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CSEC MATHEMATICS Past Paper Solution – May 2015 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)

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

(ii)

Q1a 1. May 2015

Q2.b(i)

Method 1: convert to improper fractions and solve : 12 4 7 72 40 105 − + – + = 5 3 2 30 30 30 137 = 30

=

4

17 30

=

so sum

=

4

2a b

= 2×4 2

=

= 32

(i) If P ml is poured from 500ml then remainder is: (500 – P)ml (ii) if q glasses of r ml is poured then: therefore remainder is :

(qr) ml (500−qr )ml

Q2.c

Whole numbers : ( 2 – 1 + 3) = 4 12 – 10+15 30

a–b+c=4-2–1=1

(i)

total poured is

Method2: Separate whole numbers and fractions: 2 1 1 – + Fractions : 5 3 2

if a = 4, b = 2 and c = -1 then

17 30

17 30

Using an LCD of 15, the expression may be re-written as: 10k – 3(2 – k ) = 15

10k – 6+3k 13k – 6 = 15 15

Q2.d (i) The pair of equations based on the information is:

4.14÷5.75 1.622 sum

Q1a(ii).

= 0.72 = 2.6244 = 3.3444

1) 2) (ii)

Q1a(iii).

2×3.142×1.25

= 7.855

Q1a(iv).

√ 2.89

= 1.7 =1 = 1.7

tan 45 0 1.7×1

so

(i) Taking a as a factor we get a (a 2 −12) (ii) From the quadratic expression shown on the past paper: a = 2, b = -5 , and c = 3

Item

Unit cost

Price ($)

(i)

3kg sugar

$3.60

$10.80

(ii)

4kg rice

(3.60 + 0.80) = 4.40

$17.60

$1.60

$3.20

2kg flour

(iii)

where: x is the cost of 1 mango, and y is the cost of 1 pear:

Q2. e Factorization

Q1b Item

4x + 2y = 24 2x + 3y = 16

p = −3

then:

Total before Tax

$31.60

Tax (10%)

$3.16

Total including Tax

$34.76

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now if two numbers p & q are such that: p×q = ac = 6 p+q = b = −5 and q = −2

The quadratic expression may then be re-written as: 2x 2−2x−3x+3 This expresion can factorized by grouping: that is: giving:

Email: [email protected]

and

2x( x −1) – 3( x−1) (2x−3)(x −1)

website: www.cxcDirect.org

2


Q3 b i) Construction details for Triangle ABC

Q3a)

Note that the question require that you use compass, ruler and pencil ONLY to contruct the triangle.

12 students play neither the guitar nor the violin

i)

ii) Expression for total # students: iii) Equation for total # students : ⇒

3x+16=40

⇒

3x=24

2x+x+4+12 2x+x+4+12=40

Steps:

iv) 16 + 8 = 24 students play the guitar

1. 2.

3.

4.

5. 6.

Draw a straight horizontal line AB= 9cm Extend line AB , and with centre B and suitable compass separation, construct two arcs to cut the horizontal line at p1 and p2. Now with centre p1 and suitable compass separation, construct two arcs above and below the horizontal line. With centre p2, and the same compass separation, construct two additional arcs to intersect the previous arcs at p3 and p4. Draw line CB = 6cm which passes through points p3 and p4. Draw line AC to complete the triangle ABC.

ii) Measured angle

∢BAC=41.8 0

iii) – Point D is shown such that ABCD is a parallelogram




3


Q 4:

4. d) i)

The values of x for which the function is zero are x = 3, and x = - 1

ii)

The minimum value of the function is – 4

iii)

The equation of the line of symmetry is x = 1




4


5a)

Q6 a) – Measurements

For this question, only one equation need be distance remembered, and that is : Speed = time

For a cylindrical tank of radius r and height h Volume of tank = π×r 2 ×h

The other equations for distance and time can be found by transposing the speed equation. i) if speed is 54 km/hr and time is 2.25 hrs then

Distance

= =

speed x time 54 x 2.25 = 121.5 km

ii) Converting speed to m/s.. 54×5 54km/hr = = 15m/s 18 so time to travel =

5b) i)

distance speed

for tank B ,

ii) now area of tank A =

π r2

so radius (r)

√ πA .... transposition √ 314π = 10m

ie

=

iii) Volume of tank A

=

8×Vol tank B

=

8×251.2 = 2009.6m 3

5 c) Now a scale of 1: 2000 , means that 1cm on the map represents an actual distance of 2000cm = 20m

so height of tank A

area PQRS =

2009.6 2 π×10

=

6.4m

From the past paper , we can see that:

12+6 = 18cm 2

iv) Now if 1cm represents 20m then 1cm2 represents 20m x 20m =

400m2

18cm2 represents an actual area of 18×400 = 324m 2

=

6 = 2 3

iii) P'Q' = k PQ = 2 √ 13 iv) The area of P'Q'R is where area PQR =


P'R' PR

ii) The object and the image are on opposite sides of the center of enlargement so the scale factor k will be negative.

