost per tablet. The company imports 1000 tablets from China. It sells 600 of them at €10540 each (i.e. at a profit of 24%) and the remainder at a profit of 15%. (iii)



Find the overall profit, in euro, made by the company.

(c) (i)

10 marks I Cost in euro

= =

* *

696 87 €80

II €1 = 87 元 Given €1 ÷ 87 = 元 ·114942528 ·114942528 × 696 = €80 € symbol not necessary in answer (€ 1 ÷ 8·7) × 696 is worth 4m

7m 10m

3m 4m 7m 10m


Correct answer, no work shown  Decimal error Incorrect numerator Incorrect denominator Mathematical error Inversion Fails to finish

Slips (-1) S1 Numerical error to a max of -3 S2 Early rounding of decimal if it affects final answer Misreadings (-1) M1 Misreads a digit provided it doesn’t oversimplify the question Attempts (3 marks) A1 Any relevant step Worthless (0) W1 Incorrect answer no work shown W2 No work of merit

Page 14

Att 3

(c) (ii) I 124% = €105·40 1% = 105·40 ÷ 124 = ·85 100% = ·85 × 100 = €85 €85 − €80 = €5 shipping cost

5 marks

II (€80 + x) (1·24) = €105·40 99·20 +1·24 x = 105·40 1·24 x = 105·40 − 99·20 1·24 x = 6·20 x = 6·20 ÷ 1·24 x = €5 III Each tablet with profit = €80 × 124% = €99·20 €105·40 − €99·20 Shipping charge = €6·20 ÷ 1·24 = €5 IV Profit + 124% shipping charge = €105·40 - €80 = €25·40 Profit = €25·40 – 124% shipping charge Shipping charge = x Profit = €25·40 – 1·24 x % Profit = × 100 24 =

·

– ·

100

24 = (25·40 – 1·24 x) × 24 ÷

= 25·40 – 1·24%

24 × = 25·40 – 1·24 x 19·20 = 25·40 – 1·24 x 19·20 − 25·40 = − 1·24 x − 6·20 = − 1·24 x 124% shipping charge = €6·20 Shipping charge = €6·20 ÷ 1·24 = €5 *

Accept candidate’s figure from (c)(i)


Correct answer, no work shown  Decimal error Incorrect operation Distribution error Mathematical error Incorrect value for cost price based on previous figures Fails to finish

Page 15

Att 2

Slips (-1) S1 Numerical errors to a max of -3 Misreadings (-1) M1 Misreads a digit provided it doesn’t oversimplify the question Attempts (2 marks) A1 Shows some knowledge of % profit A2 Any relevant step Worthless (0) W1 Incorrect answer no work shown W2 No work of merit

Page 16

(c) (iii) 600 × 105·40 400 × 85 × 1·15 Total 85 × 1000 Overall profit * *

5 marks = = = = =

63,240 39,100 102,340 85,000 €17,340

Accept candidate’s values from (c)(i) and (ii) Candidates may use other variations in calculating the overall profit

Blunders (-3) B1 B2 B3 B4 B5

Correct answer, no work shown  Decimal error Incorrect operation Mathematical error Fails to finish

Slips (-1) S1 Numerical errors to a max of 3 Misreadings (-1) M1 Misreads a digit provided it doesn’t oversimplify the question Attempts (2 marks) A1 Multiplies by 1000 A2 Multiplies by 600 A3 Multiplies by 400 A4 Finds 15% or states 15% = A5 Any relevant step Worthless (0) W1 Incorrect answer no work shown W2 No work of merit

Page 17

Att 2



Att 3 Att (2,3,2) Att (2,2,2,2)

Part (a)

10 marks

Att 3



Given that 1 billion is a thousand million, find the sum of €36 billion and €700 million. Give your answer in the form a × 10n where n  ℕ and 1  a < 10.

(a)

10 marks I 36  109 + 700  106

=

Att 3

36  109 + 07  109 4m

=

43  109

II €3,600,000,000 + €700,000,000 = €4,300,000,000

4m

=

43  109

III €3·6 billion + €0·7 billion = €4·3 billion

4m

=

43  109

IV 36  109 + 7  108 (or equivalent) = 3,600,000,000 + 700,000,000 = 4,300,000,000

4m

=

43  109


Correct answer, no work shown  Decimal error Answer not given in correct form Index error Mathematical error

Misreadings (-1) M1 Misreads a digit provided it doesn’t oversimplify the question Attempts (3 marks) A1 Indicates some knowledge of indices e.g. gets 109 A2 Converts either expression to a whole number and stops A3 Writes 0·7 billion and stops A4 Any relevant step Worthless (0) W1 Incorrect answer, no work shown W2 Attempts to add and shows no knowledge of indices, but note A1 W3 No work of merit

Page 18

Part (b)

(i) (ii) (iii)

20 (5,10,5) marks



Simplify

 

6 x 2  17 x  12 . 3x  4

Factorise

4c 2  3d  2cd  6c .

Express in its simplest form:

(b) (i) I (3 x  4)(2 x  3) 3x  4

Att (2,3,2)

5 3  . x3 x2

5 marks

=

2x – 3

II

2x − 3______ 3x – 4 6x2 – 17x + 12 6x2 − 8x − 9x + 12 − 9x + 12 0 III (6x2 – 17x + 12) ÷ (3x – 4) (6x2 – 9x −8x + 12) ÷ (3x – 4) [3x(2x – 3) – 4(2x – 3)] ÷ (3x – 4) [(3x – 4)( 2x – 3)] ÷ (3x – 4) = 2x − 3 IV (6x2 – 17x + 12) ÷ (3x – 4) (6x2 – 8x −9x + 12) ÷ (3x – 4) [2x(3x – 4) – 3(3x – 4)] ÷ (3x – 4) [(2x – 3)( 3x – 4)] ÷ (3x – 4) = 2x − 3

*

2 3 3 4 and continues is one blunder - B4. It will also incur B6 or B7. All other attempts to factorise apply B2, B3 and/or B4.


Correct answer, no work shown  Incorrect factors of 6 in method I Incorrect factors of +12 in method I Incorrect factors leading to an incorrect middle term in method I Mathematical error Incorrect cancellation Fails to finish i.e. no cancellation in method I

Page 19

Att 2

Slips (-1) S1 Numerical errors to a maximum of -3 Attempts (2 marks) A1 Some effort at factorising A2 Sets up division A3 Multiplies instead of dividing, with at least one correct term A4 Finds guide number (72) in methods III and IV and stops A5 Quadratic with some correct substitution A6 Sets up quadratic and identifies a, b or c A7 Uses quadratic formula and stops at correct roots (x = and x = A8 Any relevant step

)

Worthless (0) W1 Incorrect answer, no work shown W2 ( ) ( ) W3 Work of no merit

(b) (ii) 10 marks Att 3 I II 4 3 2 6 Given 4 3 2 6 Given 4 6 2 3 3m 4 2 6 3 3m 2 2 3 2 3 7m 2 2 3 2 7m 2 3 2 2 2 3 10m 10m * Accept any of the following for full marks with work (with or without brackets): (2c − d) and (2c + 3) [the word 'and' is written down] (2c − d) or (2c + 3) [the word 'or' is written down] (2c − d) , (2c + 3) [a comma is used]

Blunders (-3) B1 B2 B3 B4 B5 B6

Correct answer, no work shown  Error in factorising any pair of terms, apply once if consistent. 3 Incorrect last step e.g 2 2 3 2 Incorrect common factor and continues e.g. 2 2 3 2 Incorrect common factor and continues e.g. 2 2 also apply. 3 2 3 or similar Fails to finish, stops at 2 2

Slips (-1) 2 S1 2 S2

2 2

3 3

Attempts (3 marks) A1 Some effort at factorising e.g. groups or attempts to pair Worthless (0) W1 Incorrect answer, no work shown W2 No work of merit

Page 20

3 3 . B3 or B6 will

(b) (iii)

= = =

5 marks 5 3  x3 x2 5( x  2)  3( x  3) ( x  3)( x  2) 5 x  10  3x  9 ( x  3)( x  2) 2x  1 ( x  3)( x  2)

2m 2m 5m


Correct answer, no work shown  Incorrect common denominator or mishandles common denominator Mishandles numerator Distribution error Mathematical error Fails to combine like terms in final answer Reads as and continues

Slips (-1) S1 Numerical slips to a max of -3 Attempts (2 marks) A1 Correct common denominator and stops A2 No denominator used A3 Any relevant step Worthless (0) W1 Incorrect answer, no work shown W2 W3 No work of merit

etc

Page 21

Att 2

Part (c) 20 (5,5,5,5) marks Att (2,2,2,2) Roisín cycled from Wicklow to Bray, a distance of 30 km. She left Wicklow at 10:30 and arrived in Bray at 12:20, having stopped in Greystones for 20 minutes. Greystones is 22 km from Wicklow. (i) Roisín’s average speed between Wicklow and Greystones was x km/h. Write an expression in x for the time taken for this part of her journey. (ii) Her average speed for the second part of her journey, between Greystones and Bray, was 6 km/h slower than her speed between Wicklow and Greystones. Write an expression in x for the time it took to complete the second part of her journey. (iii) Write an equation in x to represent the above information. (iv)



Solve the equation to find Roisín’s speed for each part of the journey.

(c) (i)

5 marks

Time (1) *

=

Att 2

22 x


Blunders (-3) B1 Inversion Attempts (2 marks) A1 Writes 22 and/or x A2 Speed = Worthless (0) W1 Incorrect answer, no work shown (c) (ii)

5 marks

Time (2) *

=

8 x6


Blunders (-3) B1 Inversion B2 Uses x + 6 B3 Incorrect operation Attempts (2 marks) A1 Any combination of two of the following x, 6, 8 A2 Speed = A3 30 – 22 or 8 A4 Any relevant step Worthless (0) W1 Incorrect answer, no work shown; note A3 Page 22

Att 2

(c) (iii) 5 marks 12:20 – 20 minutes – 10:30 = 1 hour 30 minutes 2m = 1·5 hours 2m 22 8 Total time = 15 5m  x x6 * Accept candidates’ expressions from (c)(i) and (c)(ii)

* *

Accept correct answer for full marks. No work required, no  If no work, or no work of merit, at parts (i) and/or (ii) but states above, award Att 2 and Att 2 from parts (c)(i ), and/or (c)(ii) here

Blunders (-3) B1 Sign error in setting up equation e.g. has B2 B3

Att 2

=

Expression not equal to 1·5 or , but note S1 Uses 1·3

Slips (-1) S1

= 90

Attempts (2 marks) A1 Incorrect expression but uses data from (c) (i ) and (c) (ii) A2 Constructs an equation or expression using at least two of the following: , answer (c) (i ), answer (c) (ii) A3 Attempt to subtract times A4 Any relevant step Worthless (0) W1 1·3 only W2 No work of merit

Page 23

(c) (iv)

5 marks

I 22 1.5

II

= 1.5 132 8 1·5 39 132 0 26 4

x=4

but

*

Att 2

= 9

88 0 22 0 x = 22

x = 22 x=4

 

2m

44x – 264 + 16x = 3x2 − 18x 3x2 − 78x + 264 = 0 x2 − 26x + 88 = 0 4 22 0 x=4 x = 22 speed (1) speed (1)

= 22  speed (2) x − 6 = 22 – 6 = 16 = 4  speed (2) x – 6 = − 2 not possible 5m

Accept candidate’s equation from (c) (iii)


Correct answer, no work shown  Sign error Distribution error Transposition error Mathematical error Correct factors and stops Incorrect factors Errors using quadratic formula

Slips (-1) S1 Numerical errors to a max of -3 S2 Does not (or cannot) conclude that speed of – 2 is not possible S3 Doesn’t find speeds between Greystones and Bray for second part of the journey Attempts (2 marks) A1 Linear equation merits attempt at most A2 Any correct relevant step A3 Quadratic formula with some correct substitution A4 Attempt at factorising Worthless (0) W1 Incorrect answer and no work shown W2 ( )( ) W3 No work of merit

Page 24



Att 3 Att (2,2,3) Att (3,2,2)

Part (a)

10 marks

Att 3



Graph on the number line the solution set of 4 – x ≥ 2x – 5, x  ℕ.

(a)

10 marks I 4–x 4+5 9 9 3 3

2x–5 2x+x 3x x x

7m

II 4–x – x − 2x –3x 3x x x

2x–5 – 5− 4 –9 9 9÷3 3

7m

x

3  {1,2,3}

7m

≤

10m Blunders (-3) B1 B2 B3 B4 B5 B6 B7

Correct answer no work shown  Transposition error Mishandles inequality x  R or x Z indicated Mathematical error No number line drawn Values outside of range graphed, note S2

Slips (-1) S1 Numerical errors to a maximum of -3 S2 Includes 0 Misreadings (-1) M1 Excludes equals in inequality P

25

Att 3

Attempts (3 marks) A1 Tests any value in the inequality and stops A2 Draws a number-line A3 No inequality, solves equation to get x = 3 A4 Any relevant step Worthless (0) W1 List given with no correct value W2 No work of merit

Part (b) 20 (5,5,10) marks Att (2,2,3) Electricity is charged to a consumer at a day rate and at a night rate. Day rate units are charged at 14 cent per unit and night rate units are charged at 7 cent per unit. A consumer uses a total of 1100 units for a billing period, at a cost of €12950. (i) By letting x equal the number of day rate units used and y equal the number of night rate units used, write two equations to represent the above information. (ii)



Solve these equations to find the number of each type of unit used.

(b) (i)

10 (5,5)marks

x + y = 1100 014x + 007y = 12950 * * * *

Att 2,2 5m

or

14x + 7y = 12950

5m

Two equations to mark in (b)(i) Each equation is marked separately Each equation is worth 5marks, attempt 2 Equations sufficient, no  in question

Blunders (-3) B1 Incorrect term B2 Decimal error Attempts (2,2 marks) A1 Mentions x or y or 14x or 7y or ·14x or ·07y A2 Effort at creating an equation equal to 1100 or 129·50 or 12950 A3 Any relevant step

Page 26

(b) (ii)

10 marks

Att 3

I x+y 14x + 7y

= 1100 (−7) = 12950

−7x – 7y = − 7700 14x + 7y = 12950 7x = 5250 x = 5250 7

x + y = 1100 (−14) 14x + 7y = 12950

x + y = 1100 (−1) 14x + 7y = 12950 (÷ 7)

−14x – 14y = −15400 14x + 7 y = 12950 −7y = −2450 y=

− x – y = −1100 2x + y = 1850 x = 750

x = 750

y = 350

x + y = 1100

x + y = 1100 750 + y = 1100 y = 1100 −750 y = 350

x + y = 1100 x + 350 = 1100 x = 1100 − 350 x = 750

750 + y = 1100 y = 1100 −750 y = 350

x = 750 y = 350 II x  y  1100 y  1100  x 14 x  7(1100  x)  12950 14 x  7700  7x  12950 7x  12950  7700 7x  5250 x  750 y  350

*1 *2 *3 *4 *5 *6

x  y  1100 x  1100  y 14(1100  y)  7y  12950 15400 14 y  7y  12950 7y  12950 15400 7y  2450 7y  2450 y  350 x  750

Accept candidate’s answers from part (i) provided oversimplification does not occur Apply only one blunder deduction B1 or B2 to any errors in establishing the first equation Finding the second variable is subject to a maximum deduction of 3 marks Correct values of x and y without algebraic work, both verified in both equations merits full marks Correct values of x and y without algebraic work not verified or not fully verified in both equations merits attempt mark only Equations may also be solved by substituting x = or y =

Blunders (-3) B1 Error(s) in establishing the first equation in terms of x i.e. (7x  5250) through elimination by cancellation or elimination by substitution B2 Error(s) in establishing the first equation in terms of y i.e. (7y  2450) through elimination by cancellation or elimination by substitution B3 Distribution error B4 Transposition error B5 Mathematical error B6 Fails to find second variable Page 27

Slips (-1) S1 Numerical errors to a max of 3 Misreadings (-1) M1 Misreads a digit provided it doesn’t oversimplify, apply each time to a max of 3 Attempts (3 marks) A1 Any correct manipulation of either equation and stops Worthless (0) W1 Incorrect answer, no work shown W2 Trial and error, but note *4 and *5 W3 No work of merit

Page 28

Part (c)

20 (10,5,5) marks

(i)



(ii)



(iii)



Solve the equation x2 – 6x + 4 = 0, giving your answer in the form of a  b , where a, b  ℕ. Hence, or otherwise, find two values for p for which (3 + p)2 – 6(3 + p) + 4 = 0. Show that the sum of the two values of p is zero.

