Sec 3 E-Math Exam Paper 1: free PDF with answers

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Mathematics

Sec 3 E-Math · Exam Paper 1

Algebraic manipulation + Quadratic equations + Inequalities + Indices, compound interest and standard form + Coordinate geometry + Graphs of functions + Further trigonometry + Applications of trigonometry + Arc length and sector area + Congruence and similarity + Area and volume of similar figures + Circle properties

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

1 hour 30 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 60.

For Examiner's Use

/ 60

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)Expand and simplify

(2x + 2)(4x + 1)

Answer (a)[1]

(b)Factorise

7x^{2} + 39x + 20

Answer (b)[1]

(c)Factorise

6h^{2} + 13hk + 5k^{2}

completely.

Answer (c)[2]

(a)The diagram shows the curve

y = (x - a)(x - b)

. Find the equation of the curve.

Answer (a)[2]

(b)The length of a rectangle is

9

cm more than its width. Its area is

133

cm². Let the width be

x

cm. Find the perimeter of the rectangle, correct to 2 decimal places.

Answer (b)[3]

(a)Find the largest integer value of

x

that satisfies

3x - 5 < -8

Answer (a)[2]

(b)Solve

\dfrac{x - 1}{4} > \dfrac{3x + 3}{3}

Answer (b)[2]

(a)Write

7.02 \times 10^{-2}

as an ordinary number.

Answer (a)[1]

(b)Simplify

(5a^{3})^{2}

Answer (b)[2]

(c)Given that

a = 2^{x}

, express

8^{x + 2}

in terms of

a

Answer (c)[2]

Find the equation of the line passing through

A(-2, -3)

and

B(-1, -2)

Answer[3]

Triangle

ABC

has vertices

A(1, -2)

B(3, 5)

and

C(-2, -4)

. Find the perimeter of the triangle, correct to 3 significant figures.

Answer[3]

(a)The diagram shows the speed–time graph of a lorry that starts from rest. The maximum speed,

V

, of the lorry is 54 km/h. Find the acceleration, in m/s², of the lorry in the first 12 seconds.

Answer (a)m/s²[2]

(b)The diagram shows the graph of

y = \dfrac{4}{x} + 1

and the line

y = -3

. Use the graph to solve the equation

\dfrac{4}{x} + 1 = -3

Answer (b)[2]

The diagram shows the curve

y = -x^{2} + 2x + 3

and the tangent to the curve at

x = 3

. Use the two points marked on the tangent to find the gradient of the curve at

x = 3

Answer[2]

In triangle

ABC

BC = 8

cm,

AC = 6

cm and

AB = 11

cm. Find angle

BAC

, correct to 1 decimal place.

Answer°[3]

Given that

\sin x = 0.75

and

x

is obtuse, find

x

, correct to 1 decimal place.

Answer°[3]

(a)From a point

60

m from the foot of a flagpole on level ground, the angle of elevation of the top is

41^\circ

. Find the height of the flagpole, correct to 3 significant figures.

Answer (a)m[2]

(b)The diagram shows a cuboid with

AB = 10

cm,

BC = 6

cm and

CG = 3

cm. Find the angle between

AG

and the base

ABCD

, correct to 1 decimal place.

Answer (b)°[3]

A sector of a circle has radius

8

cm and angle

0.8

radians. Find its area.

Answercm²[2]

AB

is a chord of a circle with centre

O

and radius

9

cm, and angle

AOB = 1.8

radians. Find the area of the shaded segment, correct to 3 significant figures.

Answercm²[3]

Triangles

ABC

and

PQR

are congruent, with the equal parts marked. Complete the statement

\triangle ABC \equiv \triangle \ldots

Answer[2]

In the diagram,

D

lies on

AB

and

E

lies on

AC

, with

DE

parallel to

BC

. Find

DE

, correct to 3 significant figures.

Answercm[3]

(a)Triangles

ABC

and

PQR

are similar, with areas

15

cm² and

29.4

cm² respectively.

