Bee Math Tutor CSEC Booklet (May/June 2014)

Bee Math Tutor

PAST CSEC
MATHEMATICS

PAPER II
QUESTIONS & SOLUTIONS

May/June 2014

OVERVIEW OF THE CSEC MATHEMATICS
PAPER II EXAMINATION

The Mathematics Paper II General Proficiency
Examination is a 2 hours and 40 minutes paper. It consists
of two sections:
Section1 consists eight (8) compulsory questions.
Section II consists of three (3) questions. Candidates are
required to do any TWO questions from this section.
Candidates are allowed to use silent non-programmable
calculators. Candidates are also asked to carry along a
Geometry set.
A formulae sheet is provided in the examination which the
candidate can refer to at any time.

1

Volume of V  Ah where A is the area of the cross-
prism section and h is the height or length of the
prism
(Volume of a cylinder = r2h )

Volume of 1 Ah where A is the area of the base and h
pyramid 3

Circumference is the height of the pyramid. (Volume of
of a circle
cone = 1   r 2 h )
3

C  d or 2r where r is the radius (d is the

diameter)

Area of a circle A  r 2

Area of a A 1 a bh where a and b are the
trapezium
2

parallel side and h is the perpendicular

distance between the parallel sides.

Root of a If ax2  bx  c  0 , then
quadratic
equation − ± √2 − 4
= 2

2

Trigonometric sin  opposite Hypotenuse
ratios
hypotenuse Opposite
cos  adjacent θ

hypotenuse Adjacent

tan  opposite

adjacent

Area of 1. = 1 ℎ, where b is the base length
triangle 2

Sine Rule and h is the perpendicular height.

2. A  1 absinC
2

3. A  s(s  a)(s  b)(s  c)

s  a b c
2
where

B

abc a
sin A sin B sinC c

A C

a 2  b2  c2  2bcCosA b

Cosine Rule

3

MAY/JUNE 2014

4

SECTION I

Answer ALL questions in this section

All working must be clearly shown.

1. (a) Using a calculator, or otherwise, determine the

value of

(i) 5.25 ÷ 0.015 (1 mark)

(ii) 6.5025 (1 mark)

(iii) 3.142 x 2.2362 (correct to 3 significant

figures). (2 marks)

SOLUTIONS

(i) 5.25 ÷ 0.015 = 5250 ÷ 15 = 350

(ii)
(iii) 3.142 x 2.2362 = 3.142 x 4.999696 = 15.7 (3 s.f.)

(b) Concrete tiles are made using buckets of cement, sand
and gravel in the ratio 1 : 4 : 6.

(i) How many buckets of gravel are needed for 4

buckets of cement? (2 marks)

(ii) If 20 buckets of sand are used, how many buckets

of each of the following will be needed?

a) Cement

b) Gravel

(3 marks)

5

SOLUTIONS

(i) Ratio of cement to gravel = 1 : 6.
Therefore number of buckets of gravel needed

for 4 buckets of cement =
(ii) (a) Ratio of cement to sand = 1 : 4.

Therefore number of buckets of cement needed

for 20 buckets of sand =
(iii) (b) Ratio of sand to gravel = 4 : 6 or 2 : 3.

Therefore number of buckets of gravel needed

for 20 buckets of sand =

(c) The cash price of a laptop is $1299. It can be bought
on hire purchase by making a deposit of $350 and 10
monthly instalments of $120 each.

(i) What is the TOTAL hire purchase price of the

laptop? (2 marks)

(ii) How much is saved by buying the laptop for

cash? (1 mark)

(i) H.P. price = $350 + 10 x $120 = $350 + $1200 = $1550
(ii) Amount saved =$1550 - $1299 = $251

6

2. (a) Write as a single fraction in its LOWEST terms.

x 2  x 1 (3 marks)
34

(b) Write an equation in x to represent EACH statement
below. DO NOT solve the equation.
(i) When 4 is added to a certain number, the result
is the same as halving the number and adding
10. (1 mark)
(ii) Squaring a number and subtracting 6 gives the
same result as doubling the number and adding
9. (1 mark)

(i)
(ii) x2 – 6 = 2x + 9

7

(c) John drew the diagram below to show what he was
thinking.

