cxcDirect May2008

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CSEC MATHEMATICSPast Paper Solution – May 2008

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

Q1a. s3.90.27=4.17

0.6724=0.82

so: 4.170.82=4.99

b)simplifying the numerator :

212−

45

= 52−

45

= 1710

. Divide Numerator by denominator:

⇒1710

÷34

⇒1710

×43

= 3415

= 2415

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

1.b $1Can = J$72.50

⇒ $Can 250.00 = J $250 x 72.5 = J$18,125.00

Credit limit = $J 30,000 = $Can (30,000/72.5) = $Can 413.79

Credit card balance = $413.79 – $250.00 = $Can 163.79

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

Question 2

a = 2, b = -1, c = 3

a(b+c) = 2(-1 +3) = 4

4b2 – 2acabc

= 4 −1

2 – 2×2×32−13

=4 – 12

4= − 2

b)

i) 4 x5

ii) ab16

c)

i) 15 – 4x = 6x +2

⇒ 15−2=10x 4x

⇒ 13=10 x

⇒ x=1.3

ii) 6a2 b312a 4 b=6a2 b b2

2a 2

iii) 2m29m – 5 a = 2, b = 9, c = -5

let pq = ac ; p+q = b

⇒ pq = -10; p+q = 9

⇒ p = 10, q = -1

⇒ 2m29m – 5 = 2m2

10m – m−5

= 2m m5−m5

= 2m – 1 m5

© cxcDirect Institute cxcDirect.org; Math club : 2


Q3

Career # Angle

Doctor 105 35

Teacher 189 63

Artist 216 72

Lawyer 240 80

Sales 330 110

Total 1080 360

't = 1080 - ( 240 + 189 + 216 + 330) = 105

Converting to a pie chart angle:

⇒ Doctor = 3600×

1051080

= 350

⇒ Teacher = 3600×

1891080

= 630

and so on...

Pie Chart

Q4.

M = prime numbers = 17, 19, 23N = Even numbers = 16, 18, 20, 22, 24

M ∪N ' = 15, 21, 25

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

b.

Construction details.

1. Draw straight line AB = 7cm2. Construct 600 ∢ BAD using compass3. Measure AD = 7cm4. Set compass to a separation of 7cm and with centre

D, construct an arc above B. 5. Set compass to a separation of 7 cm and with centre

B, construct a second arc to intersect the first arc. The intersection of the two arcs is the point C.

6.Measure AC = 12.1cm


Doctor

( 350)

Teacher

( 630)

Artist

( 720)Lawyer

( 800)

Sales

( 1100)

N M

U

16, 18, 20, 22, 24

17, 19, 23

15,21,25

A B600

D C

7 cm

7 cm 12.1 cm


Q5.

The diagrams below highlight the main areas of Q5 (see past paper for original).

1.) RS = 8 – 2 = 6m

2.) x = 10 – RS = 4m

Perimeter equal distance around the two objects.Since A and B are joined at RS, then RS is not taken into account so:

Perimeter A = 2 ( 5 + 10 ) - RS = 30 – 6 = 24m

Perimeter B = 2 ( 8 + 3 ) - RS = 22 – 6 = 16m

Perimeter (A+B) = 24 + 16 = 40m

Area ( A + B ) = 10 x 5 + 8 x 3 = 74m2

Area of floor board = 1m x 20cm

= 1m x 0.2m ( 20cm = 0.2m)

= 0.2m2

# floor boards needed to cover A

= Area A

area of floor board

= 10×5

0.2= 250


A

10m

5m

B

8m

3m

2mRS = 6m

R S

RS = 6m

R Sx=4m


Q6 The diagrams below highlight the main areas of Q6

If GH = 12m

Then tan 320=

GHHJ

⇒ HJ = GH

tan 320 = 12

0.6249= 19.2m

similarly tan 270=

GHHK

⇒

so HK = GH

tan 270 = 12

0.5095= 23.55m

Now JK = HK – HJ = 23.55 – 19.2 = 4.35m

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

Question 6b – Transformation From the diagram given on the Past paper, we observe the following:

1. Size and Orientation of Object and Image are maintained

2. P 4,2 Q 6,−5 3. P 7,2Q 9,−5

where : Q

6−5 -

P

42 = 2

−7

and : Q

9−5 -

P

72 = 2

−7

⇒ The object is transformed by a translation T where :

T = 2−7

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

6 b contd.