AB = 5

cm. Find

PQ

Answer (a)cm[2]

(b)An inverted cone of height

12

cm has a capacity of

480

cm³. It contains

60

cm³ of water. Find the depth of the water.

Answer (b)cm[2]

In the diagram,

A

B

C

and

D

lie on the circle with centre

O

. Find

x

Answer°[2]

In the diagram,

AB

is a chord of the circle with centre

O

, and

OM

is perpendicular to

AB

AB = 14

cm. Find

x

, correct to 3 significant figures.

Answercm[3]

END OF PAPER

Answers and worked solutions

Question 1

(a)Answer:

8x^2+10x+2

[1]

8x^2+10x+2

(b)Answer:

(7x+4)(x+5)

[1]

(7x+4)(x+5)

(c)Answer:

= (2h + k)(3h + 5k)

[2]

10 \times 3 = 30

and

10 + 3 = 13

6h^{2} + 13hk + 5k^{2} = 6h^{2} + 10hk + 3hk + 5k^{2}

= 2h(3h + 5k) + k(3h + 5k)

= (2h + k)(3h + 5k)

Question 2

(a)Answer:

y = -5

, as on the diagram.[2]

The curve meets the

x

-axis at

x = -5

and

x = 1

a = -5,\ b = 1

y = (x - 1)(x + 5)

x = 0

y = (-1)(5)

y = -5

, as on the diagram.

(b)Answer:

= 49.52

cm (2 d.p.)[3]

length × width = area

x(x + 9) = 133

x^{2} + 9x - 133 = 0

a = 1,\ b = 9,\ c = -133

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \dfrac{-9 \pm \sqrt{(9)^2 - 4(1)(-133)}}{2(1)}

x = \dfrac{-9 \pm \sqrt{613}}{2}

x = 7.8794\ldots

x = -16.8794\ldots

x = 7.88

x = -16.88

(2 d.p.)

x = -16.88

is rejected because a length cannot be negative. So

x = 7.88

.Perimeter

= 2(7.8794\ldots + 7.8794\ldots + 9)

= 49.52

cm (2 d.p.)

Question 3

(a)Answer: In

x < -1

, the largest integer is

-2

.[2]

3x < -3

x < -1

x < -1

, the largest integer is

-2

(b)Answer:

x < -\dfrac{5}{3}

(dividing by a negative number reverses the sign)[2]

3(x - 1) > 4(3x + 3)

3x - 3 > 12x + 12

-9x > 15

x < -\dfrac{5}{3}

(dividing by a negative number reverses the sign)

Question 4

(a)Answer:

7.02 \times 10^{-2} = 0.070\,2

[1]

A negative power of 10 makes the number smaller: the point moves left.Number of places the point moves:

2

7.02 \times 10^{-2} = 0.070\,2

(b)Answer:

25a^{6}

[2]

Numbers:

5^{2} = 25

a

: multiply the indices,

3 \times 2 = 6

25a^{6}

(c)Answer:

= 64a^{3}

[2]

8 = 2^{3}

8^{x + 2} = 2^{3(x + 2)}

= 2^{3x} \times 2^{6}

= (2^{x})^{3} \times 64

= 64a^{3}

Question 5

Answer:

y = x - 1

[3]

m = \dfrac{y_2 - y_1}{x_2 - x_1}

m = \dfrac{-2 - (-3)}{-1 - (-2)}

\phantom{m} = 1

-3 = 1(-2) + c

c = -1

y = x - 1

Question 6

Answer:

= 21.2

units (3 s.f.)[3]

AB = \sqrt{4 + 49}

\phantom{AB} = 7.2801\ldots

BC = \sqrt{25 + 81}

\phantom{BC} = 10.2956\ldots

CA = \sqrt{9 + 4}

\phantom{CA} = 3.6055\ldots

Perimeter

= 21.1812\ldots

= 21.2

units (3 s.f.)