324m 2


=

Q6 b) – Transformation ( Enlargement)

Actual distance = 6 x 20m = 120m = 0.12km

so area PQRS =

V 2 π×r

h =

i) Size of scale factor k =

hence:

π×r 2 ×h

now if volume of tank V = then height

ii)

= r

ii) 2: 6 is the same as 1:3 therefore:1cm =3m represents a scale of 1 : 300

From the map of the past paper: i) distance QR = 6cm

8 = 4m 2

so: Volume of water in tank B = 3.14×4×4×5 = 251.2m3

315 = = 21 sec 15

A scale for 1 millimeter = meter is 1:1000

r=

h = 5m and

3×2 2

k 2 or 4 times the area of PQR =


3 units 2 5


Q7.a)

Q-8 a) Fourth pattern in the sequence:

(Sales in thousands of $) Month Jul Aug Sep Oct Nov Dec Total Sales

13

20

36

25

15

21

130

i) Greatest increase in sales occured between August and

September.

b)

ii) Smallest increase in sales occurred between July and August. iii) The slope of the line on the graph indicates the rate of the increase or the rate of decrease in sales. Once the slope is positive, the steeper ( greater) slope will indicate the greater increase in sales.

Fig

# Triangles

# unit sides

4

16

30

12

144

234

25

625

975

n

n2

(3n) (n+1) / 2

c) Now sales for period Jul to Nov Average ( mean) monthly sales is:

d i) Sales for period Jul – Dec

= $109,000 109,000 = 5 = $21,800 = $130,000

so sales for month of Dec = 130,000 – 109,000 = $21,000

i) If n is the figure number , and n = 4, then # triangles = 4 2 = 16 =

ii) If # triangles

= 144.

then n =

ii)

(3×4)(4+1) = 30 2

and # sides

√ 144

and # sides =

= 12

(3×12)(12+1) = 234 2

iii) If n = 25, then # Triangles = and # sides =

(3×25)(25+1) 2

25 2

= 625 = 975

iv) From the table shown on the past paper: The number of triangles is the square of the figure number. ie: # triangles = n 2 where n is the figure number The number of sides can be derived form the equation: (3×n)(n+1) # sides = where n is the figure number 2




6


SECTION II Q9 a (i)

i) 60 marks =

60 ×100 120

ii) 120 marks =

120×100 =100% 120

= 50%

(ii) Finding:

f

−1

step1: y = 3x + 2 step2: x = 3y + 2 x–2 step3: y = 3 f

−1

( x) =

.. replace f(x) with y .. exchange x with y .. make y the subject x−2 3

iii) From graph, 95 marks corresponds to 79%

result:

iv) From the graph, the minimum mark needed to be awarded a Grade A ( 85% or more) , is 102 marks.

(ii) Finding: gf(x) gf(x)

9b- Functions : Given f(x) and g(x) as shown on the past paper. i) g(5) =

(5)2 – 1 = 3


24 3

=8


( x)

⇒

put f into g

=

(3x +2)2 – 1 3

=

9x2 +12x+4 – 1 3

=

3x2 +4x+1

=

3( x+1 /3) ( x+1)


=

(3x+1)( x+1)

7


10.a 10 b - Bearings P

L

12m 400

N

T

80km

0400 K

120km

81.50 78km

42o

M

35o

B

60m

R

(ii) Finding

∢KLM

Now since point M is due south of L, then lines KN and ML are parallel. now BP = and BT =

So

0

60 tan 42 60 tan 350

=

∢NKL

∢KLM

=

400

.. alternate angles

(iii) Finding Length KM If TP represents the length of the flagpole then:

Using the cosine rule: ( KM )2 =802 +1202 −2×80×120×cos 400 giving KM = 78km

TP

= BP - BT = 60 tan 42 0 - 60 tan 35 0 = 54−42 = 12m

(iv) Finding

∢LKM

Using the sine rule:

78 120 0 = sinK sin40 so

sinK =

giving K =

120 ×sin40 78

∢LKM =

= 0.9889

sin−1 (0.09889)

=

81.50

(iv) Finding Bearing of M from K ∢NKM Bearing of M from K =



40+81.5 =


121.50

8


Q11. A )

which may be written as

(i) AB =

( ) ( ) ( )

(ii) BA =

( ) ( ) ( )

so AB

1 1 2 3

1 2 0 1

≠

A

where: and

so

1 2

=

1 3

R⃗ S = O⃗ S - O⃗ R

1 3 2 7

5 2

=

−1

=

7 3

now

(

(

is the vector going from S to T

S T⃗

where but

S T⃗ ⃗ SO

⃗ + O T⃗ = SO = −O ⃗S

giving

S T⃗

=

S T⃗ =

)

3 −1 −2 1

−O ⃗ S + O T⃗

( ) - () = ( ) 5 −2

1 1

4 −3

ii) Vectors RS and ST are parallel because

R⃗ S

)

3 −1 −2 1

=

S T⃗

RST is a straight line because the two parallel vectors

R⃗ S and

(iv) Finding x note that Determinant M can be written as So if ∣M ∣ = 0 then giving

4 −3

−3 4

S T⃗ = O T⃗ - O ⃗ S

Determinant A = (3 x 1 – 2 x 1) = 1

A−1 =

1 1

which may be written as

Adjoint A Determinant A

Adjoint A =

() - ( ) = ( )

=

BA

now

(iii)

1 2 0 1

S T⃗

share a common point S as seen below.

∣M ∣ R

2x×3 – 2×9=0 x=3

S

Q11. b O

( )

−3 O⃗ R= 4

a) If (i)

∣O ⃗ R∣ =

, then :

√(−3)2 +4 2

T

= 5

ii) now

R⃗ S

where but

R⃗ S ⃗ RO

⃗ + O⃗ = RO S ⃗ = −O R

so

R⃗ S

=

is the vector going from R to S ----END---

−O ⃗ R + O⃗ S




9

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