FDM June 2015 Examination Marking Scheme - Final

8/17/2019 FDM June 2015 Examination Marking Scheme - Final

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Foundation Mathematics

23 June 2015

Marking Scheme

This marking scheme has been prepared as a guide only to markers. This is not a set ofmodel answers, or the exclusive answers to the questions, and there will frequently bealternative responses which will provide a valid answer. Markers are advised that, unless aquestion specifies that an answer be provided in a particular form, then an answer that iscorrect (factually or in practical terms) must be given the available marks.

If there is doubt as to the correctness of an answer, the relevant NCC Education materialsshould be the first authority.

Throughout the marking, please credit any valid alternative point.

Where markers award half marks in any part of a question, they should ensure thatthe total mark recorded for the question is rounded up to a whole mark.


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Page 2 of 16 Foundation Mathematics © NCC Education Limited 2015

Answer ALL questions

MarksQuestion 1

a) Simplify:

i) × 3 ÷ 1 ii) × 1

1

Acc ept either − or iii) 4 1

b) Simplify:

i) 7 + 1 + 2 3 1= + ii) 12

4 1

= iii) 23 × 2 1

= =

c) Factorise:

i) 10 + 5 2+ 1 mark for correct work ing, 1 mark for co rrect answ er

ii) + 1 0 + 2 1 2 + + 1 mark for correct work ing, 1 mark for co rrect answ er


3/16

Marks


d) Simplify:

i) 13 ÷

24

2

=

×

= = 1 mark for correct work ing, 1 mark for co rrect answ er

ii) 25 +

13

2

=

+

= 1 mark for correct work ing, 1 mark for co rrect answ er

e) Transpose the following formula to make the subject.

2 = 3 8

2

=

+ 1 mark for correct work ing, 1 mark for co rrect answer

f) Solve the following equation and find the value of . 75 + 3 = 2

2

=

1 mark for correct work ing, 1 mark for co rrect answer

g) Solve the following quadratic equation by factorising.

1 4 + 3 3 = 0 2

= So, = or = 1 mark for correct work ing, 1 mark for c orrect answer

Total 20 Marks


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Marks


Question 2

a) Solve the following quadratic equation by using the Quadratic Formula:

9 + 2 3 = 0 (You may leave your answer in surd form.)

2

= ± √ Part a): 1 mark for correct working , 1 mark for correct answer

b) Solve graphically the simultaneous equations:

= 3 2 and = 1 0 for 0 ≤ ≤ 5 Use the graph paper provided.

7

Make a table of values for each equ ation

0 1 2 3 4 5

= -2 1 4 7 10 130 1 2 3 4 5

= 10 - 10 9 8 7 6 5

Draw accurate graphs fo r both equ ations on on e set of axes.

The point of intersect ion is = 3, = 7

So the solu tion is = 3, = 7

Part b):2 marks for each co rrect table, 1 mark for each correct ly plo tted graph, 1

mark fo r corr ect value of x and y. If only o ne of x o r y is correct award ½ mark.

-4

-2

0

2

4

6

8

10

12

14

0 1 2 3 4 5 6

y

x

y = 3x 2

y=10 x


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Marks


c) Calculate the gradient of the following curves at the point where = 1 by usingdifferentiation.

i) = 2 + 3

Dif ferent iat ing gives:

=

+

When = =

+ = + =

So, when = the gradient is – ii) = 2 + 1 3

Dif ferent iat ing gives:

= 2

1

When = = = = So, when = the gradient is Parts c i) and i i) : 2 marks for correct dif ferent iat ion o f curv e, 1 mark for

correct g radient .


6/16

Marks


d) A particle moves metres in seconds so that = 2 + 9 1 i) Find the velocity, , after 2 seconds 3

= = + When = = + = + = So, after 2 secon ds the veloc ity is 19 m/s

Part d i): 2 marks for cor rect work ing, 1 mark for c orrect answer.

ii) Find the acceleration, , after seconds. 2 = = + Part d) i i ): 1 mark for c orrect wo rkings , 1 mark for correct answer.

I f answ er incorrect only because of error in part i) award ful l m arks.