(c) (i) I x2 – 6x + 4 = 0

10 marks

√

a=1

=

=

b=−6

c=4

√ √

Att (3,2,2)

=

√

= 3

√5

II x2 – 6x + 4 = 0 x2 – 6x + 9 − 5 = 0 (x – 3)2 – (√5)2 = 0 (x – 3 – √5) (x – 3 + √5) = 0 x – 3 – √5 = 0 or x – 3 + √5 = 0 x = 3 + √5 or x = 3 __ √5

B1 B2 B3 B4

Correct answer no work shown  Error in quadratic formula once only Error in substitution once only Error when applying quadratic formula once only

B5

Invalid cancelling or stops at

B6 B7 B8

Error in completing the square in method II Error in factors in method II Error in establishing roots or incorrect format for roots in method II

√

or similar

Slips (-1) S1 Each numerical error to a max of -3 Attempts (3 marks) A1 Identifies a, b or c correctly and stops A2 Some attempt at factorising e.g. (x ) (x

)

Worthless (0) W1 Incorrect answer without work W2 No work of merit

Page 29

Att 3

(c) (ii) I (3 + p) 2 – 6(3 + p) + 4 = 0 From c (i) x = 3 + p So 3 + p = 3 √5 p = √5

5 marks

Given 2m 5m

II (3 + p) 2 – 6(3 + p) + 4 = 0 Given 9 + 6p + p2 – 18 – 6p + 4 = 0 2m 2 p –5=0 (p − √5 ) (p + √5 ) = 0 p = √5 and p = − √5 5m *Accept candidate’s answers from part (c)(i)


Correct answer no work shown  Sign error Transposition error Mathematical error Distribution error Finds one solution only

Slips (-1) S1 Numerical errors to a max of -3 Attempts (2 marks) A1 States x = 3 + p and stops A2 Some use of answer from part (i) A3 Some correct multiplication in II A4 Any relevant step Worthless (0) W1 Incorrect answer without work W2 No work of merit

Page 30

Att 2

(c) (iii)

* *

5 marks

Att 2

Sum of roots = 5 + ( 5 ) = 0 Accept candidate’s answers from part (c)(ii) above If candidate’s p1 + p2 ≠ 0 and candidate acknowledges this with work, award full marks

Blunders (-3) B1 States √5 - √5 ≠ 0 B2 Incorrectly states that candidate’s p1 + p2 = 0 B3 Decimal error B4 Fails to finish Slips (-1) S1 Numerical errors to a max of -3 Attempts (2 marks) A1 Some use of candidate’s answers from part (ii) A2 Some relevant step Worthless (0) W1 0 only or = 0 only W2 No work of merit

Page 31

QUESTION 5 Part (a) Part (b) Part (c) Part (a)



10 marks 10 marks 30 marks 10 marks

Given that 4d 

(a)

Att 3 Att (2,2) Att (3,3,2,2) Att 3

2c a  , write a in terms of c and d. 3 5 10 marks

I 4

Given

4

7m

5 4

or

a=

or a = 20d −

II 4 15 15(4d) = 15( ) + 15( ) 60

10

3

4m

60

10

3

7m

or a = 20d −

or

= 5 4

Other methods may be used


10m

Given 3m 3m

a= *

Att 3

Correct answer no work shown  Mishandles numerator Incorrect LCM in II (any multiple of 15 acceptable) Transposition error Mathematical error Fails to finish

Slips (-1) S1 Numerical errors to a max of -3 Misreadings (-1) M1 Misreads a digit provided it doesn’t oversimplify the question Attempts (3 marks) A1 Correct LCM only A2 Any relevant step Worthless (0) W1 Incorrect answer, no work shown W2 No work of merit Page 32

10m

Part (b) (i)



(ii)



10 (5,5)marks 4 2 Find the value of 3 x 2  5 x  , when x = . x 3 5x  2 x 1 Solve the equation   1. 3 4 5 marks

(b) (i) I 2 4 2 2 3   5   3 3 2   3 4 10 3 3 6 6 3 2 6 4

=

2m

12 2

5m OR

II 5

3 3 3

5 2 3

4 4

5

2 3 8 20 9 4 9 2 3 12 4 9

2 3

4

2 3 12 36 9 9 2 3 24 9 2 3 3 24 2 9 4

Page 33

Att (2,2)

Att 2


Correct answer no work shown  Incorrect substitution e.g. x = · 6, gives 4·74, but see S2 Incorrect handling of fractions Drops denominator or mishandles denominator Mathematical error Mishandles numerator Distribution error Fails to finish

Slips (-1) S1 Numerical errors to a max of -3 S2 If decimal is used and answer can be rounded to 4 correctly, but is not; otherwise B2 Misreadings (-1) M1 Misreads a digit provided it doesn’t oversimplify the question Attempts (2 marks) A1 Some correct substitution A2 No denominator used II A3 Any relevant step Worthless (0) W1 Incorrect answer, no work shown W2 No work of merit

Page 34

(b) (ii) I

5 marks 1

5

3

2

1

4

Given

4(x −1) – 3(5x + 2) = 12 (1) 4x −4 – 15 x – 6 = 12 − 11 x −10 = 12 − 11 x = 12 + 10 − 11 x = 22 x=−2 II 4

1

3 5

12 4

2

11 12

15 12 10

6

5m 1 1

12 4

2m

1

1

−11 x – 10 = 12 (1) −11 x = 12 + 10

* *

−11 x = 22 x=−2 Other methods may be used x = −2 verified is worth 5 marks

5m


Correct answer no work shown  Sign error Incorrect denominator Mishandles numerator Mathematical error Transposition errors Drops denominator or mishandles denominator Fails to finish

Slips (-1) S1 Numerical error to a max of 3 Attempts (2 marks) A1 Correct denominator only A2 No denominator, oversimplified A3 Some relevant step Worthless (0) W1 Incorrect answer no work shown W2 No work of merit Page 35

Att 2

Part (c)

30 (10,10,5,5) marks

Att (3,3,2,2)

Let f be the function f : x  10 – x – 2x2. (i)  Draw the graph of f for – 3  x  3, x  ℝ. (ii) Use your graph to estimate the maximum value of f (x). (iii) Use your graph to estimate the values of x for which f (x) = 6.

(c) (i) Function f

20(10,10) marks

Att (3,3)

I x

–3

–2

–1

0

1

2

3

f (x)

–5

4

9

10

7

0

– 11

II f  x  10  x  2 x 2

f  3 10  (3)  2(3) 2  10  3  2(9) 10  3  18  5   3,5

f  2   10  (2)  2 2   10  2  2( 4)  10  2  8  4   2,4  2

f  1  10  ( 1)  2 1  10  1  2(1)  10  1  2  9   1,9  2

f 0   10  (0)  20   10  0  2(0)  10  0  0  10  0,10  2

f 1  10  (1)  21  10  1  2(1)  10  1  2  7  1,7  2

f 2   10  ( 2)  22   10  2  2( 4)  10  2  8  0  2,0  2

f 3 10  (3)  23 10  3  2(9)  10  3  18  11  3,11 2

III

x −3 −2 −1 0 1 2 3 10 10 10 10 10 10 10 10 x 3 2 1 0 −1 −2 −3 −2x2 −18 −8 −2 0 −2 −8 −18 −5 4 9 10 7 0 −11 f x  Points  3,5  2,4  1,9 0,10 1,7 2,0 3,11 * Table is worth 10 marks, graph is worth 10 marks * Middle lines of table do not have to be shown * Consistent error(s) in each row/column attract a maximum deduction of 3 * Points might not be listed, mark on position on graph * Graph constitutes work in this question * Candidates may choose not to use a table * If graph is correct award 20 marks

Page 36

Points

 3,5  2,4

 1,9

0,10

1,7

2,0

3,11

10 8 6 4 2 -3

-2

-1

1

2

3

-2 -4 -6 -8 -10

*

Accept candidate’s values from table when plotting graph.

Blunders (-3) B1 Error in calculating – 2x2, once if consistent, note A2 B2 Error in calculating – x, once if consistent B3 Adds in domain row when evaluating f (x) B4 Each incorrect point without work B5 Each point plotted incorrectly, once if consistent B6 Each missing point B7 Axes reversed B8 Scale error, apply once B9 Points not joined to form curve or joined in incorrect order, apply once B10 Graph not extended to include full domain Slips (-1) S1 Numerical errors to a max of -3 Misreadings (-1) M1 Incorrect digit provided it does not oversimplify the question Attempts (3,3 marks) A1 Draws axes with some indication of scaling A2 Errors leading to a linear graph A3 Some correct substitution A4 Some relevant step Page 37

(c) (ii)

* *

5 marks

Att 2

Max at f (x) or y = 101 Accept answer consistent with candidate’s graph Tolerance =  0·2 of candidate’s graph

Blunders (-3)) B1 Maximum indicated on graph but no value stated B2 States or indicates x co-ordinate of maximum point Slips (-1) S1 Writes maximum point instead of maximum value Attempts (2 marks) A1 Some relevant work

(c) (iii)

* *

5 marks f (x) = 6  x = 12 or – 17. Accept answer consistent with candidate’s graph Tolerance =  0·2 of candidate’s graph

Blunders (-3) B1 One value only if two available, see *1 Misreadings (-1) M1 Misreads a digit providing it does not oversimplify the question M2 Solves f(x) = − 6 Attempts (2 marks) A1 Some indication on graph at y = 6 A2 States y = 6 Worthless (0) W1 Incorrect answer(s), no work shown W2 No work of merit

Page 38

Att 2

QUESTION 6 Part (a) Part (b) Part (c) Part (a) x–3

Let g be the function g : x  2




Att 3 Att (2,2,2,2) Att (3,2,2)

10 marks

Att 3

10 marks

Att 3

.

Find the value of g (3).

(a) I g (3)

= 23-3

4m

= 20

7m

= 1 II g : x  2x – 3

10m

=

3m

=

4m

=

7m

= 1

10m


Correct answer no work shown  Mishandles 20 Mishandles indices, once if consistent 23-3 = 23 − 23 and continues correctly Fails to finish e.g. stops at 20

Attempts (3 marks) A1 x = 3 and stops A2 8 Worthless (0) W1 Incorrect answer no work shown, note A2 W2 2 × 3 = 6 W3 No work of merit

Page 39

Part (b)

20 (5,5,5,5) marks

Att (2,2,2,2)

Let f be the function f : x  x 2  3 x . (i) (ii)

 

Express f (t) and f (2t + 1) in terms of t. Hence, find the values of t for which f (t) = f (2t + 1).

(b)(i) f : x  x 2  3x

f (t)

=

f (2t + 1) = * * * *

10(5,5) marks

Given t2 – 3t

5m

(2t + 1)2 – 3(2t + 1)

5m

2 parts to mark f (t) and f (2t +1) 5m each Accept t 2 – 3t for 5 marks Accept (2t + 1) 2 – 3 (2t +1) for 5 marks Ignore notation if consistent

Blunders (-3) B1 Substitution error B2 (2t + 1) 2 + 3 (2t +1) Misreadings (-1) M1 Misreads a digit providing it does not oversimplify the question Attempts (2,2 marks) A1 Some correct substitution A2 States x = t A3 States x = 2t +1 A4 Any relevant step Worthless (0) W1 No work of merit

Page 40

Att (2,2)

(b) (ii) t2–3t = t2–3t = 4t2 – 2t – 2 4t2 – 2t – 2 3t2 + t – 2

2

4t + 4t + 1 −6t −3 4t2 – 2t – 2 − (t2 – 3t) = 0 − t2 + 3t = 0 =0

(5,5) marks 2m

5m

and I (t + 1) (3t – 2) = 0 

t = – 1, t =

2 3

5m

or II 3t2 + t – 2 = 0 3t2 + 3t −2t – 2 = 0 3t (t + 1) −2( t + 1) = 0 (t + 1) (3t – 2) = 0 2  t = –1,t = 3

5m

or III √

1

a = 3, b = 1, c = − 2

24

1 6

1

√25 6

1 5 6 6 6 4 6



* *

t = –1,t =

2 3

5m

Mark in 2 parts, 5m for establishing an equation and 5m for solving Accept candidate’s answers from (b)(i), but note A1

Page 41

Att (2,2)

Blunders (-3) B1 B2 B3 B4 B5 B6 B7 B8 B9 B10 B11 B12

Correct answer no work shown  Sign error Transposition error Squaring error Mathematical error Substitution error Error in quadratic formula Distribution error (2 t 2 + 1) + 3(2 t + 1) and continues correctly, if not already penalised in (b)(i) Error in solving Finds one solution only Fails to finish

Slips (-1) S1 Numerical error to a max of -3 Misreadings (-1) M1 Misreads a digit providing it does not oversimplify the question Attempts (2,2 marks) A1 Linear equation merits Att 2, Att 2 at most A2 Attempt to form equation A3 Attempt to solve A4 Some use of answer(s) from (b)(i) A5 Any relevant step Worthless (0) W1 Incorrect answer(s), no work shown W2 t = 2t + 1 and stops W3 No work of merit

Page 42

Part (c)

20 (10,5,5) marks

Att (3,2,2)

The diagram below shows part of the graphs of the function f : x  x2 – 2x – 8, x  ℝ.

A

B

C (i)

(ii)

The graph intersects the x axis at A and B and the y axis at C.



Find the co-ordinates of A, B and C.

Hence, write down the range of values of x for which x2 – 2x – 8 ≤ 0.

(c) (i)

15(10,5) marks

Intersects x axis:





I x2 – 2x – 8 = 0 (x – 4) (x + 2) = 0 x = 4, x = − 2

3m 7m 9m

A (– 2, 0) B (4, 0)

10m

II x2 – 2x – 8 = 0 x2 – 4x + 2x – 8 = 0 x(x – 4) + 2(x – 4) = 0 (x – 4) (x + 2) = 0 x = 4, x = − 2

3m 3m 4m 7m 9m

A (– 2, 0) B (4, 0)

10m

Page 43

Att (3,2)

III x2 – 2x – 8 = 0 4 √ 2 2 4 4 1 2 2 √4 32 2 2 6 2 x = 4, x = − 2  A (– 2, 0) B (4, 0)

3m

8

9m 10m

Intersects y axis: f (0) = (0)2 – 2(0) – 8 2m f (0) = – 8 4m  C (0, – 8) 5m * Mark in two separate parts. 10m for finding where graph intersects x axis, 5m for where graph intersects y axis * Correct answer fully verified is full marks Blunders (-3) B1 Correct answer no work shown  B2 Sign error B3 Incorrect operation B4 Incorrect factors B5 Mathematical error B6 Error in quadratic B7 Error in substitution B8 Transposition error B9 Square root error B10 Finds one solution only ( i.e. x = 4 or x = − 2)

Slips (-1) S1 Does not write co-ordinate S2 Does not label A or B, or labels them incorrectly. Apply once. Misreadings (-1) M1 Misreads a digit providing it does not oversimplify the question Attempts (3,2 marks) A1 Effort to solve equation (= 0) or any indication of y = 0 A2 Effort to substitute (x = 0) A3 Quadratic with some correct substitution A4 x = 4 and/or x = − 2 only with no work A5 Some relevant step Worthless (0) W1 Incorrect answer(s), no work shown W2 No work of merit

Page 44

(c) (ii)

5 marks

Range

* *

–2  x  4

Accept candidate’s x values from (c)(i) Accept “from − 2 to 4 inclusive” or similar for full marks

Blunders (-3) B1 Reversed inequalities B2 2 x 4, minus sign omitted Slips (-1) S1 Does not include equals in inequalities Attempts (2 marks) A1 Some identification on graph A2 f (x) 0 A3 Some relevant step Worthless (0) W1 Incorrect answer(s) with no work of merit W2 No work of merit

Page 45

Att 2

JUNIOR CERTIFICATE EXAMINATION 2011 MARKING SCHEME MATHEMATICS HIGHER LEVEL PAPER 1

Page 1

MARKING SCHEME JUNIOR CERTIFICATE EXAMINATION 2011 MATHEMATICS -HIGHER LEVEL- PAPER 1 GENERAL GUIDELINES FOR EXAMINERS 1.


2.


3.


4.


5.


6.


7.


8.


9.


10.


11.


12.

Do not penalise the use of a comma for a decimal point, e.g. €5·50 may be written as €5,50.

Page 2



Att 3 Att 3,3 Att 2,2,2,2

Part (a)

10 marks

Att 3

1 1 Peter and Anne share a lotto prize in the ratio 3 to 2 . 2 2 Peter’s share is €35 000.



What is the total prize fund?

(a) I 1 1 : 2 3 2 2 1 1 3 +2 2 2 =6 €35,000 ÷ 3

10 marks

Att 3

1 2

€10,000 × 6

or € 10,000 ×2½ = €25,000 €35,000 + €25,000

The total prize fund

=

€60,000

II 3

1 2

: 2

1 2

7 :5 €35,000 ÷ 7 = €5,000 7 + 5 = 12 €5,000 × 12 or €5,000 × 5 = €25,000 €35,000 + €25,000 The total prize fund * *

=

€60,000

Other methods may be used €10,000 is 4 marks; €25,000 is 7 marks; €20,416·67 is 4 marks (B3 & B4 with work)


Correct answer no work shown.  Fails to finish - stops at €10,000 ×6, or €5,000 × 12 or €35,000 + €25,000 or equivalent Incorrect divisor Incorrect multiplier Decimal error

Page 3

Slips (-1) S1 Numerical errors to a maximum of −3 Misreadings (-1) M1 Incorrect digit which does not oversimplify the question

Attempts (3 marks) A1 3½ + 2½ and stops A2 6 or 5,000 or 25,000 or 10,000 or 12 with no work shown A3 Any relevant step

Worthless (0) W1 Incorrect answer no work shown, but note A2 W2 Work of no merit

Page 4

Part (b) (i)

20 (10,10) marks

Att (3,3)

The diameters of Venus and Saturn are 1  21  10 4 km and 1  21  10 5 km .



What is the difference between the diameters of the two planets? Give your answer in the form of a 10 n where n ℤ and 1 a< 10. (ii)



Write

(b) (i) I 1·21 × 10,000 12,100

3  27 in the form of 3n where n ℚ. 2 3 10marks

Att3

3m ↨

1·21 × 100,000 121,000

4m

121,000 – 12,100 108,900

7m

Difference: 1·089 × 105

10m

Interchangeable

II (1·21 × 105) – (1·21 × 104) 3m 108,900 7m 1·089 × 105 10m * One correct calculation 3 marks, two correct calculations 4 marks, subtraction 7 marks, conversion 10 marks * 108,900 only is B1 and B5 Blunders (-3) B1 B2 B3 B4 B5

Correct answer no work shown.  Reversed subtraction Index error Misplaced decimal Answer not given in required form

Slips (-1) S1 Numerical errors to a maximum of −3 Misreadings (-1) M1 Incorrect digit provided it doesn’t oversimplify the question Attempts (3 marks) A1 10,000 or 100,000 or 12,100 or 121,000 A2 Digits 1089 with incorrectly placed decimal or index without work A3 Any relevant step Worthless (0) W1 Incorrect answer, no work shown W2 1·21 × 101 or 1·21 × 109 W3 Work of no merit Page 5

(b) (ii) I

10 marks 3  27 32

(3½ × 33) ÷ 32 37/2 ÷ 32

3

3

2

II 1·732050808 × 27 9 46·7653718 9 5·196152423 *

3m

5·196152423 = 33/2 10m 1·5 1½ Accept 3 or 3 for full marks.

