CSEC Model Exam 1 Papers

CSEC Model Exam 1 Paper 2
CSEC MODEL EXAMINATIONS

CSEC MODEL EXAMINATION 1
MATHEMATICS
Paper 1
90 minutes
Answer ALL the questions
NEXT

1. The decimal fraction 0.625 written as a common
fraction, in its lowest terms, is
ANSWER
(C)(A) (D)(B)
Model Exam 1 Paper 1

1. Multiply the decimal fraction with three
decimal places by , which is 1, to make
the decimal fraction a common fraction.
Divide both the numerator and the
denominator by their common factor 25.
This is the common fraction written in its
lowest terms.

2. The number 8.150 46 written correct to 3 decimal
places is
ANSWER
(C) 8.151(A) 8.149 (D) 8.152(B) 8.150

2. 8.150 46 = 8.150 4 6
= 8.150
The digit in the 4th decimal place is 4, which is
less than 5, so we do not add 1 to the digit in the
3rd decimal place. The digit in the 3rd decimal
place remains unchanged.

3. The exact value of 0.615 0.07 is
ANSWER
(C) 4.305
(A) 0.043 05
(D) 43.05
(B) 0.430 5

3.
0.615 0.07 = 0.043 05
3 dp + 2 dp = 5 dp

4. The exact value of 7 (0.01)2 is
ANSWER
(C) 7 000
(A) 0.000 7
(D) 70 000
(B) 0.000 07

4.
Invert the product of fractions which is the divisor and
multiply instead of divide.

5. If $450 is divided into two portions in the ratio
4:5, then the smaller portion is
ANSWER
(C) $200(A) $50 (D) $250(B) $150

5. The number of equal parts = 4 + 5 = 9
the smaller portion = of $450
= $450
= 4 $50
= $200

6. If 40% of a number is $70, what is the number?
ANSWER
(C) $175(A) $110 (D) $200(B) $150

7. What is the least number of cherries that can be
shared equally among 5, 10 or 15 children?
ANSWER
(C) 60(A) 30 (D) 75(B) 45

7. 2 5, 10, 15
3 5, 5, 15
5 5, 5, 5
1, 1, 1
LCM = 2 3 5 = 30
The answer to this question is the LCM of 5, 10
and 15.

8. What is the greatest number that can divide
exactly into 12, 20 and 60?
ANSWER
(C) 6(A) 2 (D) 8(B) 4

8. 2 12, 20, 60
2 6, 10, 30
3, 5, 15
HCF = 2 2 = 4
The answer to this question is the HCF of 12,
20 and 60.
2 is a common factor of the
three numbers.
three numbers.

9. The exact value of 85 104 is
ANSWER
(C) (85 100) (85 4)
(A) (85 100) + 4
(D) (85 100) + (85 4)
(B) (85 100) – (85 4)

9. 85 104 = 85 (100 + 4)
= 85 100 + 85 4
= (85 100) + (85 4)
Using the
distributive
law.

10. The value of the digit 3 in 736.2 is
ANSWER
(C) 3 tens
(A) 3 tenths
(D) 3 hundreds
(B) 3 ones

10. Hundreds Tens Ones Tenths
7 3 6 2
The value of the digit 3 in 736.2 is 3 tens.

11. The simple interest earned on $600 at 5% per
annum for 3 years is given by
ANSWER
(C)
(A)
(D)
(B)

11. The simple interest,
P = $600
R = 5 %
T = 3 years

12. A woman bought a sheep for $800 and sold it
for $1200. Her gain as a percentage of the cost
price is
ANSWER
(C) 45%(A) (D) 50%(B) 40%

13. An insurance salesman is paid 4% of his sales as
commission. His sales for March were $5 025.
How much commission was he paid?
ANSWER
(C) $201.00
(A) $50.25
(D) $402.00
(B) $100.50

14. If the simple interest on $700 for 4 years is $168,
then the rate of interest per annum is
ANSWER
(C) 7%(A) 5% (D) 8%(B) 6%

14. The rate of interest per annum,
I = $168
P = $700
T = 4 years

15. The exchange rate for one United States
dollar (US $1.00) is six dollars and thirty-
four cents in Trinidad and Tobago currency
(TT $6.34). What is the value of US $50 in
TT currency?
ANSWER
(C) $264(A) $634 (D) $134(B) $317

16. A sales tax of 10% is charged on an article. How
much does a customer pay for an article marked
as $75?
ANSWER
(C) $80.00
(A) $82.50
(D) $79.50
(B) $82.00

17. Samuel invested $800 for 3 years at 5% per
annum. Marina invested $600 at the same rate.
If they both earned the same sum as simple
interest, how many years did Marina invest her
money?
ANSWER
(C) 4(A) 6 (D) 3(B) 5

17. Samuel‘s interest,
P = $800
R = 5%
T = 3 years
The time,
I = $120
P = $600
R = 5%

18. A discount of of the marked price is offered
for cash. What is the discount on a dress with a
marked price of $150?
ANSWER
(C) $37.50
(A) $25.00
(D) $40.50
(B) $30.00

19. If P = {2, 3, 5, 7, 9}, Q = {3, 7, 8} and
S = {7, 8, 9}, then
ANSWER
(C) {7}
(A) { }
(D) {2, 3, 5, 7, 8, 9}
(B) {2}

20. U = {integers}
N = {natural numbers}
Zn = {negative integers}
Which of the Venn diagrams given below
illustrates the statement:

―No natural numbers are negative integers‖?
ANSWER
(C)
(A)
(D)
(B)

20. No natural numbers are negative integers:

21.
ANSWER
(C) 6(A) 2 (D) 8(B) 4
In the Venn diagram shown above, n(L) = 8,
n(M) = 10 and
What is

22.
ANSWER
The two circles above represent set X and set Y.
If X = {factor of 8} and Y = {factor of 12}, then
the shaded region represents
(C) {2, 4, 6, 8}
(A) { }
(D) {4, 6, 8, 12}
(B) {1, 2, 4}

23. The scale on a map is stated as 1:500 000. The
distance between two towns as measured on
the map is 1.8 cm. What is the actual distance
between the two towns?
ANSWER
(C) 9.0 km
(A) 0.9 km
(D) 50 km
(B) 1.8 km

24. The number of kilometres travelled by a vehicle
in t hours at a rate of s km per hour is
ANSWER
(C)(A) (D)(B) st

24. Speed,
Distance, d = st
Formula
Multiply both sides by t.
d is the subject of the formula.

25. A cuboid with dimensions 12 cm, 10 cm and
5 cm occupies space of volume
ANSWER
(C) 81 cm3
(A) 27 cm3
(D) 600 cm3
(B) 54 cm3

25. The volume of the cuboid,
The formula for
the volume of a
cuboid.

26. A cylindrical block of cheese 8 cm thick has a
volume of 500 cm3. A student cuts a uniform
slice of 2 cm thickness. What volume of the
cheese did the student take?
ANSWER
(C) 100 cm3
(A) 50 cm3
(D) 125 cm3
(B) 75 cm3

27.
ANSWER
The figure above, not drawn to scale, shows the
sector of a circle with centre O. The length of the
minor arc PQ is 7 cm. The length of the
circumference of the circle is
(C) 56 cm
(A) 21 cm
(D) 63 cm
(B) 28 cm

28. The distance around the edge of a circular table
top is 352 cm. The radius of the table top, in
centimetres, is
ANSWER
(C)(A) 88 (D)(B) 352

29. A plane left Guyana at 21:00 h. The next
day, the plane arrived at its destination in the
same time zone at 02:30 h. How many hours did
the flight take?
ANSWER
(C)(A) (D)(B) 3

30. An aircraft leaves airport A at 07:30 h and arrives
at airport B at 12:30 h, the same day, in the same
time zone. The distance between the two airports
is 3 600 kilometres. What was the average speed
of the aircraft for the flight?
ANSWER
(C) 480 km/h
(A) 180 km/h
(D) 720 km/h
(B) 288 km/h

31. Each of the letters of the word ‗PERFORM‘ is
written on a piece of paper. One piece of paper
is drawn at random. What is the probability that
a letter ‗R‘ is drawn?
ANSWER
(C)(A) (D)(B)

32. A die is tossed twice. What is the probability
that a ‗2‘ followed by an odd number turns up?
ANSWER
(C)(A) (D)(B)

33.
ANSWER
The bar chart shows the number of students who liked one
of five stated colours. How many students took part in the
survey?
(C) 80(A) 5 (D) 125(B) 45

33. The number of students = 20 + 10 + 5 + 20 + 25
= 80

34.
ANSWER
The pie-chart above represents the fruit a group
of students ate. If 16 students ate mandarin, then
the total number of students in the group is
(C) 128(A) 125 (D) 135(B) 45

35. The lowest weekly wage of a group of employees
is $520.60. What is the wage of the highest paid
employee, if the range of the wages is $63.20?
ANSWER
(C) $520.60
(A) $63.20
(D) $583.80
(B) $457.40

35. The range = The highest weekly wage −
The lowest weekly wage
$520.60 = The highest weekly wage − $63.20
The highest weekly wage = $520.60 + $63.20
= $583.80

36.
ANSWER(C) 8.5 and 14.5
(A) 0 and 2.5
(D) 14.5 and 20.5
(B) 2.5 and 8.5
The lengths of the pencils of 40 students were measured,
to the nearest cm, and the information collected is shown
in the frequency table above.
The least and greatest length of the class interval 15–20
are
Frequency 9 17 14
Length of pencil (cm) 3–8 9–14 15–20

37. (5a)2 =
ANSWER
(C) 10a2(A) 10a (D) 25a2(B) 25a

37. Meaning of a square.
Expanding the term.
Grouping like values.
Multiplying like values.
Simplifying.

38. 2x3 3x2 =
ANSWER
(C) 6x6(A) 5x5 (D) 36x5(B) 6x5

38.
Expanding the term.
Grouping like values.
Multiplying like values.
Adding the indices.
Simplifying.

39. ( 6a) ( 3b) =
ANSWER
(C) 9ab(A) 9ab (D) 18ab(B) 18ab

39.
Expanding the
term.
Grouping like
values.
Multiplying like
values.
Simplifying.

40. 4(3x y) − 2(5y 3x) =
ANSWER
(C) 6x 14y
(A) 18x 14y
(D) 18x 6y
(B) 8x 4y

40. 4(3x y) 2(5y 3x)
= 4 3x + 4 ( y)
2 5y 2 ( 3x)
= 12x 4y 10y 6x
= 12x 6x 4y 10y
= 18x 14y
Using the distributive
law.
Simplifying each term.
Grouping like terms.
Adding like terms.

