Sec 4 E-Math O-Level style Paper 2: free PDF with answers

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Mathematics

Sec 4 E-Math · O-Level style Paper 2

All topics

Practice paper in the same style as the school and national exam papers. Not a copy of any past year paper.

2 hours 15 minutes

Name

Class

Index number

Candidates answer on the Question Paper.

Read these instructions first

Write your name, class and index number in the spaces at the top of this page.

Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs.

Do not use staples, paper clips, glue or correction fluid.

Answer all questions.

If working is needed for any question it must be shown with the answer. Omission of essential working will result in loss of marks.

The use of an approved scientific calculator is expected, where appropriate.

If the degree of accuracy is not specified in the question, and if the answer is not exact, give the answer to three significant figures. Give answers in degrees to one decimal place.

For π, use either your calculator value or 3.142, unless the question requires the answer in terms of π.

The number of marks is given in brackets [ ] at the end of each question or part question.

The total of the marks for this paper is 90.

For Examiner's Use

/ 90

Mathematical Formulae

Compound interestTotal amount

= P\left(1 + \dfrac{r}{100}\right)^{n}

MensurationCurved surface area of a cone

= \pi r l

Surface area of a sphere

= 4\pi r^{2}

Volume of a cone

= \dfrac{1}{3}\pi r^{2} h

Volume of a sphere

= \dfrac{4}{3}\pi r^{3}

Area of triangle

ABC = \dfrac{1}{2}ab\sin C

Arc length

= r\theta

, where

\theta

is in radiansSector area

= \dfrac{1}{2}r^{2}\theta

, where

\theta

is in radians

Trigonometry

\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}

a^{2} = b^{2} + c^{2} - 2bc\cos A

StatisticsMean

= \dfrac{\sum fx}{\sum f}

Standard deviation

= \sqrt{\dfrac{\sum fx^{2}}{\sum f} - \left(\dfrac{\sum fx}{\sum f}\right)^{2}}

(a)Express

676

as a product of its prime factors. Hence find

\sqrt{676}

without using a calculator.

Answer (a)[2]

(b)(i)Find the highest common factor (HCF) of

4

10

and

14

Answer (b)(i)[3]

(b)(ii)It is given that

396 = 2^{2} \times 3^{2} \times 11

and

1176 = 2^{3} \times 3 \times 7^{2}

. Find the highest common factor (HCF) of

396

and

1176

, giving your answer in index notation.

Answer (b)(ii)[1]

(c)(i)Two bus services leave an interchange together at 7.00 a.m. One leaves every

10

minutes and the other every

15

minutes. Find the number of minutes before they next leave together.

Answer (c)(i)[3]

(c)(ii)A cuboid has a volume of

1716

^3

. Its length, breadth and height are whole numbers of centimetres, all different, and each is greater than

6

cm. Find its length, breadth and height.

Answer (c)(ii)cm[2]

(a)The graph shows the length,

y

cm, of a spring when a mass of

x

kg hangs from it. Use the graph to find the length of the spring when a mass of

5

kg hangs from it.

Answer (a)[1]

(b)Complete the table of values for

y = 2x - 3

Answer (b)[3]

(c)(i)Find the

y

-intercept of the line

-6x - y = -8

Answer (c)(i)[2]

(c)(ii)Find the gradient of the line

4x - 5y = -9

Answer (c)(ii)[2]

(a)A ladder

8

m long leans against an upright wall. Its foot is on level ground,

2

m from the wall. Find the height,

h

m, that the ladder reaches up the wall, correct to 3 significant figures.

Answer (a)m[2]

(b)In triangle

PQR

\angle PQR = 90^\circ

PR = 9.6

cm and

QR = 6.5

cm. Find the length of

PQ

, correct to 3 significant figures.

Answer (b)cm[3]

(c)(i)In triangle

ABC

AB = 15

cm,

BC = 17

cm and

AC = 8

cm. Use the converse of Pythagoras' theorem to find which angle of triangle

ABC

is a right angle.

Answer (c)(i)[3]

(c)(ii)In triangle

ABC

AB = 8

cm,

BC = x

cm and

AC = y

cm, where

BC

is the longest side. It is given that

(x + y)(x - y) = 65

. Determine whether triangle

ABC

is right-angled. Choose the answer with its reason.