Blunders (-3) B1 B2 B3 B4

Correct answer no work shown. Each index error Incorrect operation Fails to finish



Slips (-1) S1 Numerical error to a maximum of −3 Misreadings (-1) M1 Misreads a digit provided it doesn’t oversimplify Attempts (3 marks) A1 27 = 33 A2 √3 = 3½ A3 1·732 A4 32= 9 A5 √ = power of ½ A6 5·196152423 A7 3√3 A6 Some relevant step Worthless (0) W1 Incorrect answer with no work W2 Work of no merit

Page 6

Att3

Part (c) (i)

20 (5,5,5,5) marks



Att 2,2,2,2

By rounding to the nearest whole number estimate the value of

(7  17) 2  14  59 8  29  1  64  2  23 . (7  17) 2  14  59 Then evaluate , correct to one decimal place. 8  29  1  64  2  23

(ii)

Úna and Conor were travelling to South Africa. They bought 5760 rand in the bank. The bank charged them €630, which included a 5% service charge.



What was the value of the euro in rand (the exchange rate) on that day?

(c) (i) Estimate

5 marks

(7  17)2  14  59 8  29  1  64  2  23 15 √7 8− 2 × 2 √49 15 8−4 64 4 8 4 = 2 Blunders (-3)

B1 B2 B3 B4 B5 B6 B7 B8

Correct answer no work shown.  Precedent error (i.e. incorrect order) Mishandles square root Incorrect squaring Incorrect use of indices Decimal error Mathematical error Calculates first, then rounds (i.e. 1·8, rounded to 2)

Slips (-1) S1 Numerical errors to a maximum of −3 S2 Incorrect rounding to a max of −3 if it affects answer Misreadings (-1) M1 Misreads a digit, provided it doesn’t oversimplify the question

Page 7

Att2

Attempts (2 marks) A1 Some correct rounding A2 Any correct step without rounding Worthless (0) W1 1·8 without work W2 Work of no merit (c) (i) Evaluate

5 marks

Att 2

(7  17)  14  59 8  29  1  64  2  23 2

√51 · 4089 14 · 59 8·29 − 3·6572 √65 · 9989 4·6328

8·123970704 4·6328 1·7535

=

1·8

Blunders (-3)

B1 B2 B3 B4 B5 B6

Apply once in (c) (i). Correct answer with no work shown. Mishandles square root Incorrect squaring Precedent error Incorrect use of indices Decimal error



Slips (-1) S1 Numerical error to a maximum of −3 S2 Incorrect or no rounding, apply once if it affects final answer Misreadings (-1) M1 Misreads a digit, provided it doesn’t oversimplify the question Attempts (2 marks) A1 Any relevant step e.g. squaring, multiplying, square root etc. Worthless (0) W1 Incorrect answer no work shown W2 Work of no merit

Page 8

(c) (ii) I 105% = €630 1% = €630 ÷ 105 1% = €6 100% = €600

10 (5,5) marks

5m

€600 = 5760 rand €1 = 5760 ÷ 600 €1 = 9·6 rand

5m

Att 2,2

Value of euro in rand: €1 = 9·6 rand II 5760 rand × 105% = 6048 rand

5m

€630 = 6048 rand €1 = 6048 ÷ 630 €1 = 9·6 rand

5m

* *

Accept final answer €1 = 9·6 rand with some work for 10 marks Two parts in marking this question; dealing with the 5% and the conversion in any order. 5 marks each (but note first *)

Blunders (-3)

B1 B2 B3 B4

Correct answer no work shown. Incorrect operation Inverted division Decimal error



Slips (-1) S1 Numerical error to a maximum of −3 Misreadings (-1) M1 Misreads a digit provided it doesn’t oversimplify the question Attempts (2,2 marks) A1 105% or 100% + 5% A2 105% = €630 and stops A3 Any relevant step Worthless (0) W1 Incorrect answer no work shown W2 Work of no merit

Page 9

QUESTION 2 Part (a) 10 marks Part (b) 25 marks Part (c) 15 marks Part (a) 10 marks A computer salesperson is paid an annual salary of €30 000. He is also paid a commission of 4% on sales. Last year the salesperson earned €38 000.



Att 3 Att 3,3,2 Att 2,2,2 Att 3

Calculate the value of the sales.

(a)

10 marks

I €38,000 − €30,000 = €8000 Commission of 4% = €8000 1% = €8,000 ÷ 4 1% = €2,000 100% = Sales = €200,000 II €30,000 + 4% of Sales = €38,000 €30,000 + ·04 Sales = €38,000 ·04 Sales = €38,000 − €30,000 ·04 Sales = €8,000 Sales = €8,000 ÷ ·04 Sales = €200,000 Blunders (-3)

B1 B2 B3 B4 B5 B6 B7 B8

Correct answer no work shown.  Decimal error Percentage error Incorrect transposition Mathematical error Expresses % as an incorrect fraction and continues In Method I, stops at €2,000 Fails to finish

Slips (-1) S1 Numerical errors to a maximum of −3 Misreadings (-1) M1 Incorrect figure if it doesn’t oversimplify the question e.g. uses €36,000 Attempts (3 marks) A1 Indicates some knowledge of percentages e.g. 4% = 4/100 or ·04 A2 Subtraction involving €38,000 and €30,000 or €8,000 mentioned without work A3 Any relevant step Worthless (0) W1 Incorrect answer no work shown W2 Adds €30,000 and €38,000 W3 Adds or subtracts 4 and €30,000 or 4 and €38,000 W4 Work of no merit Page 10

Att 3

Part (b)

25 (10,10,5) marks

Att 3,3,2

Aoife is single and earned €40 000 last year. Aoife’s tax credits are listed below. Single Person Tax Credit €1830 PAYE Tax Credit €1830 Rent Allowance Tax Credit €400 Trade Union Payment Tax Credit €70 (i)  Calculate Aoife’s total tax credits.

The standard rate cut-off point for a single person was €36 400. The standard rate of income tax was 20% and the higher rate was 41%. (ii)  Calculate the tax paid by Aoife on her income. Aoife also had to pay a 2% income levy on her gross income. (iii)  Calculate Aoife’s net income after all deductions had been made.

(b) (i) €1,830 + €1,830 + €400 + €70 =

Total tax credits = €4,130

10 marks 7m

Att 3

10m

Blunders (-3)

B1 B2 B3

Correct answer no work shown.  Omits one tax credit Addition indicated but fails to complete

Slips (-1) S1 Numerical error to a maximum of −3 Misreadings (-1) M1 Incorrect number written e.g. €1,380 etc. provided it doesn’t oversimplify the question Attempts (3 marks) A1 Adds two numbers together from list A2 Any relevant step Worthless (0) W1 Incorrect answer no work shown W2 Work of no merit

Page 11

(b) (ii) €40,000 − €36,400 = €3,600 €36,400 × 20% = €7,280

* *

10 marks

↨ Interchangeable €3,600 × 41% = €1,476 4m €7,280 + €1,476 = €8,756 7m €8,756 − €4,130 7m The tax paid = €4,626 10m Accept candidate’s tax credit figure from (b)(i)

If candidate gets 41% of €36,400 (€14,924) and 20% of €3,600 (€720) and continues correctly, this is one blunder (Total tax €15,644 minus tax credits €4,130 = €11,514 is worth 7 marks)

Blunders (-3)

B1 B2 B3 B4 B5 B6 B7 B8

Att 3

3m 3m

Correct answer no work shown.  Decimal error Percentage error 20% of an incorrect figure but note *2 41% of an incorrect figure but note *2 Mishandles tax credits Mathematical error Fails to finish

Slips (-1) S1 Numerical error to a maximum of −3 Misreadings (-1) M1 Uses 21% M2 Uses 40% or 42% Attempts (3 marks) A1 Finds 20% or 41% of any number and stops A2 Writes 20% as 20/100, 1/5 or ·2 without any further work of merit A3 Writes 41% as 41/100 or 0·41 without any further work of merit A4 Some knowledge of tax paid e.g. writes tax paid = total tax – tax credits A5 €7,280 or €1,476 with or without work A6 Any relevant step Worthless (0) W1 Incorrect answer no work shown, but note A2, A3 and A5 W2 Work of no merit

Page 12

(b) (iii) €40,000 × 2% = €800 €4,626 + €800 = € 5,426 €40,000 − € 5,426 Net income = €34,574

5 marks

* Accept candidate’s “tax paid” figure from (b) (ii) Blunders (-3)

B1 B2 B3 B4 B5 B6 B7

Correct answer no work shown.  Decimal error Percentage error Finds 2% of incorrect figure and continues Mathematical error Fails to finish Ignores “tax paid” figure when calculating net income

Slips (-1) S1 Numerical errors to a maximum of −3 Attempts (2 marks) A1 Finds 2% of any number and stops A2 2% = ·02 or 2/100 or 1/50 A3 Demonstrates some knowledge of net income e.g. Net income = Gross – Tax A4 Demonstrates some knowledge of income levy e.g. Levy = Gross × % A5 Any relevant step Worthless (0) W1 Incorrect answer no work shown, but note A2 W2 Work of no merit

Page 13

Att 2

Part (c) 15 (5,5,5) marks U is the universal set and P and Q are two subsets of U. # U = 30, #P = 16 and # Q = 6.

Att 2,2,2

(i)



Find with the aid of a Venn diagram the minimum value of # (PQ)΄.

(ii)



Find with the aid of a Venn diagram the maximum value of # (PQ)΄.

# U = u, #P = p, # Q = q and # (PQ)΄= x. (iii)



Show with the aid of a Venn diagram, that if p>q and x is a maximum, then u = p + x.

(c) (i) 16 + 6 = 22 30 −22 = 8

5 marks

Att 2

U (30) P

Q

16

6 8

Minimum value of # (PQ) ΄ = 8 * *

Accept correct Venn diagram for full marks Ignore notation

Blunders (-3)

B1 B2 B3

Correct answer no work shown.  Incorrect operation Venn diagram correct but no minimum included or stated

Slips (-1) S1 Correct value of 8 for minimum with work and no Venn diagram or incorrect Venn diagram Attempts (2 marks) A1 Venn diagram A2 16 + 6 and stops A3 30 – 6 or 30 – 16 and stops A4 Any relevant step Worthless (0) W1 Incorrect answer no work shown W2 Work of no merit

Page 14

(c) (ii) 16 – 6 = 10 30 – 16 = 14

5 marks

Att 2

U (30)

P

Q 10

6 14

Maximum value of # (PQ) ΄= 14 * *

Accept correct Venn diagram for full marks Ignore notation

Blunders (-3)

B1 B2 B3

Correct answer no work shown.  Incorrect operation Venn diagram correct but no maximum included or stated

Slips (-1) S1 Correct value of 14 for maximum with work and no Venn diagram or incorrect Venn diagram Attempts (2 marks) A1 Venn diagram A2 16 – 6 A3 Any relevant step Worthless (0) W1 Incorrect answer no work shown W2 Work of no merit

Page 15

(c) (iii) 0 in Q only for x to be a maximum u=p–q+q+x u=p+x Venn diagram to give:

5 marks

Att 2

u=p+x

U I

Q

P q

p-q

0 x

II U

P q x

III

U P

x Q

* *

Accept correct Venn diagram for full marks with conclusion u = p + x Ignore notation

Blunders (-3) B1 Omits term from equation B2 Fails to finish B3 # Q/P ≠ 0 B4 #P/Q ≠ p – q B5 # (P Q) ≠ q Slips (-1) S1 u = p –q +q + x S2 No conclusion

u = p +x only. No Venn diagram.

Page 16

Attempts (2 marks) A1 Venn diagram A2 p – q A3 Any relevant step Worthless (0) W1 Incorrect answer no work shown W2 u = p + x only (Given) W3 No work of merit

Page 17



Att 3 Att 3,3 Att 2,2,2,2

Part (a)

10 marks

Att 3



Given that t 2 – s = r, express t in terms of r and s.

(a)

10 marks t2 –s = r t2 = r+s t  rs

* * *

Att 3

Given 7m

10m Two steps in this question, transposition and square root t –s = r and continues correctly to get t = r +s is worth 6 marks - Misread and B2. Finds s correctly in terms of t and r is 6 marks - Misread and B2. (s = t2 − r)

Blunders (-3)

B1 B2 B3 B4

Correct answer no work shown.  Mishandles or fails to get square root Incorrect operation e.g. May attempt to square everything Transposition error

Misreadings (-1) M1 Note *2 and *3 M2 t 2 + s = r and continues correctly to get t = √ Attempts (3 marks) A1 Effort at square root A2 t 2 – s − r (with or without = 0) A3 Any relevant step Worthless (0) W1 Incorrect answer no work shown W2 Work of no merit

Part (b)

20 (10,10) marks

(i)



Divide 3x2 + 5x – 28 by x + 4.

(ii)



Solve the equation

4x  2 5



6x 3

Page 18

Att (3,3)

 5 .

(b) (i) I 3x2 + 5x – 28 ÷ x + 4

10 marks

(3x – 7)(x + 4) x+4 = 3x – 7

II 3x2 + 5x – 28 ÷ x + 4 3x2 + 12x −7 x – 28 ÷ x + 4 3x ( x + 4) + 7 ( x+ 4) ÷ x + 4 (3x – 7)(x + 4) ÷ x + 4 = 3x – 7

III Division to give answer 3x – 7

3x – 7 x+4 3 ²

5 – 28

2

3x + 12x − 7x − 28 − 7x − 28 0

*

(3x+7)(x – 4) and continues is one blunder (B4), will also incur B5 or B6. All other attempts to factorise apply B2, B3 and/or B4.

Blunders (-3)

B1 B2 B3 B4 B5 B6 B7

Correct answer no work shown.  Incorrect factors of 3x2 in method I Incorrect factors of – 28 in method I Incorrect factors leading to an incorrect middle term in method I Fails to finish i.e. no cancellation in method I Incorrect cancellation Mathematical error, once if consistent

Slips (-1) S1 Numerical errors to a maximum of −3

Page 19

Att3

Attempts (3 marks) A1 Some effort at factorising A2 Sets up division A3 Multiplies instead of dividing, with at least one correct term A4 Finds guide number (− 84) in method II and stops A5 Quadratic with some correct substitution A6 Sets up quadratic and identifies a,b or c A7 Uses quadratic formula and stops at correct roots (x = − 4 and x = 7/3) A8 Any relevant step Worthless (0) W1 Incorrect answer no work shown W2 ( )( ) W3 Work of no merit

Page 20

(b) (ii) I 4x  2 6x   5 5 3 3(4x + 2) – 5(6 – x) = −5 15 3(4x + 2) – 5(6 – x) = 15 (−5) 12x + 6 – 30 + 5x = – 75 17x – 24 = − 75 17x = −75 + 24 17x = − 51 x = − 51 ÷ 17 x=–3

10 marks

II 4x  2 6x   5 5 3 (4x + 2)(3)(5) – (6 – x)(3)(5) = −5(3)(5) 5 3 3(4x + 2) – 5(6 – x) = 15 (−5) 12x + 6 – 30 + 5x = – 75 17x – 24 = − 75 17x = −75 + 24 17x = − 51 x = − 51 ÷ 17 x=–3

*

III 4x  2 6x   5 5 3 3(4x + 2) – 5(6 – x) = (5)(3) (−5) 12x + 6 – 30 + 5x = – 75 17x – 24 = − 75 17x = −75 + 24 17x = − 51 x = − 51 ÷ 17 x=–3 x = − 3, by trial and error or similar, fully verified merits 10 marks

Blunders (-3)

B1 B2 B3 B4 B5 B6 B7

Correct answer no work shown.  Distribution error, once if consistent Incorrect common denominator or mishandles denominator Transposition error, once if consistent Mathematical error Mishandles numerator Combines unlike terms and continues

Page 21

Att3

Slips (-1) S1 Numerical errors to a maximum of −3 Attempts (3 marks) A1 Common denominator and stops A2 Oversimplified but some correct work A3 Cross multiplies A4 Any relevant step Worthless (0) W1 Incorrect answer no work shown W2 Adds or subtracts terms incorrectly e.g. 5 +3, or 4x +2 ± 6 – x etc. W3 Work of no merit

Page 22

Part (c) 20 (5,5,5,5) marks A car park can accommodate cars and mini-buses. On a particular day there were x cars and y mini-buses in the car park, giving a total of 520 vehicles. The parking area for a car is 7 m2 and the parking area for a mini-bus is 12 m2. On that day a total area of 3840 m2 was occupied by cars and mini-buses. (i) Write down two equations to represent the above information. (ii)



(iii)



Att 2,2,2,2

Solve these equations to find the number of cars and the number of mini-buses in the car park on that day. There is a flat rate charge per day for parking. The flat rate for mini-buses is 3 times that for cars. On that day €3000 was taken in. What is the flat rate for cars?

(c) (i) x+y 7x + 12y

* * * *

= =

520 3840

10 (5,5) marks 5m 5m

Att 2,2

Two equations to mark in (c) (i) Each equation is marked separately Each equation is worth 5 marks, attempt 2

Answer is sufficient for full marks (No  in question)

Blunders (-3) B1 Incorrect term Attempts (2,2 marks) A1 x or y or 7 x or 12 y A2 Effort at creating an equation equal to 520 or 3,840 A3 Any relevant step (c) (ii) I x+y 7x + 12y

5 marks

= =

520 (−7) 3840

−7x – 7y = − 3640 7x + 12y = 3840 5y = 200 y = 200 ÷ 5 y = 40 x+y x + 40 x x x = 480

= = = = and

520 520 520 – 40 480 y = 40.

