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CSEC MATHEMATICS Past Paper Solution – May 2012 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.

Selling Price

= cost Price + Profit = cost Price + 25% Cost Price

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

= cost Price ( 1+0.25)

Q1a. May 2012 Note that you must give your answer as a fraction so you must work using fractions.

= so: cost Price

step 1. Simplify the numerator 1 3 5

–

2 3

=

16 5

–

2 3

=

48 – 10 15

so:

=

=

2

4 5

then EC $1.00 =

38 15

=

= TT $2.5

= EC $216.00

TT $1.00 = EC $0.40

Then: TT $648 = 648 x 0.4 = EC $259.20

38 14 ÷ 15 5

Invert the denominator and multiply ⇒

TT $ 1.00 0.4

Then: US $80 = 80 x 2.7

step3. Divide the numerator by the denominator

38 5 × 15 14

= $80

ii) If US$ 1.00 = EC $2.70

iii) If ⇒

100 1.25

i) If : TT $1.00 = EC $0.40

14 5

=

selling Price 1.25

1. c

step 2. Simplify the denominator so:

=

1.25×Cost Price

19 21

If :

US$ 1.00 = EC $2.70

then:

EC $1.00 =

US $ 1.00 2.7

= US$ 0.37037

so : EC $259.20 = 259.2 x 0.37037 = US$ 96.00

1.b $ Cost Price $ Selling Price $ Profit or Loss

% Profit or Loss

55

44

-11

20% ( loss)

80

100

20

25.00%

i) Loss % =

Cost Price – selling Price cost Price

55−44 ×100 = 20% Loss 55

=


×100

-

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2


2a)

2 c) 3

2

2

i)

2x y +6x y

2

ii)

9x2 −4 =

(3x)2 – (2)2 =

=

2

iii)

4x +8xy – xy−2y

(3x+2)(3x−2)

4x ( x +2y) – y ( x+2y)

=

(4x− y )( x+2y)

2b) from the question on the Past paper Step 1: Simplify the LHS 2x – 3 3

+

step 1: Multiply equation 1 by 2 Multiply equation 2 by 3

2

=

⇒

From the simultaneous equation shown on the past paper

2x y ( x +3y)

⇒ { 3x – 2y = 10 } x 2

eqn1

⇒ { 2x +5y = 13 } x 3

eqn2

⇒

6x – 4y = 20

eqn1

⇒

6x +15y = 39

eqn2

Step 2:

5– x 2

Subtract eqn2 from eqn1

⇒

0 – 19y=−19

⇒

y=1

=

2 (2x – 3)+3(5 – x) 6

=

4x – 6+15 – 3x 6

⇒ 6x – 4(1) = 20

=

x +9 6

⇒

Step 3: Substitute (y = 1) into equation 1 ⇒ 6x = 24 x=4

Step 2: Equate the LHS with the RHS ⇒

x +9 =3 6

⇒

x+9=18

⇒

x=9


-



3


Q 3b) ( i)

Q3. (a)

N

The information given on the Past Paper is used to produce the Venn diagram below:

E

W

36 S

V

P

T

530

x

9x

30 - 9x

PR = 25km 15km

4

Q

R 20km

Where V represents students who play volleyball and T represents students who play Tennis.

The simplified expression is:

x + 30 + 4

iii) the Equation is: ⇒

x + 30+ 4 = 36 x=2

Finding distance PR

ii)

ii) The expression in x that represents the total number of students is: x + 9x + (30 – 9x) + 4

Right angled Triangle so we use pythagoras: where: so:

( PR)2 PR

iii)

Finding

Let

∢QPR =

where so

tan θ θ

= 202 +152 = 625 = √ 625 = 25 km

∢QPR θ

=

20 15

=

tan−1 1.3333

= 1.3333 =

53.1 0

= 530 ( nearest whole number)


-



4


Q4. (a)

Q 4 b)

i) volume of Prism = area x length (20cm) = 28.9×20 = 577.5 cm3

20cm

ii) Mass of prism

= volume x density

Density of tin

= 7.3 kg / cm3

Mass of tin

=

577.5×7.3 = 4,215.75 kg

= 4,216 kg ( nearest kg)