41. If
ANSWER
(C) 10(A) (D) 3(B)

41. Substituting the value for
p and for q in the formula.
Simplifying the two
terms.
Subtracting.

42. If 40 − 3x = x + 8, then x =
ANSWER
(C) 8(A) 4 (D) 29(B) 8

42.
Grouping like terms.
Adding like terms.
Dividing both sides by −4.
Simplifying.

43.
ANSWER
(C)
(A)
(D)
(B)

43. Use the distributive
law to remove the
brackets.
Adding the middle
terms.

44. If
ANSWER
(C)(A) 5 (D)(B) 5

44.
Substituting the value for v in the
formula.
Squaring and subtracting values.
Dividing.

45. Yuri‘s age is ten years less than twice that of
Christine‘s age. If Christine‘s age is x years, then
Yuri‘s age, in years, is
ANSWER
(C) x 10
(A) 2(x 5)
(D) 2x 5
(B) 2(x 10)

46. Which of the equations stated below represents
the equation of a straight line?
ANSWER
(C) y = 5x2
(A) y = 3x
(D) y = 4x3
(B)

46. The equation of a straight line is y = mx + c
If c = 0, then y = mx
So y = −3x is the equation of a straight line.

47. The gradient of the straight line 2y = 4 5x is
ANSWER
(C) 4(A) 5 (D) 2(B)

47.
Writing the terms on the RHS
in the form mx + c.
Dividing each term by 2.
It is in the form y = mx + c.
So the gradient,

48. If
ANSWER
(C) 11(A) 1 (D) 17(B) 7

48. Substitute −3 for x.
Simplifying.
Adding.

49.
ANSWER
The relation diagram shown above represents a
function. Which of the following equations best
describes the function?
(C) f (x) = x 2
(A) f (x) = x
(D) f (x) = 2(x 1)
(B) f (x) = y

50. Which of the following diagrams is not the graph
of a function?
(A)

50. (D)
ANSWER

50.
Using the vertical line test for a function, it
can be seen that The graph represents a
one-to-many relation and it is therefore not a
function.

In the graph above, when y = 2, the values of
x are:
ANSWER
(C) 1.4(A) 1.2 (D) 1.5(B) 1.3

51.
From the construction on the graph, when
y = 2, then x = 1.4 and x = 1.4, that is
x = 1.4.

52.
ANSWER
The half-lines BA and CD are parallel. If angle
BCD is 65 , then angle ABC is
(C) 130(A) 65 (D) 145(B) 115

52. Interior angles are
supplementary.
Substitute the
value of angle
BCD.
Subtract 65 from
both sides.
Subtracting.

53.
ANSWER
AC and DE are straight lines that intersects at B.
Angle ABE = 127
The size of angle ABD is
(C) 127(A) 53 (D) 233(B) 74

53. The sum of
angles on a
straight line.
Substitute the
value for angle
ABE.
Subtract 127
from both sides.
Subtracting.

54.
ANSWER
The line segment PQ is mapped onto the line
segment P′ Q′ by a translation. The matrix that
represents this translation is
(C)(A) (D)(B)

The shaded triangle is rotated through an angle
of 90 in a counter-clockwise direction about the
point P. Which of the four triangles represent the
image of the shaded triangle?
ANSWER
(C) C(A) A (D) D(B) B

In the diagram above, the line segment PQ is the
image of LM after
ANSWER
(C) a reflection in the x-axis
(A) an enlargement of scale factor 1
(D) a rotation through with centre O
(B) a translation by vector

Mx means a reflection in the x-axis.
56.

The point P shown in the graph above is reflected
in the x-axis. What are the co-ordinates of the
image of P?
ANSWER
(C) (2, 3)
(A) (3, 2)
(D) ( 2, 3)
(B) (3, 2)

58. In a triangle ABC, if angle A = 2x° and angle
B = 3x°, then angle C =
ANSWER
(C) (180 5x)
(A) 36
(D)
(B) 72

58. The sum of
the angles of a
triangle.
Substitute the
value for angle A
and for angle B.
Add the xs.
Subtract 5x from
both sides.
Simplifying.

59.
ANSWER
In the right-angled triangle, tan θ =
(C)(A) (D)(B)

59. Definition of the tangent of an
angle.
Using the capital letters
notation.
Substituting the length for
each side.

The diagram above, not drawn to scale, shows
that the angle of depression of a point A on the
ground from T, the top of a tower, is 40 . A is
25 m from B, the base of the tower. The height,
TB, of the tower, in metres, is
ANSWER
(C) 25 tan 40
(A) 25 sin 40
(D) 25 sin 60
(B) 25 cos 40

Alternate
angles.
Definition of
the tangent of
an angle.
Substitute the
length of AB.
Multiply both
sides by 25 m.

MATHEMATICS
Paper 2
2 hours 40 minutes
SECTION I
Answer ALL the questions in this section
All working must be clearly shown
NEXT

1. (a) Using a calculator, or otherwise, calculate
the EXACT value of
ANSWER
(i)
(3 marks)giving your answer as a common fraction

1. (a) (i)

(ii)
ANSWER
(3 marks)giving your answer in standard form.
1. (a) Using a calculator, or otherwise, calculate
the EXACT value of

1. (a) (ii)
Standard form

The basic wage earned by a factory worker for a
40-hour week is $640.00.
(i) Calculate her basic hourly rate.
For overtime work, the factory worker
is paid one and a half times the basic
hourly rate.
1. (b)
ANSWER
(1 mark)

1. (b) (i) The basic hourly rate

The basic wage earned by a factory worker for a
40-hour week is $640.00.
(ii) Calculate her overtime wage for 15 hours
of overtime.
1. (b)
ANSWER
(2 marks)

1. (b) (ii) The overtime hourly rate = The overtime rate
The basic hourly rate
The overtime wage = The overtime
hourly rate The number of
hours worked overtime
= $24 15
= $360

1. (b) The basic wage earned by a factory
worker for a 40-hour week is $640.00.
(iii) Calculate the total wages earned
by the factory worker for a
60-hour week.
ANSWER
(3 marks)
Total 12 marks

1. (b) (iii) The number of hours
worked overtime = (60 40) hours
= 20 hours
The overtime wage = $24 20
= $480
The total wages
earned
= The basic wage
The overtime wage
= $(640 480)
= $1120

Factorise completely:2. (a)
(i) 8px 5py 8qx 5qy
ANSWER
(2 marks)

2. (a) (i) 8px 5py 8qx 5qy
= p(8x 5y) q(8x 5y)
= (8x 5y) (p q)
Factorise pairwise
Factorise
using 8x 5y
as a common
factor.

(ii) 4x2 36
ANSWER
(2 marks)

2. (a) (ii) 4x2 36
= 4(x2 9)
= 4(x2 32)
= 4(x 3)(x 3)
Factorise using 4 as the HCF.
Write as the difference
of two squares.
Factorise as the difference
of two squares.

(iii) 5x2 6x 8
ANSWER
(2 marks)

2. (a) (iii)
Factorise pairwise
Factorise using
x 2 as a common
factor.

2. (b) One cup of yogurt costs $x and one granola
bar costs $y.
One cup of yogurt and three granola bars cost
$32.00, while two cups of yogurt and two
granola bars cost $30.00.
(i) Write a pair of simultaneous
equations in x and y to represent
the given information above.
ANSWER(2 marks)

2. (b) (i) The cost of one cup of yogurt = $x
The cost of one granola bar = $y
The first equation is:
x 3y = 32 (in dollars)
The second equation is:
2x 2y = 30 (in dollars)
The pair of simultaneous equations in x and y:
x 3y = 32
2x 2y = 30

2. (b)
ANSWER
One cup of yogurt costs $x and one
granola bar costs $y.
One cup of yogurt and three granola
bars cost $32.00, while two cups of
yogurt and two granola bars cost
$30.00.
Solve the equations to find the
cost of one cup of yogurt and
the cost of one granola bar. (4 marks)
Total 12 marks
(ii)

Hence, the cost of a yogurt is $6.50 and
the cost of a granola bar is $8.50.
2. (b) (ii)
So

In a survey of 85 students,
25 played drums
20 played tassa
x played drums and tassa
3x played neither.
Let D represent the set of students in the
survey who played drums, and T the set of
students who played tassa.
Copy and complete the Venn diagram
below to represent the information
obtained from the survey.
(i)
ANSWER
3. (a)
(2 marks)

3. (a) (i)
The Venn diagram is shown above.
The students who played drums only,
The students who played tassa only,

25 played drums
20 played tassa
3x played neither.
Let D represent the set of students in
the survey who played drums, and T
the set of students who played tassa.
Write an expression in x for the total
number of students in the survey.
(ii) ANSWER
(1 mark)
3. (a)

The total number of students in the
survey, n(U ) = 25 x + x + 20 x + 3x
= 2x + 45
3. (a) (ii)

3. (a)
ANSWER(2 marks)Calculate the value of x.
25 played drums
20 played tassa
3x played neither.
Let D represent the set of students in
the survey who played drums, and T
the set of students who played tassa.
(iii)

3. (a) (iii) n(U ) = 85 and n(U ) = 2x + 45
so we have the following equation:
Hence, the value of x is 20.
Subtract 45 from
both sides.
Divide both sides
by 2.