Page 23

Att 2

II x+y 7x + 12y

= =

520 (−12) 3840

−12x – 12 y = − 6240 7 x + 12y = 3840 −5x = − 2400 5x = 2400 x = 2400 ÷ 5 x = 480 480 + y y y

= = =

520 520 − 480 40

x = 480 and y = 40. ___________________________________________________________________________ III x = 520 − y 7(520 – y) + 12y = 3840 3640 – 7y + 12 y = 3840 −7y + 12 y = 3840 −3640 5y = 200 y = 200 ÷ 5 y = 40 x+y x + 40 x x

= = = =

520 520 520 – 40 480

x = 480 and y = 40. __________________________________________________________________________ IV y = 520 – x 7x + 12(520 – x) = 3840 7x + 6240 – 12 x = 3840 7x – 12 x = 3840 −6240 − 5 x = − 2400 5 x = 2400 x = 2400 ÷ 5 x = 480

480 + y y y

= = =

520 520 − 480 40

x = 480

and y = 40

Page 24

*1

Equations may be also solved by substituting y =

or x =

*2

Accept candidate’s equations from (c ) (i) provided oversimplification does not occur

*3

Apply only one blunder in establishing the first equation in terms of x only or the first equation in terms of y only.

*4

Finding the second variable is subject to a maximum deduction of 3 marks

*5

Correct values of x and y without algebraic work, both verified in both equations merits full marks

*6

Correct values of x and y without algebraic work not verified or not fully verified in both equations merits attempt mark only

Blunders (-3) B1 Finds one variable only B2 Distribution error B3 Mathematical error B4 Incorrect substitution when finding second variable, but note M1 B5 Transposition error in solving first variable B6 Transposition error in solving second variable B7 Error(s) in establishing the first equation in terms of x (−5x = − 2400) only or the first equation in terms of y (−5y = −200) only through elimination by cancellation I and II B8 Error(s) in establishing the first equation in terms of x (−5x = − 2400) only or the first equation in terms of y (5y = 200) only through elimination by substitution III and IV Slips (-1) S1 Numerical errors to a max of −3 Misreadings (-1) M1 Misreads digits, providing it doesn’t oversimplify Attempts (2 marks) A1 Any correct manipulation of either given equation and stops A2 Some correct partial substitution and stops A3 Any relevant step Worthless (0) W1 Incorrect answer, no work shown W2 Trial and error, but see *5 and *6 above W3 Work of no merit

Page 25

(c) (iii) 3x : x

5 marks

480x + 3x(40) = 3,000 480x + 120x = 3,000 600x = 3,000 x = 3,000 ÷ 600 x=5 *

Flat rate for cars = €5 Accept candidate’s answers from (c ) (ii)

Blunders (-3)

B1 B2 B3

Correct answer no work shown.  Mathematical error Uses ratio Buses: Cars 1:3, but note M1

Slips (-1) S1 Numerical errors to a maximum of −3 Misreadings (-1) M1 Misreads values for buses and cars (i.e. uses 40 cars and 480 buses – note B3) Attempts (2 marks) A1 States ratio 3:1 and stops A2 Some correct partial substitution and stops A3 3000 ÷ 4 or 750 or 3 + 1 A4 Some relevant step e.g. divides 3,000 Worthless (0) W1 Incorrect answer no work shown W2 Trial and error with incorrect value(s) W3 Work of no merit

Page 26

Att 2



Att 3 Att 2,3,2 Att 2,2,2,2

Part (a)

10 marks

Att 3



Graph on the number line the solution set of – 2x + 1 > – 7, x ℕ.

(a)

10 marks I – 2x + 1 > – 7 – 2x > – 7 − 1 – 2x > – 8 2x < 8 x – 7 1+7>2x 8>2x 4>x x − 7 solved to get x > − 4 with correct graph

Page 27

Att 3

Slips (-1) S1 Numerical errors to a max of −3 S2 Includes 4 on graph S3 Each incorrect or missing number to a maximum of 3 Misreadings (-1) M1 Includes equals in inequality Attempts (3 marks) A1 Tests any value in inequality and stops A2 Draws any number line A3 Any relevant step Worthless (0) W1 List given with no correct value W2 Work of no merit

Page 28

Part (b) (i) (ii)

20 (5,10,5) marks

Att 2,3,2

2

Factorise x – 1.



Factorise fully ax– 3 – a + 3x.

(iii) Factorise 6x2 + x – 35

(b) (i) x2 – 1 = x2 – 12 = (x – 1)(x + 1)

* *

*

5 marks Given 2m 5m

Answer is sufficient for full marks (No  in question) Accept also (with or without brackets) for full marks any of the following: (x −1) and (x + 1) [The word 'and' is written down] (x −1) or (x + 1) [The word 'or' is written down] (x −1), (x + 1) [a comma is used] Quadratic equation method is subject to slips and blunders

Blunders (-3) B1 Incorrect factors of x2 B2 Incorrect factors of −1 B3 (1 – x) ( 1 + x ) B4 Answer left as roots x = ± 1 Slips (-1) S1 (x – 1) + (x + 1) S2 (x – 1) − (x + 1) Attempts (2 marks) A1 x2 – 12 A2 Correct factors of x2 only A3 Correct factors of 1 or − 1 only A4 ± x or ± 1 A5 x2 – 1 = x × x – 1 × 1 A6 Difference of two squares mentioned A7 √1 A8 √ A9 Correct quadratic formula with some correct substitution Worthless (0) W1 Combines terms incorrectly and stops e.g. – 1 x2 W2 Work of no merit

Page 29

Att 2

(b) (ii) I a x – 3 − a +3 x Given ax–a+3x–3 3m a ( x – 1) + 3 (x – 1) 7m (a + 3) (x – 1) 10m

10 marks

Att 3

II a x – 3 − a +3 x Given ax + 3x – a – 3 3m x ( a + 3 ) – 1 ( a + 3 ) 7m (a + 3) (x – 1) 10m

*

Accept also (with or without brackets) for full marks any of the following with work (a + 3) and (x − 1) [The word 'and' is written down] (a + 3) or (x − 1) [The word 'or' is written down] (a + 3), (x − 1) [A comma is used]

Blunders (-3)

B1 B2 B3 B4 B5

Correct answer no work shown.  Failure to complete last step e.g. stops at a(x – 1) + 3(x – 1) Error in factorising any pair of terms, apply once if consistent Incorrect last step e.g. 3a (x −1) or (a + 3) ( – 1x) Incorrect common factor and continues e.g. x ( a + 3 ) + 1 ( −a −3). B4 will also apply.

Slips (-1) S1 (a + 3) + (x − 1) S2 (a + 3) − (x − 1) Attempts (3 marks) A1 Pairing off matching terms, or indication of common factors and stops A2 Correctly factorises any pair and stops A3 Any relevant step Worthless (0) W1 ( )( ) W2 Pairing of terms with nothing in common e.g. a x – 3 and no further work of merit W3 Work of no merit

Page 30

(b) (iii) I 6x2 + x – 35 (2x + 5) (3x – 7)

5 marks

II 6x2 + x – 35 6x2 – 14 x +15 x – 35 2 x (3 x – 7) + 5 (3 x – 7) (2x + 5) (3x – 7) III 6 x 2 + x – 35 6x2 +15 x – 14 x – 35 3x (2 x + 5) − 7 x (2 x + 7) (2x + 5) (3x – 7)

*


* *

x = − , x = and continues Quadratic may be used to solve 6x2 + x – 35 = 0 Accept also (with or without bracket) for full marks any of the following (2x + 5) and (3x – 7) [The word 'and' is written down] (2x + 5) or (3x – 7) [The word 'or' is written down] [A comma is used] (2x + 5), (3x – 7)

Blunders (-3) B1 Incorrect factors of 6x2 B2 Incorrect factors of − 35 B3 Factors leading to an incorrect middle term B4 Substitution error in quadratic B5 Uses quadratic to get roots and stops Slips (-1) S1 (2x + 5) + (3x – 7) S2 (2x + 5) − (3x – 7) Attempts (2 marks) A1 Some correct factors A2 Identifies a,b or c for quadratic A3 Quadratic with some correct substitution A4 Any correct relevant step Worthless (0) W1 Incorrect answer no work shown but note attempts W2 ( )( ) W3 Quadratic formula only W4 Work of no merit

Page 31

Att 2

Part (c) 20 (5,5,5,5) marks Att 2,2,2,2 The new Lansdowne Road stadium has seating capacity for 200 journalists. It was decided initially that this seating would be in x rows of equal value. (i) Write, in terms of x, the number of seats per row required to accommodate the 200 journalists. During the construction it was decided to have 3 fewer rows to accommodate the 200 journalists. (ii) Write, in terms of x, the number of seats per row now required. It was found that 15 extra seats per row were required compared to the initial plan. (iii)



Write an equation using the above information and solve for x.

(c) (i)

5 marks

Att 2

200 x *


Misreadings (-1) M1 Uses letter other than x Blunders (-3) B1 Inversion Attempts (2 marks) A1 Effort at forming expression using 200 and x Worthless (0) W1 x only or 200 or similar (c) (ii)

5 marks

200 x 3 *


Misreadings (-1) M1 Uses letter other than x if not penalised already Blunders (-3) B1 Inversion, but do not penalise if already blundered in (c ) (i)

B2 B3

± 3 (Linear in (c )(iii) and subject to further penalty there)

Attempts (2 marks) A1 Forms an incorrect expression with at least two of the following x, 3, 200

Page 32

Att 2

(c) (iii) 10 (5,5) marks Establish equation 200 200 – 2m = 15 x 3 x 200 x – 200 (x – 3) = 15 x (x −3) 2m x (x −3) 200 x − 200 x + 600 = 15x2 – 45 x 2m

600 = 15x2 – 45 x 15x2 – 45 x − 600 = 0 x2 – 3 x − 40 = 0

5m

(x – 8)(x + 5) = 0 → 8 and − 5 Solution: x = 8

2m 4m 5m

Att 2,2

Solve

* *

Mark in two parts: 5 marks for equation and 5 marks for solving Accept candidate’s expressions from (c ) (i) and (ii). Linear merits Att 2,2 at most

Blunders (-3)

B1 B2 B3 B4 B5 B6 B7 B8 B9

Correct answer no work shown.  Distribution error - apply each time but once if consistent Transposition error - apply each time but once if consistent Mathematical error in forming equation e.g. line 1 Incorrect factors Correct factors and stops, will also incur S2 Error in quadratic formula Each error in grouping terms or fails to group Error in establishing equation e.g. line 1

Slips (-1) S1 Numerical errors to a maximum of 3 S2 Stops at x = 8, x = − 5 or concludes x = − 5 Attempts (2,2 marks) A1 Linear equation merits attempt marks at most A2 Trial and error merits attempt at most A3 Any correct relevant step Worthless (0) W1 Incorrect answer no work shown W2 ( )( ) W3 Substitution of any number other than 8 or −5

Page 33



Att 3 Att 3,3 Att 2,3,2

Part (a)

10 marks

Att 3



Given that f(x) = 3x – 4 and that f (k) = 11, find the value of k.

(a)

10 marks

3k – 4 = 11 3k = 11 + 4 or 15 k = 15 ÷ 3 k = 5 * *

Att 3

3m 4m 7m 10m

Ignore notation (e.g. x = 5 or answer = 5 for full marks) k = 5 fully verified = 10 marks

Blunders (-3)

B1 B2 B3 B4

Correct answer no work shown.  Transposition error 3k – 4 = 0 and continues correctly Incorrect operation

Slips (-1) S1 Numerical error to a maximum of 3 Attempts (3 marks) A1 Fills in k and stops A2 Attempt to divide by 3 A3 Tests values e.g. k = 11 to get answer of 29 A4 Any relevant step Worthless (0) W1 Incorrect answer with no work shown Part (b)

20 (10,10) marks 2

Let f be the function f : x  7x – x .



Draw the graph of f for 0  x  7, xℝ.

Page 34

Att 3,3

(b) Function f

20 (10,10) marks

Att 3,3

12 10 8 6 4 2 1 x 7x −x2

0 0 0

1 7 −1

2 14 −4

2

3

4

5

3 21 −9

4 28 −16

6

7

5 35 −25

6 42 −36

7 49 −49

x

0

1

2

3

4

5

6

7

f (x)

0

6

10

12

12

10

6

0

10

12

12

(3, 12)

(4,12)

f(x) or

0

6

10

6

0

or f : x  7x – x2. f : x  7(0) –02 = 0 – 0 = 0 f : x  7(1) – 12= 7 – 1 = 6 f : x  7(2) – 22 = 14 – 4 = 10 f : x  7(3) –3 2 = 21 – 9 = 12 f : x  7(4) – 42 = 28 – 16 = 12 f : x  7(5) – 52 = 35 – 25 = 10 f : x  7(6) – 62 = 42 – 36 = 6 f : x  7(7) – 72 = 49 – 49 = 0 (0.0)

(1,6)

(2, 10)

(5, 10)

* * * *

Table is worth 10 marks, graph is worth 10 marks Middle lines of table do not have to be shown Candidates may choose not to use a table Points might not be listed, mark on position on graph

*

Graph constitutes work in this question 

Page 35

(6, 6)

(7,10)

Blunders (-3) B1 Error in calculating 7x, once if consistent B2 Error in calculating – x2, once if consistent, but note A1 B3 Error in calculating last line of table, once if consistent B4 Each incorrect point without work B5 Point plotted incorrectly, once if consistent B6 Each missing point B7 Axes scaled incorrectly, once only B8 Reversed axes B9 No curve between (3,12) and (4,12) on graph B10 Points not joined, most probably incurs B9 also Slips (-1) S1 Numerical error to a maximum of −3 Attempts (3,3 marks) A1 Error leading to a linear graph A2 Some correct substitution A3 Draws axes, with some indication of scaling Part (c) 20 (5,10,5) marks Att2,3,2 The formula for the height, y metres, of a golf ball above ground level x seconds after it is hit, is given by 7x – x2. Use your graph from part (b): (i)

 

to find the maximum height reached by the golf ball

(ii) to estimate the number of seconds the golf ball was more than 2 metres above the ground.

The graph below represents the flight of another golf ball. The flight of the golf ball is given by the formula ax – x2, xℝ. 9 8 7 6 5 metres 4 3 2 1 1 (iii)



2

3 4 seconds

Find the value of a.

Page 36

5

6

(c) (i)

*

5 marks

Att 2

Maximum height  12·25 m Accept answer consistent with candidate’s graph, tolerance ± ·2

Blunders (-3)

B1 B2 B3 B4 B5

Correct answer no indication on graph.  Maximum indicated on graph but no value given Outside of tolerance States x co-ordinate of maximum point Fails to use graph

Slips (-1) S1 Accept maximum as point if y value is correct i.e. (3·5, 12·25) Attempts (2 marks) A1 Reads maximum from table A2 Some relevant substitution in effort to find maximum A3 Uses graph of (c ) (iii) to find maximum of 9 A4 x = 3· 5 indicated on graph Worthless (0) W1 Incorrect answer no work, but note attempts (c) (ii)

*

10 marks More than 2 metres above the ground→ 6·7 – 0·3= 6·4 secs Accept values from candidate’s graph with a tolerance of ± ·2

Blunders (-3) B1 No subtraction B2 Value(s) not consistent with candidate’s graph B3 No indication on graph B4 Indication on graph but no value given each time. B1 also applies B5 Outside of tolerance, each time B6 ·3 − 6·7 = − 6·4 or candidate’s equivalent Slips (-1) S1 Numerical error Attempts (3 marks) A1 Correctly solves f(x) = 2 by formula; graph not used A2 f (2) found (answer = 10, or candidate’s equivalent) A3 Uses graph of (c )(iii) to find answer (5·7 − ·3 = 5·4) Worthless (0) W1 Incorrect answer, no work shown

Page 37

Att 3

(c) (i)

5 marks

Fills in any of the following points (1,5) (2,8) (3,9) (4, 8) (5, 5) (6,0) or any other correct points to solve equation ax – x2 = y I e.g. (1, 5) a (1) – (1)2 = 5 a–1=5 a=5+1 Value of a = 6 II ax – x2 = y x(a – x) = y Fill in e.g. (1, 5) 1(a – 1) = 5 a −1 = 5 a=5+1 Value of a = 6

Blunders (-3)

B1 B2 B3 B4 B5 B6 B7

Correct answer no work shown.  Co-ordinates reversed when substituting Incorrect squaring Transposition error Substitutes for a instead of x Fails to finish Incorrect factors

Slips (-1) S1 Numerical error to a maximum of −3 Attempts (2 marks) A1 Some relevant substitution A2 ax – x2 = f(x) or ax – x2 = y A3 Writes down any point on the curve A4 Attempt to factorise Worthless (0) W1 Incorrect answer no work shown W2 No work of merit

Page 38

Att 2

QUESTION 6 Part (a) Part (b) Part (c) Part (a)



10 marks 20 marks 20 marks 10 marks a5 a4 1 When a = , find the value of .  4 3 2

(a)

10 marks I

= II

a5 a4  3 2 ¼+5 − ¼+4 3 2 5¼ − 4¼ 3 2 1¾ − 2⅛ 3 8 a5 a4  3 2 2( a + 5) – 3 ( a + 4) 6 2a +10 – 3a − 12 6 −a–2 6

a=¼ −¼ − 2 6 −2¼ 6 −3 8

*

Accept answer in decimal format (− ·375) or equivalent fraction

Page 39

Att 3 Att 3,2,2 Att 3,2,2 Att 3

Att 3

Blunders (-3)

B1 B2 B3 B4 B5 B6 B7 B8 B9

Correct answer no work shown.  Incorrect denominator Mishandles denominator Mishandles numerator Mathematical error Distribution error, once if consistent Fails to combine like terms (if it affects final answer) Combines unlike terms Incomplete step (e.g. − 2¼/6 or similar)

Slips (-1) S1 Numerical errors to a maximum of −3 Attempts (3 marks) A1 Some correct substitution A2 No denominator used A3 Any correct relevant step Worthless (0) W1 Incorrect answer no work shown W2 No work of merit

Page 40

Part (b)

20 (10,5,5) marks

(i)



(ii)



Att 3,2,2

Express in its simplest form: 5 4  . x 1 x2 Hence, or otherwise, solve the equation: 4 5 3   , x 1 x2 2 giving your answers correct to one decimal place.

