Csec Math Jan2008

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CSEC MATHEMATICSPast Paper Solution – Jan 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. Jan 2008

step 1. simplifying the numerator :

117−

34

=87−

34

=32−21

28 =

1128

step 2. Simplifying the denominator:

⇒ 212×

15

= 52×

15

= 5

10 =

12

step 3. Divide Numerator by denominator:

⇒1128

÷12

⇒1128

×21

= 1114

(ii)

2−0.240.15

= 2−0.24×1000.15×100

= 2−2415

= 30−24

15=

615

= 25

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

1.b

Cash Price = $319.95

Total hire purchase price = $ 6910×$ 28.50 = $354.00

Difference in Cash & Hire purchase price

= 354.00 – 319.95 = $34.05

Difference as a percentage of cash price

= 34.05319.95

×100 = 10.64%

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

Q2.a

if 3 – 2x7

⇒ −2x<7 – 3 ..Mov 3 to RHS

−2x<4 ..simplify

⇒ −x < 2 ..div both sides by 2

⇒ x >−2 ..multiply both sides by (-1)

and reverse sign

Q2a(ii)

The smallest whole number that satisfies the inequality is x = 0

Q2. b

1. x2 – 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

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

From the table given:

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

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2

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Q3

1. S ∪T = l,m,k, p,q2. 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

48

48B D

alternate angles

A

C

42

B48

C

A

48B D

48

84

A

3



Q4.

Time to travel from A to B

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

Distance from A to B = 410km

Avg, Speed = Distance

time= 410 over 6.8333 = 60 km/h

b)

Note the following:The unshaded portion represents a quarter(1/4) of a circle:Shaded portion = area of square – unshaded area

i) Area of circle of radius ( r = 3.5cm) = 3.52 = 38.5 cm2

unshaded area = ¼ ( circle) = ¼ (39.5) = 9.625 cm2

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

iii) Shaded Area = area of square – unshaded area

= 12.25 – 9.625 = 2.625 cm2

Q5 a). Frequency Table

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

0 2

1 6

2 17

3 8

4 3

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

c) Modal # Books read = 2 ( most (17) boys read this amount of books)

d) Total books read = ∑ f ×x =

(2×0) + (6×1) + (17×2) + (8×3) + (3×4) = 76

e) Mean number of books read

=Total books # boys

=7636

= 2.1

f) Prob that a boy read 3 or more books = P( books >= 3)

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

=83

36=

1136


A

410km

7:30 14:20 time

6 hrs + 50 min

B

r = 3.5cm

4



Question 6 – Transformation

Note from the diagram above that if object P rotated by (900

+900)=1800 about the origin in a clockwise direction,

it will be mapped unto object Q.

Q6 a.(i)From the diagram shown on the past paper, BCL is therefore mapped unto FHL by: A clockwise rotation of

1800 about the origin L.

So:

R(−180) : B (x , y)→ F (−x ,− y)

Q6 a.(ii)

Note from the diagram above that if object P Translated by 1 unit down and 4 units to the right, then this is represented by a Translation T = (4,-1). which would map P onto Q.

So: BCL is mapped unto HFG by a translation T

represented by: T = (4,-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


W X7 cm

5.5 cm

600

Z Y

10.9 cm

5

P

Q

P

Q

4

1



Q 7 a)

Q7 C(ii)

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

Q7 C(iii)

The curve and the line intersects at the point where:

x 2−4x=2

so x2−4x−2 = 0 will give the roots ( -0.4, and 4.4)

The equation is therefore: x2−4x−2 = 0 or

f x= x2 – 4x – 2

Question 8

n Series Sum Formula

3 1+2+3 6 12(3 )(3+1)

6 1+2+3+4+5+6 21 12(6)(6+1)

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

n Series sum Formula

3 1+2+3 6 123 31

3 1323

33 62

(12

(3)(3+1))2

n 1323

33 ..+ n3

[12(n)(n+1)]

2

8 1323

33 ...+ 83 362

128 81

2

12 1323

33 ..+ 123 782

1212 121

2

Where : 1323

33 ...+ 123

= 1212 121

2

= 782 = 6084

n.b: the value of 6024 shown on the CXC website is incorrect


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 )

6



Section II

Inverse variation so:

V ∝1P

V =kP

( k is a constant)

finding k

now: 12.8=k

500 ⇒ k = 12.8×500=6,400

so if v = 480 P=6,400480

= 13.3

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

b)

For a right angle triangle h2=x2

y2 .. ( Pythgoras theorem)

so if h=a1 , x=a , y=a−7

⇒ a1 2=a 2a−7

2

⇒ a 22a1=a 2

a 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

Let x represents the # balls and y represents the # bats :

Where:

1. x y ≤ 30 ( no more than 30 items)

2. 6x24y ≤ 360 ( budget is $360

The Graphs of these two inequations are shown below:

n.b.watch video Lesson1: graphing linear inequations:

watch video Lesson2: solving a linear program

Profit is $1 per ball and $3 per bat

Given Profit: P = x + 3 y

The profit depends on which option ( vertex) is usedThe following are the choices (options):

Vertex # Balls (x) # Bats (y) P =x +3y

1 0 15 $45

2 20 10 $50

3 30 0 $30

Vertex 2 ( 20 Balls and 10 Bats) gives the maximum profit of $50,


7

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


http://www.youtube.com/watch?v=_Aya8KUk1ig

http://www.youtube.com/watch?v=hskCFeU5f10


Q 11. Geometry & Trig

Please see the past paper for the actual drawing:

The drawings shown below are general drawings use to explain the theorems.