Q5 a Constructing the

600 angle at point Q

0

270 3600

i) The cross section shown represents

or (3 / 4) of a

full circle. So: Length of arc ABC = =

(3/ 4)×2 π r

P2

3 22 ×2× ×3.5 4 7

= 16.5 cm P1

ii) Perimeter OABC

= length ABC + r + r

P

60

0

Q

8 cm

= 16.5 + 3.5 + 3.5 = 23.5cm

iii) Area of sector OABC

= (3/ 4)×π r 2 =

3 22 2 × ×3.5 4 7

=28.9

Steps: 1. Draw a straight line PQ = 8 cm 2. With centre Q, and a suitable compass separation, construct a very wide arc above line PQ. The arc should be wide enough to cut line PQ at point P1 . 3.

cm 2 4.


-


With Centre P 1 , and the same compass separation, construct a second arc above line PQ to intersect the first arc at point P 2 The straigt line through points Q and P 2 will make an angle of 60 0 with line PQ as shown.


5


Q5 a ii) Constructing the 450 angle at point P The 450 is constructed in two steps Step 1: is to construct a 900 Step 2 : is to bisect ( devide) the 900

Bisecting the

90 0 to get a

450

angle at point P R

to get the

450

90 0 angle at point P

Constructing the

P8 P6 450 P2

60

P7

P

0

Q 8 cm

P5

• SR

P1

P3

P4

P

60

0

Q

8 cm

Steps: • Extend line PQ a suitable distance to point S • With center P, and a suitable compass separation, construct two arcs to cut line PQ at P 3 and P4 • With centre P 3 , draw an arc above point P. • With center P_4 draw a second arc above point P to intersct the first arc at P 5 . • The line connecting P and P 5 is 90^0 to PQ


-


• •

•

With centre P and suitable compass separation, construct two arcs to bisect the the two lines shown at P 6 and P 7 With Centre P 7 construct an arc above line PQ With centre P 6 construct a second arc to intersect the first arc at P 8 . 0 The line through P and P 8 is at 45 to line

Complete the triangle PQR • Extend the 450 line and the intersect at point R. • Complete the triangle PQR.

600

line to

Measure Line RQ = 5.8cm


6


Q5 b)

Q6 a

Coordinates of point S = (6, 6) Coordinates of point T = (0, -2) so if : and:

y 2 =6 , and x 2 =6 , and

y1 =− 2 x1 =0

If Line L passes through these two points, then

i) Gradient of L =

y2 – y 1 x 2 −x 1 6−(−2) = (6−0 )

=

8 6

=

4 3

ii) Equation of the line is y = mx + c Using the coordinates of S( 6,6) in the equation (a) Finding the centre of enlargement 4 ( )6+c 3

⇒

6=

⇒ ⇒

6= 8 + c c = -2

so equation is:

y =

Using a ruler and pencil, draw straight line through the the object points and the corresponding image points. That is: ( R to N), (Q to M) and P to L. The Centre of enlargement is found at the intersection of the three line ( 1, 5)

4 x −2 3

iii) The midpoint coordinates ( x m , y m ) of line TS is:

so:

PQ LM

b) The scale factor ( k) =

xm

=

( x 2 +x 1 ) (6+0) = 2 2

= 3

ym

=

( y 2+ y 1 ) (6+(−2)) = 2 2

= 2

c) Area of Image = so:

=

6 3

= 2

k 2×Area of object

Area of image Area of object

=

k2 =

22 = 4

mid point of TS = ( 3 , 2)

iv) Length of TS

=

√( y − y ) +( x −x )

=

√(6−(−2))2+(6−0)2 √(8)2 +(6)2

=

2

2

1

2

2

1

= 10


-



7


d)

y

B ( -1, 4)

A ( -4, 4)

L ( 4,4) P

C ( -1, 2) N(2,1)

M(4,1)

x

O

Q R

The object point LMN are mapped onto ABC as follow: L(4, 4)

→

A(- 4, 4)

