cxcdirect jan2009

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CSEC MATHEMATICSPast Paper Solution Jan 2009

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Q1. Jan 2009

step 1. Converting the numerator from a mixed number :

3 34 =

154

step 2. Simplifying the denominator:

2 13 5

6= 145

6 = 96 =

32

step 3. Divide Numerator by denominator:

154 3

2

154 2

3 = 3012 = 2.5

AlternativelyThe LCM of 3, 6 and 4 = 12so we multiply the numerator and the denominator by 12 giving:

3 34

2 13 5

6

x 1212

= 452810 = 4518 = 2.5

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

b) B$ 2000 = EC$ 2700

so: B $ 20002000 = EC $ 2700

2000 B$1 = EC$ 1.35

now: $EC 432.00 = $BD ?The first question to ask is whether you should get more dollars, or less dollars. Note that when you convert from $B to $EC you get more dollars so from $EC to $B the reverse will happen and you should get less dollars.

If you expect less dollars then you must divide by the conversion rate of 1.35

so $EC 432.00 over 1.35 = $B 320.00

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

If I deposit $24,000 in an account and it is compounded at a rate of 8% for two years. Then after the first year ,

Total amount (yr1) = principal + Interest = 24, 000 + 8% of 24000

= 24000 + 240008100= 24000 ( 1+ 0.08)

[Total amt ( yr1)] = 24000 ( 1+ 0.08)

Now in year 2, Interest is applied to the total amount from year 1. so:

total amt (yr2) = [total amt (yr1)] (1+ .08)

= [24,000 (1+.08)] (1 + 0.08)= 24,000 1.082

= $27.996.60

Alternatively:Using the compounding formula:

S=P 1 r n

where s = total amountP = Principal amount = $24,000r = interest rate = 8% = 0.08n = Compounding interval = 2 years = 2

S = 24,000 10.082

= $27.996.60********************************************************Q2The LCM of n and 3n = 3nso placing under a common denominator we get:

6m 5m3n =

m3n

5 * 2 = 522 = 23

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

Factorizing 1st two terms 3x 6y = 3 x 2y and factorizing 2nd two terms x 22xy = x x2y

so : (x 2y) is a common in both terms, Giving: x2y x3

***********************************************d) Let the length of the first piece be xso length of second piece = x 3 (3cm shorter)and length of third piece = 2x ( twice the first piece)

All three pieces add to 21cm so: x x 32x=21 4x 3 = 21 4x = 24 x = 6cm

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2

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Q3

The total Students = 92so: (54-x) + x + (42-x) + 6 = 92 54+42+6 x = 92 x = 10

where x = # students studying both Music and Pys. Ed*********************************************************

b) This question is a test of your knowledge of trig. ratios and Pythagoras theorem. From the diagram given:

Finding MK:MK is the opposite side of right angled triangle MKN, so:

sin 300 = Opposite (MK)hypotenuse (KN) =

MK10

MK = 10 sin 300 = 5m

Finding JK:now JK = JM - MKwhere JM is the vertical side of right angled triangle JLM

Using pythagoras JM 252=132

so: JM = 13252 =12( we note also that this is a pythagorean triplet ( 5,12,13)

therefore: JK = 12 5 = 7m

Q4

From the graph given: the two points on the straight line have coordinates of: P (0, 3) and Q ( -2 , 0)

Gradient PQ:

so Gradient of line PQ = m=y2 y1x2x1

(tip: the graph has a positive slope so we expect m to be positive)

where y2=0, y1=3 x2=2, x1=0

so m= 0 320 =

32 = 1.5

Equation PQ:

From the graph, The y-intercept = c = 3

so from the general equation of a straight line: y = mx + c y=1.5x3

Finding t :

Point ( -8 , t ) lie on PQ

substituting these co-ordinates in the equation of PQ: t=1.5 8 3

so t = - 9

Finding the equation of a perpendicular line AB

If lines are perpendicular the product of their gradients = -1

so: m AB mPQ=1

where: m AB = gradient of line AB


3

P = 54 M= 42

x

N = 6

U

(54 - x) (42 - x)

K

J

ML

5

13

7

5

K

M N

5300

10

y

P( 0 , 3)

oQ (- 2 , 0)

( 6 , 2)

A

B

x

Q (- 8 , - 9 )

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and m PQ = gradient of line PQ

m AB =1mPQ

= 11.5 = 23

Now AB passes through coordinates (6 , 2)

so substituting these coordinates into the general equation:

6=2 /3 2c

so: c = 64 /3 = 223

therefore yAB=23

x 223

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

Q5.

Volume of large box = 25 x 8 x 36 = 7,200 cm3

Total Surface area :

= 2 x {area of front + area of side + area of top}

= 2 x { ( 25 x 36) + (36 x 8 ) + (8 x 25) } = 2,776 cm2

Large box can fill 6 small boxes:

volume of small box = 7,200/6 = 1,200 cm3

If height of small box is 20cm, then:

Area of small box = 1,200/20 = 60cm2

Two possible dimensions for the area are

1. 12cm x 5 cm = 60cm2 or:

2. 10cm x 6 cm = 60cm2

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

Question 6 a Construction

Construction of a Rectangle PQRS where PQ = 7.0cm and QR = 5.5cm

Construction details:

Steps:1. Draw a straight line PT of suitable length and

measure PQ = 7cm

Use a compass and ruler to construct a 900 at Q as follows:

With Centre Q, and and suitable compass separation, draw two arcs to cut line PT at P1 and P2 as shown.