(i) Using a ruler, a pencil, and a pair
of compasses, construct the
kite PQRS accurately. ANSWER(4 marks)
The diagram below, not
drawn to scale, shows a
kite, PQRS, with the
diagonal PR = 6 cm,
3. (b)

3. (b) (i)
Draw a horizontal line greater than 6 cm. Mark a point P to the left of
the line. Set your compasses to a separation of 6 cm using a ruler.
Place the steel point of the compasses at point P and construct an arc
to intersect the horizontal line at point R. PR = 6 cm.

Set your compasses to a separation of
that is, 3.25 cm. With point P as centre, construct an arc above PR. With
point R as centre and the same compasses separation, construct another
arc to intersect the previous arc at point Q.
Set your compasses to a separation of 5 cm. With centres P and R, construct
two arcs below PR to intersect at point S.
PS = RS = 5 cm.
Use a ruler and pencil to draw the four sides of the kite PQRS.

(ii) Join QS. Measure and state, in
centimetres, the length of QS.
ANSWER
(2 marks)
Total 11 marks
The diagram below, not drawn to scale,
shows a kite, PQRS, with the diagonal
PR = 6 cm,
3. (b)

Draw a straight line from Q to S.
Take a divider and open it from point
Q to point S. Measure the separation
of the divider using a ruler.
3. (b) (ii)
The length of

The table below shows two readings
from the records of a train.
4. (a)
Town Time Distance travelled (km)
X 07:20 538
Y 09:50 773
Calculate
(i) the number of hours taken for the
journey from town X to town Y
ANSWER
(1 mark)

4. (a) (i) The number of hours taken for the
journey from town X to town Y,

4. (a)
Calculate
(ii) the distance travelled, in kilometres,
between the two towns ANSWER(1 mark)
X 07:20 538
Y 09:50 773

4. (a) (ii) The distance travelled between
the two towns,

4. (a)
ANSWER(2 marks)
Calculate
(iii) the average speed of the train
in km/h
X 07:20 538
Y 09:50 773

4. (a) The average speed of the train,(iii)

4. (b) The map shown below is drawn
to a scale of 1:500 000.
ANSWER(2 marks)
(i) Measure along a straight line and state,
in centimetres, the distance on the map
from P to Q.

Open your divider from P to Q, then
measure the separation using a ruler.
4. (b) (i)
The distance on the map
from P to Q = 5.8 cm

(ii) Calculate the actual distance, in
kilometres, from P to Q
ANSWER
(2 marks)

The scale is 1: 500 0004. (b) (ii)
The actual distance
from P to Q = 5.8 5 km
= 29.0 km

(iii) The actual distance between two places is 8.5
km. Calculate the number of centimetres that
represent this distance on the map ANSWER(3 marks)
Total 11 marks

4. (b) (iii) 5 km is represented by 1 cm
1 km is represented by
8.5 km is represented by 8.5
= 1.7 cm

ANSWER
(1 mark)
5. (a) Given that f (x) = 4x − 7 and g(x) = x2 − 15,
calculate the value of
(i) f (−3)

5. (a) (i)
Substitute −3 for x.

ANSWER
(2 marks)
(ii) gf (2)

5. (a) (ii)
Substitute 2 for x.
Substitute 1 for x.

Substitute f (x)
into g(x) for x.
Or

ANSWER
(2 marks)
(iii) f −1(−1)

5. (a) (iii)
Defining equation for f(x).
Interchanging x and y.
Adding 7 to both sides.
Dividing both sides by 4.
Defining equation for f −1(x).

ANSWER
(2 marks)
5. (b) (i) Given that y = x2 + x − 6, copy and
complete the table below.
x −4 −3 −2 −1 0 1 2 3
y 6 −4 −6 −6 −4 6

5. (b) (i)

The completed table is shown below.
x −4 −3 −2 −1 0 1 2 3
y 6 0 −4 −6 −6 −4 0 6

ANSWER
(5 marks)
Total 12 marks
(ii) Using a scale of 2 cm to represent 1 unit
on the x-axis and 1 cm to represent
1 unit on the y-axis, draw the graph of
y = x2 + x − 6 for −4 ≤ x ≤ 3.
5. (b) Given that y = x2 + x − 6, copy and
complete the table below.
x −4 −3 −2 −1 0 1 2 3
y 6 −4 −6 −6 −4 6

5. (b) (ii)
Using the given scales, the graph of y = x2 + x − 6 for −4 ≤ x ≤ 3
was drawn on graph paper as shown above.

ANSWER
6. The diagram below shows trapeziums A, B and C.
The line y = −x is also shown.

ANSWER
(3 marks)
6. (a) Describe, fully, the single transformation
which maps trapezium A onto
(i) trapezium B

6. (a) (i) The single transformation which maps
trapezium A onto trapezium B is a translation
with vector
Each point on trapezium A is moved
3 units horizontally to the right, then 6
units vertically downwards.

ANSWER
(3 marks)6. (a) (ii) trapezium C

6. (a) (ii) The single transformation which maps
trapezium A onto trapezium C is a
reflection in the line y = −x .

ANSWER
(4 marks)
Total 10 marks
6. (b) State the coordinates of the vertices of
trapezium D, the image of trapezium B
after a reflection in the line y = −x.

6. (b) The coordinates of the vertices of trapezium D are:
(3, 0), (1, 2), (1, 4) and (3, 4).

7. The waiting time, to the nearest minute, experienced by
100 people to catch a bus is shown in the table below.
Waiting Time
(minutes)
Number of
Students
Cumulative
Frequency
1 – 5 9 9
6 – 10 12 21
11 – 15 15 36
16 – 20 19
21 – 25 22
26 – 30 16
31 – 35 4
36 – 40 3

ANSWER
(2 marks)
7. (a) Use the table given above to construct a
cumulative frequency table.

7. (a) Interval
(minutes)
Cumulative
Frequency
< 5.5 9
< 10.5 9 + 12 = 21
< 15.5 21 + 15 = 36
< 20.5 36 + 19 = 55
< 25.5 55 + 22 = 77
< 30.5 77 + 16 = 93
< 35.5 93 + 4 = 97
< 40.5 97 + 3 = 100
The cumulative frequency table is shown above.

ANSWER
(4 marks)
7. (b) Use the values from your table to draw a
cumulative frequency curve.

7. (b)
The completed cumulative frequency curve is shown above.

ANSWER
(2 marks)
7. (c) Use your graph to estimate
(i) the median for the data

7. (c) (i) Half of the total frequency,
From the graph, the waiting
time corresponding to a total
frequency of 50, Q2 = 19 minutes
Hence, the median for the data is
19 minutes.

ANSWER
(2 marks)
7. (c) (ii) the number of people who waited less
than 23 minutes

7. (c) (ii) From the graph, the
number of people who
waited less than 23 minutes = 65

ANSWER
(2 marks)
Total 12 marks
7. (c) (iii) the probability that a person, chosen at
random from the group, waited for at
least 18 minutes

7. (c) (iii) From the graph, the
number of people who
waited less than 18 minutes = 45
The number of people who
waited for at least 18 minutes
P(x ≥ 18 minutes)

8. The first three diagrams in a sequence are shown below.
Diagram 1 has a single circle, which can be considered as a
square pattern formed by a single circle.
Diagram 2 consists of a square of side two circles with two
triangles formed at the ends as shown.
Diagram 3 consists of a square of side three circles with two
triangles formed at the ends as shown.
Diagram 1 Diagram 2 Diagram 3

Diagram
Number
Number of
Circles Forming
the Square
Number of Additional
Circles in Two
Triangles
Pattern for Calculating
the Total Number of
Circles in the Diagram
1 12 1(0) 12 + 1(0)
2 22 2(1) 22 + 2(1)
3 32 3(2) 32 + 3(2)
(i) 4 42 — —

(ii) — — 8(7) —

(iii) n — — —

ANSWER
(2 marks)8. (a) Draw Diagram 4 in the sequence.

8. (a)
Diagram 4
Diagram 4 in the sequence is shown above.

ANSWER
(8 marks)
Total 10 marks
8. (b) Complete the table by inserting the
appropriate values at the rows
marked (i), (ii) and (iii).

(b) Diagram
Number
Number of
Circles Forming
the Square
Number of Additional
Circles in Two
Triangles
the Total Number of
Circles in the Diagram
1 12 1(0) 12 + 1(0)
2 22 2(1) 22 + 2(1)
3 32 3(2) 32 + 3(2)
(i) 4 42 4(3) 42 + 4(3)

(ii) 8 82 8(7) 82 + 8(7)

(iii) n n2 n(n 1) n2 + n(n 1)
8.
The completed table is shown above.

SECTION II
Answer TWO questions in this section
NEXT

ANSWER
(4 marks)
9. (a) Solve the pair of simultaneous equations
y = 1 − 2x
y = 2x2 + 5x − 3

Either 2x – 1 = 0
i.e. 2x = 1
Or x + 4 = 0
x = – 4

ANSWER
(3 marks)
9. (b) Express in the form
where a, h and k are real
numbers

9. (b) Factorise out the
coefficient of x2
i.e. 2.
Write as a perfect square.
The LCM of 2 and 16 is 16.
Adding the fractions.

Simplifying the fraction.
Multiplying the fraction by 2.
It is in the form where a, h and k are 2, respectively.

