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CSEC MATHEMATICS Past Paper Solution – Jan 2008 cxcDirect Institute

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

Q2.a

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

if

Q1a. Jan 2008

=

11 28

..Mov 3 to RHS

−2x< 4

..simplify

⇒

−x < 2

..div both sides by 2

⇒

x > −2

..multiply both sides by (-1) and reverse sign

step 2. Simplifying the denominator: 1 1 2 × 2 5

−2x< 7 – 3

⇒

step 1. simplifying the numerator : 1 3 8 3 32 −21 1 − = − = 7 4 7 4 28

⇒

3 – 2x 7

=

5 1 × = 2 5

5 10

=

Q2a(ii)

1 2

The smallest whole number that satisfies the inequality is x = 0 Q2. b

step 3. Divide Numerator by denominator: 11 1 ⇒ ÷ 28 2 11 2 × = 28 1

⇒

11 14

x – xy= x  x− y

2.

a – 1= a− 1 a 1 

3.

2 2p – 2q – p  pq =

2

=

2  p− q  – p  p− q 

2 − p  p − q

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

(ii)

From the table given:

2−

0.24 = 0.15

2−

=

2

1.

24 = 15

2−

0.24× 100 0.15×100

30− 24 6 = = 15 15

Money collected for sponge cake Money collected for Chocolate cake Money collected for fruit cake 2 5

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

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

1.b

= 2 (k +5) = 10k = 4 × 2k =8k

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

Cash Price = $319.95 Total hire purchase price =

$ 69 10×$ 28.50

= $354.00

Difference in Cash & Hire purchase price = 354.00 – 319.95 = $34.05 Difference as a percentage of cash price =

34.05 × 100 319.95

= 10.64%

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

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2


Q3 Step 3.

U

A

T

S

A 84

l, m

B

q

k, p

48 42

48

48

B

D

C

n, r

∢ ABC S ∪T = l,m,k, p,q S' = n, r, q

1. 2.

=

48 42

=

0

90

Triangle ABD is isoceles so base angles are equal = Hence

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

∢ ABC

=

180−48 48 = 84

48

0

0

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

b) The following diagrams are drawn to demonstrate the theorems: B

42

D

48

90 C

step1. Find angle BDC Angles in a triangle = 180 ⇒

∢ DBC =

180−90 42

=

48

step2. Find angle ABD A

B

48

alternate angles 48

D

C

Angle ABD =∢ CDB

=

48

( alt. Angles)


3


Q4.

Q5 a). Frequency Table B

410km

A 7:30

# Boys = Frequency (y)

0 1 2 3 4

2 6 17 8 3

b) Total Boys =

time

14:20

# Books (x)

∑f

= (2 + 6 + 17 + 8 + 3) = 36

6 hrs + 50 min

c) Modal # Books read = 2

Time to travel from A to B =

( most (17) boys read this amount of books)

14:20 – 7 :30 = 6hrs + 50 min = 6.833 hrs Distance from A to B = 410km d) Total books read =

Avg, Speed =

Distance = 410 over 6.8333 = 60 km/h time

∑ f ×x

=

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

= 76

b) e) Mean number of books read r = 3.5cm

=

Total books # boys

=

76 36

= 2.1

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

Note the following: The unshaded portion represents a quarter(1/4) of a circle: Shaded portion = area of square – unshaded area 2

i) Area of circle of radius ( r = 3.5cm) =

 3.5

unshaded area = ¼ ( circle)

= ¼ (39.5) =

ii) Area of square of side 3.5 cm

=

3.5

2

=

# boys reading >= 3 books Total Boys 83 = 36

11 36

= 38.5 cm2 9.625 cm

2

= 12.25 cm 2

iii) Shaded Area = area of square – unshaded area =

12.25 – 9.625

= 2.625 cm 2


4


Question 6 – Transformation

6.b Z

P

Y

5.5 cm

10.9 cm

Q

60

Note from the diagram above that if object P rotated by 0 0 0 (90 +90 )=180 about the origin in a clockwise direction, it will be mapped unto image Q. Q6 a.(i) From the diagram shown on the past paper, BCL is therefore mapped unto FHL by: A clockwise rotation of 0 180 about the origin L.

So:

0

7 cm

W

X

Construction details. 1. 2. 3. 4. 5.

R(−180) : B ( x , y) → F (− x ,− y) 6.

