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CSEC MATHEMATICS Past Paper Solution – Jan 2009 cxcDirect Institute

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step 1. Converting the numerator from a mixed number : 15 3 3 = 4 4

Total amount (yr1) = principal + Interest = 24, 000 + 8% of 24000 24000×8 = 24000 + 100 = 24000 ( 1+ 0.08)

step 2. Simplifying the denominator:

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

Q1. Jan 2009

1 5 14− 5 2 − = 3 6 6

⇒

=

9 = 6

3 2

Now in year 2, Interest is applied to the total amount from year 1. so: total amt (yr2) = [total amt (yr1)] (1+ .08)

step 3. Divide Numerator by denominator: 15 3 ⇒ ÷ 4 2 15 2 × = 4 3

⇒

30 12

= 2.5

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

3

3 4

1 5 2 − 3 6 45 28−10

=

12 12

x

Alternatively: Using the compounding formula: n S = P 1 r  where s = total amount P = Principal amount = $24,000 r = interest rate = 8% = 0.08 n = Compounding interval = 2 years = 2 ⇒ S = 24,000 1 0.082 = $27.996.60 ********************************************************

Q2

45 = 2.5 18

=

= [24,000 (1+.08)] (1 + 0.08) = 24,000 1.082 = $27.996.60

The LCM of n and 3n = 3n so placing under a common denominator we get: 6m – 5m m = 3n 3n

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

5 *2=

b) B$ 2000 = EC$ 2700 so: ⇒

B $ 2000 2000

B$1

= =

EC $ 2700 2000 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 **********************************************************

2 5 −2 = 23

************************************************ Factorizing 1st two terms 3x – 6y = and factorizing 2nd two terms x 2− 2xy =

3  x – 2y  x  x− 2y 

so : (x – 2y) is a common in both terms, Giving:  x− 2y  x 3 *********************************************** d) Let the length of the first piece be x so 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 – 3 2x= 21 ⇒

4x – 3 = 21

⇒

4x = 24

⇒

x = 6cm

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 , © cxcDirect Institute Email: [email protected] website: www.cxcDirect.org

2


Q3

Q4

U

From the graph given: the two points on the straight line have coordinates of: P (0, 3) and Q ( -2 , 0) M= 42

P = 54

Gradient PQ: (54 - x)

(42 - x)

x

m=

so Gradient of line PQ =

y2− y1 x2− x1

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

N=6

where so

y2 = 0, y 1= 3

m=

0– 3 −2− 0

x 2=−2, x1 =0 −3 −2

=

= 1.5

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

x = 10 A

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

y ( 6 , 2)

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

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

B P( 0 , 3)

Finding MK: MK is the opposite side of right angled triangle MKN, so: Opposite (MK) MK 0 = sin 30 = hypotenuse (KN) 10 ⇒

MK =

10 sin 30

0

Q (- 2 , 0) o

x

= 5m Q (- 8 , - 9 )

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

Equation PQ: From the graph, The y-intercept = c = 3 so from the general equation of a straight line: y = mx + c

2

 JM  5 2 =13 2

Using pythagoras

y=1.5x 3

⇒

so:

JM =

 13 −5  =12 2

2

( we note also that this is a pythagorean triplet ( 5,12,13)

Finding t :

therefore: JK = 12 – 5 = 7m

Point ( -8 , t ) lie on PQ

J

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

⇒

7 13

5 M

5

t=-9

10 5

L

so

K

K

M

300

Finding the equation of a perpendicular line AB N

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

m AB m PQ =−1

where: © cxcDirect Institute Email: [email protected] website: www.cxcDirect.org

m AB

= gradient of line AB 3


m PQ

and

m AB =

⇒

Question 6 a – Construction

= gradient of line PQ −1 mPQ

−1 1.5

=

−2 3

=

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

Now AB passes through coordinates (6 , 2) so substituting these coordinates into the general equation: 6=−2 /3 × 2 c

so: c =

6 4 / 3

=

R

S

22 3 5.5cm

C

therefore

y AB =−

2 22 x 3 3

90o

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

7cm

P

P1

Q

T

P2

Q5. Volume of large box = 25 x 8 x 36 =

7,200 cm

3

Construction details: Steps:

Total Surface area :

1.

