cxcDirect-May2013

© cxcDirect Institute

CSEC MATHEMATICS Past Paper Solution – May 2013 cxcDirect Institute

© All rights reserved. No part of this document may be reproduced without the written consent of the Author

cxcDirect Institute Email: [email protected] Website: www.cxcDirect.org All Help videos are free, and can be viewed from our website



** Please see the original past paper for the questions. Only the answers will be provided as per copyright obligations.

Q1. b

**********************************************************

To assess which size carton of orange is the better buy, we must determine the cost per ml .

Q1a. (i)May 2013

So for carton (350ml):

Since the exact value for the expression is requested, do not convert to decimal.

4 5

=

9 5

=

27 – 5 15

=

22 15

–

1 3

–

1 3

= 1.14 cents per ml

Q1. c Principal (P) = $9,600 Interest (r) = 8% after first year Time (T) = 1 year Interest after first year = 8% of $9600

2 5 12 5 2

=

Step 3: Invert denominator and multiply by numerator 5 22 × 12 15

=

= 1.2 cents per ml

The better buy is therefore carton size (450 ml) as it costs less than carton size(350ml)

(LCM = 15)

Step 2: Simplify the denominator viz:

=

$4.2 350

and for carton (450ml) : $5.13 cost per ml = 450

Step1: Simplify the numerator viz: 1

cost per ml =

=

1 22 × 12 3

=

9600×

8 100

= $768

Total amount owing after first year = Principal + Interest = (9600 + 768) = $10,368 Amount repaid at end of first year = $4, 368 so amount still owing at beginning of second year is: = 10,368 – 4,368 = $6,000

1 11 × 6 3

11 18

Interest due after second year is 8% of $6000 8 6000× = $480 100

1.a (ii)

√( 1.5625)

= 1.25

and

2

(0.32)

= 0.1024

so: 1.25 + 0.1024 = 1.3524 ( do not round your answer)


-

Email: [email protected]

website: www.cxcDirect.org

2


2 a) (i)

2b (i)

The first thing to note is that we can factor 2x from both terms in (i) to get: 2x ( x2 – 4)

To make C the subject:

The term in bracket is now the familiar difference of two squares , where: 2

x – 4=( x+2 )( x−2)

Step

Action

1

Move (32) to the LHS

2

so the the completely factorized espression is:

Result

Mul both sides by

3

F-32 =

5 9

Swap RHS with LHS

2x ( x+2)( x−2 )

9 C 5

5 ( F −32 ) 9

C=

=C

5 ( F −32 ) 9

2 a) (ii) From the question paper, we can see that the coefficients of the quadratic expression are : a = 3,

C=

b = -5 and c = -2

we must now identify two numbers whose product = ac and whose sum = b now : and

2b (ii) If F = 113 then

ac = 3×−2 = b = −5

5 (113−32) = 45 9

2c (i) Total tickest sold = 500 so if x tickets were sold for $6 each then

−6

a) (500 – x) tickets were sold for $10 each

The challenge is now to find two numbers whose product is (- 6) and whose sum is (-5 )

b) total money collected =

= 6x + 5000 – 10x

Condition 1 The pair of numbers must multiply to give negative 6, so one number must be negative. Condition 2 The pair of numbers must add (sum) to give negative 5, so the bigger number must be negative.

6x+10(500− x)

= 5000 - 4x 2c(ii) If amount collected is known to be $4108 then:

Pairs of numbers are: (- 6) and (+1) product = - 6 and sum = -5 (-3) and (+2) product = - 6 and sum = -1

4108 = 5000 – 4x so 4x = 5000 – 4108

clearly the first pair of numbers that match our criteria is the first pair of ( - 6 ) and ( +1)

giving:

x =

5000 – 4108 4

= 223 tickets were sold for $6 each.

Now we must rewrite the original quadratic equation, and replace the coefficient of the middle term ( which is -5) with the pair of numbers that we just found. ( - 6, and +1) Note that:

-6+1 =-5

Rewriting we get : = =

2

3x −6x+ x – 2

3x ( x− 2)+( x−2)

= (x – 2) (3x +1) © cxcDirect Institute

-

grouping factor ( x - 2)



3


3 a. (i) Sets

3b (ii)

U=30

C M

C 4x

x

E 20-x

6

2

A

8 - 3.2 = 4.8

From the information given on the past paper, the ven diagram is produced showingthe number of each student in each group in terms of x.