Draw straight line WX = 7cm ∢ ZWX using compass Construct 600 Measure WZ = 5.5cm Set compass to a separation of 7cm and with centre Z, construct an arc above X. 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. Measure WY = 10.9cm

Q6 a.(ii) P 1

Q 4

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)


5


Q 7 a)

Question 8

y 5

4

x y

-1 5

0 0

1 -3

2 -4

3 -3

4 0

5 5

n

Series

Sum

Formula

3

1+2+3

6

1 (3 )(3 +1) 2

6

1+2+3+4+5+6

21

1 (6)(6+1) 2

n

1+2+3+..........+ n

n

Series

sum

3

1+2+3

6

1  n n 1 2

3

y=2 2

( 4.4, 2 )

( - 0.4, 2 ) 1

0

-1

1

2

3

4

5

Formula 1 3 3 1 2

x 3

3

3

3

3

3

..+

3

3

3

...+ 8 3

36

3

3

3

..+ 123

78

3

1  2 3

n

1  2 3

8

1  2 3

12

1  2 3

6

2

-1

n

3

-2 2

-3 2

-4

Q7 C(ii)

Where :

3

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

Q7 C(iii)

3

1  2 3



3

( [  

2

) ] 

1 (3) (3+ 1) 2

1 ( n)( n+1) 2 1 8 8 1 2

2

2

2



1 12 121  2

...+ 123 2



1 12 121  2

= 782 = 6084

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

The curve and the line intersects at the point where: 2

x −4x=2

so

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

The equation is therefore:

2 x −4x− 2 = 0

or

2

f  x= x – 4x – 2


6


Section II

Question -10 Linear programming

Inverse variation so:

Let

V∝

1 P

V=

k P

x

represents the # balls and y represents the # bats :

Where: ( k is a constant)

1.

x y ≤ 30

( no more than 30 items)

2.

6x 24y ≤ 360

( budget is $360

finding k now:

12.8=

k 500

P=

so if v = 480

k=

⇒

6,400 480

12.8×500=6,400

The Graphs of these two inequations are shown below: n.b. watch video Lesson1: graphing linear inequations:

= 13.3

watch video Lesson2: solving a linear program **************************************************

b) 2

For a right angle triangle

2

h =x  y

2

Profit is $1 per ball and $3 per bat .. ( Pythgoras theorem) Bats

30

h = 13 y=5

x + y = 30 25

x = 12 20

so if

,

h =a 1  2

x= a ,

y=a − 7

2

2

⇒

a 1  =a a −7 

⇒

a  2a1 =a a −14a 49

⇒

0 = a – 16a  48

⇒

0 = a − 4  a −12

⇒

2

2

(0, 15)

(20, 10) Max profit = $50

2

2

a = 4,

15

or a = 12

10

6x + 24y = 360 x> = 0

5 y> = 0

(30, 0)

since y must not be negative, we will choose a = 12 the result is shown in the diagram.

10

0

Given Profit:

20

30

40

50

60

ball

P =x+3y

The max profit is found at one of the vertices of the shaded area. 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, © cxcDirect Institute Email: [email protected], website: www.cxcDirect.org , Math club

7


Q 11. Geometry & Trig

Q11 b)

Please see the past paper for the actual drawing:

N

N 125m

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

A

B

80o

190o

Y 134.1 m

X

O

50o

75m

100o C

W

Since the bearing C from B is Theorem1.

Then

∢ ABC +190

so:

∢ ABC =80

0

190

=

270

0

0

0

Angle at the center = twice angle at the circumference i.e

∢ WOY =2×∢WXY

so

∢ WOY =2 (50)=100

Finding AC: 0

Use the cosine rule since we know 2 sides + included angle Now AB = c = 125m; 2

so

⇒

Triangle OWY is isocesles so, base angles are equal

⇒

0

2

2

2

2

b =125  75 − 2 12575 Cos80

b = 134.1m = AC

Finding Bearing C from A

o

N

40 O

∢ ABC =80

b = a c −2ac.CosB

Q11 a(ii)

Y

2

BC = a = 75m ;

100o

N 90o

125m

A

o

40

x

80o

o

B 190o

W 134.1 m

⇒

∢ OWY

=

∢ OYW

=

½ 180−100

=

40

75m

0

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

C

Bearing C from A =

0

90 + x

0

0

using the sine rule:

⇒ sin x =

⇒ x=

⇒

75 × sin80 =0.54914 134.1

−1 sin 0.54914 =

therefore Bearing C from A = © cxcDirect Institute Email: [email protected], website: www.cxcDirect.org , Math club

0

sin x sin 80 = 75 134.1

0

33.4

90 33.4=123.4

0

8


Q12. Angle of Elevation

13 . Vectors & matrices (i)