Answer (c)(ii)[2]

(a)A bag contains

7

red,

5

blue,

3

green and

5

yellow balls. A ball is taken at random. Find the probability that it is blue.

Answer (a)[2]

	Bus	Walk	Total
Sec 3	6	14	20
Sec 4	13	7	20
Total	19	21	40

(b)The two-way table shows how some Sec 3 and Sec 4 students travel to school. One of the Sec 3 students is chosen at random. Find the probability that this Sec 3 student takes the bus.

Answer (b)[2]

(c)(i)A bag contains

9

red balls and some blue balls. The probability of taking a red ball is

\dfrac{1}{2}

. Find the number of blue balls in the bag.

Answer (c)(i)[2]

(c)(ii)A bag contains

7

red balls and

5

blue balls.

x

red balls are taken out of the bag. A ball is then taken at random. The probability that it is red is

\dfrac{2}{7}

. Find the value of

x

Answer (c)(ii)[3]

The diagram shows the points

A(-2, 2)

and

B(-1, 4)

and the straight line through them.

O

is the origin.

(a)Find the gradient of

AB

Answer (a)[1]

(b)Find the equation of the line

AB

Answer (b)[2]

(c)The line

AB

meets the

x

-axis at

P

. Find the coordinates of

P

Answer (c)[1]

(d)The line

AB

meets the

y

-axis at

Q

. Find the area of triangle

OPQ

Answer (d)units²[2]

(e)The point

R

lies on the line

AB

, above the

x

-axis, and the area of triangle

OPR

3

units². Find the coordinates of

R

Answer (e)[2]

In triangle

PQR

PQ = 13

cm,

QR = 7

cm and angle

QPR = 24^\circ

. Angle

PRQ

is obtuse.

(a)Find angle

PRQ

, correct to 1 decimal place.

Answer (a)°[3]

(b)Find angle

PQR

, correct to 1 decimal place.

Answer (b)°[1]

(c)Find the length of

PR

, correct to 3 significant figures.

Answer (c)cm[2]

(d)Find the area of triangle

PQR

, correct to 3 significant figures.

Answer (d)cm²[2]

In the diagram,

AB

is a diameter of the circle with centre

O

C

is a point on the circle such that angle

CAB = 35^\circ

T

is a point on

AB

produced such that angle

CTB = 38^\circ

E

is the point on

AB

such that

CE

is perpendicular to

AB

(a)(i)Find angle

ACB

, giving a reason.

Answer (a)(i)°[1]

(a)(ii)Find angle

COB

, giving a reason.

Answer (a)(ii)°[1]

(a)(iii)Find angle

OCT

Answer (a)(iii)°[2]

(a)(iv)Explain why

CT

is not a tangent to the circle at

C

[1]

(a)(v)Explain why

E

lies on the circle with

OC

as its diameter.

[2]

\xi = \{x : x \text{ is an integer and } 1 \le x \le 30\}

A = \{x : x \text{ is a factor of } 16\}

and

B = \{x : 4x - 3 > 26\}

. In the Venn diagram, part of

\xi

is shaded.

(b)(i)Find

n(B)

Answer (b)(i)[2]

(b)(ii)List the elements of

A \cap B

Answer (b)(ii)[1]

(b)(iii)List the elements of

A' \cap B'

Answer (b)(iii)[1]

(b)(iv)Which set is shaded in the Venn diagram?

Answer (b)(iv)[1]

(b)(v)

C = \{x : x \text{ is a multiple of } 3\}

. Which one of these statements is true?

Answer (b)(v)[1]

Two fair six-sided dice are thrown. The possibility diagram shows the outcomes.