Question 7

(a)Answer:

\phantom{\text{Acceleration}} = 1.25 \text{ m/s}^2

m/s²[2]

V = \dfrac{54 \times 1000}{3600}

\phantom{V} = 15 \text{ m/s}

\text{Acceleration} = \dfrac{15}{12}

\phantom{\text{Acceleration}} = 1.25 \text{ m/s}^2

(b)Answer:

x = -1

[2]

Left-hand side:

y = \dfrac{4}{x} + 1

, the curveRight-hand side:

y = -3

, the horizontal lineWhere the curve and the line meet, both sides have the same

y

-value for the same

x

, so the equation is true there.Number of points where they meet:

1

x = -1

Question 8

Answer:

\phantom{m} = -4

[2]

A = (2, 4)

B = (4, -4)

m = \dfrac{y_2 - y_1}{x_2 - x_1}

m = \dfrac{-4 - 4}{4 - 2}

\phantom{m} = -4

Question 9

Answer:

\phantom{\theta} = 45.2^\circ

(1 d.p.) °[3]

\text{Cosine rule}

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

a = 8

(the side opposite the angle),

b = 6

c = 11

\cos \theta = \dfrac{6^2 + 11^2 - 8^2}{2(6)(11)}

\phantom{\cos \theta} = 0.7045\ldots

\theta = \cos^{-1}(0.7045\ldots)

\theta = 45.2071\ldots^\circ

\phantom{\theta} = 45.2^\circ

(1 d.p.)

Question 10

Answer:

\phantom{x} = 131.4^\circ

(1 d.p.) °[3]

\sin x = 0.75

Acute angle

= \sin^{-1}(0.75)

= 48.5903\ldots^\circ

\sin x = \sin(180^\circ - 48.5903\ldots^\circ)

x = 180^\circ - 48.5903\ldots^\circ

\phantom{x} = 131.4096\ldots^\circ

\phantom{x} = 131.4^\circ

(1 d.p.)

Question 11

(a)Answer:

\phantom{h} = 52.2

m (3 s.f.)[2]

\tan 41^\circ = \dfrac{h}{60}

h = 60 \tan 41^\circ

h = 52.1572\ldots

\phantom{h} = 52.2

m (3 s.f.)

(b)Answer:

\phantom{\theta} = 14.4^\circ

(1 d.p.) °[3]

AC^2 = 10^2 + 6^2

AC^2 = 136

AC = \sqrt{136}

The angle between

AG

and the base is

\angle GAC

(

G

is directly above

C

\tan \theta = \dfrac{CG}{AC}

\tan \theta = \dfrac{3}{\sqrt{136}}

\theta = \tan^{-1}\left(\dfrac{3}{\sqrt{136}}\right)

\theta = 14.4264\ldots^\circ

\phantom{\theta} = 14.4^\circ

(1 d.p.)

Question 12

Answer:

A = 25.6

cm²[2]

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

\mathstrut A = \dfrac{1}{2} \times 8^2 \times 0.8

A = 25.6

cm²

Question 13

Answer:

\phantom{\text{Segment}} = 33.5

cm² (3 s.f.)[3]

\mathstrut \text{Sector} = \dfrac{1}{2} \times 9^2 \times 1.8

\text{Sector} = 72.9

\text{Triangle} = \dfrac{1}{2} \times 9 \times 9 \times \sin 1.8

\text{Triangle} = 39.4408\ldots

\mathstrut \text{Segment} = \text{Sector} - \text{Triangle}

\text{Segment} = 33.4591\ldots

\phantom{\text{Segment}} = 33.5

cm² (3 s.f.)

Question 14

Answer: RQP[2]

\angle C

(

90^\circ

) matches

\angle \text{P}

AB

(

13

cm) matches

\text{RQ}

BC

(

11

cm) matches

\text{QP}

A

matches

R

B

matches

Q

C

matches

P

\triangle ABC \equiv \triangle \text{RQP}

Question 15

Answer:

\phantom{DE} = 3.33

cm (3 s.f.)[3]

\angle ADE = \angle ABC

(corresponding angles)

\angle A

is common, so the triangles are similar (AA).