Theorem1.

Angle at the center = twice angle at the circumference

i.e ∢WOY =2×∢WXY

so ∢WOY =2 (50)=1000

Q11 a(ii)

Triangle OWY is isocesles so, base angles are equal

⇒ ∢OWY = ∢OYW = ½ 180−100 = 400

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

Q11 b)

Since the bearing C from B is 1900

Then ∢ABC +190 0 = 2700

so: ∢ABC=800

Finding AC:

Use the cosine rule since we know 2 sides + included angle

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

so b2=a2

c2−2ac.CosB

⇒ b2=1252

752−2 12575Cos80

⇒ b = 134.1m = AC

Finding Bearing C from A

Bearing C from A = 900+ x0

using the sine rule: ⇒sin x0

75=

sin 800

134.1

⇒ sin x = 75

134.1×sin80 =0.54914

⇒ x = sin−1 0.54914 = 33.40

therefore Bearing C from A = 9033.4=123.40


8

X

Y

W

50o O 100o

Y

W

O 100o

40o

40o

N

190o B A

C

N

125m

75m

80o

134.1 m

N

190o B A

C

N

125m

75m

80o

134.1 m

90o

xo



Q12. Angle of Elevation

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

13 . Vectors & matrices

(i)

(ii)Finding A⃗C

To find vector : A C⃗Imagine that you are at Point A , and you need to get to point C. You will need to first travel from A to B, and then from B to C. Note the direction of the arrows:

So : A C⃗ = A B⃗ + B C⃗

But we know that :

A⃗B = 2x

and B C⃗ = 3y

so: A C⃗ =2x+3y

⇒ A C⃗ =2(x +1.5y)


2.5

5

5D

A B28.58

20

2.5

B

E

C

28.58

10.4

12.9m

9

B

A

C

P

Q

A

B

C

A B⃗

B C⃗

A C⃗



Finding P Q⃗

To get from P to Q we must first go from P to B and then from B to Q

So : P Q=P BB Q

but P⃗B = 12

A⃗B = 12

(2x) = x

and B⃗Q = 12

B⃗C = 12

(3y) = 1.5y

⇒ P Q=x1.5 y

(iii) Now recall that :A C⃗ = 2 (x+15 y)

= 2 P Q⃗

so: P Q⃗ = 12

AC⃗

13.b

To find R⃗T :

The three position vectors O S⃗ , O R⃗ vector triangle equation is:

Now the vector RT⃗ is the vector from Point R to pointT

So if you once again imagine that you are at R and need to get to T, you must first go from R to O , anf then from O to T

so: R T⃗ = R O⃗+O T⃗

Now recall that RO⃗ = − O R⃗

This simply means that going from R to O is the reverse of going from O to R

so given O R⃗ = 34 , and OT⃗ = 5

−2⇒ R O⃗ = (−3

−4)hence R⃗T = R O⃗ + OT⃗

R⃗T = (−3−4) + 5

−2 =

2−6

To find S⃗R : Consider Triangle OSR

Similarly, to get from S to R, we first go from S to O and then O to R

so: S R⃗ = S O⃗ + O R⃗

but O S⃗ = −16

⇒ S O⃗ = − O S⃗ = ( 1−6) ( reverse the signs)

so : S R⃗ = ( 1−6) + 3

4 = 4−2


10

B

A

C

A⃗CP

Q

x

x

1.5 y

1.5 y

P⃗Q

O S⃗

O R⃗

OT⃗

O

S( -1, 6 )

T( 5, - 2 )

R( 3, 4 )

RT⃗

S R⃗

O S⃗

O R⃗

OT⃗

O

S( -1, 6 )

T( 5, - 2 )

R( 3, 4 )



n.b: The value of (−42 ) shown on the official CXC website is

(incorrect)

13b ii (a) Position Vector O⃗F

To find : O F⃗ we can go from O to R , then from R to F

so: O F⃗ = O R⃗ + R F⃗

But R F⃗ = F T⃗ = 12

R T⃗

so O⃗F = O R⃗ + 12

R T⃗

= 34 +

12 ( 2

−6) = 3

4 + ( 1−3)

so: O F = 41

b) The coordinates of point F are found directly from the position vector, = (4, 1)


11

O S⃗ O R⃗

OT⃗

O

S( -1, 6 )

T( 5, - 2 )

R( 3, 4 )

F ( 4, 1)O F⃗

R F⃗

F T⃗



14 – Matrices

if AB = C

⇒A

(2 1 )

B

(1 xy −2)

C= (5 6)

⇒ 2 + y = 5 and 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 −11 3

−1

07 = x

y

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

⇒ 1/7 3 1−1 2 0

7 = xy

⇒ x = 1/7 (0 + 7) = 1

⇒ y = 1/7 ( 0 + 14) = 2

⇒ x = 1, and y = 2

Note :

Additional Past Paper Type practice questions and answers are available in our: Vectors and Matrices Practice WORKBOOK for CSEC students:Available for immediate download at www.cxcDirect.org


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Csec Math Jan2008