With center P1, construct an arc above Q With center P2, construct a second arc above Q to

intersect the first arc at Point C. Use a ruler and pencil to draw a straight line

through points Q and C. Measure QR = 5.5cm.

Complete the Rectangle

2. With Centre R and compass separation = 7 cm, draw an arc above the point P

3. With center P and compass separation = 5.5cm, draw a second arc above P to intersect the first arc. The Point S is located at the intersection of the two arcs.

4. Use a ruler and pencil to complete the rectangle PQRS.


4

90o

P QP1P2

S

5.5cm

7cm

R

T

C

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Question 6 b Transformation

Expressing the translation as a column vector T=xyWe note that : T: P(2, 3) P' (-2, -1)

This means that the object point P maps to its corresponding image point under the translation T.

In column vector format, this is written as:

T

(xy) +P

(23) =P '

(21)

so: T

(xy) =P '

(21) -P

(23) = T

(44)

that is: T=(44)

Note that we could have found the translation vector T, using

any of the other two points Q or R.

Q6 b ii (a)To find the centre of enlargement ( point G)

Use a ruler to draw straight lines connecting the object and corresponding image points. The intersection of the lines will be the center of enlargement

Q6 b ii (b )Centre of elargement = G( - 5, 0 )

Q6 b ii (c)

Scale factor k = vertical height of Nvertical height of L = 63 = 2


5

P(2,3)

M

L

P'(-2, -1)

Q(4,3)

Q'(0, -1)

R'(-2, -4)

R(2,0)

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Question -7 Statistics

Marks # Students ( f ) upper class boundaryCumulative Frequency

1 - 10 2

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Question 8 Pattern Recognition

n d rule l1 5 2 x 5 - 4 62 8 2 x 8 - 4 123 11 2 x 11 - 4 184 14 2 x 14 - 4 245 17 2 x 17 - 4 306 20 2 x 20 - 4 367 23 2 x 23 - 4 42.. .. .. ..

d 2 x d - 4 2 x d - 4.. .. .. ..30 92 2 x 92 - 4 180.. .. .. .... .. .. ..62 188 2 x 188 - 4 372

Section 2

See past paper for question To make t the subject:

Step Action Result

1 Square both sides P2 2

= trg 2 Multipy both sides by g g P2

2

= tr

3 Move r to LHS g ( P2 )2

r= t

4 Swap LHS and RHS t = g ( P2 )2

r

(b)If ax2+bx+c is expressed in the form a xh 2k

then: h=b2a and k=4ac b2

4a

From the past paper we obtain,

a=2 , b=4 , and c=13

so h=422 = 1

and k=42(13) 42

42= 15

so: a xh 2k = 2 x1 215

2. The values of x at which the graph cuts the x-axis are the roots of the equation: so we need to solve the equation:

so if : a=2 , b=4 , and c=13

Using the quadratic formula:

x = bb2 4ac

2a

= 4(4)2 4 (2 )(13)

2 (2 )

= 416 (104)4

= 41204

= 41204= 12.74

that is: x=3.74 or x=1.74

3. Now the function is a parabola with a minimum value. The interval for which f(x) is less than zero is:

1.74 x3.74

4. Now given that f x=2 x1215The minimum value occurs when f (x)=kso f min(x) = - 15

The value of x at the minimum = hso x= (1) = 1


7

1 2 3 4

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Question 10: - FunctionsWe can write the equations given in the form:1 f x= x 32 g x=x21

so: f 6 =6 3=3

and: f 1 x = inverse of f x

f 1 x = x3 ( reverse the operation)

3. now fg(x) g(x) into f(x) fg x= x21 3 = x24

so: fg 2=224=0

and: fg 2=224=0***********************************************************

b.

Train travels 100 km from A B in 40 mins: so

1. Average speed = distancetime =100

40 /60 = 150 km/hr

2. Train remains at rest for ( 60 40) = 20 minutes

3. Time from B to C = Distance BCspeed BC

= 50 km60 = 5/6 hrs

= 5660 = 50 mins


8

speed = 100/40

0.833333

60time (min)

Distance (km)

A

B (100km )

0 40

50 min

(0 km)60

C (150km )Train

110

20 min

At rest (0 km/hr)

150 km/hr

60 km/hr

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Question 10: - Trig

when y=0.7 , x=540


9

x 0 10 20 30 40 50 60Tan x 0.2 0.4 0.6 0.8 1.2 1.71/2.Tan x 0.09 0.18 0.29 0.42 0.60 0.87

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Question 11. C

1. Angle at radius and tangent OAE=OBE=900

2.now: OAEOBE=1800 ( both are right angles)then for quadrilateral OAEB, AOB AEB=1800 since angles in a quadrilateral = 3600