\mathstrut AB = 3 + 6

AB = 9

\dfrac{DE}{BC} = \dfrac{AD}{AB}

\mathstrut \dfrac{DE}{10} = \dfrac{3}{9}

DE = \dfrac{3}{9} \times 10

DE = 3.3333\ldots

\phantom{DE} = 3.33

cm (3 s.f.)

Question 16

(a)Answer:

PQ = 7

cm[2]

\dfrac{l_1}{l_2} = \sqrt{\dfrac{A_1}{A_2}}

\mathstrut \dfrac{PQ}{5} = \sqrt{\dfrac{29.4}{15}}

\dfrac{PQ}{5} = \dfrac{7}{5}

PQ = \dfrac{7}{5} \times 5

PQ = 7

(b)Answer:

d = 6

cm[2]

\dfrac{d}{H} = \sqrt[3]{\dfrac{\text{volume of water}}{\text{volume of cone}}}

\mathstrut \dfrac{d}{12} = \sqrt[3]{\dfrac{60}{480}}

\dfrac{d}{12} = \dfrac{1}{2}

d = \dfrac{1}{2} \times 12

d = 6

Question 17

Answer:

x = 100

°[2]

\mathstrut \angle ADC = \dfrac{160^\circ}{2}

(∠ at centre = 2 ∠ at circumference)

\angle ADC = 80^\circ

\mathstrut x = 180^\circ - 80^\circ

(∠s in opp. segments)

x = 100

Question 18

Answer:

\phantom{x} = 7.14

cm (3 s.f.)[3]

AM = MB

(⊥ from centre bisects chord)

AM = 7

OM^2 = OA^2 - AM^2

\mathstrut x^2 = 10^2 - 7^2

x^2 = 51

x = \sqrt{51}

x = 7.1414\ldots

\phantom{x} = 7.14

cm (3 s.f.)

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Sec 3 E-Math Exam Paper 1: free PDF with answers

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MugCornerT-Leng Tuition · mugcornersg.com

Mathematics

Sec 3 E-Math · Exam Paper 1

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

1 hour 30 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 60.

For Examiner's Use

/ 60

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)Expand and simplify

(2x + 2)(4x + 1)

Answer (a)[1]

(b)Factorise

7x^{2} + 39x + 20

Answer (b)[1]

(c)Factorise

6h^{2} + 13hk + 5k^{2}

completely.

Answer (c)[2]

(a)The diagram shows the curve

y = (x - a)(x - b)

. Find the equation of the curve.

Answer (a)[2]

(b)The length of a rectangle is

9

cm more than its width. Its area is

133

cm². Let the width be

x

cm. Find the perimeter of the rectangle, correct to 2 decimal places.

Answer (b)[3]

(a)Find the largest integer value of

x

that satisfies

3x - 5 < -8

Answer (a)[2]

(b)Solve

\dfrac{x - 1}{4} > \dfrac{3x + 3}{3}

Answer (b)[2]

(a)Write

7.02 \times 10^{-2}

as an ordinary number.

Answer (a)[1]

(b)Simplify

(5a^{3})^{2}

Answer (b)[2]

(c)Given that

a = 2^{x}

, express

8^{x + 2}

in terms of

a

Answer (c)[2]

Find the equation of the line passing through

A(-2, -3)

and

B(-1, -2)

Answer[3]

Triangle

ABC

has vertices

A(1, -2)

B(3, 5)

and

C(-2, -4)

. Find the perimeter of the triangle, correct to 3 significant figures.

Answer[3]

(a)The diagram shows the speed–time graph of a lorry that starts from rest. The maximum speed,

V

, of the lorry is 54 km/h. Find the acceleration, in m/s², of the lorry in the first 12 seconds.