= 2 x {area of front + area of side + area of top} = 2 x { ( 25 x 36) + (36 x 8 ) + (8 x 25) } =

2,776 cm

2

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

Large box can fill 6 small boxes: volume of small box = 7,200/6 =

1,200 cm

3

• If height of small box is 20cm, then: Area of small box = 1,200/20 =

60cm

2

•

• Two possible dimensions for the area are 2

1.

12cm x 5 cm

=

2.

10cm x 6 cm

= 60cm 2

60cm

or:

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

Draw a straight line PT of suitable length and measure PQ = 7cm

•

0

90

at

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 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. 2.

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4


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

Question 6 b – Transformation

P(2,3)

Q(4,3) M

R(2,0)

P'(-2, -1)

Q'(0, -1) L

R'(-2, -4)



x Expressing the translation as a column vector T =

y

We 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 x y

+

T x y

=

P 2 3

P' −2 −1

() () ( ) so:

that is:

P' −2 −1

=

P 2 3

() ( ) () -

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 )

=

T −4 −4

( )

Q6 b ii (c) Scale factor k =

vertical height of N vertical height of L

=

6 =2 3

( )

T=

−4 −4

Note that we could have found the translation vector T, using any of the other two points Q or R.


5


Question -7 Statistics Marks

1 - 10 11 - 20 21 - 30 31 - 40 41 - 50 51 - 60 61 - 70 71 - 80 Graph:

# Students ( f )

upper class boundary

Cumulative Frequency

2 5 9 14 16 12 8 4

<10.5 <20.5 <30.5 <40.5 <50.5 <60.5 <70.5 <80.5

2 7 16 30 46 58 66 70

4.) The probability that a student scores less than or equal to 30 marks: P  <= 30 =

# students scoring ≤ 30 6

Now reading directly from the cumulative frequency table, # students scoring less than or equal to 30 = 16 so

P  <= 30 =

16 70

1.) A plot of cumulative frequency against Upper class boundary is drawn because we want to produce a ( less than graph). 2.) The coordinate (0,0) is assuming that no student gets a score of zero, Or a score less than 1 3) Because this is a “less than” graph, It can be seen directly from the graph that 40 students scored less than the pass mark of 47, and therefore 40 failed the test. The Number who passed = 70 – 40 = 30 students © cxcDirect Institute Email: [email protected] website: www.cxcDirect.org

6


Question 8 – Pattern Recognition

1

2

3

(b) If ax2 +bx+c h=

then:

4

2

is expressed in the form

b 2a

and

k=

4ac – b 4a

a  x h  k 2

From the past paper we obtain,

n 1 2 3 4 5 6 7 ..

d 5 8 11 14 17 20 23 .. d .. 92 .. .. 188

.. 30 .. .. 62

rule 2x5-4 2x8-4 2 x 11 - 4 2 x 14 - 4 2 x 17 - 4 2 x 20 - 4 2 x 23 - 4 .. 2xd-4 .. 2 x 92 - 4 .. .. 2 x 188 - 4

a=2 ,

l 6 12 18 24 30 36 42 .. 2xd-4 .. 180 .. .. 372

so

−4 h= 2×2

and

k=

so:

b=−4 , and

c=−13 =

4× 2×(−13) – 4 4×2 2

2

= −15 2

=

a  x h  k

−1

2  x−1 −15

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:

a=2 ,

so if :

b=−4 , and

c=−13

Using the quadratic formula: =

−b± √ b – 4ac 2a

=

4± √(−4) – 4 (2 )(−13) 2 (2 )

=

4± √ 16 – (−104) 4

2

x

2

Section 2 See past paper for question Step 1 2

To make t the subject:

Action Square both sides Multipy both sides by g

Move r to LHS

4

Swap LHS and RHS

4± √ 120 4

=

4± √ 120 4

=

1± 2.74

Result 2

    () P 2

g

=

P 2

tr g

2

= t r

2

3

=

P g − r= t 2 2

t=g

()

P − r 2

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≤ x≤3.74

4. Now given that f  x= 2 x −12 − 15 The minimum value occurs when f ( x)=k so f min ( x) = - 15 The value of x at the minimum = so x= − (−1) = 1