M(4, 1) → B(- 1, 4) N(2, 1)

→ A(- 1, 2)

This describes the Transformation R90 : P ( x , y)→ Q(−y , x) which is an anti-clockwise rotation of origin


-

900

about the



8


7a)

c)

Q 2 ) age for the data is the age 1 th n person .. where n= 50 corresponding to the 2

i) The medium (

The median age is therefore the age of the person in the grouped data = 64

© cxcDirect Institute - 876 469-2775 Email: [email protected] website:

25

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th

ii) The Probability that a person visiting the clinic is 75 years or younger : P ( x≤75) =

# persons <= 75 = Total persons

43 50

( This value is read directly from the curve).

9


8 a)

Section 2 - Question 9 To solve the simultaneous equations shown in Q9. Substitute equation 1 into into equation 2 and simplify: that is: y=8–x

( equation1)

By substituting equation 1 into equation 2 :

Fig

Area of Triangle

# Pins on Base

1

1

2 x 1+1 = 3

2

4

2 x 2+1 = 5

3

9

2 x 3+1 = 7

4

16

2 x 4+1 = 9

.. .. 10

100

2 x1 0+1 = 21

.. .. 20

400

2 x 20+1 = 41

n2

2×n+1

.. .. n

then : becomes:

2x 2+ xy=−16 ..... equation 2 2 x 2 + x ( 8 – x ) = -16

⇒

2x 2+8x−x2 =−16

⇒

x2 +8x +16=0

⇒

( x+4 )( x+4)=0

⇒

x=−4

so using equation 1 y = 8 - (-4) = 12 so solution is : x = - 4, y = 12 The significance of this result is that the straight line that is represented by equation1, will just touch the curve which is described by equation 2, when x = - 4. The double root indicates that the line touches the curve at one point only This means that y = 8 – x is a tangent to the curve 2x 2+ xy=−16 . see diagram below

By observation we can see that: and :

twice ( a double root)

−7≤x≤−0.5 ( not to scale ) y

2

Area of the triangle

=

n

# Pins on Base

=

2n+1

2

2x + xy=−16

y=8 – x

x

-4


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10


Question 9 b (i)

iii) Coordinates of the vertices of the shaded region

let # roses = x

Vertex

# roses (x)

# orchids(y)

A

2

10

Description

B

8

4

# orchids must be at least ½ the number of roses

C

2

1

let #orchids = y (i) In equations 1 y≥ x 2

iv) Profit Equation and Maximum profit

x≥2

At least 2 roses

x+ y ≤12

No more than 12 flowers

The proft is $3 for a rose and $4 for an orchid The Profit equation is therefore:

P = 3x + 4y

Maximum profit To find the maximum profit, we must apply the profit equation at each of the three vertices, and choose the one which gives the highest result. From the table shown on the graph below, Vertex A which represents a bouquet of 2 roses and 10 orchids will yield the maximum profit of $46.

(ii) Graph


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11


Q10 a) From the diagram shown n the past paper, we can see that for triangle QRS, we have two sides and a non included angle so we need to apply the sine rule to find RS.

Q10b) Circle Theorem:

Finding RS

a) Finding

The following are observations from the diagram shown on the past paper. ∢OUZ

Q 480

Z

7cm

O 70 0

X

600

S

R

? cm

U

Applying the sine rule to find the side RS RS sin

RS 0 sin 48

⇒

so:

QS sin

=

7 0 sin 60

=

RS =

7×

Note the following: 1)

ZOX is a straight line passing through center O

⇒ so :

sin48 = 6cm sin60

∢ZOU + 70 0 = 1800 (angles on a straight line) ∢ZOU = 180 – 70 = 110

Z 35 0

O 110 0

Finding

∢QTS =

(∢T ) shown shaded T

8cm

70 0

350

Q

U

10cm 7cm

2) Triangle ZOU is isosceles because OZ = OU = radius

⇒ base angles are equal = ½ (180-110) = 350

S

Since we do not know any of the angles in the triangle, we must use the cosine rule. So:

t2

=

s2 +q 2 – 2 s q CosT

⇒

72

=

82 +102 – 2×8×10cos T

⇒

49

=

164−160CosT

⇒

160cost T =

164 – 49

=

115

⇒

cos T

=

115/160

=

0.71875

=

−1

=

440

⇒

∢T


-

cos (0.71875)


so:

∢OUZ = 350


12


b) Finding

c) Finding

∢UVY

∢UWO

shown shaded

Y

O

Z

70 0

35 0

O 90 0

X

W

V

U Tangent

V

U

OUW is a right angled triangle because the angle between the radius OU and the tangent UW = 900

Note the following: 1) Triangle UYV is a right angled triangle because angle

so:

900 +700 +

because angles in a triangle =

Y

hence :

35 0

O

1800 .

∢UWO = 180 – 90 – 70 =

Z 35 0

∢UWO = 1800

200

Q9b (ii) a) Triangle ZOU is congruent to triangle YOX

X 90 0

b) Triangle YXU is congruent to triangle ZUX

550

V

U Tangent

between radius and tangent =

900

2) Angle UYV = UZX 35 0 = because they are subtended from the same chord (UX) . 3) ⇒

∢UVY +90 0 +35 0 = 1800 .( ∢' s in Triangle UVY) ∢UVY


= 550

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13


⃗ Finding Vector B C

Q11 (a) Vectors

B(3,4)

Finding Vector B ⃗ A

b

⃗ OB

B(3,4)

B⃗ A

⃗ OB

⃗ BC A(6, 2)

⃗ OA

O

⃗ OC

O

C(12, - 2)

⃗ OC C(12, - 2)

A Diagram showng the position vectors is shown above. Now B ⃗ A is the vector going from B to A that is: First go from B to O and then from O to A so:

B⃗ A

O ⃗A =

so

() 6 2

⃗ BO

Now

⃗ BO

giving:

⃗ + BO

=

,

O⃗ B=

O⃗ A

() 3 4

( )

, and O C ⃗ = 12 −2 ⃗ OB

( ) ( ) () ( ) −3 −4

+

⃗ i.e : B C

=

so:

=

⃗ BC

where:

is the opposite of vector −3 = −O ⃗ B = −4

B⃗ A =

⃗ , Similarly: to find Vector B C we must go from B to O and then from O to C

6 2

=

3 −2

⃗ BO −3 −4

⃗ OC 12 + −2

+

( ) ( ) ( ) ( ) ( ) ( ) 9 −6

=

(ii) Now we note that :

⃗ = BC

9 −6

which may be written as:

⃗ = BC

3

but: recall that:

B⃗ A =

so we can write

3 −2

3 −2

⃗ in terms of B ⃗ BC A ⃗ = BC

giving :

One geometric relationship is that three times B ⃗ A

3. B ⃗ A ⃗ BC

is equal to

iii) The relative positions of A, B and C are shown below B(3,4)

⃗ OB ⃗ OA

A(6, 2)

O

⃗ OC C(12, - 2)


-



14


Q11b ) Matrices

nb:

To solve for a, and b we must expand the matrix shown on the past paper, to get: 2a - 4 = 2 2 +b=0

⇒ ⇒

Inverse of

Download CSEC Maths Practice workbooks at : www.cxcDirect.org

so a = 3 so b = - 2

(

2 −4 1 −3

) (

2 −4 1 −3

=

−1

)

=

1 2 (−3) – 1 (−4)

=

−1 2

(

−3 4 −1 2

(

−3 4 −1 2

)

)

iii) now to solve the simultaneous equation: we can rewrite using the inverse matrix equation i.e:

(

2 −4 1 −3

) ( ) () 12 7

(

2 −4 1 −3

where:

so

−1

(

−1 −3 4 2 −1 2

x y

=

−1

)

(

−1 −3 4 2 −1 2

=

)

) ( ) () 12 7

x y

=

By expanding the above we get: ( - ½) [( -3 x12) + (4x7) ] = x ( - ½) [( -1 x12) + (2x7) ] = y giving x = (- ½) (-8)

=

4

and

=

-1

y = (- ½) (2)

-END-


-



15

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