G

270320

J KH

1

y = x

2

1

y

x0 3 4

2

3

4

5 6 7

5

6

7

( 7 , 2 )P

S

( 6 , 4 )

( 4 ,2 )

( 4 ,2 )

( 4 ,6 )

( 2 ,4 )

( 2 ,7 )


Question 7

General equation = y = mx + c

c = y intercept = 7

m = slope = y2 – y2

x2−x1

= 0 – 72 – 0

= −3.5

1. so equation of line = y= −3.5 x7

2.

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

For a Point ( -2, k ) on this line ⇒ k = y, and x = - 2

substituting these values in the equation :

⇒ y= −3.5 x7

⇒ k =−3.5 −2 7

⇒ k =14

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

If a second line y = x – 2, intersects the first line, the point of intersection is found by making the two equations equal:

⇒ x – 2 = −3.5 x7

by solving this equation we get

⇒ x= 2, and y = 0

so coordinates of intersection = ( 2, 0)


0

A(0, 7)

B(2, 0)

mid(1, 3.5)

(-2, k) … w here k = 14

-2

2

y = x -2

-2


Question 8

$cost -stamps Quantity $Total

1 6 6

1.2 6 7.2

2.5 6 15

4 6 24

Total 24 52.2

Total cost of stamps = $52.20

Parcel postage cost = 25.70

1. Using as many $4 stamps as possible

= 5 x $4 + 1 x $2.5 + 1 x $1.2 + 2 x $1 = $25.70

2. Using all the $1 stamps

= 6 x $1 + 4 x $4 + 1 x $2.5 + 1x $1.2 = $25.70

3. Largest # stamps ⇒ use all the smallest value stamps

= 6 x $1 + 6 x $1.2 + 5 x $2.5 = $25.70

largetst # = (6+6+5) = 17 stamps, with values as shown

above

Section II

x2×x3

÷ x4 = x23−4 ) = x

a 3/2 b5/2×ab3 = a 3/2 b5/2

×a1 /2 b3 /2

= a3 /21 /2 b5/ 23 /2 = a 2 b4

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

if f x=2x – 3

then f −1 x = x3

2

so f(2) = 2(2) – 3 = 1

and f −10 =

03

2 =

32

also f −1 f x = 2x−3 3

2= x

so f −1 f 2 = 2

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

Q 9.c

1.Temperature after 15 min = 500 C

2.Rate of cooling at t = 30 min = gradient of the tangent a t

t = 30min = 1420

= 0.70 C /min


10 20

10

y

min0 30 40

20

30

40

50 60 70

50

60

70

80

15

20

14

slope = 14/20 = 0.7 degree / min


Q10.a

If y4x=27

⇒ y=27−4x ... 1

also if xy x=40 ... 2

then Substituting 2 into 1

⇒ x 27−4x x=40

⇒ 4x2−28x40=0

⇒ x2−7x10=0

⇒ x−2 x−5 =0

so x=2 ; or x=5

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

10.b

From the diagram ( see Past paper)

1. 10 boys and 5 girls falls outside the feasible region and therefore cannot be members of the club at the same time.

2. 6 boys and 6 girls falls inside the feasible region and therefore can be members at the same time.

The three equations that define the region are:1. y≥2

2. y≤−45

x12

3. y≤2x

If Profit equation = 3x + 5y, then max profit occurs at one of the three vertices.

Vertex A ⇒ 3 x 1 + 5 x 2 = $13

Vertex B ⇒ 3 x 5 + 5 x 8 = $55 ( max Profit)

Vertex C ⇒ 3 x 12 + 5 x 2 = $46

Q11 – Geometry & Trig

The figs below are used to show the theorems used for this problem. ( see past paper for original diagram)

Now: Lines OS and PR are parallel ⇒ alt. Angle as are equal:

⇒ ∢RPS =∢OSP=260

2. Triangle OPS is isosceles ⇒ Base angles are equal

⇒ ∢OSP=∢OPS =260

⇒ ∢ POS=180 – 2 26 =1280

Angle at centre = twice angle at circumference

⇒ ∢ PTS=128

2=640

Angle between Radius (OP) and Tangent (PQ) = 900

⇒ ∢RPQ=90 – 2626 =280

Q11 – B © cxcDirect Institute cxcDirect.org; Math club : 8

B(5,8) - profit = $55

A(1 ,2) - $13 C(12, 2)- $46

boys

girls

(10 ,5)

10

5

(6, 6) 6

6

128

26

P

R

26

26

O

S

128

P

64

O

ST

2890

P

R

OS

Q

2626


The diagram above is drawn to illustrate the principles used to solve this problem, ( see past paper for original drawing)

Define x as the perpendicular bisector of AB.