−h 7


Question 10: - Functions We can write the equations given in the form: f  x= x – 3 1 2 2 g  x= x −1 so:

f 6 =6 – 3= 3

and:

f f

⇒

−1

−1

 x

= inverse of =

 x

3. now fg(x)

⇒

f  x

( reverse the operation)

x 3

g(x) into f(x) 2

2

=

fg  x= x −1 − 3

⇒

x −4

2

so:

fg 2= 2 − 4= 0

and:

fg −2=−2 − 4=0

2

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

b. Distance (km) Train

C (150km ) 60 km/hr

speed = 100/40

At rest (0 km/hr) B (100km ) 150 km/hr

(0 km)

0.833333

50 min

20 min A 0

40

60

110

time (min)

60

Train travels 100 km from A – B in 40 mins: so 1. Average speed =

distance 100 = = 150 km/hr time 40 /60

2. Train remains at rest for ( 60 – 40) = 20 minutes 3. Time from B to C =

Distance BC speed BC 50 km 60

= =

5 ×60 6

= 5/6 hrs

= 50 mins


8


Question 10: - Trig

x0 Tan x 1/2.Tan x

when

10 0.2 0.09

y=0.7 , x=54

20 0.4 0.18

30 0.6 0.29

40 0.8 0.42

50 1.2 0.60

60 1.7 0.87

0


9


Question -12 Question 11. C

U

A

40

90 w=8 V

t= 9.2

32.5

Area = 34.64

C 66

114 D

O 132

48

E

u' = 11 T

60 u= 10

W

Considering Triangle UTW:

B

( 2 sides with included angle: apply cosine rule) TU = w = 8; TW = u = 10; UW = t; ∢ UTW = 600 , 1. Angle at radius and tangent

∢ OAE=∢OBE =90

0

2. now: ∢ OAE∢ OBE= 1800 ( both are right angles) 0 then for quadrilateral OAEB, ∢ AOB∢ AEB =180 0 since angles in a quadrilateral = 360 ⇒

∢ AOB

=

0

180 – 48=132

Applying cosine rule: 2 2 2 ⇒ t =u  w −2uw.CosT 2

2

2

⇒

t =10  8 – 2×10×8 ×cos 60

⇒

2 t =164 – 160×0.5 = 84

⇒

t =  84= 9.2

0

Considering Triangle UVW: Now angle at centre = twice angle at the circumference ⇒ ∢ AOB= 2×∢ ACB 0 so ∢ ACB= 132/2=66 Now Opposite angles cyclic quadrilateral (ACBD) are supplementary, so: 66∢ ADB= 1800 0 so : ∢ ADB=180 – 66= 114

( 2 sides with excluded angle: apply sine rule) 0 UW = t = 9.2; VW = u' = 11; ∢ VUW = 40 , Applying sine rule: ⇒

u' t = sinU sinV

⇒

11 9.2 = sin40 sinV

⇒

sinV = sin40 ×

⇒

−1 V =sin 0.5376 =

9.2 = 0.5376 11 0

32.5

Considering Triangle UTW: Area = ½ uw SinT

(T is the included angle)

Area− ½ 8× 10× sin60 =34.64m


2

10


Question 13 – Vectors

Finding Vector

Q⃗ R R( 8, 9 )

Q⃗ R

R( 8, 9 ) Q( -1, 3 )

Q(-1, 3)

O⃗ R O⃗ R

O⃗ Q

P ( 3,2 )

O⃗ Q

O

⃗ OP

O

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

The Position vectors are found directly from the coordinates: so:

P ( 3,2 )

⇒

⃗ OP =

Q ( -1, 3)

⇒

O⃗ Q =

Q R =

so:

   3 2

⃗ QO

so

R(8, 9)

O⃗ R

⇒

 8 9

=

= =

and b(i)

⃗ QO

now the vector

−1 3

⃗ + QO

Q R

so:

=

O⃗ R

is the reverse of the vector

⃗ OQ

⃗ − OQ

( ) =( ) ( ) +  = () −

1 −3

1 −3

−1 3

8 9

9 6

now note that:

ii) Proof that:

O P

Q⃗ R

is parallel to

Q R=

R( 8, 9 ) ⃗ QR

Q(-1, 3) O⃗ R

P ( 3,2 )

O⃗ Q

 9 6

can be written as:

()

3 ⃗ QR = 3 2

 3 2

but

O P =

so

Q R =

3. ⃗ OP

or

Q R =

k. ⃗ OP

..( where k = 3)

This completes the proof that the lines are parallel.