Total 20 Marks

Question 3

a) i) Using differentiation, find the coordinates of the turning point on the curve:

= 3 4 4

Dif ferent iat ing gives: = Turning p oints are located where

= =

= when = When = , = So, the turnin g po int is at (-2, 7)

Aw ard 2 marks for corr ect ly dif ferent iating the curve. Award 1 mark for

wo rking and 1 mark for correct coo rdinates.


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Marks


ii) Construct a table of values for the curve = 3 4 for -5 ≤ ≤ 1 Use these values to plot a graph of the curve = 3 4 and identify theturning point found in (i) as either a maximum or minimum turning point.

5

Table of values

-5 -4 -3 -2 -1 0 1 = -2 3 6 7 6 3 -2

Graph of the curve = y

Hence, the turning point at (-2, 7) is a maximum turning point .

Aw ard 2 marks for the table of values. Award 2 marks for co rrect ly

plot t ing cu rve and award 1 mark for correct classi f icat ion of turning

point .

b) Integrate the following expression:

2 + 2√ 2

∫ ( + √ ) = ∫ ( + /) = + ×

/ / +

= + ∕ + where is a cons tant .Award 1 mark for correct working . Award 1 mark for co rrect answ er.

-3

-2

-1

0

1

2

3

4

5

6

7

8

-6 -5 -4 -3 -2 -1 0 1 2


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Marks


c) The gradient of the curve which passes through the point (1, 2) is given by3 + 2. Find the equation of the curve. 3The gradient is given b y + , hence = + Integrate to ob tain = ∫ +

= + + = + + The curve passes thro ugh the point (1, 2), so wh en = , = . Subs titutethese values b ack into th e integrated equat ion to get: = + + = + + = + So, = Therefore the equat ion o f the cur ve is = + Award 2 marks for correct wor king. Award 1 mark for correct answer.

d) Evaluate the definite integral:

7 2 + 4 ⅆ

3

+

= +

= + = + + =

Award 2 marks for correct work ing. Award 1 mark for correct answer.

e) The part of the curve = 3 between the ordinates = 1 and = 2 is rotatedabout the -axis. Calculate the volume of the solid generated. Leave your answeras a multiple of .

3

=

=

= =

= = = Award 2 marks for correct work ing. Award 1 mark for correct answer.

Total 20 Marks


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Marks


Question 4

a) The acceleration of a moving body at the end of seconds from thecommencement of motion is 82 m/s2.i) Find the velocity at the end of 4 seconds if the initial velocity is 8 m/s. 3

The equat ion for accelerat ion is = We can integrate this to ob tain an expression for the veloci ty = ∫ + Subst i tut ing in = we obtain = ∫ + = + To calculate the veloci ty we need to determine the con stant of

integratio n. The init ial velocity is the veloc ity at

= .So when

= ,

= . Subst i tute these values into the expression for to give: = + So, = The equat ion fo r veloci ty is therefore: = + After 4 s econd s, i.e. at = , the veloci ty is: = + = +

= / 2 marks for correct wo rking, 1 mark for cor rect answ erii) Find the distance travelled by the body at the end of 3 seconds. 3

Distance is given by th e equation = ∫ Integrate the equation fo r using the l imits of 0 and 3. Ass ume = when = therefore = . = +

= [ +]

= { +} = + = 2 marks for correct wo rking, 1 mark for cor rect answ er


10/16

Marks


b) The 8.15am train to Newtown is either on time or late. It is never early.The probability that 8.15am train to Newtown will be late on Monday is 0.2. Theprobability that the 8.15am train will be late on Tuesday is 0.1.

i) Draw a probability tree diagram to show all the possible outcomes. 8Draw a tree diagram, wri t ing L for late and T for o n t im e.

Monday Tuesday Outcome Probability

L L, L) 0.2 × 0.1 = 0.02

0.1

L 0.9

0.2 T L, T) 0.2 × 0.9 = 0.18

L T, L) 0.8 × 0.1 = 0.08

0.8 0.1

T

0.9

T T, T) 0.8 × 0.9 = 0.72

1 mark for co rrect workin gs for each branch, 1 mark for each co rrect

outcome probabi l i ty.

ii) Use your tree diagram to find out the probability that the train is late onONE (1) of the two days. 2

There are two po ssib le outcomes w here the tra in is late on o ne day.