(b)(i)

10 marks

Att 3

4(x +2) − 5 (x – 1) (x – 1) (x + 2) 4x+8−5x+5 (x – 1) (x + 2) − x + 13__ (x – 1) (x + 2) *

Accept common denominator as (x – 1)(x+2). Penalise incorrect multiplication in (b) (ii)

Blunders (-3)

B1 B2 B3 B4 B5 B6 B7 B8 B9

Correct answer no work shown.  Incorrect denominator Mishandles denominator Mishandles numerator Mathematical error Distribution error, once if consistent Fails to combine like terms Combines unlike terms Reads as + . Continue to apply slips and blunders

Slips (-1) S1 Numerical errors to a maximum of −3 Attempts (3 marks) A1 Identifies common denominator and stops A2 No denominator used A3 Oversimplification A4 Any relevant step Worthless (0) W1 Incorrect answer no work shown

W2 Adds or subtracts numerators and denominators e.g.

Page 41

or − or −

etc

(b)(ii)

10 (5,5) marks

 x  13 3  ( x  1)( x  2) 2 2 ( − x + 13) = 3 (x – 1)( x + 2 ) −2x + 26 = (3x – 3)(x + 2) −2x + 26 = 3x2 +3x – 6 3x2 + 5x – 32 = 0 −5 ± √25−4(3)(−32) 2(3) −5±√25 + 384 6 −5±√409 6 −5±20·22374842 6 −25·22374842 and 6

Att 2,2

Equation 5m

15·22374842 6

x = – 4·2 and x = 2·5 Solve 5m * Mark in two parts: 5 marks for equation and 5 marks for solving * Accept candidate’s expressions from (b ) (i). Linear merits Att 2,2 at most Blunders (-3) B1 Correct answer no work shown.  B2 Distribution error - apply each time but once if consistent B3 Transposition error - apply each time but once if consistent B4 Mathematical error in forming equation B5 Incorrect denominator B6 Mishandles denominator B7 Mishandles numerator B8 Fails to combine like terms B9 Combines unlike terms B10 Error in quadratic formula B11 Error in application of quadratic formula B12 Finds only one solution Slips (-1) S1 Numerical errors to a maximum of −3 S2 Fails to round or rounds incorrectly Attempts (2,2 marks) A1 Linear equation merits attempt marks at most A2 Quadratic formula with some correct substitution A3 Trial and error of correct solution(s) merits attempt at most A4 Any relevant step Worthless (0) W1 Incorrect answer no work shown W2 ( )( ) W3 Substitution of any number other than 2·5 and − 4·2 or equivalent W4 No work of merit

Page 42

Part (c)

20 (10,5,5) marks

Att 3,2,2

The diagram below shows part of the graphs of the functions f (x) = x2 – 4x + 3 and g(x) = x + k. y axis

f (x)

g(x)

x axis A

B

The graph of f(x) cuts the x axis at A and B. The graphs of f(x) and g(x) intersect at A. (i) (ii) (iii)

  

Find the coordinates of A and the coordinates of B. Find the value of k. Verify that f(x) and g(x) intersect also at the point (4, 3).

(c) (i)

10 marks

Att3

2

*

Solve x – 4x + 3 = 0 3m (x – 1)(x – 3) = 0 4m x = 1, x = 3 7m (1, 0) and (3, 0) 9m  A (1, 0) B (3, 0) 10m If trial and error is used, must be fully verified for 10 marks

Blunders (-3) B1 Correct answer no work shown.  B2 Incorrect factors of x2 B3 Incorrect factors of +3 B4 Factors leading to an incorrect middle term B5 Fails to find roots B6 Fails to list coordinates with y = 0 included B7 Uses quadratic and stops at roots B8 (0,1) and (0,3) Reversed co-ordinates. Slips (-1) S1 Fails to specify A or B or incorrectly names A and B, apply once. S2 After solving x = 1, x = 3 only states one point i.e (1,0) or (3,0). May also incur S1.

Page 43

Attempts (3 marks) A1 Some correct factors A2 Identifies a, b or c for quadratic A3 Quadratic with some correct substitution A4 Finds where graph cuts y axis (0,3) A5 f(x) = 0 or g(x) = 0 or (x , 0 ) A6 Substitution of 0,1 or 3 A7 Any correct relevant step A8 Uses graph to read answer Worthless (0) W1 Incorrect answer no work shown (c) (ii)

5 marks g(1) = 1 + k = 0

2m

 k= –1 5m * Accept (4,3) or candidate’s A co-ordinate from (c ) (i) Blunders (-3)

B1 B2 B3 B4

Correct answer no work shown.  Fails to let g(x) = 0 Fails to let x = 1 (or candidate’s equivalent value) Transposition error

Attempts (2 marks) A1 Substitutes x = 1 (or candidate’s equivalent) and stops A2 g(x) = 0 A3 Any relevant step Worthless (0) W1 Incorrect answer no work shown

Page 44

Att 2

(c) (iii)

Solve

5 marks I x2 – 4x + 3 = x – 1 x2 – 4x + 3 − x + 1 = 0 x2 – 5x + 4 = 0 ( x – 4) (x – 1) = 0 x = 1 and x = 4 x2 – 4x + 3 = y (4)2 – 4(4) + 3 = y 16 −16 + 3 = 3 = y  point (4, 3)

Att 2 2m

or x – 1 = y 4–1=y 3=y 5m

II Substitute x = 4 into f(x) = x2 – 4x + 3 and g(x) = x – 1 f(x) = x2 – 4x + 3 f(x) = (4)2 – 4(4) + 3 f(x) = 16 – 16 + 3 f(x) = 3 (4,3)

2m

g(x) = x – 1 g(x) = 4 – 1 g(x) = 3 (4,3)

*

4m

(4,3) on both lines → point of intersection 5m Accept g(x) = x + k, based on candidate’s k value from (c ) (ii)

Blunders (-3) B1 Fails to equate f(x) and g(x) in method I B2 Incorrect squaring B3 Transposition error B4 Does not substitute into second function B5 Fills in x = 1 B6 Fails to finish Slips (-1) S1 Does not conclude in method II S2 Numerical errors to a maximum of 3 Attempts (2 marks) A1 Some relevant substitution A2 Linear equation attempt at most in Method I Worthless (0) W1 Incorrect answer no work shown W2 x – 1 = 0 W3 x2 – 4x + 3 = 0 W4 Work of no merit Page 45

JUNIOR CERTIFICATE EXAMINATION 2010 MARKING SCHEME MATHEMATICS HIGHER LEVEL PAPER 1



2.


3.


4.


5.


6.


7.


8.


9.


10.


11.


12.

Do not penalise the use of a comma for a decimal point, e.g. €5.50 may be written as €5,50. Page 1


10 marks 25(10, 10, 5)marks 15(10, 5) marks

Att 3 Att(3, 3, 2) Att(3, 2)

Part (a) 10 marks st The price of a litre of petrol on the 1 of August was €1·20. The price on the 1st September was €1·17.



Calculate the percentage decrease over this period.

(a)

10 marks I €1·20 - €1·17 = €0·03

Att 3 II

or

3 cents

· ·

Percentage decrease

= 2·5%


Att 3

Correct answer no work shown. Decimal error Percentage of incorrect figure Inversion Mathematical error Sign error

Slips (-1) S1 Numerical error, max -3 Misreadings (-1) M1 Incorrect number written down, provided it doesn’t oversimplify the question Attempts (3 marks) A1 Subtracts and stops A2 Some mention of 100, 3 or ·03 Worthless (0) W1 Adds €1·20 and €1·17 and stops W2 Incorrect answer no work shown, except for numbers mentioned in A2.

Page 2

Part (b)

25(10,10, 5) marks



(i)

By rounding correct to the nearest whole number, estimate the value of . .

.

Att(3, 3, 2)

+ (2.97)3 ÷ √9 · 16 .

Then, evaluate

.

+ (2.97)3 ÷ √9 · 16

correct to one decimal place.



(ii)

By putting the largest number first, place the following numbers in order:

,

√

,

(1ּ11)2, √1 · 3456

. (b) (i) Estimate +

10 marks

33 √9=

1 + 27  3 = 1 + 9 = Estimate: 10 Blunders (-3) B1 B2 B3 B4 B5 B6 B7

Correct answer no work shown. Incorrect order i.e. precedent error Incorrect square root Incorrect use of indices Decimal error Mathematical errors Calculates first, then rounds (e.g. 9with work, rounded to 10)

Slips (-1) S1 Numerical errors S2 Incorrect rounding, to a max of 3, if it affects answer Misreadings (-1) M1 Misreads a digit, provided it doesn’t oversimplify the question Attempts (3 marks) A1 Some correct rounding A2 Any correct step without rounding Worthless (0) W1 Incorrect answer no work shown W2 96 without work

Page 3

Att 3

(b) (i) Evaluate Evaluate . .  

10 marks

+ (2.97)3 ÷ √9 · 16

898345153 + 26198073  302654919 898345153 + 8656086967  = 9·6


Correct answer no work shown. Incorrect order i.e. precedent error (e.g. 8952908561 = 90 to one decimal place) Incorrect square root Incorrect use of indices Decimal error Mathematical errors

Slips (-1) S1 Numerical errors S2 Incorrect or no rounding, apply once if it affects answer Misreadings (-1) M1 Misreads a digit, provided it doesn’t oversimplify the question Attempts (3 marks) A1 Any relevant correct step i.e. division, cubing, square root etc. Worthless (0) W1 Incorrect answer no work shown

Page 4

Att 3

(b) (ii) 5 marks 12321, 1224744871, 1166666, 116 or 123, 122, 117, 116 (1ּ11)2, * *

√

,

,

√1 · 3456

Accept correct decimals in order for full marks, at least 2 correct decimal places required. (Ignore rounding if it doesn’t affect answer). Accept candidate’s values when arranging


Correct answer no work shown. Incorrect order with relevant work shown, but note W1 and W2 Mathematical or decimal error Square root error Indices error

Slips (-1) S1 Reverses order S2 Incorrect or early rounding which affects answer, max -3 Misreadings (-1) M1 Incorrect reading of a digit provided it doesn’t oversimplify the question Attempts (2 marks) A1 Finds decimal value of any given number and stops Worthless (0) W1 Incorrect answer no work or no work of merit W2 Original list presented with no work of merit W3 √9 = 4.5 , with otherwise no work of merit

Page 5

Att 2

Part (c) 15(10, 5) marks (i) The standard rate of income tax is 20% and the higher rate is 41%. The standard rate cut-off point is €36 500. Aisling has a gross income of €47 500 and total tax credits of €1830.

 (ii)

Att(3, 2)

Calculate Aisling’s net income.

The following year Aisling’s gross income increases. The tax rates, cut-off point and tax credits remain unchanged. Her net tax now amounts to €15 105.



What is her new gross income?

(c) (i)

10 marks

Att 3

€47 500  €36 500 = €11 000 €36 500  20% = € 7300 €11 000 × 41% = €4510 €7300 + €4510 = €11 810 €11 810  €1830 = €9980 €47 500  €9980 Net income = €37 520 *

If candidate gets 41% of €36 500 and 20% of €11 000 and continues, this is one blunder only (€32 165)


Correct answer no work shown. Decimal error 20% of an incorrect figure, but note * 41% of an incorrect figure, but note * Mishandling of tax credits leading to the wrong answer Fails to get total tax Fails to complete final step Mathematical blunder

Slips (-1) S1 Numerical error to a max of 3 Misreadings (-1) M1 Incorrect reading of a digit which does not oversimplify the question M2 Uses 21% M3 Uses 40% or 42%

Page 6

Attempts (3 marks) A1 Finds 20% or 41% of a number and stops, but note W1 A2 Writes 20% as 1/5 or 20/100 or 2, without further work of merit A3 Writes 41% as 41/100 or 41 without further work of merit A4 Some knowledge of tax credits eg. Tax payable=total tax  tax credits A5 Some knowledge of net income eg. Net income = gross income  net tax A6 Any relevant step A7 Any one of the following numbers without work €7,300 or €4,510 or €11,810 or €9,980 Worthless (0) W1 Incorrect answer no work shown

(c) (ii)

5 marks

I €15 105 + €1830 = €16 935 €16 935  € 7300 = € 9635 €9635 = 41% €9635  41 = €235 €235 = 1% €235  100 = €23 500 €23 500 + €36 500 = New gross income = €60 000 II €15 105  €9980 = €5125 €5125 = 41% of increase €5,125 41 = €125 = 1% €125×100 = €12 500 €12 500 + €47 500 = €60 000

Net tax  previous net tax

Increase + previous

III x = Gross income

(€7300 + (x  €36 500)(41))  €1830 = Tax paid

*

€7300 + 41x  €14 965  €1830 = €15 105 41x  €9495 = €15 105 41x = €15 105 + €9495 41x = €24 600 x = €24 600  41 = €60 000 Accept candidate’s values from (c ) (i) in II and III

Page 7

Att 2


Correct answer no work shown. Uses 20% instead of 41% Incorrect use of tax credits Adds €7300 instead of subtracting in method I Divides by 100, and multiplies by 41 Stops at €23 500 I or €12 500 II or €24 600  41 III Mathematical blunder

Slips (-1) S1 Stops at €23 500 + €36 500 I or €12 500 + €47 500 II S2 Numerical error, max -3 Misreadings (-1) M1 Incorrect reading of a digit which does not oversimplify the question Attempts (2 marks) A1 Subtracts €7300 (as in Method I) A2 Adds €1830 (as in Method I) A3 Divides by 41 A4 Multiplies by 100 A5 Adds €36 500 to some relevant number A6 Any relevant step Worthless (0) W1 Incorrect answer no work shown

Page 8


10 marks 20(5, 10, 5) marks 20(10, 10) marks

Att 3 Att(2, 3, 2) Att(3, 3)

Part (a) 10 marks P is the set of divisors of 12. Q is the set of divisors of 9.



Att 3

Using this information copy and complete the Venn diagram. Q

P

(a)

10 marks

Att 3

Divisors of 12 = {1,2,3,4,6,12} Divisors of 9 = {1,3,9} Venn diagram

*

P

2.

6.

4.

12.

1. 3.

Q 9.

Ignore notation

Slips (-1) S1 Each missing or incorrect or misplaced entry from the Venn Diagram Attempts (3 marks) A1 Any correct divisors of 12 or 9 Worthless (0) W1 Copies diagram and stops

Page 9

Part (b) 20 (5, 10, 5) marks Att (2, 3, 2) A group of 100 students were asked if they had a presence on particular social networking websites A, B and C. 24 students had a presence on A only, 40 had a presence on B and 50 had a presence on C. 14 students had a presence on A and B but not on C. 18 students had a presence on A and C but not on B. 8 students had a presence on B and C but not on A. 4 students stated that they did not have a presence on any of the websites (i)



(ii)

 Hence, calculate the ratio of students with a presence on B only to the

Using x to represent the number of students who had a presence on all three websites, construct a Venn diagram and solve for x.

students with a presence on C only.

(b) (i) Venn diagram

5 marks

Att 2

U (100)

A

B (40)

24

14

x

18

8

C(50)

4

Page 10

or

U (100)

A

B

24

18 - x

14

x

18

8

24-x

C

4

or

Page 11

U (100)

A

B

24

40−(x+14+8)

14

x

18

8

50−(x+8+18)

C

4

Slips (-1) S1 Each incorrect or misplaced or missing entry from the Venn Diagram, to a max of -3 S2 Universal box not drawn on diagram S3 U = 100 not indicated on diagram Attempts (2 marks) A1 Any correct entry A2 Draws a Venn Diagram of 3 intersecting circles and stops Worthless (0) W1 Diagram of one or two circles, but note A1

Page 12

10 marks (b) (i) Value of x I 24 + 14 + 18 + x + 18  x + 8 + 24  x + 4 = 100 110  x = 100  x =  10 x = 10

Att 3

II 24 + 14 + 18 + x + 18  x + 8 + 24  x = 96 106  x = 96  x =  10 x = 10 *Accept candidate’s work from previous part, provided it does not oversimplify the question Blunders (-3) B1 B2 B3 B4 B5 B6

Correct answer, no work shown  Transposition error Mathematical error U not equal to 100 in I Each missing or incorrect element from previous work in forming equation, to a max of 2; but if equation is oversimplified Att 3 applies Distribution error

Attempts (3 marks) A1 Any correct term in forming equation and stops A2 Any effort to combine terms from Venn Diagram A3 Oversimplification (e.g. uses 40 and 50 to get x =  46 etc.) Worthless (0) W1 Incorrect answer, no work shown

Page 13

(b) (ii) 5 marks Ratio B only: C only 18  x : 24  x 18 10: 24 10 8 : 14= 4:7 * Accept candidate’s value of x from previous part * Accept 8 : 14 (or equivalent) for full marks with work

Att 2

Blunders (-3) B1 B2 B3 B4

Correct answer, no work shown  Value of x from previous part not used Mathematical error Fails to find ratio, stops at 18  10 : 24  10

Slips (-1) S1 Finds ratio C only: B only ( 14 : 8 or 7:4 ) S2 Numerical errors to a max of 3 S3 Two numbers correct in correct order, with no ratio sign Attempts (2 marks) A1 States 18  x or 24  x or candidate’s equivalent from (b)( i) and stops A2 Mention of 8 or 14 in this section, or candidate’s equivalent. This may appear on the diagram A3 Ratio of two numbers Worthless (0) W1 Produces the same diagram as in (b)(i) without relevant work

Page 14

Part (c) 20(10, 10) marks Att(3, 3) €2000 was invested at r % for 2 years compound interest. A tax of 25% was deducted each year from the interest gained. At the end of the first year the investment amounted to €2030, after tax was deducted. (i) (ii) (c) (i)

 

Calculate the rate of interest r %.