ANSWER
(1 mark)
9. (c) Using your answer from (b) above, or
otherwise, calculate.
(i) the minimum value of

9. (c)
(i) The minimum value of

ANSWER
(1 mark)
9. (c) Using your answer from (b) above, or
otherwise, calculate.
(ii) the value of x where the minimum
occurs

9. (c) (ii) The minimum occurs where the value of x

ANSWER
(4 marks)
9. (d) Sketch the graph of y = 2x2 + 5x − 3, clearly
showing
the coordinates of the minimum point.
the value of the y-intercept.
the values of x where the graph cuts the
x-axis.

9. (d) The coordinates of the minimum
point are
y = 2x2 + 5x − 3 the value of
the y intercept, c = 3
y = 2x2 + 5x − 3 and y = 0 on the x-axis,
so 0 = (2x − 1)(x + 3) by factorising the
expression.

Hence, x = and x = −3 are the values
of x where the graph cuts the x-axis.
A sketch of the graph of y = 2x2 + 5x − 3 is
shown below.

ANSWER
(2 marks)
Total 15 marks
9. (e) Sketch on your graph of y = 2x2 + 5x − 3,
the line which intersects the curve at the
values of x and y as calculated in (a) above.

9. (e) A sketch of the line y = 1 − 2x which intersects the curve
y = 2x2 + 5x − 3 at the points (−4, 9) and is shown below.

ANSWER(1 mark)(i) ABC
10. (a) The diagram following, not drawn to scale,
shows a circle, centre O. The line DCE is a
tangent to the circle. Angle ACE = 46 and angle
OCB = 25 .
Calculate:

10. (a) (i) in alternate
segment

ANSWER
(1 mark)(ii) AOC
OCB = 25 .
Calculate:

10. (a) (ii) at centre =2· at
circumference

ANSWER
(1 mark)
(iii) BCD
OCB = 25 .
Calculate:

10. (a) (iii) between
radius and
tangent at
point of
tangency.

ANSWER
(1 mark)
(iv) BAC
OCB = 25 .
Calculate:

10. (a) (iv) in alternate
segment.

ANSWER(1 mark)
(v) OAC
OCB = 25 .
Calculate:

10. (a) (v)
ΔOAC is
isosceles,
since OC =
OA = r.

ANSWER(1 mark)(vi) OAB
OCB = 25 .
Calculate:

10. (a) (vi)

10. (b) The diagram below, not drawn to scale,
shows the positions of two ships, P and Q,
relative to a point O. P is on a bearing of
045° from O and the distance OP = 500 km.
Q is on a bearing of 080° from P and the
distance PQ = 800 km.

ANSWER
(2 marks)
10. (b) (i) Copy the diagram above. On you diagram
indicate the angles that represent the
bearings of 045 and 080 .

10. (b) (i)
A copy of the diagram is shown above.
The angles that represent the bearings of
045 and 080 are indicated. The
distances are also indicated.

ANSWER
(7 marks)
Total 15 marks
10. (b) (ii) Calculate
a) OPQ
b) the distance OQ, to the nearest
kilometre
c) the bearing of Q from O

10. (b) (ii) a)

Interior
angles are
supplementary,
NO//NP.

S
at a
point.
Hence, OPQ is 145 .

10. (b) (ii) b)

Hence, the distance OQ is 1243 km, to the
nearest kilometre.
Considering ΔOPQ and using the cosine rule:

10. (b) (ii) c)

Hence, the bearing of Q from O is 066.7 .
Considering ΔOPQ and using the sine rule:

ANSWER
(4 marks)
11. (a) The value of the determinant of
is −36.
Calculate the values of x.

11. (a)
Hence, the values of x are +4 and −4.

ANSWER(2 marks)
11. (b) The transformation R is represented by the
matrix
The transformation S is represented by the
matrix
(i) Write a single matrix, in the form
to represent the combined
transformation S followed by R.

11. (b) (i) The combined transformation S followed by R,
Hence, the single matrix that represents the
combined transformation S followed by R is

ANSWER
(3 marks)
(ii) Calculate the image of the point P(−7, 4)
under the combined transformation S
followed by R.
11. (b) The transformation R is represented by the
matrix
The transformation S is represented by the
matrix

11. (b) (ii) RS P P′
P′ (7, 4)
Hence, the image of the point P (−7, 4) under
the combined transformation S followed by
R is P′ (7, 4).

ANSWER
(2 marks)
11. (c)
(i) Determine the inverse matrix of N.

11. (c) (i)

(4 marks)
Total 15 marks
(ii) Hence, calculate the value of x and the
value of y for which
11. (c)
ANSWER

11. (c) (ii)

Hence, x = 2 and y = −3.

MATHEMATICS
Paper 1
90 minutes
NEXT

1. ( 1)3 + ( 3)2 =
ANSWER
(A) 4
(B) 9
(C) 8
(D) 10

1. ( 1)3 + ( 3)2 =
= ( 1) ( 1) ( 1)
+ ( 3) ( 3)
= 1 ( 1) + 9
Use the meaning of a
square and a cube.
The product of two
negative signs is a
positive sign.
The product of a positive
sign and a negative sign is
negative.
= 1 + 9
= 8 Subtracting.

2. Express as a decimal correct to 3 significant figures.
ANSWER
(A) 5.27
(B) 5.28
(C) 5.29
(D) 5.30

2.
The digit after the 3rd
significant figure is 5, so
we add 1 to the digit 8.

3. The decimal fraction 0.016 expressed as a common
fraction in its lowest terms is
ANSWER
(A)
(B)
(C)
(D)

Write the decimal fraction
as an equivalent common
fraction.
Divide both the numerator
and the denominator by
their common factor 8.
This is the common fraction
written in its lowest terms.
3.

4. In standard form, 8 504 is
ANSWER
(A) 8.504 102
(B) 8.504 103
(C) 8.504 10 2
(D) 8.504 10 3

4. 8 504 = 8.504 1 000
= 8.504 103
The first number must have a
value between 1 and 10.
That is,
1 < first number < 10

5.
ANSWER
(A)
(B) 12
(C)
(D)

6. If 70% of a number is 80, then the number is
ANSWER
(A) 10
(B) 56
(C) 80
(D)

7. The multiplicative inverse of –5 is
ANSWER
(A) 5
(B) 5
(C)
(D)

7. Definition
Divide both sides by –5.
A positive value divided by a
negative value is a negative value.

8. The HCF of 15, 30 and 60 is
ANSWER
(A) 3
(B) 5
(C) 15
(D) 45

8. 3 15, 30, 60
5 5, 10, 20
1, 2, 4
Each of the numbers 15, 30
and 60 is divisible by 15.
The HCF = 3 5 = 15

9. If 2n is an even number, which of the following
is an odd number?
ANSWER
(A) 2n 1
(B) 2(n + 1)
(C) 2n 2
(D) 2(n + 3)

9. Even number = 2n
Odd number = 2n 1

10. The next term in the sequence 5, 2, 1, 4 is
ANSWER
(A) 5
(B) 6
(C) 7
(D) 8

10. 5, 5 3 = 2,
2 3 = 1,
1 3 = 4,
4 3 = 7
A term in the sequence is obtained by subtracting 3
from the term just to its left (the preceding term).

11. A butcher bought a car for $2 500 and sold it for
$3 000. His profit as a percentage of the cost price is
ANSWER
(A) 5%
(B) 10%
(C) 15%
(D) 20%

11. The profit = $(3 000 2 500)
= $500
The percentage profit =
= 20%

12. A boutique gives 10% discount for cash. What is the
cash price of a dress with a marked price of $350?
ANSWER
(A) $35
(B) $315
(C) $340
(D) $360

12. (100 10)% of $350 = 90% of $350
= $315

13. If J $90.00 is equivalent to US $1.00, then
J $5 400.00 equivalent to
ANSWER
(A) US $6.00
(B) US $60.00
(C) US $600.00
(D) US $540

14. The freight charges on a parcel is $150 plus custom duties of
20%. What amount of money was paid to collect the parcel?
ANSWER
(A) $160
(B) $170
(C) $180
(D) $190

15. A man pays $0.25 for each unit of electricity used
up to 400 units and $0.31 for each unit of electricity
used in excess of 400 units. How much does he pay for
consuming 1 200 units of electricity?
ANSWER
(A) $56
(B) $324
(C) $348
(D) $672

15. The cost for the first 400 units = $0.25 400
= $25 4
= $100
The cost for the remaining = $0.31 800
800 units = $31 8
= $248
The electricity bill = $(100 + 248)
= $348

16. The table below shows the rates charged by
an insurance company for home insurance.
ANSWER
(A) $2 100 (B) $4 500
(C) $4 020 (D) $6 600
House $4.50 per $1 000
Contents $2.10 per $1 000
A house is valued at $800 000 and the contents at $200 000.
How much will the owner pay for home insurance?