1st \ 2nd	1	2	3	4	5	6
1	(1, 1)	(1, 2)	(1, 3)	(1, 4)	(1, 5)	(1, 6)
2	(2, 1)	(2, 2)	(2, 3)	(2, 4)	(2, 5)	(2, 6)
3	(3, 1)	(3, 2)	(3, 3)	(3, 4)	(3, 5)	(3, 6)
4	(4, 1)	(4, 2)	(4, 3)	(4, 4)	(4, 5)	(4, 6)
5	(5, 1)	(5, 2)	(5, 3)	(5, 4)	(5, 5)	(5, 6)
6	(6, 1)	(6, 2)	(6, 3)	(6, 4)	(6, 5)	(6, 6)

(a)(i)Find the probability that the sum of the two numbers is

8

Answer (a)(i)[1]

(a)(ii)Find the probability that the product of the two numbers is greater than

15

Answer (a)(ii)[1]

(a)(iii)Find the probability that the two numbers are the same.

Answer (a)(iii)[1]

The matrix

P = \begin{pmatrix} 15 & 5 & 15 \\ 14 & 8 & 9 \end{pmatrix}

shows the numbers of adult, child and senior tickets sold on Friday and Saturday (one row for each day). The matrix

Q = \begin{pmatrix} 12 \\ 8 \\ 6 \end{pmatrix}

shows the prices, in dollars, of an adult, child and senior ticket.

(b)(i)Find

PQ

Answer (b)(i)[2]

(b)(ii)State what the elements of

PQ

represent.

[1]

(b)(iii)Find the total amount, in dollars, from all the tickets sold.

Answer (b)(iii)[1]

Relative to the origin

O

, the position vectors of the points

A

and

B

are

\overrightarrow{OA} = \begin{pmatrix} 6 \\ -4 \end{pmatrix}

and

\overrightarrow{OB} = \begin{pmatrix} -6 \\ -12 \end{pmatrix}

(a)Find

\overrightarrow{AB}

Answer (a)[1]

(b)Find

|\overrightarrow{AB}|

, correct to 3 significant figures.

Answer (b)[2]

(c)The point

P

lies on

AB

such that

AP : PB = 1 : 3

. Find the position vector of

P

Answer (c)[2]

(d)The point

E

has coordinates

(18, h)

, and

A

B

and

E

lie on a straight line. Find the value of

h

Answer (d)[2]

The Ong family is going to a resort on Bintan island. They drive

18

km from home to the ferry terminal at an average speed of

60

km/h. At the terminal, they must check in at least

30

minutes before their ferry departs. The table shows the times the ferries depart from Singapore, in Singapore time.

Ferry	Departs (Singapore time)
A	0740
B	0900
C	1030
D	1200
E	1320
F	1450

(a)Find the time taken for the drive to the ferry terminal, in minutes.

Answer (a)min[2]

(b)The family leaves home at 1140. Which is the earliest ferry the family can take?

Answer (b)[2]

(c)The ferry travels

60

km at an average speed of

36

km/h. Find the time taken for the crossing, in minutes.

Answer (c)min[2]

(d)The local time on Bintan is 1 hour behind Singapore time. A shuttle bus to the resort leaves the Bintan ferry terminal at these local times: 1235, 1325, 1415, 1505 and 1555. Using your answers to parts (b) and (c), find how long the family waits at the Bintan terminal for the next shuttle bus, in minutes.

Answer (d)min[3]

END OF PAPER

Answers and worked solutions

Question 1

(a)Answer:

= 26

[2]

\def\arraystretch{1.8}\begin{array}{r|l} 2 & \underline{\;676\;} \\ 2 & \underline{\;338\;} \\ 13 & \underline{\;169\;} \\ 13 & \underline{\;13\;} \\ & \;1 \end{array}

676 = 2^{2} \times 13^{2}

\sqrt{676} = 2 \times 13

= 26

(b)(i)Answer:

= 2

[3]

\def\arraystretch{1.8}\begin{array}{r|l} 2 & \underline{\;4\;} \\ 2 & \underline{\;2\;} \\ & \;1 \end{array}

\def\arraystretch{1.8}\begin{array}{r|l} 2 & \underline{\;10\;} \\ 5 & \underline{\;5\;} \\ & \;1 \end{array}

\def\arraystretch{1.8}\begin{array}{r|l} 2 & \underline{\;14\;} \\ 7 & \underline{\;7\;} \\ & \;1 \end{array}

4 = 2^{2}

10 = 2 \times 5

14 = 2 \times 7

Take only the primes common to every number, each with its lowest power.HCF

= 2

= 2

(b)(ii)Answer:

2^2*3

[1]

Take only the primes common to every number, each with its lowest power.