B 8

Triangle ABC and ADE are similar because their corresponding angles are equal.:(see diag. on past paper)

The equation for the total students in the survey is:

1. 2. 3.

3 a. (ii) 30 = 4x + x + 20-x +2 30 – 20 – 2 = 4x + x - x 8 = 4x x=2

⇒ ⇒ ⇒

D

∢ AED=∢ ACB ∢ ADE=∢ ABC

Both Triangles share a common angle in ∢CAB

3 b (iii) Finding Length DE

3 a. (iii)

Since the triangles are similar then:

Number of students using only cameras = 4x = 8

6 8

3b The diagram on the past paper should be used here in conjunction with the following:

=

so DE =

DE 4.8 4.8×6 8

= 3.6

(i) Finding length BC Q4a.

C

Length DE = ∢ ECD

10

6

=

Perimeter = DE + EC + CD = Area (CDE) =

A 8

1/ 2 (CD )× EG

=

B

From the diagranm of the past paper, we can find BC for this right angled trialgle as follows: Let length BC = x then:

82 +x 2=10 2 x=

√( 102 −82 )

giving: x =

BC = 6

so


-

( pythagoras)



4


Let the equation of the perpendicular bisector be: y = m 2 x +c

(m AB )

4b i) Finding Gradient of line AB Given A(-1, 4) and B(3,2) then

m1 =

Let the gradient of Line AB =

where and so

:

and

so

m1

x1, y1 x 2, y 2

x 2= 3

y 2 =2

and

2–4 3−(−1)

=

Point A …

xm

=

x 1 +x 2 2

=

y1+ y 2 2

ym

so coordimates of mid point (

−1 m1

=

− 1 −0.5

=2

−2 4

=

−1 2

=

3−1 2

=

4+2 2

=

−0.5

( xm , ym )

=

1

Finding the intercept c recall also that the perpendicular bisector passes through the mid point of line AB, so one point on the perpendicular bisector is the mid point coordinates ( 1, 3) of line AB So by substituting the coordinates (1, 3) , and the gradient m 2 = 2 into the equation for the perpendicular bisector we get can find the intercept c.

So:

⇒ and

m2 =

m 1 ×m 2 =−1

.. Point B

4b ii) Finding Mid point coordinates of AB now

is the gradient , and c is the imtercept

gradient such that

⇒ y1= 4

m2

Finding the gradient Now recall that the perpendicular bisector must have a

are the coordinates of point A are the coordinates of point B

x 1 =−1 and

=

where

y2 – y 1 x2 −x 1

xm , y m )

= =

Giving :

3

y = m 2 x +c

..

3 = 2 (1) + c c=3–2=1

so the equation of the perpendicular bisector is

(1, 3 )

y = 2x +1 4b ii) Finding the equation of the perpendicular bisector of Line AB Lin e2 is perpendicular bisector

A(-1,4) M id p oint o f line AB

B(3,2)


-



5


5. a(i) R2

Now if A is proportional to Then we may write this as:

⇒

A∝ R2 A=k R 2

....

36 =

hence k =

Action

Result

1

Replace f(x) with y

2

Interchange x and y

3

Make y the subject

where k is a constant

now from the table when A = 36, R = 3 so

Step

4×(5 2 ) = 100

A=

and when A = 196, R =

x=

2y+1 3

3x = 2y +1

3x – 1 =y 2

=4

also , when R = 5 ,

2x+1 3

3x – 1 = 2y

k (32 ) 36 2 (3 )

y=

√

A k

=

√

196 4

y=

so: 4

= 7

Replace y with

f −1 ( x )

f

−1

3x – 1 2

( x) =

3x – 1 2

we may now complete the table as shown: A

36

100

196

R

3

5

7

So

f −1 (3)

=

3(3)– 1 = 4 2

5. b(i) Finding fg(2) Now given f(x) and g(x) as shown on the past paper: to find fg(2) we must first find g(2) by substituting 2 into the function g(x). ie: g(2) = 4(2) + 5 = 13 now since g(2) = 13, then fg(2) is the same as f(13) so by substituting 13 in the function f(x) we get: fg(2) is the same as f(13) =

f −1 (3)

5. b(ii) Finding

Now to find

2 (13)+1 = 9 3

f −1 (3) , we must first find

f −1 ( x )

and then replace x with the number 3


-



6


6a i)

T M '' 0 = 6 −3 1

N 5 + 6

T N'' 0 = 5 −3 3

() ( ) ()

Now to convert the speed of a car from 54 km/h to m/s we can do the following:

() ( ) ()

now so

1km = 1000m 54km/h = 54,000m/h

also so

1 h = 3600 sec 54,000m/h means that the car will travel 54,000 54,000 = 15 meters in meters in 3600 sec or = 3600 1 sec

⇒

M 6 + 4

The fig below shows the image after the translation

54km/hr = 15m/s

Now Distance = Speed x Time so if the speed if 15 m/s, then after 20 seconds : Distance =

15×20

= 300m

6.b) mirror line

6biii)

Y d

Y

d

C

C'

object

image

A

B

0

B'

5

N '( 9, 6)

A'

P'

4

X

M' (8, 4)

The diagram above shows an object triangle ABC and its image A'B'C'. Note that each image point is a reflection of the object point in a mirror line at x = 5. Now compare the above with the diagram shown on the past paper, where it can be clearly seen that the image L'M'N' is a reflection of the triangle LMN in the line x = 7 6b ii)

( )

0 has the effect of moving the object 3 −3 units down on the y-axis. This is described mathematically as : Object point + Translation (T) = Image point

A translation

so:

T=

L 1 + 4

T L'' 0 = 1 −3 1

() ( ) ()


-


L' (13, 4)

N'' (5, 3)

1

P M'' (6, 1)

L" (1, 1)

X

Triangle P may be mapped unto P' by the following combined transformations: 1) A translation T = 0 followed by a reflection in 3 the line x = 7

()

2) A reflection in the line x = 7 followed by a translation 0 T= 3

()


7


Q 7a) ( i) Statistics

Q 7c.i) 1. The median Q2 correspond to the 1/2 n th ranked student, where n = 40 So the median amount spent corresponds to the amount spent by the 20th ranked student on the curve. ⇒ median = $31 Q 7c.ii)

The amount of students that spend less than $23 is read from the curve as 9 9 so the probability that a student spent less than $23 = 40

© cxcDirect Institute - 876 469-2775 Email: [email protected] website:

www.cxcDirect.org

8


Q8.a)

Finding a rule relating B and N

The fourth diagram is the sequence is shown below

The formula or rule that relates the number N with the number of balls B may be found using general equation for a straight line , since the relationship is linear. Ie

B1 =m 2 N 1 +c B2 =m 2 N 2+c

The completed table is shown below

(i) (ii)

N 1 2 3 4 20

# Wires 12 20 28 36 164

# Balls 8 12 16 20 84

hence

m2

=

B2 – B1 = N 2 −N 1

12 – 8 2−1

=4

substituting this gradient into the first equation 8 = 4 (1) +c so c=4 Therfore the equation for the number of balls B is:

Finding a rule relating W and N The formula or rule that relates the number N with the number of wires W may be found using general equation for a straight line , since the relationship is linear. Ie

B=m 2 N +c ( m2 = gradient, c = intercept) so using diagrams 1 and 2

W = m N + c ( m = gradient, c = intercept) so using diagrams 1 and 2

B = 4N +4

so

if N = 4,

then

B = (4 x 4) + 4 = 20

and

if N = 20,

then

B = (4x 20) + 4 = 84

W 1 =m N 1 +c W 2=m N 2 +c hence

m=

w 2 – w1 = N 2 −N 1

20 – 12 2−1

=8

substituting this gradient into the first equation 12 = 8 (1) +c so c=4 Therfore the equation for the number of wires W is: W = 8N +4 so

if N = 4,

then

W = (8 x 4) + 4 = 36

and

if N = 20,

then

W = (8 x 20) + 4 = 164


-



9


The new form is therefore: SECTION II

y

=

2

3( x− 2) – 4

Q 9. a) Linear programming Question: The graph is shown below:

Establish the variables: Person Plans to buy x oranges and y mangoes

y

Conditions: i) If Bag can only hold 6 fruits then x + y <= 6 ii)If we must have at least 2 mangoes, then y >= 2 iii) The inequality y <= 2x implies that the number of mangoes must Not be more than two times the number of

8

oranges. v) The Graph of the solution set is shown below

0

2

x

-4

(2, - 4) The y – intercept is shown = 8 the coordinates of the minimum point is (2, - 4)

b) Given a quadratic function in the form ax 2 +bx+c we can rewrite this function in the form a ( x+h )2+ k where

h=

b 2a

;

and

k=

4ac−b 4a

2

The function given on the past paper is in the form : 2 f ( x)=ax +bx+c , with a = 3, b = -12 and c = 8

so : h =

−12 2 (3)

and k =

4(3)(8)−(−12) 4(3)

=

−2

=

−4

2


-



10


Q 10 a) Circle Theorem

(iii) Finding ∢ AFB F

The following observations are made from the diagram on the past Paper: #

Observation

Applicable theorem(s)

1

one Tangents

angle between tangent and chord = angle in the alternate segment

50 O 100

B

Angle between radius and Tangent = 90 0 2

Triangle BOA is isosceles

Base angles are equal

3

Angle O at the center

Angle at center is twice that at the circumference

A

From observation #2, 100 ∢AFB= = 50 0 because angle at the centre is 2 twice the angle at the circumference.