P(train l ate on one day) = P(L, T) + P(T, L)

= 0.18 + 0.08

= 0.26

1 mark for correct work ing, 1 mark for co rrect answ er

iii) Use your tree diagram to find the possibility that the train is on time at leastonce.

2

P(train on tim e at least on ce) = 1 – P(L,L)

= 1 – 0.02

= 0.98

1 mark for correct work ing, 1 mark for co rrect answ er


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Marks


c) How many ways can we arrange three letters from the word ‘pencil’ if repetitionsare not allowed and different orders of the same letter count as the samearrangement?

2

= ! ! ! =!

!! = × × × × =

1 mark for clear calculat ions, 1 mark for correct answ er. Candidates m ay

use different notation.

Total 20 Marks


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Marks


Question 5

a) A five-sided spinner is thrown 30 times with these results:

1 2 1 4 2

4 4 5 2 3

1 3 1 5 4

4 1 2 2 5

3 4 3 5 2

2 5 3 4 3

i) Summarise this data as a frequency distribution table. 2Scores obtained wh en throwin g a f ive sided spinn er 30 t imes.

Score Frequency

1 5

2 73 6

4 7

5 5

1 mark for t i t le and column headings, 1 mark for c orrect frequencies

ii) The mean of the data is 3. Calculate the mean deviation. You may leaveyour answer as a fraction.

3

Since ̅ =

| ̅ |

| ̅ |

1 5 2 10

2 7 1 7

3 6 0 0

4 7 1 7

5 5 2 10 = | ̅ |= 34 = | ̅ |∑

= = =

2 marks for c orrect work ing, 1 mark for correct answer as either a

mix ed number or imp roper fract ion or decimal, simp l i f ied or not.


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Marks


b) The heights of 60 plants are given in the table below.

Height (cm) 0 < 4 4 < 8 8 < 12 12 < 16 16 < 20

Frequency 8 10 12 16 14

i) Construct a histogram to illustrate this data. 4

2 marks for histog ram with h eight on horizontal axis and frequency on

vert ical axis. 2 marks for accurate plot t ing of bars.

ii) Within which class interval will the modal value lie? 1Within the 12 < 16 class.

iii) Calculate the mode. You may leave your answer as a fraction. 2The formula for the mode we obtain:

=+ ×

= + ×

= + =

So the mode is 1 mark for correct wor kings , 1 mark for correct answer.


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Marks


iv) Calculate the mean. You may leave your answer as a fraction. 4

Weight

(kg)

Midpoint Frequency 0 < 4 2 8 16

4 < 8 6 10 60

8 < 12 10 12 120

12 < 16 14 16 224

16 < 20 18 14 252 = = ̅ = ∑

=

= = 3 marks for correct wo rking, 1 mark for cor rect answ er.

Be aware that candidates may not sim pl i fy the fract ion completely but

fu l l credit should be given.

c) The following data set is recorded.

105 117 113 123 118 120 101

i) Find the lower quartile and the upper quartile. 3First put the data into ascending or der:

101 105 113 117 118 120 123

The lower q uart i le is: + =

=

So, the lower q uart i le is 105

The upper qu art i le is:

+ = = So, the upper quart i le is 120

1 mark for pu tt ing data into ascending order. 1 mark for co rrect lower

quart i le value, 1 mark for c orrect up per qu art i le value.


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Marks


ii) Calculate the quartile range. 1The quart i le range is g iven by:

upper q uart i le – lower quart i le = 120 – 105 = 15

Total 20 Marks

End of paper


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Learning Outcomes matrix

Question Learning Outcomesassessed

Marker can differentiatebetween varying levels ofachievement

1 1, 2 Yes

2 2, 3, 4 Yes

3 4, 5 Yes

4 5, 7 Yes

5 2, 3, 6 Yes

FDM June 2015 Examination Marking Scheme - Final

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