(c)(i)Answer:

= 30

[3]

\def\arraystretch{1.8}\begin{array}{r|l} 2 & \underline{\;10\;} \\ 5 & \underline{\;5\;} \\ & \;1 \end{array}

10 = 2 \times 5

\def\arraystretch{1.8}\begin{array}{r|l} 3 & \underline{\;15\;} \\ 5 & \underline{\;5\;} \\ & \;1 \end{array}

15 = 3 \times 5

The smallest number that every amount divides into: the LCM.LCM

= 2 \times 3 \times 5

= 30

(c)(ii)Answer: The cuboid measures

11

cm by

12

cm by

13

cm (the only way with every edge greater than

6

cm and all different).[2]

\def\arraystretch{1.8}\begin{array}{r|l} 2 & \underline{\;1716\;} \\ 2 & \underline{\;858\;} \\ 3 & \underline{\;429\;} \\ 11 & \underline{\;143\;} \\ 13 & \underline{\;13\;} \\ & \;1 \end{array}

1716 = 2^{2} \times 3 \times 11 \times 13

1716 = 11 \times 12 \times 13

The cuboid measures

11

cm by

12

cm by

13

cm (the only way with every edge greater than

6

cm and all different).

Question 2

(a)Answer: Length:

22

cm[1]

Length:

22

(b)Answer:

y = 3

[3]

When

x = -1

y = 2(-1) - 3

y = -5

When

x = 0

y = 2(0) - 3

y = -3

When

x = 3

y = 2(3) - 3

y = 3

(c)(i)Answer:

c = 8

[2]

-y = 6x - 8

y = -6x + 8

c = 8

(c)(ii)Answer:

m = \dfrac{4}{5}

[2]

-5y = -4x - 9

y = \dfrac{4}{5}x + \dfrac{9}{5}

m = \dfrac{4}{5}

Question 3

(a)Answer:

= 7.75

m (3 s.f.)[2]

The hypotenuse is the ladder, opposite the right angle at the foot of the wall.

h^2 = 8^2 - 2^2

= 60

h = \sqrt{60}

= 7.7459\ldots

= 7.75

m (3 s.f.)

(b)Answer: Check:

7.06 < 9.6

, shorter than the hypotenuse. cm[3]

The hypotenuse is

PR

, opposite the right angle at

Q

PR^2 = PQ^2 + QR^2

PQ^2 = 9.6^2 - 6.5^2

= 92.16 - 42.25

= 49.91

PQ = \sqrt{49.91}

= 7.0647\ldots

= 7.06

cm (3 s.f.)Check:

7.06 < 9.6

, shorter than the hypotenuse.

(c)(i)Answer: ∠BAC[3]

AB^2 = 15^2

= 225

BC^2 = 17^2

= 289

AC^2 = 8^2

= 64

The longest side is

BC

64 + 225 = 289

BC^2 = AC^2 + AB^2

, so by the converse of Pythagoras' theorem the angle opposite

BC

90^\circ

\angle BAC = 90^\circ

(c)(ii)Answer: No[2]

(x + y)(x - y) = x^2 - y^2

, so

x^2 - y^2 = 65

8^2 = 64

x^2 - y^2 = 65

, but

8^2 = 64

, so

x^2 \neq y^2 + 8^2

: triangle

ABC

is not right-angled.

Question 4

(a)Answer:

P(\text{blue}) = \dfrac{1}{4}

[2]

\mathstrut \text{Total} = 7 + 5 + 3 + 5

\text{Total} = 20

\mathstrut P(\text{blue}) = \dfrac{5}{20}

P(\text{blue}) = \dfrac{1}{4}

(b)Answer:

P = \dfrac{3}{10}

[2]

From the table:

6

6 + 14 = 20

\mathstrut P = \dfrac{6}{20}

P = \dfrac{3}{10}

(c)(i)Answer:

\text{Blue} = 9

[2]