Answer (a)m/s²[2]

(b)The diagram shows the graph of

y = \dfrac{4}{x} + 1

and the line

y = -3

. Use the graph to solve the equation

\dfrac{4}{x} + 1 = -3

Answer (b)[2]

The diagram shows the curve

y = -x^{2} + 2x + 3

and the tangent to the curve at

x = 3

. Use the two points marked on the tangent to find the gradient of the curve at

x = 3

Answer[2]

In triangle

ABC

BC = 8

cm,

AC = 6

cm and

AB = 11

cm. Find angle

BAC

, correct to 1 decimal place.

Answer°[3]

Given that

\sin x = 0.75

and

x

is obtuse, find

x

, correct to 1 decimal place.

Answer°[3]

(a)From a point

60

m from the foot of a flagpole on level ground, the angle of elevation of the top is

41^\circ

. Find the height of the flagpole, correct to 3 significant figures.

Answer (a)m[2]

(b)The diagram shows a cuboid with

AB = 10

cm,

BC = 6

cm and

CG = 3

cm. Find the angle between

AG

and the base

ABCD

, correct to 1 decimal place.

Answer (b)°[3]

A sector of a circle has radius

8

cm and angle

0.8

radians. Find its area.

Answercm²[2]

AB

is a chord of a circle with centre

O

and radius

9

cm, and angle

AOB = 1.8

radians. Find the area of the shaded segment, correct to 3 significant figures.

Answercm²[3]

Triangles

ABC

and

PQR

are congruent, with the equal parts marked. Complete the statement

\triangle ABC \equiv \triangle \ldots

Answer[2]

In the diagram,

D

lies on

AB

and

E

lies on

AC

, with

DE

parallel to

BC

. Find

DE

, correct to 3 significant figures.

Answercm[3]

(a)Triangles

ABC

and

PQR

are similar, with areas

15

cm² and

29.4

cm² respectively.

AB = 5

cm. Find

PQ

Answer (a)cm[2]

(b)An inverted cone of height

12

cm has a capacity of

480

cm³. It contains

60

cm³ of water. Find the depth of the water.

Answer (b)cm[2]

In the diagram,

A

B

C

and

D

lie on the circle with centre

O

. Find

x

Answer°[2]

In the diagram,

AB

is a chord of the circle with centre

O

, and

OM

is perpendicular to

AB

AB = 14

cm. Find

x

, correct to 3 significant figures.

Answercm[3]

END OF PAPER

Answers and worked solutions

Question 1

(a)Answer:

8x^2+10x+2

[1]

8x^2+10x+2

(b)Answer:

(7x+4)(x+5)

[1]

(7x+4)(x+5)

(c)Answer:

= (2h + k)(3h + 5k)

[2]

10 \times 3 = 30

and

10 + 3 = 13

6h^{2} + 13hk + 5k^{2} = 6h^{2} + 10hk + 3hk + 5k^{2}

= 2h(3h + 5k) + k(3h + 5k)

= (2h + k)(3h + 5k)

Question 2

(a)Answer:

y = -5

, as on the diagram.[2]

The curve meets the

x

-axis at

x = -5

and

x = 1

a = -5,\ b = 1

y = (x - 1)(x + 5)

x = 0

y = (-1)(5)

y = -5

, as on the diagram.

(b)Answer:

= 49.52

cm (2 d.p.)[3]

length × width = area

x(x + 9) = 133

x^{2} + 9x - 133 = 0

a = 1,\ b = 9,\ c = -133

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \dfrac{-9 \pm \sqrt{(9)^2 - 4(1)(-133)}}{2(1)}

x = \dfrac{-9 \pm \sqrt{613}}{2}

x = 7.8794\ldots

x = -16.8794\ldots

x = 7.88

x = -16.88

(2 d.p.)

x = -16.88

is rejected because a length cannot be negative. So

x = 7.88

.Perimeter

= 2(7.8794\ldots + 7.8794\ldots + 9)

= 49.52

cm (2 d.p.)