(i) Use the information from the diagram to write a

formula for y in terms of x. (1 mark)

(ii) If the number 4 is the input, what number would be

the output? (1 mark)

(iii) If the number 8 was the output, what number was

the input? (1 mark)

(iv) Reverse the formula written at (c)(i) above to write

x in terms of y. (1 mark)

SOLUTIONSs

(i) y = 3x + 5
(ii) y = 3(4) + 5 = 12 + 5 = 17
(iii) 8 = 3x + 5

8 – 5 = 3x
3 = 3x
1=x

(iv) y = 3x + 5
y – 5 = 3x

8

(d) Solve the following simultaneous equations

2x + 3y = 9 (3 marks)
3x – y = 8

2x + 3y = 9 (equa. 1)
3x – y = 8 (equa. 2)
Substitution method:
By rearranging equa. 2 to make y the subject we get:

y = 3x – 8
Substituting equa. 2 into equa. 1 we get:

2x + 3(3x – 8) = 9
2x + 9x – 24 = 9

11x = 33
x=3

Substitute x = 3 into the rearranged equa. 2 we get:
y = 3(3) – 8 = 9 – 8 = 1

Therefore x = 3 and y = 1

9

3. (a) The universal set, U, is defined as the set of integers
between 11 and 26. A and B are subsets of U such that:
A = {even numbers}
B = {multiples of 3}

(i) How many members are in the universal set, U?

(1 mark)

(ii) List the members of the subset A. (1 mark)

(iii) List the members of the subset B. (1 mark)

(iv) Draw a Venn diagram to show the relationships

among A, B and U. (3 marks)

(i) U = {12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23,
24, 25}. Therefore n(U) = 14

(ii) A = {12, 14, 16, 18, 20, 22, 24}
(iii) B = {12, 15, 18, 21, 24}
(iv)

10

(b) (i) Using a ruler, a pencil and a pair of compasses, construct

(a) a triangle PQR in which PQ = 8 cm, PR = 6 cm and

angle P = 60o. (3 marks)

(b) the line segment RX which is perpendicular to PQ and

meets PQ at X. (2 marks)

[Note{ Credit will be given to clearly constructed lines]

(ii) Measure and state the size of angle QRX. (1 mark)

(i) (a) & (b)

(ii) 44o

11

4. The diagram below shows the map of an island drawn on
a grid of 1-cm squares. The map is drawn to a scale of
1:150,000.

(a) Copy and complete EACH of the following

sentences:

(i) 1 cm on the map represents __________ cm

on the island. (1 mark)

12

(ii) An area of 1 cm2 on the map represents an

area of _______ cm2 on the island. (1 mark)

(iii) Given that 1 km = 100,000 cm, a distance of

1 cm on the map represents a distance of

________ km on the island. (1 mark)

(i) 1 cm on the map represents 150,000 cm on the
island.

(ii) An area of 1 cm2 on the map represents an area of
2.25 x 1010 cm2 on the island.
NB. Ratio of area = (1 cm)2 : (150,000 cm)2
= 1 cm2 : 22,500,000,000 cm2

(iii) Given that 1 km = 100,000 cm, a distance of 1 cm on
the map represents a distance of 1.5 km on the island.

NB. 1 cm represents 150,000 cm on island.

and 150,000 cm = (150,000/100,000) km = 1.5 km

(b)(i) L and M are two tracking stations. State, in

centimetres, the distance LM on the map.

(1 mark)

(ii) Calculate the ACTUAL distance, in kilometre,

from L to M on the island. (2 marks)

13

(i) 8 cm

(ii) 8 x 150,000 cm = 1,200,000 cm
(NB. 1 km = 100,000 cm)
Therefore 1,200,000 cm = (1,200,000/100,000) km
= 12 km

(c) (i) The area shaded on the map is a forested reserve.

By counting the squares estimate, in cm2, find

the area of the forest reserve as shown on the

map. (2 marks)

(ii) Calculate, in km2, the ACTUAL area of the

forest reserve. (2 marks)

(i) 15 cm2
(ii) Ratio of area = (1 cm)2 : (1.5 km)2

= 1 cm2 : 2.25 km2
Therefore actual area = 15 x 2.25 km2

= 33.75 km2

14

5. (a) Triangle ABC has coordinates A(1, 2), B(4, 2) and
C(1, 0).
(i) On the answer sheet provided, draw
triangle A’B’C’, the image of triangle ABC,
under enlargement, center O and scale factor
2. (3 marks)
(ii) Triangle A”B”C” is the image of triangle
ABC, under transformation, M. Describe
completely the transformation M. (3 marks)

15

(i)

(ii)

16

(b) The diagram below, not drawn to scale, shows the
position of two ships, P and Q, at anchor. FT is a
vertical face of a cliff jutting out of the water. P and Q
are 118 m apart. FT = 80 m and angle FPT = 40o.