16. The cost for insuring the house = $4.50
= $4.50 800
= $450 8
= $3 600
The cost for insuring the contents = $2.10
= $2.10 200
= $210 2
= $420

The cost for the home insurance = $(3 600 + 420)
= $4 020

17. A student bought 12 blue pens at $15 each and 13 green
pens at $10 each. What is the mean cost per pen?
ANSWER
(A) $12.40
(B) $12.50
(C) $12.60
(D) $12.70

17. The cost for the 12 blue pens = $15 12
= $180
The cost for the 13 green pens = $10 13
= $130
The total cost for the 25 pens = $(180 + 130)
= $310
The mean cost per pen =
= $12.40

18. A woman invested a sum of money at 6% per annum for
2 years. If she collected $ 300 as simple interest, what
was the sum of money that she invested?
ANSWER
(A) $2 500
(B) $2 700
(C) $2 800
(D) $10 000

18. I = $300
R = 6%
T = 2 years

19.
ANSWER
(A) (P Q)
(B) (P Q)
(C) P Q
(D) P Q
In the Venn diagram above, the shaded region represents

19. The unshaded region represents
P Q P or Q
The shaded region represents
(P Q) Not P or Q

20. If U = {2, 3, 5, 7, 11, 13} and A = {5, 11}, then n(A ) =
ANSWER
(A) 2
(B) 4
(C) 6
(D) 8

20. A = {2, 3, 7, 13}
n(A ) = 4

21.
ANSWER
(A) {3, 9, 15, 18, 24}
(B) {3, 6, 9, 12, 15}
(C) {6, 12, 18, 24}
(D) {6, 12}
In the Venn diagram, set L and set M are represented by two
intersecting circles. If L = {multiples of 3 less than 16} and
M = {multiples of 6 less than 25}, then the shaded region
represents

22. Which of the following pairs of sets are equivalent?
ANSWER
(A) {2, 3} and {a, b, c}
(B) { } and {1, 2, 3}
(C) {a, b, c} and {2, 4}
(D) {1, 2, 3} and {a, b, c}

22. {1, 2, 3} {a, b, c}
n{1, 2, 3} = 3 n{a, b, c} = 3
The number of elements in each of the sets is 3,
therefore the sets are equivalent.
Or
1 a
2 b
3 c
There is a 1 1 correspondence between the elements
of the two sets, therefore the sets are equivalent.

23. The volume of a cube with edges of length 1 cm is
ANSWER
(A) 1 cm3
(B) 12 cm3
(C) 16 cm3
(D) 24 cm3

23. The volume of the cube,
V = l3
= (1 cm)3
= 1 cm3
The formula for the
volume of a cube.

24. Expressed in millimetres, 470 centimetres is
ANSWER
(A) 4.7
(B) 47
(C) 4 700
(D) 47 000

24. 1 cm = 10 mm
470 cm = 10 470 mm
= 4 700 mm

25. The lengths of the sides of a triangle are x, 2x and 3x
centimetres. The perimeter of the triangle is 30 centimetres.
What is the value of x?
ANSWER
(A)
(B) 5
(C) 10
(D) 15

25. The perimeter = (x + 2x + 3x) cm
= 6x cm
Equating the values for the perimeter:
6x = 30
= 5

26. If Usain Bolt runs the 100 metres race in 9.6 seconds,
what was his average speed in metres per second?
ANSWER
(A)
(B)
(C)
(D) 96

26. The average speed,
d = 100 m and
t = 9.6 s

27. Forty students each drank 2 bottles of sweet drink.
Each bottle held 250 millilitres of sweet drink.
How many litres of sweet drink were used?
ANSWER
(A) 20
(B) 80
(C) 500
(D) 20 000

27. The number of bottles used = 40 2
= 80
The number of millilitres used = 250 80
= 20 000
The number of litres used =
= 20

28. The length of a rectangle is three times that of its width.
If the area of the rectangle is 108 cm2, then its width, in cm, is
ANSWER
(A) 6
(B) 26
(C) 27
(D) 36

l = 3w cm28.
A = 108 cm2 b = w cm
The area of the rectangle,
A = lb
= (3w w) cm2 Substitute 3w for l and w for b.
= 3w2 cm2
Equating the values for the area:

29. A student leaves home at 06:25 h and arrives at school at 07:45 h.
The student travels non-stop at an average speed of 60 km/h.
What distance, in kilometres, is the student‘s home from school?
ANSWER
(A) 40
(B) 50
(C) 70
(D) 80

29. The time taken,
t = (07:45 – 06:25) h
= 1 h 20 min
The distance, s = 60 km/h
t

30.
ANSWER
(A) (B)
(C) (D)
The diagram above shows a sector POQ with
sector angle POQ = 45° and radius OQ = r units.
The area of the sector POQ is

Items 31 – 34 refer to the following frequency distribution.
The distribution shows the mass of parcels, in kilograms,
sent to a skybox by an individual.
Mass of parcel (kg) Number of parcel
2 3
3 7
4 2
5 1

31. The mode, in kilograms, of the distribution is.
ANSWER
(A) 2
(B) 3
(C) 4
(D) 5

31. Mode = 3 kg 7 (highest frequency)

32. What is the median, in kilograms, of the distribution?
ANSWER
(A) 4
(B) 3.5
(C) 3
(D) 2

32. The total frequency = 3 + 7 + 2 + 1
= 13
So the middle value is in the 7th ordered position.
The 7th parcel in ascending or descending order has a
mass of 3 kg.
So the median of the distribution has a mass of 3 kg.

33. The total mass, in kilograms, of all the parcels sent
to the skybox by the individual is
ANSWER
(A) 13
(B) 14
(C) 40
(D) 182

x(kg) f fx
2 3 6
3 7 21
4 2 8
5 1 5
fx = 40
33.

34. The mean, in kilograms, of the distribution is
ANSWER
(A)
(B)
(C)
(D)

35.
ANSWER
(A) 400
(B) 300
(C) 200
(D) 100
The pie chart shown above represents the ways in which a
school of 600 children watched a movie. The number of
children who watched the movie at a cinema is approximately

36. The volume, in millilitres, of five sizes of bottled
orange juice are 500, 250, 2 000, 750, 1 000.
The range, in millilitres, is
ANSWER
(A) 250
(B) 500
(C) 1 000
(D) 1 750

36. The range
= The greatest volume The least volume
= (2 000 250) ml
= 1 750 ml

37. 5(x 2) =
ANSWER
(A) 5x 2
(B) 5x + 2
(C) 5x 10
(D) 5x + 10

37. 5(x 2) = 5 x
5 ( 2)
= 5x + 10
Use the distributive
law to remove the
brackets.
The product of a
positive and a negative
sign is a negative
sign. The product of
two negative signs is a
positive sign.

38. 4(2x 1) 3(x 5) =
(A) 5x 11
(B) 5x +11
(C) 5x – 6
(D) 5x + 6
ANSWER

38. 4(2x 1) 3(x 5)
= 8x 4 3x + 15
= 8x 3x + 15 4
= 5x + 11
Use the distributive law
on the terms in each pair
of brackets.
Group like terms.
Add like terms.

39. For all x, 4x(x + 3) 2x(5x 1) =
ANSWER
(A) 6x2 + 14x
(B) 6x2 14x
(C) 4x2 10x + 4
(D) 4x2 10x 4

39. 4x(x + 3) –2x(5x 1)
≡ 4x2 + 12x 10x2 + 2x
≡ 4x2 10x2 + 12x + 2x
≡ 6x2 + 14x
Use the distributive law twice to
remove the two pairs of brackets.
Group like terms.
Add like terms.

40.
ANSWER
(A) 1
(B) 1
(C)
(D)

40. State the given formula.
Substitute the value for p and for q.
Use the meaning of a square root.
Dividing.
Subtracting.

41. If a = 2 and ab = 10, then (a + b)2 (a2 + b2) =
ANSWER
(A) 20
(B) 20
(C) 78
(D)

41.
Substitute 2 for
a and 5 for b.

42.
ANSWER
(A)
(B)
(C) 3x
(D)

42. The common denominator is 9x.
Simplify the values in the numerator by subtracting.
Divide both the numerator and denominator by their
common factor 3.

43. The statement ―When 2 is added to five times a number n,
the result is 40.‖ May be represented by the equation
ANSWER
(A) 2(5n) = 40
(B) 2 5n = 40
(C) 5n + 2 = 40
(D) 5n 40 = 2

43. Five times a number n = 5n
2 added to five times a number n = 5n + 2
The equation is: 5n + 2 = 40

44. If x and y are numbers with x greater than y, then the statement.
―The square of the difference of two numbers is always positive.‖
May be represented as
ANSWER
(A) (x y)2 > 0
(B) x2 y2 > 0
(C) 2(x y) > 0
(D) (x + y)2 > 0

44. The difference of the two numbers = x y
The square of the difference of the two numbers = (x y)2
The statement is:
(x y)2 > 0
A positive number is greater than zero.

45. Given that 3x + 8 29, then the range of values of x is
ANSWER
(A) x 7
(B) x > 7
(C)
(D)

45. Subtract 8 from both sides.
Subtracting.
Divide both sides by 3.
Dividing.
So

46.
ANSWER
(A) y is greater than x
(B) x is a factor of y
(C) x is less than y
(D) x is a multiple of y
The arrow diagram above describes the relation

46. 2 4 = 8
4 2 = 8
3 3 = 9
2 5 = 10
Hence, x is a multiple of y.

47. Which of the following relation diagrams illustrates a function?
ANSWER
(A) (B)
(C) (D)

47.
Each element in the domain is mapped onto one
and only one element in the range.
This relation diagram represents a function.

48. If f(x) = x2 + x 1, then f( 3) =
ANSWER
(A) 5
(B) 5
(C) 7
(D) 13

48. Substitute 3 for x.
Simplify each term.
Subtracting.

49. Which of the following sets is represented
by the relation f: x x2 3?
ANSWER
(A) {(0, 3), (1, 2), (2, 1), (3, 6)}
(B) {(0, 3), (1, 2), (2, 1), (3, 0)}
(C) {(0, 3), (1, 6), (2, 9), (3, 12)}
(D) {(0, 3), (1, 3), (2, 3), (3, 4)}

49. f(x) = x2 3
f(0) = 02 3 = 3 (0, 3)
f(1) = 12 3 = 1 3 = 2 (1, 2)
f(2) = 22 3 = 4 3 = 1 (2, 1)
f(3) = 32 3 = 9 3 = 6 (3, 6)
The set is {(0, –3), (1, 2), (2, 1), (3, 6)}

50.
ANSWER
(A) y = ax2 + bx
(B) y = bx ax2
(C) y = ax2 + bx + c
(D) y = c + bx ax2
If a, b and c are constants with a > 0, then the
equation of the graph could be

a A maximum turning point
c 0 (y-intercept)
Equation is: y = c + bx ax2
50.