Question 3

(a)Answer: In

x < -1

, the largest integer is

-2

.[2]

3x < -3

x < -1

x < -1

, the largest integer is

-2

(b)Answer:

x < -\dfrac{5}{3}

(dividing by a negative number reverses the sign)[2]

3(x - 1) > 4(3x + 3)

3x - 3 > 12x + 12

-9x > 15

x < -\dfrac{5}{3}

(dividing by a negative number reverses the sign)

Question 4

(a)Answer:

7.02 \times 10^{-2} = 0.070\,2

[1]

A negative power of 10 makes the number smaller: the point moves left.Number of places the point moves:

2

7.02 \times 10^{-2} = 0.070\,2

(b)Answer:

25a^{6}

[2]

Numbers:

5^{2} = 25

a

: multiply the indices,

3 \times 2 = 6

25a^{6}

(c)Answer:

= 64a^{3}

[2]

8 = 2^{3}

8^{x + 2} = 2^{3(x + 2)}

= 2^{3x} \times 2^{6}

= (2^{x})^{3} \times 64

= 64a^{3}

Question 5

Answer:

y = x - 1

[3]

m = \dfrac{y_2 - y_1}{x_2 - x_1}

m = \dfrac{-2 - (-3)}{-1 - (-2)}

\phantom{m} = 1

-3 = 1(-2) + c

c = -1

y = x - 1

Question 6

Answer:

= 21.2

units (3 s.f.)[3]

AB = \sqrt{4 + 49}

\phantom{AB} = 7.2801\ldots

BC = \sqrt{25 + 81}

\phantom{BC} = 10.2956\ldots

CA = \sqrt{9 + 4}

\phantom{CA} = 3.6055\ldots

Perimeter

= 21.1812\ldots

= 21.2

units (3 s.f.)

Question 7

(a)Answer:

\phantom{\text{Acceleration}} = 1.25 \text{ m/s}^2

m/s²[2]

V = \dfrac{54 \times 1000}{3600}

\phantom{V} = 15 \text{ m/s}

\text{Acceleration} = \dfrac{15}{12}

\phantom{\text{Acceleration}} = 1.25 \text{ m/s}^2

(b)Answer:

x = -1

[2]

Left-hand side:

y = \dfrac{4}{x} + 1

, the curveRight-hand side:

y = -3

, the horizontal lineWhere the curve and the line meet, both sides have the same

y

-value for the same

x

, so the equation is true there.Number of points where they meet:

1

x = -1

Question 8

Answer:

\phantom{m} = -4

[2]

A = (2, 4)

B = (4, -4)

m = \dfrac{y_2 - y_1}{x_2 - x_1}

m = \dfrac{-4 - 4}{4 - 2}

\phantom{m} = -4

Question 9

Answer:

\phantom{\theta} = 45.2^\circ

(1 d.p.) °[3]

\text{Cosine rule}

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

a = 8

(the side opposite the angle),

b = 6

c = 11

\cos \theta = \dfrac{6^2 + 11^2 - 8^2}{2(6)(11)}

\phantom{\cos \theta} = 0.7045\ldots

\theta = \cos^{-1}(0.7045\ldots)

\theta = 45.2071\ldots^\circ

\phantom{\theta} = 45.2^\circ

(1 d.p.)

Question 10

Answer:

\phantom{x} = 131.4^\circ

(1 d.p.) °[3]

\sin x = 0.75

Acute angle

= \sin^{-1}(0.75)

= 48.5903\ldots^\circ

\sin x = \sin(180^\circ - 48.5903\ldots^\circ)

x = 180^\circ - 48.5903\ldots^\circ

\phantom{x} = 131.4096\ldots^\circ

\phantom{x} = 131.4^\circ

(1 d.p.)