(i) Finding ∢ EBF F

(iv) Finding ∢OAF F 50

O E

55

R

35

O B

35 40

From Observation 2, radius and tangent = Therefore:

0

∢ EBO=90 0 90 0

because angles between

B

where and and so

∢B=40+35=750 ∢F =50 0 ∢A=40+∢OAF

75+50+40+∢OAF

hence

O 40

A

∢B+∢F +∢A=1800

(ii) Finding ∢ BOA

100

40

Considering triangle BFA:

= 900 = 90 -35 = 550

∢ EBF +35 ∢ EBF

25

=

1800

∢OAF =250

A

40 B

From observation 2: Base angles are equal in an isoscelecs triangle so ∢OBA=∢OAB = 40 0 Therefore: ∢BOA=1000


-

.. angles in a triangle = 1800



11


Q10.b i) Triangle RFT

Finding angle of elevation T from S considering triangle TFS

T

T

25m

25m 270

F

x0

S

R

43.3m

F

Triangle TFS Let x represent the angle of elevation T from S:

T

so:

25m

Tan x =

giving: x =

S F

43.3m

25 43.3

= 0.57736

tan−1 (0.57736 ) =

30 0

Triangle SFR

R

F 43.3m

S

Finding RF 0

tan 27 =

considering Triangle RFT: so

RF =

25 0 tan 27

=

25 0.5095

25 RF

= 49.1m

Finding SR considering triangle RFS 49.1m F

R

43.3m

S

(SR)2 =(SF )2 +( RF )2 so SR

=

√ 43.32 +49.12


-

= 65.5m Email: [email protected]


12


Q11 (a) Vectors The following diagrams are to be used in conjunction with the diagram on the past paper.

Q11a.ii)

Finding Vector A ⃗ B

1) OB is parallel to PQ because

Two geometrical relationships are;

⃗ O⃗ B=2 P Q

2) Triangle OBA is similar to Triangle PQA because the

B

corresponding angles are equal. ⃗

A

O

B

2b

⃗

B

Q11. b 2a

If

A

O

M=

⃗

O

A

then

Now A ⃗ B is the vector going from A to B that is: First go from A to O and then from O to B so:

A⃗ B

where:

O ⃗A=2a , and

Now so giving: or

⃗ AO ⃗ AO ⃗ AB ⃗ AB

=

⃗ ⃗ A O+O B

M

−1

=

( ) 2 1 4 3

Adjoint M Determinant of M

where Adjoint M =

O⃗ B=2b

is the negative of vector = −O ⃗A = −2a

(

3 −1 −4 2

)

and Determinant of M = 3 x 2 – 4 x 1 = 2 ⃗ OA

M −1

so

=

= −2a+2b = 2 (b−a ) now

M ×M −1 =

Finding Vector P ⃗ Q

(

1 3 −1 2 −4 2

)

( )(

2 1 3 −1 4 3 −4 2 2

)

( ) ( )

1 2 0 2 0 2

B

1 0 0 1

=

Q 11 b.iii) now if [M][P] = [Q] then

Q

M −1

[P] =

(b - a)

[Q]

A

O

a

P

so Now P is midpoint OA so

P ⃗A =

1/ 2 (O ⃗A)

( ) r t

s u

so:

)( )

1 3 −1 2 1 × 2 −4 2 4 −1

=

( )

= a

and Q is the midpoint of AB ⃗ = 1/ 2( A ⃗ so A Q B) = (b - a)

(

=

1 2 0 −3

r = 1, s = 2, t = 0, u = -3

⃗ is the vector going from P to Q Now P Q that is: First go from P to A and then from A to Q ⃗ ⃗ so: = P ⃗A+A Q PQ

----END---

= a + ( b – a) =b © cxcDirect Institute

-



13

cxcDirect-May2013

Recommend Documents