51. Which of the following diagrams is the graph of a function?
ANSWER
(A) (B)
(C) (D)

51.
Using the vertical line test for a function:
x1 y1
x2 y2
The graph represents a 1 1 relation which is a function.

52.
ANSWER
(A) x = y
(B) x < y
(C) x + y = 180
(D) x + y > 180
In the figure above, AB and CD are parallel. The
relation between x and y is

52. x + y = 180 The interior angles are supplementary.

53. Which of the following plane shapes has no line of symmetry?
ANSWER
(A) (B)
(C) (D)

53.
Each of these three
plane figures has a line
of symmetry.
This figure has no line of symmetry.

54.
ANSWER
(A) 6 8 (B) 6 10
(C) 8 10 (D) 6 16
The area of PQR, in cm2, is given by

54.
The area of PQR, A = bh
= 6 cm 8 cm
= 6 8 cm2

55.
ANSWER
(A) 28
(B) 56
(C) 102
(D) 124
In ABC, angle ABC = xand angle BAC = 28.
What is the value of x?

55.
Δ ABC is isosceles since AB = CB.
Also angle BCA = angle BAC = 28°
So x+ 28+ 28= 180
i.e. x+ 56= 180
 x= 180– 56
= 124
 x = 124
Sum of the angles of a
triangle.

56.
ANSWER
(A) 640 m
(B) 160 m
(C)
(D)
In the diagram above, not drawn to scale, TB represents a
hill which is 320 m high, and S is the position of a ship.
The angle of elevation of S from T is 30°. The distance of
the ship from the top of the hill is

57.
ANSWER
(A) DAB = 90
(B) ADB = ACB
(C) CAB = ACB
(D) ACB + ABD = 90
In the diagram above, not drawn to scale, BOD
is a diameter of the circle centre O. Which of the
four statements below is false?

57. DAB = 90
ADB = ACB
CAB + ADB = 90
CAB = ACB
DAB = 90
ADB = ACB
Each of these three
statements is true.
ADB + ABD = 90
ACB + ABD = 90 since ADB = ACB
This statement is false.
The angle in a semicircle is 90º
Angles at the circumference
standing on the same arc.

58.
ANSWER
(A) (B)
(C) (D)
In the triangle shown above, tan M is

58.
Definition of the tangent of an angle.
Using the capital letters notation.
Substitute the length of each side.

59. A ship sailed 75 km due east from A to B. It then sailed 50 km
due south to C. Which of the diagrams below best represents
the path of the ship?
ANSWER
(A) (B)
(C) (D)

59. This diagram best
describes the path
of the ship.

60.
ANSWER
(A) x = 0
(B) y = 0
(C) y = x
(D) x = –y
In the diagram shown, if the line y = –x is rotated about 0
through a clockwise angle of 90°, then its image is

60.
The image is the line y = x.

MATHEMATICS
Paper 2
2 hours 40 minutes
SECTION I
Answer ALL the questions in this section
All working must be clearly shown
NEXT

1. (a) (i) Using a calculator, or otherwise,
determine the exact value of
(2 marks)
ANSWER
(ii) Express as a single fraction
(3 marks)

1. (a) (i)

1. (a) (ii) Use the mixed number
Function, , to simplify
the numbers in the
numerator. Use the mixed
number function, , to
divide the mixed number
in the numerator by the
mixed number in the
denominator.
(single fraction)

1. (b) In this question, use CAN $1.00 = GUY
$164.00.
(2 marks) ANSWER
(i) While vacationing in Canada, Robert
used his credit card to buy a camcorder
for CAN $450.00.
How many Guyanese dollars is
Robert owing on his credit card
for this transaction?

1. (b) (i) CAN $1.00 = GUY $164.00
CAN $450.00 = GUY $164.00 450
= GUY $73 800.00
Hence, Robert is owing GUY $73 800.00
on his credit card for the transaction.

(3 marks)
Total 10 marks
ANSWER
(ii) Robert‘s credit card balance is GUY
$102 500.00. After buying the camcorder,
how many canadian dollars does he have
left on his credit card for spending?
1. (b) In this question, use CAN $1.00 = GUY
$164.00.

1. (b) (ii) The credit card balance after
the transaction
Now GUY $164.00 = CAN $1.00
So GUY $1.00 = CAN
GUY $28 700.00 = CAN 28 700.00
= CAN $175.00
Hence, Robert has CAN $175.00 left on
his credit card for spending.

2. (a) Find the value of each of the following
algebraic expressions when a 3, b 1
and c 2
(1 mark)
ANSWER
(i) a (b c)

2. (a) (i) Substitute the
values for a,
b and c into
the algebraic
expression, then
simplify.

(2 marks)
ANSWER
(ii)
2. (a) Find the value of each of the following
algebraic expressions when a 3, b 1
and c 2

2. (a) (ii) Substitute the
values for a,
b and c into
the algebraic
expression,
then simplify
according to
the arithmetic
rules

2. (b) Change the following statements into
algebraic expressions:
(1 mark)
ANSWER
(i) Seven times the sum of x and 3.

2. (b) (i)

(2 marks)
ANSWER
(ii) Fifteen more than the product
of p and q.
2. (b) Change the following statements into
algebraic expressions:

2. (b) (ii) The product of p and q p × q pq
Fifteen more than the
product of p and q pq+15

2. (c) Solve the equation
3(2x + 1) 4x 1 (2 marks)
ANSWER

2. (c) Use the distributive law
Group like terms
Add like terms
Divide both sides by 2
Simplify
So

2. (d) Factorise completely
(2 marks)
ANSWER
(i) 8a3b4 − 16a6b2

2. (d) (i) Factorise using
8a3b2 as the HCF

ANSWER
(ii) 3m2 + 11m − 4 (2 marks)
Total 12 marks
2. (d) Factorise completely

2. (d) (ii)
Factorise pairwise.
Factorise using the
common factor (3m – 1).

3. Students taking part in a community project were
surveyed to find out the type of movies that they were
most likely to view. Each student choose only one type of
movie and 1 260 students were surveyed. The results are
shown in the table below.
Movie
Number of
Students
Horror 168
Detective 210
Romance r
War 182
Musical 462

3. (a) Calculate the value of r, the number of
students who were most likely to view
romance movies.
ANSWER
(2 marks)

3. (a)
Hence, 238 students were most likely to view
romance movies.
So

(i) The data collected in the table are to be represented
on a pie chart. Calculate the size of the angle in
each of the five sectors of the pie chart.
ANSWER(4 marks)
Movie
Number of
Students
Horror 168
Detective 210
Romance r
War 182
Musical 462
3. (b)

3. (b) (i)

The sector angle
representing
romance movies
The sector angle
representing
romance movies
The sector angle
representing
musical movies

(ii) Using a circle of radius 4.5 cm, construct
a pie chart to represent the data.
ANSWER
(4 marks)
Total 10 marks
Movie
Number of
Students
Horror 168
Detective 210
Romance r
War 182
Musical 462
3. (b)

3. (b) (ii)
The constructed pie chart with radius 4.5 cm is shown above.

4. (a) A universal set, U, is defined as
U = {25, 26, 27, 28, 29, 30, 31, 32, 33, 34,
35, 36, 37, 38}.
Sets P and E are subsets of U such that
P = {Prime Numbers} and E = {Even
Numbers}.
(5 marks)
ANSWER
(i) Draw a Venn diagram to represent
the sets P, E and U.

4. (a) (i) U = { 25, 26, 27, 28, 29, 30, 31, 32, 33,
34, 35, 36, 37, 38},
P = {29, 31, 37} and
E = {26, 28, 30, 32, 34, 36, 38}.
The Venn diagram representing the sets P, E
and U is shown above.

(ii) List the elements of the set (1 mark)
ANSWER
4. (a) A universal set, U, is defined as
U = {25, 26, 27, 28, 29, 30, 31, 32, 33, 34,
35, 36, 37, 38}.
Sets P and E are subsets of U such that
P = {Prime Numbers} and E = {Even
Numbers}.

4. (a) (ii) The elements of the set
= {25, 27, 33, 35}.

4. (b) (i) Using only a pair of compasses and a
pencil, construct parallelogram ABCD in
which AB = 5 cm, AD = 8 cm and the
angle BAD is 60º. (5 marks)
ANSWER

4. (b) (i)
The constructed parallelogram ABCD
with AD = BC = 8 cm, AB = DC =
5 cm and BAD = 60 .

4. (b) (ii) Measure and state the length of the
diagonal AC.
ANSWER
(1 mark)
Total 12 marks

4. (b) (ii) The length of the diagonal AC = 11.4 cm.

5. The diagram below, not drawn to scale, represents the floor
plan of a house. The broken line PS, divides the floor plan
into a semi-circle, A, and a rectangle, B.
Use as
(a) Calculate the radius of the semi-circle PST. ANSWER(1 mark)

(b) Calculate the perimeter of the entire floor plan.
ANSWER
(3 marks)
Use as

5. (b) The length of the arc PTS
The perimeter of the entire
floor plan PQRST = (12 + 7 + 12 +11) m
= 42 m

(c) Evaluate the area of the entire floor plan.
ANSWER
(4 marks)
Use as

5. (c) The area of the
semi-circle PST
The area of the rectangle PQRS = lb
= 12 7 m2
= 84 m2

The area of the
entire floor plan

(d) Section B of the floor is to be covered with floor
tiles measuring 1 m by 50 cm. How many floor
tiles are needed to just completely cover Section B?
ANSWER
(4 marks)
Total 12 marks
5. The diagram below, not drawn to scale, represents the floor plan of
a house. The broken line PS, divides the floor plan into a semi-
circle, A, and a rectangle, B.
Use as

5. (d) The area of a floor tile
The number of floor
tiles needed to just
completely cover Section B

6. (a) In the diagram below, not drawn to scale, TB is a vertical
lantern post standing on a horizontal plane. B, P and Q
are points on the horizontal plane.
TB = 10 metres and the angles of depression from the top of the
pole T to P and Q are 35º and 29º respectively.
(i) Copy the diagram and insert the angles
of depression. (1 mark)
ANSWER

6. (a) (i)
The angles can be seen inserted in the diagram above.
TBP = TBQ = 90 Vertical post
standing on
horizontal
ground.

