csec math jan2008

© Omega Education Unit

© Club Omega

CXC MATHEMATICSClub Omega

Solutions – Jan 2008

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

Quote: “Luck is the strategic management of risk”

Omega Education UnitMandeville, JamaicaEmail: [email protected]: www.cxcdirect.schools.officelive.com Math Club: http://cxclink.50.forumer.com/

Telephone: 876 469-2775, 876 860-5263

mailto:[email protected]

http://www.cxcdirect.schools.officelive.com/

http://cxclink.50.forumer.com/


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

Q1a. Jan 2008

step 1. simplifying the numerator :

1 17− 3

4= 8

7− 3

4= 32−21

28 = 1128

step 2. Simplifying the denominator:

⇒ 2 12×1

5 = 52× 1

5 = 510 = 1

2

step 3. Divide Numerator by denominator:

⇒1128

÷ 12

⇒1128

× 21 = 11

14

(ii)

2− 0.240.15 = 2− 0.24×100

0.15×100

= 2− 2415 = 30−24

15 = 615 = 2

5**********************************************************

1.b

Cash Price = $319.95

Total hire purchase price = $ 6910×$ 28.50 = $354.05

Difference in Cash & Hire purchase price

= 354.05 – 319.95 = $34.05

Difference as a percentage of cash price

= 34.05319.95

×100 = 9.62%

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

Q2.a

if 3 – 2x7

⇒ −2x7 – 4

⇒ −2x3

⇒ 2x−3 multiply both sides by (-1) and reverse sign

⇒ x−3 /2

1. x 2 – xy=x x− y

2. a 2 – 1=a−1 a1

3. 2p – 2q – p2 pq = 2 p−q – p p−q

= 2− p p−q**********************************************************

Money collected for sponge cake = 2 (k +5)Money collected for Chocolate cake = 10kMoney collected for fruit cake = 4×2k =8k

Total money collected = ( 2k + 10 + 10k + 8 k) = 20k + 10If total momey collected = $140.

⇒ 20k + 10 = 140⇒ k = (140 – 10) over 20 = 6.5

© Club Omega - 876 469-2775Email: [email protected], website: www.cxcdirect.schools.officelive.com , Math club : http://cxclink.50.forumer.com/

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Q3

1. S∪T = k, p2. S' = n, r, q

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

b)

The following diagrams are drawn to demonstrate the theorems:

step1. Find angle BDC

Angles in a triangle = 180

⇒ ∢ DBC = 180−9042 = 48

step2. Find angle ABD

Angle ABD=∢CDB = 48 ( alt. Angles)

Step 3.

∢ ABC = 4842 = 900

Triangle ABD is isoceles so base angles are equal = 480

Hence ∢ ABC = 180−4848 = 840

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


S T

k, p

U

l, m q

n, r

48D

C

90

42B

4848

B D

alternate angles

A

C

42B 48

C

A

48B D48

84

A

3





Q4.

Time to travel from A to B

= 14:20 – 7 :30 = 6hrs + 50 min = 6.833 hrs

Avg, Speed = Distancetime = 410 over 6.8333 = 60 km/h

b)

Area of circle of radius r = 3.5 cm = 3.52 = 38.5cm2

Area of square of side 3.5 cm = 3.52 = 12.25cm2

Area of ¼ ( circle) = ¼ (39.5) = 9.625 cm2

Area of shaded portion = area of square - ¼ (area of circle)

= 12.25 – 9.625 = 2.65cm2

Q5. Frequency Table

# Books (x) # Boys = Frequency (y)

0 21 62 173 84 3

1.Total Boys = ∑ f = (2 + 6 + 17 + 8 + 3) = 36

2.Modal # Books read = 2 ( most (17) boys read this amount

of books)

3.Total books read = ∑ f ×x =

2×06×117×28×33×4 = 76 books

4. Mean number of books read

= Total books # boys = 76

36=2.1

P( books >=3) = # boys reading >= 3 books Total Boys

=83

36 = 1136


A

410km

7:30 14:20 time6 hrs + 50 min

B

r = 3.5cm

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Question 6 – TransformationFrom the diagram shown on the past paper,

BCL is mapped unto FHL by: A clockwise rotation of 1800

about the origin L.

So:R L ,−180 :C −2,0 H 2 , 0

R L ,−180 : B −2,1 H 2,−1

2 BCL is mapped unto HFG by a translation T represented

by: T = 4−1

soT : C 0,0 F 4,−1

T : B 0,1 H 4, 0

T : L 2, 0 G 6,−1

6.b

Construction details.

1. Draw straight line WX = 7cm2. Construct 600 ∢ZWX using compass3. Measure WZ = 5.5cm4. Set compass to a separation of 7cm and with centre

Z, construct an arc above X. 5. Set compass to a separation of 5.5 cm and with

centre X, construct a second arc to intersect the first arc. The intersection of the two arcs is the point Y.