⇒ ∢ XOB=½ ∢ AOB

where: sin ∢ XOB=7.258.5

= 0.85294

⇒ ∢ XOB = sin−1 0.85294 = 58.50

⇒ ∢ AOB=2×58.5=1170

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

Area of triangle AOB = ½ r2 sin 1170

= ½ 8.52×0.8906 = 32.17 cm2

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

Area of shaded section

= Area of Sector AOB – Area of triangle AOB

now :area of sector AOBarea of circle

=1170

3600

so : Area of sector = r2×

117360

= ×8.52×

117360

= 73.73 cm2

⇒ Shaded area = 73.73 – 32.17=41.56 cm2

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

Length of minor arc = 2 r×117360

length of major = 2 r×360−117

360= 36cm

Q12.a

1. The completed bearing diagram is shown above.

∢RST =6833=1010

2. To find ∢RTS we use the sine rule:

⇒56km

sin ∢RTS= 75km

sin 1010

⇒ sin∢ RTS = sin 1010×

5675

= 0.7329

⇒ ∢RTS = sin−1 0.7329 = 470

3. Bearing of R From T = 1803347=2600

4. To find the distance TX, we note that Triangle XTR is a right angled triangle, and:

∢ XTR=90 – 4733 = 100

Hence: cos 100=

TX75

⇒ TX = 75Cos100 = 73.9km


58.5r = 8.5

1170

BA

O

7.25 7.25x

56 km

4733

10

33

68

68

75 cos 10 = 73.9 km

N

R

2600

S

N

X

N

75 km

112

T


Vector & matrices

a)

b)i

b ii)

Finding A⃗B

Vector A B⃗ ( shown dotted ) is the vector going from point A to point B. Imagine that you are at point A and you wish to get to Point B

⇒ First go from A to O and then from O to B

⇒ A B⃗ = A O⃗ + O B⃗

Given O A⃗ = a and O B⃗ = b

now if: O A⃗ = a then A O⃗ =− a (reverse the sign)

so A⃗B = − a + b

= b − a

now: OP : PA = 2 : 1

so ; O⃗P=23

a and P⃗A=13

a

Finding P⃗M

To go from P to M : ⇒ P to A then A to M

so : P M = P⃗A + A⃗M

but BM = MA

therefore A⃗M = 12

A⃗B = 12

(b−a )

and since: P⃗A = 13

a

then: P M = 13

a +12

(b−a )

= 13

a + 12

b −12

a

= 2a+3b−3a

6

=3b−a

6

so: P M = 16

(3b−a )


B

O⃗BO⃗A

A⃗B

O

A

b a

B

O

A

b

aP

M

B

O⃗BO⃗A

O

A

b a P

M

N

B

O⃗B

O

A

bP

M

13

a

23

a


13. c)

Collinear means: On a straight line.

If the three points P, M and N are collinear then, the two vectors P M⃗ and M N⃗ that connect these three points will also be collinear.

The proof that two vectors are collinear is that one vector must be a constant ( scalar) multiple of the other vector :

⇒ M N⃗ = k P M⃗ ( where k is a constant)

we have already found P⃗M = 16

(3b−a )

so now we need to find M⃗N

Finding M N⃗

To go from M to N : ⇒ M to B then B to N

So: M⃗N = M⃗B + B⃗N

where B⃗N = O B⃗ = b

and M⃗B = 12

A⃗B = 12

(b−a )

so M⃗N = b +12

(b−a )

= b +12

b −12

a

= 2b+b−a

2

⇒ M⃗N = 12

(3b−a )

now note that since: P⃗M = 16

(3b−a )

and M⃗N = 12

(3b−a )

then : M⃗N = 3× P⃗M

This completes the proof that :

P , M and N are on a stright line ( collinear)

nb. The proof that three point are collinear (on a straight line), is the same as the proof that the two vectors connecting the points are collinear.

Steps:• Find the two vectors connecting the three points• Show that one vector is a constant multiple of the

other vector i.e V⃗ 1=k×V⃗ 2


P⃗M

O

B

A

P

b

23

a

13

a

M

b

N

M⃗N

M⃗B A⃗M


To find the length of A⃗N :

Finding A⃗N

Now: A⃗N = A⃗O + O⃗N

where O⃗N = 2b

and O⃗A = a ⇒ A⃗O = − a

so A⃗N = −a + 2b

now given : a = 62 and b = 1

2

A⃗N = - 62 + 2 1

2= (−6

−2) + (24) = −4

2

Length = −4222 = 4.47

Q 14: Matrices

X 2Y = −2 0

5 12

4 −13 7

= −2 05 1−2 0

5 14 −13 7

= 4 0−5 14 −1

3 7 = 8 −1

−2 8

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

14 c i (a)