AOB = 1800 48=1320

Now angle at centre = twice angle at the circumference AOB=2 ACB

so ACB=132 /2=660

Now Opposite angles cyclic quadrilateral (ACBD) are supplementary, so: 66 ADB=1800so : ADB=180 66=1140

Question -12

Considering Triangle UTW:

( 2 sides with included angle: apply cosine rule)TU = w = 8; TW = u = 10; UW = t; UTW=600 ,

Applying cosine rule: t 2=u 2w 22uw.CosT t 2=10282 2108cos 60

t 2=164 1600.5 = 84

t=84=9.2

Considering Triangle UVW:

( 2 sides with excluded angle: apply sine rule)UW = t = 9.2; VW = u' = 11; VUW=400 ,

Applying sine rule:

u '

sinU= t

sinV

11

sin40= 9.2

sinV

sinV=sin40 9.211

= 0.5376

V =sin1 0.5376 = 32.50

Considering Triangle UTW:

Area = uw SinT (T is the included angle)

Area 810sin60=34.64m2


10

A

48

B

C O DE

13266

90

114

40

u' = 11

U

T

W

Vw=8

u= 10

60

t= 9.2 32.5Area = 34.64

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Question 13 Vectors

The Position vectors are found directly from the coordinates:

so: P ( 3,2 ) OP = 32 Q ( -1, 3) OQ = 13and

b(i) R(8, 9) O R = 89

ii) Proof that: O P is parallel to Q R

Now, if O P is parallel to Q R :

Then: Q R = k. O P where k is a constant

We already know that O P = 32so we must now find QR

Finding Vector Q R

To go from Q to R , we first go from Q to O and then from O to R:

so: Q R = QO + O R

now the vector QO is the reverse of the vector OQ

so QO = OQ

= (13) = ( 13)so: Q R = ( 13) + 89 = (96)now note that:

Q R=96 can be written as: QR = 3(32)but O P = 32so Q R = 3. OP

or Q R = k. OP ..( where k = 3)

This completes the proof that the lines are parallel.


11

Q(-1, 3)

O

P ( 3,2 )

R( 8, 9 )

OP

OQ

O R

QR

Q(-1, 3)

O

P ( 3,2 )

R( 8, 9 )

OP

O Q

O RO R

O

Q( -1, 3 )

R( 8, 9 )

OQ

Q R

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iii) To find the magnitude of PR

First: Find vector P R

To go from P to R: First go from P to O , then O to R

so PR = PO + O R

but P O = O P = (32) (reverse the vector)where (32) = (32) so PR = (32) + 89 = (57)

and magnitude P R = 52+72 = 8.6

Vector Q S

Point S (a, b) OS = (ab)

c i) Finding Q S in terms of a and b

To go from Q to S: First go from Q to O , then O to S

so: QS = QO + O S

but: QO = O Q = (13) = ( 13)and O S = (ab)so: QS = ( 13) + (ab) = (a+1b3)

c ii) If Q S=O P and given O P = 32then: a1b3 = 32

and solving for a, and b gives:

a +1 = 3; giving a = 2;

and b - 3 = 2 giving b = 5

O S = 25


12

Q(-1, 3)

O

P ( 3,2 )

R( 8, 9 )

OP

OQ

O R

PR

Q(-1, 3)

O

P ( 3,2 )

OP

OQ

S( a, b )

QS

OS

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Proof that OPSQ is a parallelogram:

Now if OPSQ is a parallelogram, Then :

Q S=O P and O Q=P S

We already know that Q S=O P so the remainder of the

proof is to show that: O Q=P S

Now we also know that OQ = 13so we must now find the vector PS

To Find P S

We can go from: P to O and then O to S

so PS = P O + O S

where P O = O P was found earlier as (32)

and O S was found earlier as : 25so: P S = (32) + 25 = (13)From this result we conclude that since :

Q S=O P and O Q=P S

Then OPSQ is a parallelogram.


13

O S

O

P(3, 2)

S( 2, 5)

OQ

Q( -1, 3 )

P S

O P

Q S

O S

O

P(3, 2)

S( 2, 5)

OQ

Q( -1, 3 )

P S

O P

Q S

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Question 14 Matrices

Multiplying Matrices AB :

(1 22 1)(1 32 5) = (5 134 11) 3 AB = 3 (5 134 11)

= (15 3912 33)

b)

matrix v can be written as:

V = 2 (1 00 1)The effects of this transformation on the triangle ABC is a enlargement with scale factor two (2)

The combined transformation of V followed by W :( V first then W = [W][V)])

hint: Just remeber that fir st is Right

so the transformation that is first will be on the right side

W V WV

(1 00 1)(2 00 2) = (2 00 2)

The effect of the combined transformation on Triangle ABC:

(2 00 2)A B C

(1 1 22 1 1) =A ' B ' C '

(1 1 44 2 2)

This is an enlargement followed by a reflection in the y axis.

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

The simultaneous equation can be written as:

(11 69 5)(xy) = (67) (xy) = (11 69 5)

1

(67)(1 00 1)(2 00 2)


14

(W ) (V )Firs tSecond

Right s ideLeft s ide

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