Find the amount of the investment at the end of 2 years, after tax has been deducted. 10 marks Att 3

I €2030  €2000 = €30 €30 = 75% 30 75 = 0  4 = 1% 0  4  100 = € 40 = Interest 40  2000  100 1 Rate of interest r = 2% II €2000  r100 = 20r 20 r  25100 = 500r 100 = 5r 20r - 5r = 15r 15 r = 30 r=2%

(or 20r  75100 = 15r)

* Answer of r = 15% with work is worth 4 marks * Stops at 15r or at € 40 is worth 4 marks 40 *  2000  100 1 is worth 7 marks Blunders (-3) B1 Correct answer, no work shown  B2 Mathematical error B3 Incorrect operation B4 Incorrect transposition B5 Inverted fraction B6 Expresses % as incorrect fraction and continues Slips (-1) S1 Numerical error to a max of 3 Attempts (3 marks) A1 Finds 30 and stops A2 Expresses 25% or 75% as a correct fraction and stops A3 Any relevant step Worthless (0) W1 Incorrect answer, no work shown Page 15

(c) (ii) €2030  2100 = €4060 €4060  25100 = €1015 €4060  €1015 = €3045 €2030 + €3045 = €2060·45 *

10 marks

Att 3

Interest Tax on interest Interest less tax (or €4060  75100 = €3045) Value of the investment

Accept candidate’s r from previous part


Correct answer, no work shown  Mathematical error Decimal error Different value of r from previous part (unless worked out again in this part: mark on slip and blunder) Mishandles 25% or 75% Fails to finish

Slips (-1) S1 Numerical error to a max of 3 Attempts (3 marks) A1 Some correct effort to find interest before tax A2 Some correct effort to simplify 25% or 75% A3 Any relevant step Worthless (0) W1 Incorrect answer, no work shown

Page 16


10 marks 25(10, 5, 10) marks 15(5, 5, 5) marks

Part (a)

10 marks



Att 3 Att(3, 2, 3) Att(2, 2, 2) Att 3

Write the reciprocal of 10 000 in the form 1  10 n, where n  Z.

(a)

10 marks 1

1

Att 3

104

10000 = (or 00001) 4 = 10 Answer = 1  10 - 4


Correct answer, no work shown  Not in form of 1  10 n , final answer left as 10 4 , with work Index error e.g. answer given as 1  10 4 , with work Final answer given as 1  1104 or 1 00001, with work Decimal error

Attempts (3 marks) A1 Indicates knowledge of reciprocal A2 10 4 without work Worthless (0) W1 Incorrect answer, no work shown, but note A2

Part (b) 20(10, 5, 10) marks Att(3, 2, 3) A builders’ supplier sells two types of copper pipes. One has a narrow diameter and costs €x per length. The other has a wider diameter and costs €y per length. Tony buys 14 lengths of the narrow diameter pipes and 10 lengths of the wider diameter pipes at a cost of €555. Gerry buys 12 lengths of the narrow diameter pipes and 5 lengths of the wider diameter pipes at a cost of €390. (i) Write two equations to represent the above information. (ii) pipe.



Solve these equations to find the cost of a length of each type of copper

Page 17

(b) (i) 15(10, 5) marks Att(3, 2) 14x + 10y = 555 12x + 5y = 390 * Two equations to mark in (b) (i) * Each equation is marked independently of the other; the first is worth 10 marks, with Att 3 for any correct work, the second 5 marks, Att 2. Blunders (-3) B1 Incorrect term, once per equation Attempts (3,2 marks) A1 14x or 10y or 12x or 5y A2 Effort at creating an equation equal to 555 or 390 (b) (ii) I 14x + 10y 12x + 5y

10 marks = 555 = 390 (-2)

3m

14x + 10y = 555 −24x −10y = −780 −10x = −225 x = €22·50

4m 7m

14(22.5) + 10y = 555 315+ 10y = 555 10y = 240 y = €24

10 m

II 14x + 10y 12x + 5y

= 555 ×6 = 390 ×−7

3m

86x + 60y = 3330 −84x −35y = −2730 25y = 600 y = €24·00

4m 7m

14x + 10(24) = 555 14x + 240 = 555 14x = 315 x = €22·50

10m Page 18

Att 3

III 14x + 10y 12x + 5y

= 555 = 390

3m

14x = 555 −10y x= 12

+ 5y = 390 (×14)

6660 – 120y + 70y = 5460 −50y = −1200

4m 7m

y = €24·00 x=

10m

x = €22·50

IV 14x + 10y 12x + 5y

= 555 = 390

3m

5y = 390 – 12 x y= 14x + 10

= 555

70x + 3900 – 120x = 2775 −50x = −1125

y = €24·00

4m 7m

x = €22·50 y=

(×5)

.

10m

Page 19

Notes * 1. Accept candidate’s equations from part (b)(i) provided that it does not oversimplify * 2. Apply only one blunder deduction in establishing the first equation in terms of x only or the first equation in terms of y only *3. Finding the second variable is subject to a maximum deduction of 3 marks *4. Correct values without algebraic work, both verified in both equations merits 10 marks *5. Correct values without algebraic work not verified or not fully verified merits attempt 3 marks Blunders (-3) B1 Finds one variable only B2 Distribution error B3 Mathematical error B4 Incorrect substitution when finding second variable, but note M1 B5 Transposition error in solving first variable B6 Transposition errors in solving second variable B7 Error(s) in establishing the first equation in terms of x only (10x = 225) or the first equation in terms of y only (25y = 600) through elimination by cancellation ( I and II ) B8 Error(s) in establishing the first equation in terms of x only (-50x = −1125) or the first equation in terms of y only ( −50y = −1200 ) through elimination or by substitution ( III and IV) Misreadings (-1) M1 Misreads digits, provided it doesn’t oversimplify answer Slips (-1) S1 Numerical errors to a max of −3 Attempts (3 marks) A1 Any correct manipulation of either given equation and stops A2 Some correct partial substitution and stops Worthless (0) W1 Incorrect answer, no work shown W2 Trial and error, but see *4 and *5 above

Page 20

Part (c) (i)

15(5, 5, 5)marks



Express in its simplest form:

2

3 (ii)



Att (2, 2, 2)

4 1 Hence, or otherwise, solve the equation: 2

3

4

1

√ , where a, b  N.

giving your answers in the form

(c) (i)

1 3

5 marks 3

Att 2

2 1

3

4

4

2 1

3

1 4

12 1

2

2 4

10 1

4 10 5

*

4

Accept common denominator as (x + 1)( x + 4 ). Penalise incorrect multiplication in part (c) (ii).

Page 21

Blunders (-3) B1 B2 B3 B4 B5 B6 B7 B8 B9

Correct answer but no work shown.  Incorrect denominator. Mishandles denominator. Mishandles numerator. Distribution error, once if consistent Mathematical error. Fails to combine like terms in final answer. Combining unlike terms and continues. . Reads equation as Continue to apply slips and blunders

Slips (-1) S1 Numerical errors to a max of −3. Attempts (2 marks) A1 Identifies common denominator and stops. A2 Any correct relevant step and stops. A3 No denominator used A4 Oversimplification Worthless (0) W1 Incorrect answer and no work shown. W2 W3

or

or

etc

Page 22

(c) (ii) 5 marks

Att2

10(5, 5) marks 5 marks

Att(2, 2) Att2

Establishing quadratic equation

Solving quadratic equation

I

a = 1; b = 2; c = −26

1 3

10 5

4

3(x+10) =1(x+1)(x+4)

2

2

4 1 2 1

3x + 30 =x2 + x + 4x + 4 x2 + 2x −26 = 0

2

5m

√4 2

2 1

4

1 3

104

2

√108 2

2

6√3

II 3

26

2 3(3)(x + 4) −2(3)(x + 1) = (x + 1)(x + 4)

−1 ± 3√3

or

1 ± 3√3

(see *)

9x + 36−6x−6 = x2 + x + 4x + 4 x2 + 2x – 26 = 0

5m

5m

*

Due to a typographical error, “where a ,b N” should have read “where a ,b  Z”. Accordingly, accept candidate answers −1 ± 3√3 or 1 ± 3√3 for full marks.

*

Accept candidate’s equation for solving, provided it doesn’t oversimplify

Page 23


Correct answer but no work shown.  Fails to group or groups incorrectly. Combines unlike terms and continues e.g. 3x + 30 = 33 Error in transposition. Error in quadratic formula (given in tables) Error in the application of the quadratic formula Finds only one solution.

B8 Stops at B9 Mathematical error B10 Distribution error

√

Slips (-1) S1 Numerical errors to a max of −3.

Attempts (2 marks) A1 Any correct multiplication and stops. A2 Simplification to a linear equation may merit at most Att 2 marks. A3 Solving a linear equation correctly for a single value may merit at most.Att 2 marks A4 Quadratic formula with some correct substitution A5 Any correct relevant step Worthless (0) W1 Incorrect answer and no work shown.

Page 24


10 marks 20(10, 10)marks 20(5, 5, 5, 5) marks

Part (a)

10 marks



Solve

10 marks 3(x −2 ) 5 (x − 3) = 1 3x −6  5x + 15 = 1 3x 5x = 1 + 6 −15  2x = 8 2x = 8 x=4 Accept x = 4 fully verified in equation for full marks

Blunders (-3)

B1 B2 B3 B4

Att 3

3(x – 2) – 5(x – 3) = 1.

(a)

*

Att 3 Att(3, 3) Att(2, 2, 2, 2)

Correct answer but no work shown  Mathematical error Transposition error Distribution error

Slips (-1) S1 Numerical errors to a max of −3. Attempts (3 marks) A1 Some correct relevant work. A2 Tests any value in the equation and stops but note * Worthless (0) W1 Incorrect answer and no work shown.

Page 25

Att 3

Part (b)

20(10, 10) marks

(i)



Simplify fully

(ii)



List the elements of the solution set of - 5  3x – 2 < 7, 10 marks

(b) (i)

Att(3, 3)

(3x – 4)(2x2 + 5x – 2).

(3x − 4)(2x² + 5x −2) 3x(2x² + 5x −2) − 4 (2x² + 5x − 2) 6x³ + 15x² − 6x − 8x² − 20x + 8 6x³ + 7x² −26x + 8

x  Z.

Att 3

3m 7m 10m


Correct answer but no work shown.  Mathematical error. Fails to group or groups incorrectly. Each incorrect or omitted term. Distribution error

Slips (-1) S1 Numerical errors to a max of −3. Attempts (3 marks) A1 Some correct relevant work. A2 Combining unlike terms merits at most attempt mark, subject to marks already secured Worthless (0) W1 Incorrect answer and no work shown.

Page 26

(b) (ii) I 5  3x – 2 < 7  5 +2 x < 7 + 2  3 3x < 9   1 x < 3 {1, 0, 1, 2 } II  

*

*

10 marks

7m 10m

5  3x – 2 5 +2  3x 3 3x  1  x 4m and 3x – 2 < 7 3x < 7 +2 3x < 9  x < 3 7m {1, 0, 1, 2 } 10m One correct inequality with work is worth 4m ( i.e. 1  x or x < 3 ) Two correct inequalities with work is worth 7m (i.e. 1  x and x < 3 or 1 x < 3) Correct set or list and work is worth 10 marks Accept correct answers indicated on a number line, with work


Correct answer but no work shown  Mathematical error Mishandles inequality Transposition error Fails to list. Ignores the negative value in the original (i.e. 5  3x – 2 < 7 and continues) Solves one inequality (B5 may also apply) x  R indicated

Slips (-1) S1 Numerical errors to a max of −3 S2 Includes the number 3 on the list, but note S3 S3 Each incorrect or missing number to a max of −3 Misreading (-1) M1 Reverses one or both inequality signs M2 Omits or includes “equals” incorrectly in inequality, apply once Attempts (3 marks) A1 Tests any value in the inequality and stops Worthless (0) W1 Number line drawn, no correct range or value indicated

Page 27

Att 3

Part (c) 20(5, 5, 5, 5) marks Rectangular tiles are to be placed side by side on a wall. Each tile has a length of x cm.

Att(2, 2, 2 ,2)

of these tiles are required.

(i) (ii)

(iii)

If each tile was 1 cm longer, write down an expression in x for the number of tiles that would now be required. If the longer tiles were used, the number of tiles required would decrease by 10.

 

Write an equation in x to represent this information. Solve this equation to find the value of x.

(c) (i)

5 marks 300 1

Blunders (-3) B1 Inversion B2 x + 1 only Attempts (2 marks) A1 Effort to form relevant expression e.g. A2

Att 2

1

Uses x + 1

Worthless (0) W1 or

(c) (ii)

5 marks –

*

=

10

(or equivalent, based on (c) (i))

Accept candidate’s answer from (c) (i) above

Blunders (-3) B1 Error in setting up equation = 10 or = 10 B2 Writes correct expression with required terms but no equal sign Attempts (2 marks) A1 Must construct an equation or expression using at least two of the following: 10, answer (i), Worthless (0) W1

Page 28

Att 2

(c) (iii)

(5, 5) marks

Att 2,2

= 10 = 10

=

10

300 = 10(x)( x+1 ) 300 = 10(x2+x) 30 = x2 + x

or

2m x2+ x  30 = 0

or

300 = 10x2 +10 x

5m

x2+ x  30 = 0 (x + 6) (x  5) = 0 x = 6, x = 5 4m Solution: x = 5. 5m *Mark in two parts : 5 marks for equation and 5 marks for solving *Accept candidate’s expression from (c) (ii) provided it doesn’t oversimplify the question Blunders (-3) B1 B2 B3 B4 B5 B6 B7

Correct answer, no work shown  Mathematical error in forming equation Distribution error Transposition error Incorrect factors Correct factors and stops Errors using quadratic formula

Slips (-1) S1 Numerical errors to a max of 3 S2 Stops at x = −6, x = 5 or concludes x = −6 Attempts (2 marks) A1 Linear equation in x merits attempt at most A2 Any correct relevant step Worthless (0) W1 Incorrect answer and no work shown W2 ( )( )

Page 29


5 marks 25(15, 10) marks 20(5, 15) marks

Part (a)

5 marks



Given that

Att 2

, write x in terms of c and y.

(a)

5 marks c2 = y  x c2  y =  x x = y – c2 .


Att 2 Att(5, 3) Att(2, 5)

Correct answer, no work shown  Transposition error Solves for y Square root error Mathematical error

Attempts (2 marks) A1 Some correct relevant work e.g. effort at squaring or effort to isolate x. Worthless (0) W1 Incorrect answer and no work shown W2 Interchanges x and c to get x =

Page 30

Att 2

Part (b) (i)

25(15, 10) marks



When m = 25 and n = 54, find the value of Write your answer in the form

(ii)



Att(5, 3)

Use factors to simplify 3

19

. , where a, b  N.

14 49

(b) (i)

15 marks

Att 5

I 1 2 2 5

1 2 1 3 5 4

II

=

1 4 5

1 5 3 4 4 15

5 4

1 15 4

4 15

16

59 60

75 60

=

=


Correct answer, no work shown  a Answer not in correct form b (e.g.9833333333 decimal answer). May also incur other blunders Incorrect denominator or mishandles denominator Error in multiplying fractions Error in subtracting fractions Error in dividing fractions Mathematical error Substitution error, but note M1

Misreading (-1) M1 Swaps values of m and n when substituting Page 31

Slips (-1) S1 Numerical errors to a max of 3 Attempts (5 marks) A1 Finds common denominator and stops A2 Some correct substitution A3

Effort to subtract candidate’s values of

from

Worthless (0) W1 Incorrect answer and no work shown W2 04 and/or 125, but note A2

Page 32

(b) (ii)

10 marks

Att 3

I 3 3

19

14

49 7 7

2 7

3

2 7

II 3

19

14 49

3

21 7

2

3

7 7

7 3

*1

14 2 7

7 7

3

2 7 2 7

and continues is one blunder

*2

Candidate may use quadratic to solve numerator = 0. (x7)(3x + 2) and continues. Blunders (-3) B1 B2 B3 B4 B5 B6 B7 B8

7

Correct answer, no work shown  Incorrect factors of denominator Incorrect factors 3x2 Incorrect factors of 14 Incorrect factors, leading to an incorrect middle term Fails to finish Blunder in quadratic Substitution error in quadratic

Attempts (3 marks) A1 Sets up factors or sets up quadratic, identifies a, b or c A2 Finds guide number or a relevant number II A3 Uses formula and stops at roots A4 Mentions difference of 2 squares A5 Any correct relevant step A6 Quadratic with some correct substitution. (See *2) Worthless (0) W1 Incorrect answer and no work shown W2 ( )( )

Page 33

x = 7 and x= 23



Part (c)

20 (5, 15) marks

Let f be the function

Att (2, 5)

2

f : x  - x – 4x + 5, x  R.

(i)



Find the co-ordinates of the points where the graph of f (x) cuts the x-axis.

(ii)



Solve f (x) = f (x + 1).