Determine:

(i) the angle of elevation of T from P. (1 mark)

(ii) the length of FP. (2 marks)

(iii) the angle of elevation of T from Q. (3 marks)

(i) 40o
(ii)
(iii) Length FQ = 95.34 m + 118 m = 213.34 m

Angle of elevation of T from Q = 20.6o

17

6. The graph of the quadratic function y = x2 for -4 ≤ x ≤ 4
is shown below.

(a) The coordinates of M and N are (-1, y) and (x, 9)

respectively. Determine the value of

(i) x (1 mark)

(ii) y (1 mark)

(i) y = (-1)2 = 1
(ii) 9 = x2, therefore x = the square root of 9 = 3

18

(b) Determine

(i) the gradient of the line MN. (1 mark)

(ii) the equation of the line MN. (2 marks)

(iii) the equation of the line parallel to MN, and

passing through the origin. (2 marks)

(i)

(ii) Using gradient m = 2 and point M(-1, 1)…
y - yp = m(x – xp) or y = m(x – xp) + yp
y = 2(x - -1) + 1
y = 2x + 2 + 1
y = 2x + 3

(iii) Using gradient m = 2 (because gradient are
parallel) and the origin (0, 0)
y = 2(x – 0) + 0
y = 2x

(c) (i) On the answer sheet provided, carefully draw the
tangent line to the graph y = x2 at the point (2, 4).
(2 marks)

(ii) Estimate the gradient of the tangent to the curve at

(2, 4) (2 marks)

19

(i)

20

7. A class of 30 students counted the number of books in
their bags on a certain day. The number of books in
EACH bag is shown below.
5463217453
6543762545
5754321634

(a) Copy and complete the frequency table for the data
shown above.

Number of Tally Frequency fxx
books (x) (f)
II 2 2
1 III 6
2 3
3
4
5
6
7

(4 marks)

(b) State the modal number of books in the bags of the

sample students. (1 mark)

(c) Using the table in (a) above, or otherwise, calculate

(i) the TOTAL number of books. (2 marks)

(ii) the mean number of books per bag.(2 marks)

(d) Determine the probability that a student chosen at

random has LESS THAN 4 books in his/her bag.

(2 marks)

21

(a) . Tally Frequency fxx
(f)
Number of II 2 2
books (x) III 6
IIII 3 15
1 IIII I 24
2 IIII II 5 35
3 IIII 6 24
4 III 7 21
5 4 Ʃfx = 127
6 3
7 Ʃf = 30

(b) Modal number of books = 5 books
(c) (i) Total # of books = 2 + 6 + 15 + 24 + 35 + 24 + 21

= 127 books

22

8. A number sequence may be formed by counting the
number of dots used to draw each of a set of geometric
figures. The first three figures are shown below.

(a) Draw figure 4, the next figure in the sequence.
(2 mrks)

(b) Complete the table below by inserting the missing
information at the rows numbered (i) and (ii). (4 mrks)

Figure (f) Total number of dots

1 Formula Number (n)
2
3 5x2-5 5
4
(i) 5 5x3-5 10
(ii) 6
5x4-5 15

Not required Not required

(c) Write an expression in f for the number (n) of dots

used in drawing the fth figure. (2 marks)

(d) Which figure in the sequence contains 145 dots?

(2 marks)

23

(a) .

(b) .

Figure Total number of dots
(f)
1 Formula Number (n)
2
3 5x2-5 5
4
5x3-5 10
(i) 5
(ii) 6 5x4-5 15

Not required Not required

5x6-5 25

5x7-5 30

(c) n = 5 x (f + 1) – 5
(d) 145 = 5f + 5 – 5

145 = 5f
Therefore f = 29. Hence the 29th figure has 145 dots.

24

SECTION II

Answer TWO questions in this section.