D

C

5

Q B

2.5 5 28.58

A

Now

0

tan 5 =

2.5 AB

2.5 0 Tan5

so: AB =

P

B

A

= 28.58m

⃗ AC

(ii)Finding

C

⃗ BC

*************************************************************** B

C 10.4 20 28.58

A⃗ B

12.9m

E 2.5 B

Now

0

tan 20 =

so CE =

A

CE 28.58

28.58 Tan20

and BC = BE + CE =

⃗ AC

⃗ To find vector : A C Imagine 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:

= 10.4 2.5 10.4

= 12.9m

So :

⃗ AC

=

A⃗ B +

⃗ BC

But we know that : ⃗ AB =

and

⃗ = 3y BC

so:

⃗ = 2x +3y AC

⇒


2x

⃗ = 2 ( x +1.5y) AC

9


⃗ PQ

Finding

B x

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

C

1.5 y Q

1.5 y

⃗ PQ

P

⃗ AC

x

so:

⃗ +O T⃗ R T⃗ = R O

Now recall that

⃗ RO

= − O ⃗R

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

A

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

R T⃗

Now the vector

 =P   PQ B B Q

but and

⇒

⃗ PB

=

1⃗ AB = 2

1 (2x) = x 2

=

1⃗ BC = 2

1 (3y) = 1.5y 2

⃗ BQ

=

⇒

⃗ RO

=

hence

⃗ RT

=

O T⃗ =

  5 −2

−3 −4

⃗ + RO

O T⃗

( )+  5 −2

−3 −4

=

=

2 −6

To find

⃗ SR : Consider Triangle OSR

S( -1, 6 )

S⃗ R

1 ⃗ AC 2

=

3 4

 

 = x 1.5 y PQ

⃗ PQ

  , and ( ) ⃗ RT

(iii) Now recall that : ⃗ = 2 ( x+15 y) AC ⃗ = 2PQ so:

O⃗ R

so given

R( 3, 4 )

O⃗ S

13.b

O⃗ R

S( -1, 6 ) O

⃗ OT

R( 3, 4 )

T( 5, - 2 )

O⃗ S

O⃗ R

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

R T⃗

O

⃗ OT

To find

T( 5, - 2 )

⃗ RT :

The three position vectors equation is:

O⃗ S ,

O⃗ R vector triangle


⃗ SO

so:

S⃗ R

but

O⃗ S =

 

⇒

⃗ SO

=

−O⃗ S

=

so :

S⃗ R

=

( )

+

=

O⃗ R

+

−1 6

1 −6

( )  =   1 −6 3 4

( reverse the signs)

4 −2

10


n.b: The value of

( ) −4 2

shown on the official CXC website is

(incorrect)

⃗ OF

13b ii (a) Position Vector S( -1, 6 )

R( 3, 4 )

O⃗ S

⃗ RF

O⃗ R

F ( 4, 1)

O⃗ F

F T⃗

O

O T⃗

To find : so:

O⃗ F we can go from O to R , then from R to F

O⃗ F =

O⃗ R

+ R F⃗

But

⃗ = RF

F T⃗ =

1 ⃗ RT 2

so

⃗ OF =

O⃗ R +

1 ⃗ RT 2

=

 3 4

+ =

so:

T( 5, - 2 )

 OF

=

( )  + ( ) 1 2 2 −6

3 4

1 −3

 4 1

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


11


14 – Matrices if

Note:

AB = C A

⇒

(2 1 )

(

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

B C 1 x = ( 5 6) y −2

)

⇒ 2+y =5 and 2x – 2 = 6 y=3 , and

⇒

x= 4

Matrix is singular if the determinant = 0



Det of

2 −1 1 3



= 6 +1 = 7

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



Inverse of matrix is

2 −1 1 3

−1



= 1/7



3 1 −1 2



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

matrix times its inverse is: 1/7

= 1/7



3 1 −1 2

 

2 −1 1 3

   7 0 = 1 0 0 7 0 1





= 1/7



6 1 −3 3 −2 2 16

= Identity Matrix (I)

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

The Equation is shown below in Matrix form:



  =     = 

2 −1 1 3

⇒

0 7

x y

2 −1 1 3

−1

x y

0 7

    = 

Inverse was found above as: 1/7



3 1 −1 2

0 7

⇒

1/7

⇒

x = 1/7 (0 + 7) = 1

⇒

y = 1/7 ( 0 + 14) = 2

⇒

x = 1,

and


3 1 −1 2

x y

y=2

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12

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