(ii) Calculate to one decimal place
a) the length of BP
b) the length of PQ
ANSWER
(5 marks)
6. (a) In the diagram below, not drawn to scale, TB is a vertical
lantern post standing on a horizontal plane. B, P and Q
are points on the horizontal plane.
TB = 10 metres and the angles of depression from the top of the
pole T to P and Q are 35º and 29º respectively.

6. (a) (ii)
In the diagram above:
of depression =
of elevation

Considering ΔTBP:
(to one decimal place)

Considering ΔTBQ:
(to one decimal place)
The length of

6. (b)
ANSWER
(i) The figure labelled P undergoes a transformation, such
that its image is Q. Completely describe this
transformation.
(2 marks)

6. (b)
(i) The transformation is a translation
represented by the column vector .

(ii) On graph paper, draw and label
a) the line y = −x
b) S, the image of P under a reflection in
the line y = −x.
ANSWER(4 marks)
Total 12 marks
6. (b)

(ii) a) The line y = −x can be seen drawn and labelled on graph
paper.
b) S, the image of P under a reflection in the line y = −x
can be seen drawn and labelled on graph paper.
6. (b)

(a) The equation of the line above is y = mx + c.
ANSWER(1 mark)(i) State the value of c.
7. The diagram below shows the graph of a straight line passing
through the points A and B.

7. (a)
(i) From the graph, the
intercept on the y-axis = 4.
The value of c = 4.

(ii) Determine the value of m. ANSWER
(2 marks)
7. (a) The diagram below shows the graph of a
straight line passing through the points A and B.

7. (a) (ii)
The gradient of the
line segment AB

(ii) From the graph, the slope of AB
indicates a negative gradient.
The gradient of the
line segment AB
Or

(a) (iii) Determine the coordinates of the midpoint
of the line segment AB.
ANSWER
(2 marks)
7. The diagram below shows the graph of a straight line passing
through the points A and B.

7. (a) (iii) Let the mid-point of the line segment
AB be M (x, y).
The x-coordinate of M

So the coordinates of the mid-point of
the line segment AB is .

(iii) From the construction on the graph, the
coordinates of the mid-point of the line
Or

(iii) Using A(0, 4) and
B(3, 0), the midpoint
of the line
segment AB,
Or

7.
ANSWER
(3 marks)
(b) The point lies on the line. State
the value of p.

7. (b) From the construction on the graph, when
, then y = p = −2.
So the value of p is –2.

(b) The equation of AB is
When and y = p, then
Hence, the value of p is –2.
Or

7.
ANSWER
(4 marks)
Total 12 marks
(c) Determine the coordinates of the
point of intersection of the line
y = x − 3 and the line shown previously.

Group like terms

When x = 3, then
y = x − 3
= 3 − 3
= 0
Hence, the coordinates of the point of
intersection of the line y = x − 3 and the line
shown is (3, 0)

(c) Given y = x − 3, then m = 1 and c = −3.
Using c = −3 and , the graph of the
line y = x − 3 was drawn on the same graph
paper as shown above.
The graph of the lines and
y = x − 3 intersect at B (3, 0).
Or

8. The first three diagrams in a sequence are shown below.
Diagram 1 has a single dot, which can be considered as a
triangular pattern formed by a single dot.
Diagram 2 consists of a triangle formed by three dots.
Diagram 3 consists of a triangle formed by six dots.

8. (a) Draw Diagram 4 in the sequence.
ANSWER
(2 marks)

8. (a)
Diagram 4
Diagram 4 in the sequence can be seen above.

8. (b) Complete the table by inserting the
appropriate values at the row 2 marked (i),
(ii) and (iii). (6 marks)

ANSWER
Diagram
Number
Number of
Dots Forming
the triangle
the Total Number of
Dots in the Diagram
1 1 1 (1 + 1) ÷ 2
2 3 2 (2 + 1) ÷ 2
3 6 3 (3 + 1) ÷ 2
(i) 4 — —

(ii) — 21 —

(iii) n — —

Diagram
Number
Number of Dots
Forming the
triangle
the Total Number of Dots
in the Diagram
1 1 1 (1 + 1) ÷ 2
2 3 2 (2 + 1) ÷ 2
3 6 3 (3 + 1) ÷ 2
4 10 4 (4 + 1) ÷ 2

6 21 6 (6 + 1) ÷ 2

n n (n + 1) ÷ 2
8. (b)
The completed table can be seen above.

8. (c) How many dots will be needed to form the
triangle in Diagram 100?
ANSWER
(2 marks)
Total 10 marks

8. (c) The total number of
dots in the diagram

(c) The total number of
dots in the diagram
Or

SECTION II
Answer TWO questions in this section

9. (a) Simplify
ANSWER
(1 mark)(i) x3 x4 x6

9. (a) (i)

ANSWER
(2 marks)(ii)
9. (a) Simplify

9. (a) (ii)
5

ANSWER
(1 mark)
9. (b) If f(x) = 4x − 1, find the value of
(i) f (3)

9. (b) (i)

(ii) f –1(0)
ANSWER
(2 marks)

9. (b) (ii) Given
then
So
i.e.
is the defining equation for f (x)
Interchanging x and y
Adding 1 to both sides
Dividing both sides by 4
is the defining equation for f–1(x)

(iii) f –1 f (3)
ANSWER
(2 marks)

9. (b) (iii)

(i) Using a scale of 8 cm to represent 100 years on the
horizontal axis and a scale of 4 cm to represent 100 kg on
the vertical axis, construct a mass-time graph to show
how the solid decays in the 168 years interval.
ANSWER
(4 marks)
Draw a smooth curve through all the plotted points.
9. (c) The mass, in kg, of strontium, a radioactive
material, after a number of years is given in
the table below.
t
(time in years)
0 28 56 84 112 140 168
m
(mass in kg)
400 200 100 50 25 12.5 6.25

9. (c) (i)
The points were plotted on graph
paper and a smooth curve
drawn as shown above.

(ii) Use your graph to estimate
a) the mass of the solid after 50 years
b) the rate of decay of the solid at t = 75
years.
ANSWER
(3 marks)
Total 15 marks
9. (c) The mass, in kg, of strontium, a radioactive
material, after a number of years is given in
the table below.
t
(time in years)
0 28 56 84 112 140 168
m
(mass in kg)
400 200 100 50 25 12.5 6.25

9. (c) (ii) (a) From the construction on the graph:
The mass of the
solid after 50 years = 116 years

(b) Draw a tangent to the curve at t = 75 years.
Using two points on the tangent, (0, 180)
and (112.5, 0), the gradient of the tangent
Hence, the rate of decay of the solid at t = 75 years is −1.6 kg/year.

10. (a) In the diagram below, not drawn to scale,
PQ is a tangent to the circle, centre O.
PS is parallel to OR and angle RPS = 32º.
ANSWER(2 marks)
Calculate, giving reasons for your answer,
the size of
(i) angle PQR

(i) Alternate S.
ΔOPR is isosceles since
OP = OR = r (radius)
Sum of the
angles of a Δ
at centre = 2.
at circumference
Hence, the size of angle PQR is 58º.

ANSWER(2 marks)
10. (a) In the diagram below, not drawn to scale,
PQ is a tangent to the circle, centre O.
PS is parallel to OR and angle RPS = 32º.
Calculate, giving reasons for your answer,
the size of
(ii) angle SPT

10. (a) (ii)
Hence, the size of angle SPT is 26º.
between radius and tangent
at point of tangency.

10. (b) In the diagram below, not drawn to scale, O
is centre of the circle of radius 9.5 cm and
AB is a chord of length 16.5 cm.
ANSWER
(3 marks)
(i) Calculate the value of θ to the
nearest degree.

10. (b) (i)
Considering Δ AOB and using the cosine rule:

(b) (i) Considering Δ AOB and using the cosine
rule:
Or

10. (b) (i)
Δ AOB is isosceles since AO = BO = r (radius)

Considering AOB:

ANSWER
(2 marks)
(ii) Calculate the area of
triangle AOB.

10. (b) (ii)

(ii) The semi-perimeter of ΔAOB,
Or

The area of ΔAOB, A1

ANSWER(3 marks)
(iii) Hence, calculate the area of the
shaded region. [Use = 3.14]

10. (b) (iii)
The area of the minor sector AOB, A2

The area of the shaded region,

ANSWER(iv) Calculate the length of
the major arc AB.
(3 marks)
Total 15 marks

10. (b) (iv)

The major sector angle, reflex
The length of the major arc AB,

ANSWER(2 marks)
(a) Copy the diagram and complete it to show the points
of P and M.
11.
In the diagram above, the position vectors
of A and B relative to the origin are a and b respectively.
The point P is on OA such that OP = 3 PA.
The point M is on BA such that BM = MA.

11. (a)
The diagram was copied and completed as
shown above. The points P and M are learly
shown.

ANSWER
(1 mark)
(b) OB is produced to N such that OB = 2 BN
(i) Show the position of N on your diagram.
11.

11. (b) (i)
The position of N is shown in the diagram.

ANSWER
(5 marks)
(ii) Express in terms of a and b the vectors.
(b) OB is produced to N such that OB = 2 BN

11.