(c) (i) 5 marks 2 I  x – 4x + 5 = 0 (- x - 5)(x - 1) = 0 -x=5 x –1 = 0 x = -5 x=1 Cuts the x-axis at (-5, 0) and (1, 0). II  x2 – 4x + 5 = 0 a= - 1, b = -4, c = 5 √ x = =

* * *

Att 2

=

√

=

= -5 and 1

Cuts the x-axis at (-5, 0) and (1, 0) Accept x2 + 4x - 5 = 0 solved correctly using I or II for full marks The two correct x values verified fully by trial and error = full marks Accept graph with points of intersection indicated correctly, with work


Correct answer, no work shown  Incorrect factors of  x2 Incorrect factors of 5 Incorrect factors leading to an incorrect middle term Transposition error Mathematical error Fails to find roots Graph showing correct points of intersection, no work otherwise

Slips (-1) S1 Numerical errors to a max of 3 S2 Does not state co-ordinates, apply once Attempts (2 marks) A1 Some correct relevant work A2 Tries to set up factors A3 Quadratic with some correct substitution A4 Correct x value(s) without algebraic work, not fully verified or not verified A5 (0,5) or y = 5 stated (where graph cuts the y axis)

Page 34

Worthless (0) W1 Incorrect answer and no work shown W2 Trial and error of x values other than x = -5 and x = 1 W3 States x = 5, no work W4 ( )( )

(c) (ii) f(x) = – x2 – 4x + 5

15 marks

f(x + 1) = – (x + 1 )2 – 4( x +1) + 5 – (x2 + 2x + 1) – 4x – 4 + 5 – x2 – 2x – 1 – 4x +1 – x2 – 6x f (x) = f (x + 1) – x2 – 4x + 5 = – x2 – 6x – x2 – 4x + 5 + x2 + 6x = 0 2x + 5 = 0 2x = – 5 x = – 52

*Accept x =  25 fully verified in f (x)

=

f (x + 1) for full marks


Correct answer, no work shown  Mathematical error Substitution error, provided it does not oversimplify question Distribution error Error in squaring Transposition error

Slips (-1) S1 Numerical errors to a max of 3 Attempts (5 marks) A1

Some correct relevant substitution

Worthless (0) W1 Incorrect answer and no work shown W2 f (x) = f (x + 1) W3 x = x + 1, even if continued

Page 35

Att 5


5 marks 30(10, 10, 5, 5) marks 15(5, 5, 5) marks

Part (a)

5 marks

Att 2

5 marks 2m

Att 2

Let h be the function h : x √

 

4 .

Show that h(0) > h(-4).

(a) h(0) =

√0

h(-4) =

√ 4

4

√4 4

=

=2

√0 = 0 2m 5m

h(0) > h(-4) or 2>0 Blunders (-3) B1 B2 B3 B4 B5

Att 2 Att(3, 3, 2, 2) Att(2, 2, 2)

Correct answer ( i.e 2 > 0 ), no work shown  Incorrect h(0) Incorrect h(-4) √4 = -2 leading to incorrect conclusion Fails to finish

Attempts (2 marks) A1 Some correct relevant substitution Worthless (0) W1 Incorrect answer and no work of merit W2 h(0) > h(-4) W3 0 > -4 and no work of merit

Page 36

Part (b)

30(10, 10, 5, 5) marks

Att(3, 3, 2, 2)

Let f be the function f : x  x2 + 5x and let g be the function g: x  x + 2.



Using the same axes and scales, draw the graph of f and the graph of g, for - 5  x  1, x  R.

‐5

‐4

‐3

‐2

6

6

5

5

4

4

3

3

2

2

1

1

‐1

1

‐5

‐4

‐3

‐2

‐1

1

‐1

‐1

‐2

‐2

‐3

‐3

‐4

‐4

‐5

‐5

‐6

‐6

or Page 37

(b) Function f f : x  x2 + 5x

20 (10, 10) marks

Att (3, 3)

x

-5

-4

-3

-2

-1

0

1

x2

25

16

9

4

1

0

1

+5x

-25

-20

-15

-10

-5

0

5

f(x)

0

-4

-6

-6

-4

0

6

* * * *

Table is worth 10 marks, graph is 10 marks Middle lines of table do not have to be shown Candidate may choose not to use a table. Points might not be listed, mark on position on graph

*


Blunders (-3) B1 Error in calculating x2, if consistent penalise once, but note A3 B2 Error in calculating 5x, if consistent penalise once B3 Error in calculating last line of table, once if consistent B4 Each incorrect point without work B5 Point plotted incorrectly, once only if consistent B6 Incomplete domain B7 Axes scaled incorrectly, once only B8 Reversed axes B9 No curve between (3,6) and (2, 6) on graph Slips (-1) S1 Numerical errors to a max of 3 Attempts (3 ,3 marks) A1 Some correct substitution A2 Draws axes, with some indication of scaling A3 Error leading to a linear graph

Page 38

(b) Function g g: x  x + 2.

10 (5, 5) marks

Att (2, 2)

x

-5

-4

-3

-2

-1

0

1

+2

2

2

2

2

2

2

2

g(x)

-3

-2

-1

0

1

2

3

* * * * *

Table is worth 5 marks, graph is 5 marks In g(x) function only 2 points are necessary in calculation; graph must be drawn from -5 to 1 Middle lines of table do not have to be shown Candidate may choose not to use a table. Points might not be listed, mark on position on graph

*


Blunders (-3) B1 Graphs not drawn on the same axes or scales (apply once only) B1 Error in “x” line B2 Error in “2” line B3 Error in calculating last line of table, once if consistent B4 Each incorrect point without work B5 Point plotted incorrectly, once only if consistent B6 Incomplete domain B8 Axes scaled incorrectly B9 Reversed axes, but if already blundered, do not penalise again Slips (-1) S1

Numerical errors to a max of 3

Attempts (2,2 marks) A1 Some correct substitution A2 Draws axes, with some indication of scaling

Page 39

Part (c) 15(5, 5, 5) marks Use your graphs from part (b) to estimate: (i)

 

Att(2, 2, 2)

The minimum value of f (x)

(ii) The values of x for which f (x) = g(x) (iii) The range of values of x for which f (x)  g(x).

(c) (i)

5 marks The minimum value of f (x) = -6·25

Att 2

(c) (ii)

5 marks f (x) = g(x) at x = 0·5 and x = -4·5

Att 2

(c) (iii) 5 marks The range of values of x for which f (x)  g(x). -4·5 ≤ x ≤ 0·5 * Apply marks to parts (c) (i) (ii) and (iii) *Accept “from 45 to 5” for (c) (iii) Blunders (-3) B1 Value(s) not consistent with candidate’s graph, tolerance  2 B2 Fails to use graph B3 y values given instead of x values (ii) and (iii) B4 x value only given instead of f (x) in (c) (i) Slips (-1) S1 In (c) (i) answer given as minimum point instead of minimum value S2 In (c) (iii) “” given instead of “” or states “between 45 and 5” Misreading (-1) M1 f (x) ≥ g(x) Attempts (2 marks) A1 Some correct substitution A2 Some correct indication of answer on graph

Page 40

Att 2

BONUS MARKS FOR ANSWERING THROUGH IRISH Bonus marks are applied separately to each paper as follows: If the mark achieved is 225 or less, the bonus is 5% of the mark obtained, rounded down. (e.g. 198 marks  5% = 9.9  bonus = 9 marks.) If the mark awarded is above 225, the following table applies: Bunmharc Marc Bónais Bunmharc Marc Bónais (Marks obtained) (Bonus Mark) (Marks obtained) (Bonus Mark) 11

261 – 266

5

227 – 233

10

267 – 273

4

234 – 240

9

274 – 280

3

241 – 246

8

281 – 286

2

247 – 253

7

287 – 293

1

254 – 260

6

294 – 300

0

226

Page 41

(b) (i)

5 marks

Braking distance (m) Number of Drivers

Att 2

0 – 10

10 – 20

20 – 28

28 – 34

34 – 40

2

28

10

6

4

Blunders (-3) B1

Omission of a value

Slips (-1) S1 Arithmetic slips to a maximum of (-3) Attempts (2 marks) A1 Any one value correctly filled into table A2 Indication of subtraction of frequencies A3 Addition of frequencies Worthless (0) W1 Table copied from examination paper W2 Cumulative frequency curve drawn

(b) (ii)

15 marks The mid-interval values are: 5, 15, 24, 31, 37

Mean 

25  2815  1024  631  437 

2  28  10  6  4 10  420  240  186  148  50 1004 or 20  08  50  20

*

Accept candidate’s answer from (b) (i)

Page 38

Att 5

Blunders (-3)

B1 B2 B3 B4 B5

Correct answer without work shown () Consistent incorrect mid-interval values Division by 5 Division by sum of mid-interval values Mid-interval values added to frequencies instead of multiplied

Slips (-1) S1 Arithmetic slips to a maximum of (-3) S2 Answer not rounded or incorrectly rounded Attempts (5 marks) A1 Step towards mid-interval values A2 One correct multiplication in numerator A3 Indication of division by 50 Worthless (0) W1 Sum of frequencies divided by 5

Part (c)

20(10, 5, 5) marks

Att(3, 2, 2)

An airline recorded the weight of each passenger’s baggage on a particular flight. The results are shown in the histogram

25 20 15 10 5 0

5

10

20

50

30

Weight in kg

(i)

Copy and complete the following frequency table in your answer book.

Weight in kg No. of passengers

0–5 20

5 – 10

10 – 20

20 – 30

30 – 50

[Note: 5 – 10 means 5 or more but less than 10, etc.] (ii) (iii)

How many passengers were on the plane? The airline charged an excess baggage fee of €8 for every kg over 25 kg. The airline collected €2880 from passengers in the 30 – 50 kg group.



What was the average excess baggage fee paid per passenger in the 30 – 50 kg group?

Page 39

(c) (i)

10 marks

Weight in kg No. of passengers

0–5 20

5 – 10 25

10 – 20 20

Att 3 20 – 30 50

30 – 50 40

Blunders (-3) B1 Height taken as frequency B2 Mishandling of base B3 Omission of a value, each time Slips (-1) S1 Arithmetic slips to a maximum of (-3) Attempts (3 marks) A1 Indication of work with base Worthless (0) W1 Table copied from examination paper with no further entries

(c) (ii)

* *

5 marks Number of passengers = 20 + 25 + 20 + 50+40 = 155

Accept candidate’s answer from (c) (i) Accept answer without work

Blunders (-3) B1 Omission of each number Slips (-1) S1 Arithmetic slips to a maximum of (-3) Attempts (2 marks) A1 Indication of addition of number of passengers

Page 40

Att 2

(c) (iii)

5 marks Number of passengers in 30 – 50 kg group = 40 Average excess baggage fee per passenger = €2880 ÷ 40 = €72 or

Excess baggage = €2880 ÷ 8 = 360 kg Average excess baggage per person = 360 ÷ 40 = 9kg Average excess baggage fee per passenger = 9 × 8 = €72 *

Accept candidate’s answer from (c) (i)

Blunders (-3)

B1 B2 B3

Correct answer without work shown () Incorrect number of passengers Multiplication by 40

Slips (-1) S1 Arithmetic slips to a maximum of (-3) Attempts (2 marks) A1 Number of passengers = 40 A2 €2880 ÷ 8 A3 €2880 ÷5 Worthless (0) W1 25 × 8

Page 41

Att 2

BONUS MARKS FOR ANSWERING THROUGH IRISH

Bonus marks are applied separately to each paper as follows: If the mark achieved is 225 or less, the bonus is 5% of the mark obtained, rounded down. (e.g. 198 marks  5% = 9.9  bonus = 9 marks.) If the mark awarded is above 225, the following table applies: Bunmharc Marc Bónais Bunmharc Marc Bónais (Marks obtained) (Bonus Mark) (Marks obtained) (Bonus Mark) 226

11

261 – 266

5

227 – 233

10

267 – 273

4

234 – 240

9

274 – 280

3

241 – 246

8

281 – 286

2

247 – 253

7

287 – 293

1

254 – 260

6

294 – 300

0

Page 42

JUNIOR CERTIFICATE EXAMINATION 2009

MARKING SCHEME MATHEMATICS HIGHER LEVEL PAPER 1

Page 1


Penalties of three types are applied to candidates’ work as follows: • Blunders - mathematical errors/omissions (-3) • 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, B2, B3,…, S1, S2,…, M1, M2,…etc. These lists are not exhaustive.

2.

When awarding attempt marks, e.g. Att(3), 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.

3.


4.


5.


6.


7.


8.


9.


10.


11.


12.

Do not penalise the use of a comma for a decimal point, e.g. €5.50 may be written as €5,50.

Page 2


10 marks 20(10, 10) marks 20(10, 10) marks

Att 3 Att (3, 3) Att (3, 3)

Part (a) 10 marks Att 3 In a school library, 28% of the books are classified as fiction and the remainder as non-fiction. There are 3240 non-fiction books in the library.

"

Find the number of books which are classified as fiction.

Part (a) Non-fiction = 100 − 28 = 72 %

1% of Non-fiction books = ⇒

10 marks

Att 3

3240 = 45 72

No. of books (fiction) = 45 × 28 =1260

Blunders (-3)

B1 B2 B3 B4

Correct answer but no work shown. (") Mishandles % and continues. (3240 ≠ 72%) Decimal error. Fails to finish.

Slips (-1) S1 Numerical errors to a max of 3. S2 Rounds off incorrectly. Attempts (3 marks)

A1

Works with 28% or similar and stops. (28% =

A2

100-28 and stops.

Worthless (0) W1 Incorrect answer and no work shown. W2 28 × 3240 and stops.

Page 3

28 100

)

Part (b) (i)

(ii)

"

20 (10, 10) marks

Given that x = 2 × 10 -3 and y = 7 × 10 -4 , evaluate x + 8 y. Express your answer in the form a × 10 n, where n ∈ Z and 1 ≤ a < 10.

A supermarket has a special offer on three different brands of packets of soap. The following table gives details of the offer: Brand No. of bars per packet Weight of each bar Price of packet A 3 100g €1·35 B 6 100g €2·40 C 4 125g €2·38

"

Which brand has the cheapest price per gram?

Part (b) (i) (i)

Att (3, 3)

"

10 marks

Att 3

Given that x = 2 × 10 -3 and y = 7 × 10 -4 , evaluate x + 8 y. Express your answer in the form a × 10 n, where n ∈ Z and 1 ≤ a < 10.

Part (b) (i)

10 marks

Att 3

" I −3

(

x + 8 y = 2 ×10 + 8 7 × 10

−4

)

= 2 × 10 −3 + 56 ×10 −4 = 2 × 10 −3 + 5 ⋅ 6 ×10 −3 = 7.6 × 10 −3

II 2 7 : 7 ×10 −4 = 2 ×10 −3 = 1000 10000 2 2 56 76 ⎛ 7 ⎞ = ⇒ x + 8y = + 8⎜ + ⎟ = 1000 1000 10000 10000 ⎝ 10000 ⎠ 76 ⇒ = 0 ⋅ 0076 = 7 ⋅ 6 × 10 −3 10000

III 2 ×10 = 0 ⋅ 002 : 7 ×10 = 0 ⋅ 0007 ⇒ x + 8 y = 0 ⋅ 002 + 8(0 ⋅ 0007 ) = 0 ⋅ 002 + 0 ⋅ 0056 = 0 ⋅ 0076 ⇒ 0 ⋅ 0076 = 7 ⋅ 6 × 10 −3 −3

−4

Blunders (-3)

B1 B2 B3 B4 B5 B6 B7

Correct answer but no work shown. (") Index error. Fails to add. Distribution error. Mathematical error. Final answer in decimal or fraction format. Incorrect operation.

Page 4

Slips (-1) S1 Numerical errors to a max of 3. S2 Index in wrong format e.g. 76 × 10 −4 Misreadings (-1) M1 Reads as 2 × 10 3 and 7 × 10 4 and continues to 5 ⋅ 62 × 10 5 Attempts (3 marks) A1 56 ×10 −4 or 0 ⋅ 0056 and stops. A2 Some correct work with indices and stops, (e.g. 2 × 10 −3 = 0 ⋅ 002 ) Worthless (0) W1 Incorrect answer and no work shown.

Page 5

Part (b) (ii)

10 marks

Att 3

1 (b) (ii) A supermarket has a special offer on three different brands of packets of soap. The following table gives details of the offer: Brand No. of bars per packet Weight of each bar Price of packet A 3 100g €1·35 B 6 100g €2·40 C 4 125g €2·38

"

Which brand has the cheapest price per gram?

(b) (ii)

10 marks

Att 3

" I Brand A

3 × 100

gram/pkt 300

B

6 × 100

600

C

4 × 125

500

II Brand A

B C

135 3 240 6 238 4

Price/bar 45 40

59 ⋅ 5

Price/gram 135 300 240 600 238 500 Price/gram 45 100 40 100 59 ⋅ 5 125

0 ⋅ 45 0 ⋅ 40 0 ⋅ 476

0 ⋅ 45 0 ⋅ 40 0 ⋅ 476

Blunders (-3)

B1 B2 B3 B4 B5 B6

Correct answer but no work shown. (") Incorrect multiplier. Decimal error. Incorrect division. Price /bar or gram / packet only and continues incurs 2 blunders. Each missing component.

Slips (-1) S1 Numerical errors to a max of 3. S2 No conclusion or incorrect conclusion. Misreadings (-1) M1 Indicates most expensive Page 6

B Cheapest

B Cheapest

Attempts (3 marks) A1 Indicates total grams in packet and stops. A2 Works cost per packet and stops. A3 One correct or partially correct step and stops.

Worthless (0) W1 Incorrect answer and no work shown.

Page 7

Part (c) 20 (10,10) marks A man travels from Arklow to Blanchardstown, a distance of 90 km. Arklow at 09:25 and arrives in Blanchardstown at 10:55.

Att (3,3) He leaves

(i) " Calculate his average speed for the journey. He continues from Blanchardstown to Cootehill, a distance of 112 km. He increases his average speed by 4 km/h for this section of his journey. (ii)

"

At what time does he arrive in Cootehill?

Part (c) (i) 10 marks 1(c) A man travels from Arklow to Blanchardstown, a distance of 90 km. Arklow at 09:25 and arrives in Blanchardstown at 10:55.

"

(i)

Calculate his average speed for the journey.

(c) (i)

10 marks

"

Time of journey = 10:55 - 09:25 = 1:30 = 1 ⋅ 5 hrs Dis tan ce 90 90 = = 60km/h or Average Speed = =1 km / min Time 1⋅ 5 90 Average speed for the journey = 60 km/h Blunders (-3) B1 B2 B3 B4 B5

Att 3 He leaves

Correct answer but no work shown. (") Time error. (1 hr = 100min.) Decimal error. Error in S/T/D formula. Mathematical error.

Slips (-1) S1 Numerical errors to a max of 3. Attempts (3 marks) A1 Writes correct S/T/D relationship in this part. A2 Time 1hr: 30min or 1⋅ 5 hrs and stops.