Question 11

(a)Answer:

\phantom{h} = 52.2

m (3 s.f.)[2]

\tan 41^\circ = \dfrac{h}{60}

h = 60 \tan 41^\circ

h = 52.1572\ldots

\phantom{h} = 52.2

m (3 s.f.)

(b)Answer:

\phantom{\theta} = 14.4^\circ

(1 d.p.) °[3]

AC^2 = 10^2 + 6^2

AC^2 = 136

AC = \sqrt{136}

The angle between

AG

and the base is

\angle GAC

(

G

is directly above

C

\tan \theta = \dfrac{CG}{AC}

\tan \theta = \dfrac{3}{\sqrt{136}}

\theta = \tan^{-1}\left(\dfrac{3}{\sqrt{136}}\right)

\theta = 14.4264\ldots^\circ

\phantom{\theta} = 14.4^\circ

(1 d.p.)

Question 12

Answer:

A = 25.6

cm²[2]

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

\mathstrut A = \dfrac{1}{2} \times 8^2 \times 0.8

A = 25.6

cm²

Question 13

Answer:

\phantom{\text{Segment}} = 33.5

cm² (3 s.f.)[3]

\mathstrut \text{Sector} = \dfrac{1}{2} \times 9^2 \times 1.8

\text{Sector} = 72.9

\text{Triangle} = \dfrac{1}{2} \times 9 \times 9 \times \sin 1.8

\text{Triangle} = 39.4408\ldots

\mathstrut \text{Segment} = \text{Sector} - \text{Triangle}

\text{Segment} = 33.4591\ldots

\phantom{\text{Segment}} = 33.5

cm² (3 s.f.)

Question 14

Answer: RQP[2]

\angle C

(

90^\circ

) matches

\angle \text{P}

AB

(

13

cm) matches

\text{RQ}

BC

(

11

cm) matches

\text{QP}

A

matches

R

B

matches

Q

C

matches

P

\triangle ABC \equiv \triangle \text{RQP}

Question 15

Answer:

\phantom{DE} = 3.33

cm (3 s.f.)[3]

\angle ADE = \angle ABC

(corresponding angles)

\angle A

is common, so the triangles are similar (AA).

\mathstrut AB = 3 + 6

AB = 9

\dfrac{DE}{BC} = \dfrac{AD}{AB}

\mathstrut \dfrac{DE}{10} = \dfrac{3}{9}

DE = \dfrac{3}{9} \times 10

DE = 3.3333\ldots

\phantom{DE} = 3.33

cm (3 s.f.)

Question 16

(a)Answer:

PQ = 7

cm[2]

\dfrac{l_1}{l_2} = \sqrt{\dfrac{A_1}{A_2}}

\mathstrut \dfrac{PQ}{5} = \sqrt{\dfrac{29.4}{15}}

\dfrac{PQ}{5} = \dfrac{7}{5}

PQ = \dfrac{7}{5} \times 5

PQ = 7

(b)Answer:

d = 6

cm[2]

\dfrac{d}{H} = \sqrt[3]{\dfrac{\text{volume of water}}{\text{volume of cone}}}

\mathstrut \dfrac{d}{12} = \sqrt[3]{\dfrac{60}{480}}

\dfrac{d}{12} = \dfrac{1}{2}

d = \dfrac{1}{2} \times 12

d = 6

Question 17

Answer:

x = 100

°[2]

\mathstrut \angle ADC = \dfrac{160^\circ}{2}

(∠ at centre = 2 ∠ at circumference)

\angle ADC = 80^\circ

\mathstrut x = 180^\circ - 80^\circ

(∠s in opp. segments)

x = 100

Question 18

Answer:

\phantom{x} = 7.14

cm (3 s.f.)[3]

AM = MB

(⊥ from centre bisects chord)

AM = 7

OM^2 = OA^2 - AM^2

\mathstrut x^2 = 10^2 - 7^2

x^2 = 51

x = \sqrt{51}

x = 7.1414\ldots

\phantom{x} = 7.14

cm (3 s.f.)

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