11. (b) (ii)


Hence,

ANSWER
(4 marks)
11. (c) Use a vector method to prove that P, M
and N are collinear.


Since , then the vectors are either parallel or coincident. Since the vectors
have a common point M, then P, M and N are collinear.

ANSWER
(3 marks)
Total 15 marks
11. (d) Calculate the length of AN if.

Considering ΔNOA:

Model Exam 2 Paper 2

Considering ΔMAN:


The length of
Hence, the length of AN is 5.39 units.

MATHEMATICS
Paper 1
90 minutes
NEXT

1. The decimal fraction 0.85 written as a common
fraction, in its simplest form, is
ANSWER
(A)
(B)
(C)
(D)

1. Multiply the decimal fraction with two
decimal places by , which is 1, to make
the decimal fraction a common fraction.
This is the common fraction written in
its simplest form.

2. The number 75 836 written correct to 4 significant figures is
ANSWER
(A) 80 000
(B) 76 000
(C) 75 800
(D) 75 840

2. 75 836
= 75 840 (4 s.f.)
The digit after the 4th
significant figure is 6, so
we add 1 to the digit 3. 0 is
needed as a place holder.

3. Given that 768 51.2 = 39 321.6, then 76.8 0.512 =
ANSWER
(A) 3 932.16
(B) 393.216
(C) 39.321 6
(D) 3.932 16

4.
ANSWER
(A) 0.018
(B) 0.18
(C) 1.8
(D) 18

5. y is inversely proportional to the square root of 7 may be
expressed as
ANSWER
(A)
(B)
(C) y 72
(D)

5. y is inversely proportional to means

6. One hundred thousand written as a power of 10 is
ANSWER
(A) 104
(B) 105
(C) 106
(D) 107

6. One hundred thousand = 100 000
= 105

7. By the distributive law, 74 13 + 74 12 =
ANSWER
(A) 86 87
(B) 74 25
(C) 86 + 87
(D) 74 + 25

7. 74 13 + 74 12
= 74 (13 + 12)
= 74 25
The common
factor is 74.
Adding.

8. The highest common factor of 12, 24 and 30 is
ANSWER
(A) 2
(B) 4
(C) 5
(D) 6

8.
The HCF = 2 3 = 6
three numbers.
three numbers.
2 12, 24, 30
3 6, 12, 15
2, 4, 5

9. The lowest common multiple of 5, 8 and 20 is
ANSWER
(A) 1
(B) 10
(C) 20
(D) 40

9. 2 5, 8, 20
2 5, 4, 10
2 5, 2, 5
5 5, 1, 5
1, 1, 1
The LCM = 2 2 2 5 = 40

10. The next two terms in the sequence 7, 6, 8 . . . is
ANSWER
(A) 7, 9
(B) 7, 7
(C) 7, 8
(D) 7, 6

10. 7, 6, 8, 7, 9, . . .
1 + 2 –1 + 2

11. A man‘s annual income is $60 000. His non-taxable
allowances is $15 000. If he pays a tax of 25% on his
taxable income, then the tax payable is
ANSWER
(A) $3 750
(B) $11 250
(C) $15 000
(D) $33 750

12. The basic rate of pay is $28.00 per hour.
What is the overtime rate of pay if it is
one-and-a-half times the basic rate?
ANSWER
(A) $32.00
(B) $35.00
(C) $36.00
(D) $42.00

13. Alfred saved $74 when he bought a cell phone at a sale
which gave a discount of 20% on the marked price.
What was the marked price of the cellphone?
ANSWER
(A) $370
(B) $296
(C) $222
(D) $148

14. A store offers a discount of 10% off the marked
price for cash. If the cash price of a calculator is $135,
what is the marked price?
ANSWER
(A) $13.50
(B) $121.50
(C) $148.50
(D) $150.00

15. The charge per kWh of electricity used is 35 cents.
There is also a fixed charge of $27.00. What amount is
the electricity bill if 80 kWh of electricity is consumed?
ANSWER
(A) $55
(B) $62
(C) $142
(D) $307

15. The cost for the electricity = 35¢ 80
= 2 800¢
= $28.00
The fixed charge = $27.00
the amount of the bill = $(28.00 + 27.00)
= $55.00

16. The exchange rate for US $1.00 is GUY $200.
What amount of Guyanese dollars will a tourist
receive for changing US $75.00?
ANSWER
(A) $150
(B) $1 500
(C) $15 000
(D) $150 000

16. US $1.00 = GUY $200
US $75.00 = GUY $200 75
= GUY $15 000

17. Calculate the book value of a computer valued at $3 000,
after two years, if it depreciates by 10% each year.
ANSWER
(A) $300
(B) $2 400
(C) $2 430
(D) $2 920

18. A man pays $540 as income tax. If income tax is charged
at 20% of the taxable income, what was his taxable income?
ANSWER
(A) $1 800
(B) $2 160
(C) $2 700
(D) $3 100

19. X = {a, p, e}. How many subsets has the set X?
ANSWER
(A) 3
(B) 6
(C) 8
(D) 10

19. { }, {a}, {p}, {e}
{a, p}, {a, e}, {p, e}
{a, p, e}
The number of subsets = 8.
or
The number of subsets, N = 2n X = {a, p, e}
= 23 n(X) = 3
= 8 n = 3

20. A school has 200 students. 108 students play both soccer and
basketball, 52 students play soccer only, and 15 students play
neither sport. How many students play basketball only?
ANSWER
(A) 25
(B) 40
(C) 50
(D) 77

20.
Hence, 25 students play basketball only.

21. All students in a class play chess or scrabble or both.
15% of the students play chess only, and 37% of the students
play scrabble only. What percentage of students play both games?
ANSWER
(A) 22
(B) 48
(C) 52
(D) 78

21.
Hence, 48% of the students play both games.

22.
ANSWER
(A) X Y
(B) Y X
(C) X Y = { }
(D) X Y { }
The Venn diagram above is best represented by the statement

22. Sets X and Y have no common elements, so X Y = { }.

23. 5:30 p.m. may be represented as.
ANSWER
(A) 05:30 h
(B) 17:30 h
(C) 15:30 h
(D) 18:30 h

23. 5 : 30 p.m. = (12 + 5) : 30 h
= 17 : 30 h

24.
ANSWER
(A) 8 (B) 16
(C) 24 (D) 32
The diagram above shows a circle with centre O and diameter
8 cm. The area of the circle, in cm2, is

24.
Formula for the
area of a circle.
Substitute r = 4 cm.
Squaring.

25.
ANSWER
(A) (B)
(C) (D)
In the diagram above, POQ is a minor sector of
a circle with angle POQ = 60° and OQ = r cm.
The area, in cm2, of the minor sector POQ is

26. Mark takes 35 minutes to drive to university which is 45 km
away from his apartment. His speed in km per hour is
ANSWER
(A)
(B)
(C)
(D)

27.
ANSWER
(A) (B)
(C) (D)
The diagram above, not drawn to scale, shows a cone of radius
r cm and height r cm. The volume of the cone, in cm3, is

28. The length of the edge of a cube is 20 cm. The
volume of the cube is
ANSWER
(A) 8 000 cm3
(B) 400 cm3
(C) 240 cm3
(D) 200 cm3

29. The mass of one tonne of sugar in kilograms is
ANSWER
(A) 100
(B) 1 000
(C) 10 000
(D) 100 000

29. 1 tonne = 1 000 kg

30. Robert has 0.75 kg of sweets. He has bags which
can each hold 15 g of sweets. How many bags of
sweets can he fill?
ANSWER
(A) 0.5
(B) 5
(C) 50
(D) 500

31. A bowl contains 6 green marbles and 7 yellow marbles.
A marble is picked at random from the bowl. The marble is
found to be green and it is not replaced. What is the probability
that the next ball picked at random from the bowl will be yellow?
ANSWER
(A) (B)
(C) (D)

The number of green marbles
remaining in the bowl = 6 1 = 5
The number of yellow marbles
in the bowl = 7
The total number of marbles
remaining in the bowl = 5 + 7 = 12
P(second marble is yellow) =
31.

32. The mode of the numbers 1, 2, 3, 4, 4, 5, 5, 5, 6, 7, 7, 8 is
ANSWER
(A) 4
(B) 5
(C) 6
(D) 7

32. The mode is 5, since it occurs the most number of times.

33. The median of the numbers 1, 2, 3, 4, 4, 5, 5, 5, 6, 7, 7, 8 is
ANSWER
(A) 6
(B) 5.5
(C) 5
(D) 4

1, 2, 3, 4, 4, 5, 5, 5, 6, 7, 7, 8
Two middle values
The median of the numbers,
Q2 = 533.

34. The mean of the numbers 1, 2, 3, 4, 4, 5, 5, 5, 6, 7, 7, 8 is
ANSWER
(A) 4
(B)
(C) 5
(D)

34. The sum of the numbers, x = 1 + 2 + 3 + 4 +
4 + 5 + 5 +
5 + 6 + 7 +
7 + 8
= 57
The total frequency of the number, n = 12
The mean of the numbers,

35. The scores of 100 students who took part in a shooting
competition at a May Fair is recorded in the table shown below.
ANSWER
(A) (B)
(C) (D)
Score 0 1 2 3 4 5 6 7 8 9 10
Frequency 2 4 5 7 10 31 20 12 5 3 1
The probability that a student chosen at random
from these students scored exactly 6 is

36. The mean of the five numbers 7, p, 5, 9 and 18 is 12.
The number p is
ANSWER
(A) 15
(B) 17
(C) 19
(D) 21

The sum of the numbers, p = 7 + p + 5 + 9 + 18
= p + 39
The total frequency, f = 5
The mean of the numbers,
36.

37. ( 7a) (+2b) =
ANSWER
(A) 14ab
(B) –14ab
(C)
(D)

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