Page 8

Att 3

(c) (ii) 10 marks Att 3 1(c) He continues from Blanchardstown to Cootehill, a distance of 112 km. He increases his average speed by 4 km/h for this section of his journey.

"

(ii)

At what time does he arrive in Cootehill?.

(c) (ii)

"

⇒

10 marks

New speed = 64km/h. Distance B → C = 112km. Distance 112 Time = = = 1⋅ 75 h =1hr: 45min Speed 64 Arrival Time in C = 10:55 + 1:45 = 12:40


Correct answer but no work shown (") Decimal error Error in S/T/D formula Conversion error ( 0 ⋅ 75 hrs ≠ 45 min ) (km/hr added to km/min) Mathematical error Mishandles calculation of new speed Fails to calculate arrival time.

Slips (-1) S1 Numerical errors to a max of 3 S2 Rounds off incorrectly Attempts (3 marks) A1 New speed correct and stops A2 Writes correct S/T/D relationship in this part Worthless (0) W1 Incorrect answer and no work shown.

Page 9

Att 3


10 marks 20(10, 5, 5) marks 20(5, 10, 5) marks

Att 3 Att (3, 3, 2) Att (2, 3, 2)

Part (a) 10 marks 2(a) Eight workers can build a cabin in 60 hours.

"

Att 3

How many workers are needed if the cabin is to be built in 32 hours?

(a) (a)

10 marks

Att 3

" I

II 60 : 32 = 60 x = 32 8 x 1 ⋅ 875 = 8 x = 15

Work hours to build a cabin 8 × 60 = = 15 Hours 32

Blunders (-3)

B1 B2 B3 B4

Correct answer but no work shown. (") Incorrect Ratio. Mathematical error. Incorrect operation.

Slips (-1) S1 Numerical errors to a max of 3 S2 Fails to round off correctly. Attempts (3 marks) A1 Writes 8× 60 and stops. 60 32 8 32 or or or . and stops A2 Writes 32 60 32 8 Worthless (0) W1 Incorrect answer and no work shown. 60 8 W2 Writes and stops. or 8 60

Page 10

x:8

Part (b) 20 (10, 5, 5) marks Att (3, 2, 2) A group of 49 students was asked which fruit each liked. 28 said they liked apples. 25 said they liked pears while 26 said they liked oranges. 8 said they liked all three types of fruit. 17 said they liked pears and oranges. 11 said they liked apples and oranges. 5 said they did not like any of the three types of fruit. Let x represent those students who liked apples and pears but not oranges. (i) (ii) (iii)

" " "

Represent the above information on a Venn diagram. Calculate the value of x. Calculate the percentage of students who liked one type of fruit only. Give your answer correct to the nearest whole number.

(b) (i) 2 (b) (i)

10 marks

"

Att 3

Represent the above information on a Venn diagram.

(b) (i)

10 marks

Att 3

" U = 49

P= 25

A =28

x 8

3

9 6

5

O=26

Slips (-1) S1 Each incorrect or missing entry from the Venn Diagram above. Attempts (3 marks) A1 Any single correct entry. A2 Draws a Venn diagram of 3 intersecting circles and stops.

Page 11

(b) (ii) 2 (b) (ii)

"

5 marks

Att 2

5 marks

Att 2

Calculate the value of x.

(b) (ii)

"

28 − (11 + x ) + x + 8 + 3 + 25 − (17 + x ) + 9 + 6 = 49 − 5 28 −11 − x + x + 8 + 3 + 25 − 17 − x + 9 + 6 = 44 17 − x + x + 8 + 3 + 8 − x + 9 + 6 = 44 51 − x = 44 x = 51 − 44 x=7

*

Accept candidate’s work from previous part.

Blunders (-3)

B1 B2 B3 B4 B5 B6

Correct answer but no work shown. (") Missing or incorrect element from previous work in forming equation. Transposition error. Distribution error. Mathematical error. #U ≠ 49 .

Slips (-1) S1 Numerical errors to a max of 3.

Attempts (2 marks) A1 Any correct term in forming the equation and stops. A2 Any effort to combine terms from Venn diagram.


Page 12

Part (b) (iii) 2 (b) (iii)

"

5 marks

Att 2

Calculate the percentage of students who liked one type of fruit only. Give your answer correct to the nearest whole number.

Part (b) (iii)

5 marks

"

Att 2

I

U = 49 P= 25

A =28 1

10

x 8

3

9 6

5

O=26 Total who liked one type of fruit only = 10 + 1 + 6 = 17. 17 % ×100 = 34 ⋅ 7 = 35% 49

II Total who liked one type of fruit only = 6 + (8 − x ) + (17 − x ) ⇒ 31− 2 x = 31 − 2(7 ) = 31 − 14 = 17 17 % ×100 = 34 ⋅ 7 = 35% 49

*

Accept candidate’s work from previous part.

Page 13

Blunders (-3)

B1 B2 B3 B4 B5 B6 B7

Correct answer but no work shown. (") Missing or incorrect element from previous work. Decimal error Distribution error. Mathematical error. Fails to find %. #U ≠ 49

Slips (-1) S1 Numerical errors to a max of 3. S2 Fails to round off or rounds incorrectly.

Misreadings (-1) M1 Finds % who liked two fruits.

Attempts (2 marks) A1 Any correct value/term and stops. Worthless (0) W1 Incorrect answer and no work shown.

Page 14

Part (c) 20 (5, 10, 5) marks Att (2, 3, 2) Three business partners, Aideen, Brian and Caroline, invest €30 000, €40 000 and €70 000 respectively. At the end of each year, 22⋅5 % of the profit made is placed in reserve and the remainder is divided among the partners in proportion to their investments. (i)

"

(ii)

"

(iii)

"

Given that in 2007, the profit amounted to €12 880, calculate the amount placed in reserve. In 2008, Caroline’s portion of the profit was €9331. Calculate how much Aideen and Brian each received in 2008. Calculate the amount placed in reserve in 2008.

Part (c) (i) 2 (c) (i)

5 marks

"

Att 2

Given that in 2007, the profit amounted to €12 880, calculate the amount placed in reserve.

Part (c) (i)

5 marks

Att 2

" I 12880 × 22 ⋅ 5 = €2898 100

II

12880 × 0 ⋅ 225 = €2898

Blunders (-3)

B1 B2 B3 B4 B5

Correct answer but no work shown. (") Mathematical error. Incorrect operation. Expresses 22 ⋅ 5% as incorrect fraction and continues. Profit taken as 77 ⋅ 5% and continues.

Slips (-1) S1 Numerical errors to a max of 3. S2 Rounds to 23% and continues. ( €2962 ⋅ 40 ) Attempts (2 marks) A1 Writes 77 ⋅ 5% and stops. 22 ⋅ 5 or similar and stops. A2 22 ⋅ 5 % as 100 A3 Finds 22 ⋅ 5 % of any of the given values. A4 Leaves answer as €12880 × 22 ⋅ 5 % Worthless (0) W1 Incorrect answer and no work shown. W2 2007×12880 or similar and stops. Page 15

III 9 40 9 12880 × = €2898 40 22 ⋅ 5 =

Part (c) (ii) 2 (c)(ii)

10 marks

"

Att 3

In 2008, Caroline’s portion of the profit was €9331. Calculate how much Aideen and Brian each received in 2008.

(c) (ii)

10 marks

Att 3

" I 3 4 7 Ratios : : : 14 14 14 C = €9331 ⇒ Profit for distribution = 9331× 2 = €18662 3 A = 18662 × = € 3999 14 4 B = 18662 × = € 5332 14

*

II C = €9331 9331 = 1 ⋅ 333 7000 A = 1 ⋅ 333× 3000 = € 3999 B = 1 ⋅ 333× 4000 = € 5332

III C = €9331 A: B: C = 3:4:7 (or similar). 9331 = 1333 7 A = 1333× 3 = €3999 B = 1333 × 4 = €5332

Candidates may work with other correct ratios or correct percentages.

Blunders (-3)

B1 B2 B3 B4 B5 B6

Correct answer but no work shown. (") Mathematical error. Incorrect denominator. [Note M1] Incorrect numerator. [Note M1] Finds Aideen's or Brian's value only. Premature rounding that effects the answer i.e. ( 1⋅ 333 to 1) and continues correctly.

Slips (-1) S1 Numerical errors to a max of 3. S2 Rounds off incorrectly. Misreadings (-1) M1 Reads Caroline as Aideen or Brian and continues. Attempts (3 marks) A1 Some correct work at simplifying ratios. 9331 A2 Writes or similar and stops. 7000 Worthless (0)

W1 Incorrect answer and no work shown.

Page 16

Part (c) (iii) 2 (c) (iii)

5 marks

"

Att 2

Calculate the amount placed in reserve in 2008.

(c) (iii)

5 marks

Att 2

" Distributed Profit (2008) = €9331 + €5332+ €3999 = €18 662 or €9331 × 2 = € 18 662 (or similar) I II € 18 662 = 77 ⋅ 5% 18 662 ×100 = €24 080 Total Profit = 22 ⋅ 5 ⎛ 9 ⎞ 77 ⋅ 5 ⇒ Reserve = 18 662 × ⎜ ⎟ = €5418 Reserve = €24 080 - €18 662 = €5418 77 ⋅ 5 ⎝ 31 ⎠ *

Accept candidate’s answers from part (ii).

Blunders (-3)

B1 B2 B3 B4 B5

Correct answer but no work shown. (") Mathematical error. Incorrect distributed profit. Incorrect ratio Decimal error

Slips (-1) S1 Numerical errors to a max of 3. S2 Incorrect rounding or early rounding which affects answer. Misreadings (-1) M1 Reads € 18 662 as 100% i.e. Finds 22 ⋅ 5% of € 18 662 (€ 4198 ⋅ 95 ) and continues. Attempts (2 marks) A1 Some correct work at finding total profit. A2 States € 18 662 = 77 ⋅ 5% or similar and stops. A3 Writes 100 − 22 ⋅ 5 = 77 ⋅ 5 and stops. Worthless (0) W1 Incorrect answer and no work shown.

Page 17


10 marks 20(10, 10) marks 20(10, 5, 5) marks

Part (a)

10 marks

Att 3

10 marks

Att 3

"

Att 3 Att (3, 3) Att (3, 2, 2)

Simplify: (2x – 3) (4 – 5x).

(a)

" (2 x − 3)(4 − 5 x )

2 x(4 − 5 x ) − 3(4 − 5 x )

8 x − 10 x 2 − 12 + 15 x − 10 x 2 + 23x − 12 Blunders (-3)

B1 B2 B3 B4

Correct answer but no work shown. (") Mathematical error. Fails to group or groups incorrectly. Each incorrect or omitted term.

Slips (-1) S1 Numerical errors to a max of 3.

Attempts (3 marks) A1 Some correct relevant work. A2 Combining unlike terms merits at most attempt mark subject to marks already secured.


Page 18

Part (b) (i)

20 (10, 10) marks

"

Given that x = 2t – 1 and y =

Att (3, 3)

2 t + 2 , express 3x – y + 2 3

in terms of t, in its simplest form. (ii)

"

Hence, find the value of t when 3x – y + 2 = 0.

(b) (i) 3(b) (i)

10 marks

"

Given that x = 2t – 1 and y =

Att 3

2 t + 2 , express 3x – y + 2 3

in terms of t, in its simplest form. (b) (i)

10 marks 3x − y + 2 ⎛2 ⎞ 3(2t − 1) − ⎜ t + 2 ⎟ + 2 ⎝3 ⎠ 2 6t − 3 − t − 2 + 2 3 16t − 9 16 1 t − 3 or 5 t − 3 or 3 3 3

Att 3

Blunders (-3)

B1 B2 B3 B4 B5 B6 B7 B8 B9

Correct answer but no work shown (") Mathematical error Distribution error Fails to group or groups incorrectly Incorrect substitution for x and /or y and continues Ignores the constant in the expression and continues. [see A3] Eliminates t . Answer as ( x − 3 y + 7 = 0 ) (Apply B5 also.) Answer not in simplest form Mishandles fractions

Slips (-1) S1 Numerical errors to a max of 3 Attempts (3 marks) A1 Some correct relevant work A2 Partial substitution and stops

A3

2 Ignores all the constants in each equation to give. [ 3(2t ) − t ] and stops or continues 3

Worthless (0) W1 Incorrect answer and no work shown

Page 19

(b) (ii) 3(b) (ii)

10 marks

"

Att 3

Hence, find the value of t when 3x – y + 2 = 0.

(b) (ii)

10 marks

Att 3

" 3x − y + 2 = 0 16 t −3=0 3 16t − 9 = 0 9 t = ( 0 ⋅ 5625) 16

* *

Accept candidate's answer from previous part. 1 19 Solving 3x − y + 2 = 0 and x − 3 y + 7 = 0 ( , ) followed by substitution gives correct 8 8 answer (see B7 previous part).

Blunders (-3)

B1 B2 B3

Correct answer but no work shown. (") Mathematical error. Transposition error.

Slips (-1) S1 Numerical errors to a max of 3. Attempts (3 marks) A1 Some correct relevant work. Worthless (0) W1 Incorrect answer and no work shown.

Page 20

Part (c) 20 (10, 5, 5) marks Att (3, 2, 2) A swimming pool can be filled by a large pipe operating alone in 4 hours. (i) What fraction of the pool can be filled by this pipe in 1 hour? The swimming pool can be filled by a small pipe operating alone in x hours.

"

Derive an expression in x for the fraction of the pool filled by the two pipes working together in 1 hour. It takes 3 hours for the two pipes working together to fill the pool. (ii)

(iii) (c) (i) 3(c) (i)

"

Find x.

10 marks A swimming pool can be filled by a large pipe operating alone in 4 hours. What fraction of the pool can be filled by this pipe in 1 hour?

(c) (i)

Fraction of the pool

10 marks 1 = 4

Blunders (-3) B1 Any variable divided by 4. B2 15 minutes or 900 seconds. Slips (-1) S1 Numerical errors to a max of 3. Misreadings (-1) M1 Not in required form (e.g. decimal or percentage).

M2

3 . 4

Attempts (3 marks) A1 Some correct relevant work. Worthless (0) W1 Incorrect answer and no work shown.e.g (4)

Page 21

Att 3

Att 3

(c) (ii) 5 marks The swimming pool can be filled by a small pipe operating alone in x hours. 3(c) (ii)

"

Derive an expression in x for the fraction of the pool filled by the two pipes working together in 1 hour.

(c) (ii)

5 marks

" I 1 Large pipe delivers of Volume /hour. 4 1 Small pipe delivers of Volume/ hour. x 1 1 Together they deliver + of Volume /hour 4 x II L.C.M of 4 and x is 4 x . In 4 x hours the Large Pipe will fill the pool x times In 4 x hours the Small Pipe will fill the pool 4 times In 4 x hours Both Pipes will fill the pool x + 4 times x +4 of the pool. In 1 hour Both Pipes will fill 4x

*

Accept candidate's answer from previous part.

Blunders (-3)

B1 B2 B3

Att 2

Correct answer but no work shown. (") Error in forming the expression. Mathematical error.


Page 22

Att 2

Part (c) (iii)

5 marks

Att 2

"

It takes 3 hours for the two pipes working together to fill the pool. Find x.

3(c) (iii)

(c) (iii)

5 marks

Att 2

" I

⎛1 1⎞ 3⎜ + ⎟ = 1 ⎝4 x⎠ 3 3 ⇒ + =1 4 x 3 1 ⇒ = x 4 ⇒ x = 12

II 1 1 1 + = 4 x 3 x+4 1 = ⇒ 4x 3 ⇒ 3x + 12 = 4 x ⇒ x = 12

III

3 of pool 4 1 3 hours small pipe → of pool 4 1 ⇒ pool in 3 hours 4 ⇒ 1 pool in 12 hours 3 hours large pipe →

* Accept candidate's answer from previous part unless oversimplifying applies. * Accept 4 × 3 = 12 Blunders (-3) B1 B2 B3 B4 B5 B6

Correct answer but no work shown. (") Mathematical error. Transposition error. Error in setting up equation. Mishandles fractions Mishandles common denominator.


Page 23


10 marks 20(10, 10) marks 20(10, 5, 5) marks

Part (a)

10 marks

"

Att 3 Att (3, 3) Att (3, 2, 2) Att 3

Given that y = 2 x − a , find the value of y when x = 4 and a = -1.

(a)

10 marks

Att 3

" I y = 2x − a

II y = 2x − a

y = 2(4) − (− 1)

y 2 = 2(4 ) − (− 1)

y = 8 +1

y2 =9 y=± 3

2

y= 9 y =3 *

Accept y = ± 3


Correct answer but no work shown. (") Mathematical error. Mishandles square root. Incorrect operation. Substitutes into the expression and stops. Incorrect substitution ( x = − 1; a = 4 or y = 2a − x ). May incur B3

Slips (-1) S1 Numerical errors to a max of 3. Misreadings (-1) M1 Arbitrary substitution with value of a negative. M2 Misreads coefficient of x . Attempts (3 marks) A1 Partial substitution and stops. A2 Squares both sides and stops. Worthless (0) W1 Incorrect answer and no work shown.

Page 24

Part (b)

20 (10, 10) marks

(i)

"

(ii)

"

Graph on the number line the solution set of -3 < 4x + 7 ≤ 23, x ∈ R. Solve the following simultaneous equations: 1 x = - y + 36 2 y = 2x + 12.

(b) (i) 4(b) (i)

10 marks

"

Att 3

Graph on the number line the solution set of -3 < 4x + 7 ≤ 23, x ∈ R.

(b) (i)

"

Att (3, 3)

10 marks

Att 3 II

I − 3 < 4 x + 7 ≤ 23 − 3 < 4 x + 7 and 4 x + 7 ≤ 23 and 4 x ≤ 23 − 7 − 10 < 4 x 10 16 and x≤ −

question 1

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