If H is a 2 x 2 transformation matrix , which represents an enlargement with scale factor k , then this may be written as:

H = (k 00 k)

If the point D(5, 12) → (7.5, 18)

This means that under the transformation (H ) , the point D(2,5) maps onto its image point at D' (7.5, 18). This is written in the column matrix format as :

H

(k 00 k)

D

( 512) =

D '

7.518

⇒ 5 k = 7.5

and 12 k = 18

From the first equation : k = 7.55

= 1.5

Note also that we will get the same result for k if we use the second equation;

Hence: k = 1.5

14 c i (b)

Similarly, under the transformation (H) , the image points E'

and F' , are determined as shown below:

⇒

H

(1.5 00 1.5)

E F

(2 87 4) =

E ' F '

( 3 1210.5 6 )

coordinates of E' = (3, 10.5)

coordinates of F' = (12, 6)


P⃗M

O

B

A

P

b

23

a

13

a

M

b

N

M⃗N

M⃗B A⃗M

A⃗N


14 c ii (a)

Finding a 2 x 2 transformation matrix

To Find a 2 x 2 transformation matrix that represents a clockwise rotation of 900 about the origin:

Consider the Object OPQ above with coordinates P(1, 0) and Q (0, 1)

Now if rotate this triangle through 900 in a clockwise direction about the origin. We note from the graph that

P 1,0 P ' 0,−1

and Q 0, 1Q ' 1, 0

The means that the object point P( 1, 0) maps unto its image P'(0, -1) , and the object point Q( 0, 1) maps unto its corresponding image point Q'( 1, 0)

Our task is now to find a 2 x 2 matrix , that will produce the exact result that we obtained graphically above. If we define this transformation matrix as R, then under this transformation, object points P( 1, 0) and Q( 0, 1) must map unto their corresponding image points at P'(0, -1) , and Q'( 1, 0)

So : let R = a bc d ( transformation matrix)

where a, b, c and d are unknowns

then: R

(a bc d)

P Q

(1 00 1) =

P ' Q '

0 1−1 0

We can now solve for the the unknowns a, b, c, and d

⇒ a×1+b×0 = 0 .. so a = 0

⇒ a×0+b×1 = 1 .. so b = 1

⇒ c×1+d×0 = -1 .. so c = -1

⇒ c×0+d ×1 = 0 .. so b = 0

Giving: R = ( 0 1−1 0)

This result can be obtained faster if we observe that

since P Q

(1 00 1) is the identitiy Matrix ( I )

then: R

(a bc d)

P Q

(1 00 1) =

R

(a bc d)

Hence: R

(a bc d) =

P ' Q '

0 1−1 0

Giving: R = ( 0 1−1 0)

( see matrix transformation workbook at www.cxcDirect.org for tutorial)


1

0 1x

Q( 0, 1 )

P(1, 0)

y

0 1x

Q' ( 1, 0)

y

P' ( 0 , -1)-1

http://www.cxcDirect.org/


14 c ii (b)

Under the transformation R:

The image of D', E' F' is D", E", F" :

That is:

R

( 0 1−1 0)

D ' E ' F '

( 7.5 3 1218 10.5 6 )

= D ' ' E ' ' F ' '

( (0x7.5+1x18) (0x3+1x10.5) (0x12+1x6)

(−1x7.5+0x18) (−1x3+0x10.5) (−1x12+0x6))

=D ' ' E ' ' F ' '

( 18 10.5 6−7.5 −3 −12)

so:coordinates of D'' = ( 18, -7.5 )coordinates of E'' = ( 10.5, -3 )coordinates of F'' = ( 6, -12 )

14 c ii (c)

If the symbol H is used to represent the Enlargement, and R used to represent the rotation, then two successive transformations, H first, followed R will map the object points DEF onto its image at D" E" F" .

The 2 x 2 matrix that represent the combined transformation, H first then R is:

[ R ] [ H ] = ( 0 1−1 0)(1.5 0

0 1.5)

= ( 0 1.5−1.5 0 )

END

nb:

Notes on combined transformations:

If H and R represent two separate transformations

Then the combined transformation that represents H first, followed by R is writtens as : [R][H]

And the combined transformation that represents R first, followed by H is writtens as : [H][R]

Note carefully that: [H][R] ≠ [R][H]

One way to remember the correct order, is to view the two matrices as functions, and recall that fg (x) means function g (first) followed by function f .

so RH ( ) ⇒ H first then R


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