⃗ OP

O

Now, if

O P is parallel to

Then:

Q R =

k. O ⃗ P

We already know that so we must now find

Q R :

where k is a constant

O P =

 3 2

⃗ QR


11


iii) To find the magnitude of

c i) Finding

⃗ PR R( 8, 9 )

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

so:

⃗ QS

=

⃗ QO

but:

⃗ QO

=

⃗ −O Q

and

O⃗ S

=

so:

⃗ QS

=

O⃗ S

+

⃗ PR

Q(-1, 3) O⃗ R

P ( 3,2 )

O⃗ Q

⃗ OP

O

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

⃗ PR

=

⃗ PO

but

⃗ PO

=

⃗ − OP

()

where

−

so

⃗ PR

3 2

=

−

( ) −1 3

O⃗ R

+

=

−

( ) ( ) + 

() 3 2

8 9

−3 −2

=

P⃗ R =

√ 52 +72

1 −3

1 −3

+

() a b

=

( )

Q⃗ S=O ⃗ P

then:

  a 1 b− 3

and given

⃗ OP

=



=

a +1 b −3

3 2

 3 2

(reverse the vector)

and solving for a, and b gives:

−3 −2

=

( )

=

a b

c ii) If

() 5 7

⇒

a +1 = 3;

giving a = 2;

and

b-3=2

giving b = 5

⇒ and magnitude”

() ( )

=

P⃗ R

First: Find vector

Vector

Q⃗ S in terms of a and b

O S =

 2 5

= 8.6

Q S

Point S (a, b)

⃗ OS =

⇒

() a b

S( a, b ) ⃗ QS

Q(-1, 3) ⃗ OS

P ( 3,2 )

O⃗ Q

⃗ OP

O


12


Proof that OPSQ is a parallelogram:

and O ⃗S

S( 2, 5)

was found earlier as :

Q⃗ S

so:

P⃗ S Q( -1, 3 )

O⃗ S

P⃗ S

=

( ) −3 −2

+

 2 5

 = ( ) 2 5

−1 3

From this result we conclude that since :  =P  and O Q Q⃗ S= O ⃗ P S

P(3, 2)

O⃗ Q

Then OPSQ is a parallelogram.

O⃗ P O

Now if OPSQ is a parallelogram, Then :  =P  OQ S

and

Q⃗ S=O ⃗ P

We already know that

so the remainder of the

Q⃗ S=O ⃗ P

 =P  proof is to show that: O Q S

Now we also know that

O⃗ Q =

so we must now find the vector

To Find

  −1 3

⃗ PS

P⃗ S S( 2, 5)

Q⃗ S P⃗ S Q( -1, 3 )

O⃗ S P(3, 2)

O⃗ Q O⃗ P O

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

⃗ PS

⃗ where P O

= =

⃗ PO −O ⃗ P

+

O⃗ S

was found earlier as


( ) −3 −2

13


Question 14 – Matrices

The effect of the combined transformation on Triangle ABC:

Multiplying Matrices AB : ⇒

( )( ) 1 2 1 3 2 1 2 5

⇒ 3 AB

( ) ( ) ( ) 5 13 4 11

=

(

⇒

−2 0 0 2

A B C 1 1 2 2 1 1

)(

A' B ' C ' −1 −1 −4 4 2 2

) ( =

)

5 13 4 11

=3

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

15 39 12 33

=

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

b)

The simultaneous equation can be written as:

matrix v can be written as: V=2

( )( ) = ( ) () = ( )() ( )( ) 11 6 9 5

( ) 1 0 0 1

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

⇒

x y

−1 0 0 1

x y

11 6 9 5

6 7

−1

6 7

2 0 0 2

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

hint: Just remeber that

“first is Right “

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

Firs t

(W ) (V ) Left s ide

W ⇒

(

−1 0 0 1

Right s ide

V

WV

)( ) = ( 2 0

0 2

−2 0 0 2

)


14

cxcDirect Jan2009

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