6. Measure WY = 10.9cm

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

From the diagram: A 2,1 B 1,−2

now recall that : R0,−90 : x , y y ,− , x

Graph

Coordinates of intersection = (4.4, 2) and ( - 0.4, 2)

Equation = x 2 – 4x=2 or f x= x2 – 4x – 2

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

Question 8

n Series Sum Formula

3 1+2+3 6 123 31

n 1+2+3+..........+ n - 12n n1

n Series sum Formula

3 1+2+3 6 123 31

3 132333 62

123 31

2

n 132333 ..+ n3

12n n1

2

8 132333 ...+ 83 362

128 81

2

12 132333 ..+ 123 782

1212 121

2


W X7 cm

5.5 cm

600

Z Y

10.9 cm

1- 1

y = 2

52

-2

-1

1

y

x0

- 3

- 4

x -1 0 1 2 3 4 5y 5 0 -3 -4 -3 0 5

3 4

2

3

4

5

( 4.4, 2 )( - 0.4, 2 )

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Section II

Inverse variation so:

V ∝ 1P

V = kP ( k is a constant)

finding k

now: 12.8= k500 ⇒ k = 12.8×500=6,400

so if v = 480 P=6,400480 = 13.3

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

b)

For a right angle triangle h2=x 2 y2 .. ( Pythgoras theorem)

so if h=a1 , x=a , y=a−7

⇒ a1 2=a 2a−7 2

⇒ a 22a1=a 2a 2−14a49

⇒ 0 = a2 – 16a48

⇒ 0 = a−4 a−12

⇒ a = 4, or a = 12

since y must not be negative, we will choose a = 12

the result is shown in the diagram.

Question -10 Linear programming

If x balls and y bats such that :

1. x y ≤ 30 ( no more than 30 items)

2. 6x24y ≤ 360 ( budget is $360

Re-writing equations:

y ≤ 30−x (1)

24y ≤ 360−6x

y ≤ 15−0.25x (2)

see graph below:

Profit is $1 per ball and $3 per bat

So Profit Equation is Profit = x = 3y

The maximun profit occurs at one of the vertices:

Vertex 1 ( 0,15) ⇒ Profit = 1 x 0 + 3 x 15 = $45

Vertex 2 ( 20,10) ⇒ Profit = 1 x 20 + 3 x 10 = $50

Vertex 3 ( 30,0) ⇒ Profit = 1 x 30 + 3 x 0 = $30

so Max Profit = $50 (20 balls and 10 bats)


x = 12

h = 13y = 5

10 20 30 40

5

50 60

(20, 10) Max profit = $50

Bats

0 ball

10

15

20

30

25

(30, 0)

(0, 15)

x + y = 30

6x + 24y = 360

y> = 0

x> = 0

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Q 11. Geometry & Trig

From the drawing given,

1. ∢WOY angle at centre = twice angle at circumference

= ∢WOY =2 50=1000

2. Triangle OWY is isocesles so, base angles are equal

⇒ ∢OWY = ∢OYW = ½ 180−100 = 400

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

b)

Find AC:

Now AB = c = 125m; BC = a = 75m ; ∢ ABC=800

Use cosine rule since we know 2 sides + included angle

so b2=a2c2−2ac.CosB

⇒ b2=1252752−2 125 75Cos80

⇒ b = 134.1m = AC

Bearing C from A = 900∢ BAC

Now considering Triangle ABC, and using the sine rule:

⇒a

sinA= b

SinB

where ∢ A=∢ BAC

⇒75

sinA= 134.1

sin80

⇒ sin A = 75134.1

×sin80 =0.54914

⇒ A = sin−1 0.54914 = 33.40

therefore Bearing C from A = 9033.4=123.40


33.4

b = 134.1m

a=75 m

c=125 mA

C

19080

N

B

N

90 + 33.4 = 123.40

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Q12

Now tan 50= 2.5AB

so: AB = 2.5

Tan50 = 28.58m

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

Now tan 200= CE28.58

so CE = 28.58 Tan20 = 10.4

and BC = BE + CE = 2.510.4 = 12.9m


2.5

5

5D

A B28.58

20

2.5

B

E

C

28.58

10.412.9m

8





Vectors & matrices

From Triangle laws of vectors

A BB C= A C

⇒ 2x3y=A C

⇒ A C=2 x1.5y

Considering Triangle BPQ

P Q=P BB Q

⇒ P Q=x1.5 y

Note that this is ½ A C

⇒ P Q=½ A C

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

13.b

R T =O T – O R = 5−2 - 34 = 2

−6

S R=O R – O S = 34 - −16 = 4

−2

Position Vector OF

now R F=O F – O R

⇒ O F= R FO R = ½ R T O R

⇒ O F = ½ 2−6 + 34 = 41

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


x1.5y

1.5y

x

B

Q

P

A

C

x+1.5y

2(x+1.5y)

9

R ( 3, 4)S ( -1, 6)

T ( 5, -2 )

F ( 4, 1 )

RF = 1/2 ( RT)

o





14 – Matrices

if AB = C

⇒ 2 1 1 xy −2 = 5 6

⇒ 2 + y = 5and 2x – 2 = 6

⇒ y=3 , and x=4

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

Matrix is singular if the determinant = 0

Det of 2 −11 3 = 6 +1 = 7

Therefore matrix is not Singular.***************************************************************

Inverse of matrix is 2 −11 3

−1

= 1/7 3 1−1 2

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

matrix times its inverse is:

1/7 3 1−1 2 2 −1

1 3 = 1/7 61 −33−22 16

= 1/7 7 00 7=1 0

0 1 = Identity Matrix (I)

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

The Equation is shown below in Matrix form:

2 −11 3 x

y = 07⇒ 2 −1

1 3−1

07 = xy

Inverse was found above as: 1/7 3 1−1 2

⇒ 1/7 3 1−1 2 07 = x

y⇒ x = 1/7 (0 + 7) = 1

⇒ y = 1/7 ( 0 + 14) = 2

⇒ x = 1, and y = 2


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csec math jan2008

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