\dfrac{9}{n} = \dfrac{1}{2}

\mathstrut n = 9 \times \dfrac{2}{1}

n = 18

\mathstrut \text{Blue} = 18 - 9

\text{Blue} = 9

(c)(ii)Answer:

x = 5

[3]

Red balls:

7 - x

; all the balls:

12 - x

\dfrac{7 - x}{12 - x} = \dfrac{2}{7}

7(7 - x) = 2(12 - x)

49 - 7x = 24 - 2x

5x = 25

x = 5

Question 5

(a)Answer:

2

[1]

\text{Gradient} = \dfrac{4 - 2}{-1 - (-2)}

\phantom{\text{Gradient}} = 2

(b)Answer:

y = 2x + 6

[2]

2 = 2 \times (-2) + c

c = 6

y = 2x + 6

(c)Answer:

(-3, 0)

[1]

0 = 2x + 6

x = -3

P(-3, 0)

(d)Answer:

9

units²[2]

Q(0, 6)

, so

OP = 3

and

OQ = 6

\text{Area} = \dfrac{1}{2} \times 3 \times 6

\phantom{\text{Area}} = 9

units²

(e)Answer:

(-2, 2)

[2]

\dfrac{1}{2} \times 3 \times h = 3

h = 2

, so

y = 2

2 = 2x + 6

x = -2

R(-2, 2)

Question 6

(a)Answer:

130.9^\circ

°[3]

\dfrac{\sin PRQ}{13} = \dfrac{\sin 24^\circ}{7}

\sin PRQ = 0.7553\ldots

The calculator gives

\sin^{-1}(0.7553\ldots) = 49.0575\ldots^\circ

, an acute angle.Angle

PRQ

is obtuse, so it is

180^\circ

minus this.

PRQ = 180^\circ - 49.0575\ldots^\circ

\phantom{PRQ} = 130.9424\ldots^\circ

\phantom{PRQ} = 130.9^\circ

(b)Answer:

25.1^\circ

°[1]

Angle

PQR = 180^\circ - 24^\circ - 130.9424\ldots^\circ

PQR = 25.0575\ldots^\circ

PQR = 25.1^\circ

(c)Answer:

7.29

cm[2]

\dfrac{PR}{\sin 25.0575\ldots^\circ} = \dfrac{7}{\sin 24^\circ}

PR = \dfrac{7 \sin 25.0575\ldots^\circ}{\sin 24^\circ}

= 7.2889\ldots

\phantom{} = 7.29

(d)Answer:

19.3

cm²[2]

\text{Area} = \dfrac{1}{2} \times 13 \times 7.2889\ldots \times \sin 24^\circ

= 19.2705\ldots

\phantom{} = 19.3

cm²

Question 7

(a)(i)Answer:

90^\circ

°[1]

\angle ACB = 90^\circ

(∠ in semicircle)

(a)(ii)Answer:

70^\circ

°[1]

\angle COB = 2 \times 35^\circ

(∠ at centre

= 2

∠ at circumference)

\angle COB = 70^\circ

(a)(iii)Answer:

72^\circ

°[2]

\angle OCT = 180^\circ - \angle COT - \angle CTO

(∠ sum of △)

\angle OCT = 180^\circ - 70^\circ - 38^\circ

\angle OCT = 72^\circ

(a)(iv)[1]

\angle OCT = 72^\circ

, not

90^\circ

.A tangent at

C

is perpendicular to the radius

OC

(tan. ⊥ rad.), so

CT

is not a tangent.

(a)(v)[2]

\angle OEC = 90^\circ

, as

CE

is perpendicular to

AB

OC

subtends a right angle at

E

, so

E

lies on the circle with

OC

as diameter (converse of ∠ in semicircle).