ALGEBRA AND RELATIONS, FUNCTIONS AND
GRAPHS

9. (a) Two functions are defined as follows:

g(x) = 4x + 3 f (x)  2x  7
x 1

(i) State the value of x for which f(x) is undefined.

(ii) Calculate the value of gf(5).

(iii) Find f -1(x). (7 marks)

(i) f(x) is undefined when x + 1 = 0. Therefore x = -1

25

(b) A ball is thrown vertically upwards. Its height, h
metres, above the ground after t seconds is shown in the
table below.

t(s) 0123456
h(m)
0 50 80 90 80 50 0
(i)
Using 2 cm to represent 1 second on the x-
(ii)
axis and 1 cm to represent 10 metres on the

y-axis, plot the graph to show the height of the

ball during the first 6 seconds. (4 marks)

Using your graph, determine

a) The average speed of the ball during the first

2 seconds. (2 marks)

b) The speed of the ball when t = 3s. (2 marks)

(i)

26

(b) Since the gradient of the curve at 3 seconds is 0,
then the speed at 3 seconds is 0 m/s

27

MEASUREMENT, GEOMETRY AND
TRIGONOMETRY

10. (a) The diagram below, not drawn to scale, shows a
circle with centre O. The line BC is a tangent to the
circle at B. Angle CBD = 42o, and angle
OBE = 20o.

Calculate, giving a reason for EACH step of your
answer, the measure of

(i) ∠BOE (2 marks)
(ii) ∠OED (2 marks)
(iii) ∠BFE (3 marks)

28

(i) ∠BOE = 180o – (20o + 20o) = 180o – 40o = 140o
Reason: Angles within a triangle add to give 180o.

(ii) ∠OED = 42o – 20o = 22o
Reasons:
1. The angle between a tangent (BC) and the chord
(BD) is equal to the angle that the chord (BD)
subtends to the circumference in the opposite
segment (i.e. ∠BED). Therefore ∠BED = 42o.
2. Since triangle BOE is an isosceles triangle, then
∠BEO = 20o.

(iii) ∠BFE = 180o – 70o = 110o
Reasons:
1. ∠BDE is half of ∠BOE = 140o ÷ 2 = 70o.
2. Angles BFE (180o – 70o) and BDE (70o) are
supplementary angles.

29

(b) The diagram below, not drawn to scale, shows the
positions of three ports P, Q and R.

Q is 80 km from P. R is 100 km from Q on a bearing

of 066o. ∠PQR = 54o.

Calculate:

(i) the bearing of P from Q. (2 marks)

(ii) the distance PR correct to 2 decimal places.

(3 marks)

(iii) the measure of ∠QPR to the nearest degree.

(3 marks)

30

(i) Bearing of P from Q = 066o + 54o = 120o

(ii) PR2 = 802 + 1002 - 2 x 80 x 100 x cos54o
PR2 = 6400 + 10000 – 9404.56
PR2 = 6995.44

PR = 83.64 km

(iii) sinP  sin54o  sinP  100sin54o
100 83.64 83.64

P  sin1  100 sin 54o  75o
 83.64 
 

31

VECTORS AND MATRICES

11. (a) The matrix M is defined as

M   7 2
 p 1


Determine the value of p for which the matrix M

does NOT have an inverse. (2 marks)

If matrix M does not have an inverse, then the
determinant is 0.

Therefore 7(-1) – 2(p) = 0

-7 – 2p = 0

-7 = 2p

p = -3.5

(b) Express the equations
4x – 2y = 0
2x + 3y = 4
in the form AX = B, where A, X and B are matrices.
(3 marks)

32

(c) In the diagram below, the coordinates of P and Q are
(2, 4) and (8, 2) respectively. The line segment joining

the origin (0, 0) to the point P may be written as OP .

(i) What term is used to describe OP ? (2 mark)

a
 
(ii) Write each of the following in the form  b 

a) OP (1 mark)

b) OQ (1 mark)

c) PQ (2 marks)

(iii) Given that OP = RQ , determine the

coordinates of the point R. (3 marks)

(iv) State the type of Quadrilateral formed by

PQRO. Justify your answer. (2 marks)

33

(i)
(ii)

(iii)

(iv) PQRO is a parallelogram, because OP and RQ
(opposite sides) are parallel and equal, hence
the other two sides are also parallel and equal.

34



Notes

36