(b)(i)Answer:

23

[2]

4x > 26 + 3

x > \dfrac{29}{4}

B = \{8, 9, \ldots, 30\}

n(B) = 30 - 8 + 1

\phantom{n(B)} = 23

(b)(ii)Answer:

\{8, 16\}

[1]

A = \{1, 2, 4, 8, 16\}

A \cap B = \{8, 16\}

(the factors of

16

that are at least

8

)

(b)(iii)Answer:

\{3, 5, 6, 7\}

[1]

B' = \{1, 2, 3, 4, 5, 6, 7\}

A' \cap B' = \{3, 5, 6, 7\}

(the elements of

B'

that are not factors of

16

)

(b)(iv)Answer:

A'

[1]

The shaded part is

A'

(b)(v)Answer:

A \cap C = \varnothing

[1]

C = \{3, 6, 9, 12, 15, 18, 21, 24, 27, 30\}

A = \{1, 2, 4, 8, 16\}

B = \{8, 9, \ldots, 30\}

A \cap C = \varnothing

is the true statement.

Question 8

(a)(i)Answer:

\dfrac{5}{36}

[1]

\text{Outcomes with sum } 8: 5

P = \dfrac{5}{36}

\phantom{P} = \dfrac{5}{36}

(a)(ii)Answer:

\dfrac{11}{36}

[1]

\text{Outcomes with product greater than } 15: 11

P = \dfrac{11}{36}

\phantom{P} = \dfrac{11}{36}

(a)(iii)Answer:

\dfrac{1}{6}

[1]

P = \dfrac{6}{36}

\phantom{P} = \dfrac{1}{6}

(b)(i)Answer:

\begin{pmatrix} 310 \\ 286 \end{pmatrix}

[2]

\text{Row } 1: 15 \times 12 + 5 \times 8 + 15 \times 6 = 310

\text{Row } 2: 14 \times 12 + 8 \times 8 + 9 \times 6 = 286

PQ = \begin{pmatrix} 310 \\ 286 \end{pmatrix}

(b)(ii)[1]

The top element is the total amount, in dollars, from tickets sold on Friday.The bottom element is the total amount, in dollars, from tickets sold on Saturday.

(b)(iii)Answer:

596

[1]

310 + 286 = 596

Question 9

(a)Answer:

\begin{pmatrix} -12 \\ -8 \end{pmatrix}

[1]

\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}

= \begin{pmatrix} -6 \\ -12 \end{pmatrix} - \begin{pmatrix} 6 \\ -4 \end{pmatrix}

= \begin{pmatrix} -12 \\ -8 \end{pmatrix}

(b)Answer:

14.4

units[2]

|\overrightarrow{AB}| = \sqrt{(-12)^2 + (-8)^2}

= \sqrt{208}

= 14.4

(c)Answer:

\overrightarrow{OP} = \begin{pmatrix} 3 \\ -6 \end{pmatrix}

[2]

\overrightarrow{AP} = \dfrac{1}{4}\overrightarrow{AB}

= \begin{pmatrix} -3 \\ -2 \end{pmatrix}

\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP}

= \begin{pmatrix} 6 \\ -4 \end{pmatrix} + \begin{pmatrix} -3 \\ -2 \end{pmatrix}

= \begin{pmatrix} 3 \\ -6 \end{pmatrix}

(d)Answer:

h = 4

[2]

\overrightarrow{AE} = k\overrightarrow{AB}

\begin{pmatrix} 12 \\ h - (-4) \end{pmatrix} = k\begin{pmatrix} -12 \\ -8 \end{pmatrix}

k = -\dfrac{12}{12}

= -1

h - (-4) = -1 \times (-8)

h = 4

Question 10

(a)Answer:

18

minutes[2]

\text{Time} = \dfrac{18}{60} \text{ h}

\phantom{\text{Time}} = \dfrac{18}{60} \times 60 \text{ min}

\phantom{\text{Time}} = 18 \text{ min}

(b)Answer: Ferry E, departing at 1320[2]

At the terminal: 1140 + 18 min = 1158Check in at least 30 min before: the ferry must depart at 1228 or later.The earliest such ferry is ferry E, at 1320.

(c)Answer:

100

minutes[2]

\text{Time} = \dfrac{60}{36} \text{ h}

\phantom{\text{Time}} = \dfrac{60}{36} \times 60 \text{ min}

\phantom{\text{Time}} = 100 \text{ min}

(d)Answer:

15

minutes[3]

Arrival on Bintan: 1320 + 100 min = 1500 (Singapore time)Local time: 1500 − 1 h = 1400The next shuttle leaves at 1415: from 1400 to 1415 is 15 min.

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