Question 1 percentages Question 2 solve equations Question 3 area of parallelogram

1

2008 practice (June 06) 43005 H2

Question 1 percentages

Question 2 solve equations

Question 3 area of parallelogram

Question 4 given the circumference, calculate the diameter

Question 5 area of a photo frame

Question 6 solve an equation using trial and improvement

Question 7 find the length of a side in a triangle

Question 8 solve an inequality

Question 9 construct an angle bisector

Question 10 multiply out and simplify

Question 11 find an angle in a RAT; calculate a side in a RAT

Question 12 find an angle in a circle; give reasons

Question 13 solve a quadratic equation

Question 14 use y = mx+c to identify parallel and perpendicular lines

Question 15 solve a non-right angled triangle

Question 16 from surface area of a sphere, calculate its volume

Question 17 solve an equation involving algebraic fractions

Grade boundaries

2


1. (a) Calculate 36% of £420

2 marks

36% = 0.36 as a decimal

0.36 x 420 = 151.2

1 (b) Mike says that 18% of £840 is equal to 36% of £420

Explain why Mike is correct. Do not calculate 18% of £840.

1 mark

1 (c) 9% of £x is equal to 36% of £420

Work out the value of x

2 marks

Answer: the percentage has been halved; the amount has been doubled; the effect is to leave the answer the same

the percentage has been halved again so the amount must be doubled again

double 840 = 1680

Answer = £151.20

Answer x = 1680

3


2 Solve the equations

(a) 4(z + 3) = 8

(b) 3t + 4 = 19 – 2t

divide by 4

z + 3 = 2 z = -1 3 marks

OR multiply out the bracket by 4

4z + 12 = 8 4z = -4 z = -1

add 2t to both sides 5t + 4 = 19

subtract 4 5t = 15

divide by 5 t = 3 3 marks

- 2t

4


3 XYZT is a parallelogram

The base XT = 11.6cm and the perpendicular height is 7.7 cm

Calculate the area of the parallelogram.

Area of parallelogram = base x height

2 marks

Area of parallelogram = 11.6 x 7.7

= 89.32 = 89.3 cm2 to 3 sig figs

Y Z

X T

7.7 cm

11.6 cm

5


2 marks

4 The circumference of a circle measures 26.7 cm

Work out the length of the diameter of the circle.Circumference = π x diameter

so 26.7 = π x diameter

so diameter = 26.7 ÷ π

= 8.498873961 = 8.50 cm to 3 sig figs

6


5 A rectangular photo is surrounded by a frame which is 4cm wide.

The outer measurements of the frame are 21.5cm by 17.2cm.

Calculate the area of the frame. This area is shaded in the diagram.

photo

21.5 cm

17.2cm

4 cm

4cm

Area of frame and photo = 21.5 x 17.2

= 369.8

Size of photo is 8cm less than size of frame

Area of photo = 13.5 x 9.2

= 124.2

Area of frame = 369.8 – 124.2

= 245.6

= 246 cm2 (to 3 sig figs)

5 marks

7


3.85 3.853 + 4x3.85 = 72.466 too large

3.9 3.93 + 4x3.9 = 74.919 too large

3.8 3.83 + 4x3.8 = 70.072 too small

3.7 3.73 + 4x3.7 = 65.453 too small

3.5 3.53 + 4x3.5 = 56.875 too small

3 marks

6 Liam is using trial and improvement to find a solution to the equation x3 + 4x = 72

The table shows the first two trials

Continue the table to

find a solution to the

equation.

Give your answer to

1 decimal place.

x x3 + 4x comment

3 33 + 4x3 = 39 Too small

4 43 + 4x4 = 80 Too large

x = 3.8 to 1 decimal place

8


4 marks

Find the RAT – use Pythagoras’ rule

h2 = 182 - 92

7 The diagram shows an equilateral triangle of side 18cm.

Calculate the height of the triangle (marked h in the diagram)

h2 = 243

h = √243

h = 15.58845727

= 15.6 cm (to 3 sig figs)

18 cm 18 cm

18 cm

h18 cm 18 cm

18 cm

h

9 cm

9


8 (a) Solve the inequality 5x + 3 18

Subtract 3 from both sides 5x 15

Divide by 5

-3 y 0

3 marks

x 3

(b) y is an integer

Write down all the solutions of the inequality -6 2y 0

Divide by 2

The list of solutions is -3, -2 and -1

2 marks

10


9 Using a ruler and compasses only, construct the bisector of angle DEF

D

E F

D

E F

D

E F

2 marks

11


(b) Expand and simplify (p + 7)(p + 2)

2 marks

times p 7

p

2

p2

7p

2p 14

= p2 + 9p +14

10 (a) Multiply out 2p2q3 x 3p5q

2p2q3 x 3p5q

2 x 3 = 6

p2 x p5 = p7 q3 x q = q4

Answer = 6p7q4

2 marks

12


10 (c) solve the equation x +13

= 5 – 2x

Multiply by 3

x + 1 = 15 – 6x

Add 6x 7x + 1 = 15

Subtract 1 7x = 14

Divide by 7 x = 2

3 marks

13


11 (a) In triangle ABC, angle A = 900, AB = 16cm and AC = 7cm.

Calculate the value of x.

3 marks

7 cm

16 cm

x

RATs with angles means

SOHCAHTOA

opposite

adjacent

T

O

A

tan x = opposite ÷ adjacent tan x = 16 ÷ 7 x = tan-1(16 ÷ 7)

x = 660 (to the nearest degree)

14


11 (b) In triangle PQR, angle Q = 900, angle R = 370 and QR = 12.6cm.

Calculate the length of PR.

RATs with angles means

SOHCAHTOA

37°

12.6 cm

P

Q R C

A

H

hypotenuse

adjacent

hypotenuse = adjacent ÷ cosine of the anglePR = 12.6 ÷ cos 37PR = 15.77690929

PR = 15.8 cm (to 3 sig figs)

3 marks

15

2008 practice (June 06) 43005 H212 In the diagram, A, P, Q and B are points on the circumference of the circle.

Angle APB = 470.

Find the value of x.

Give a reason for your answer.

P

Q

B

A

47°

x

angle APB = angle AQB

These are equal angles in the same segment, standing on the same chord AB

x = 470 2 marks

16


13 Solve the equation y2 – 3y - 14 = 0

Give your answer to 2 decimal places. You must show your working.

Compare y2 – 3y - 14 = 0 with ax2 + bx + c = 0

a = 1, b = -3 and c = -14

… helpful formula on page 2

y = – (–3) ± (–3)2

– 4x 1 x (–14)2

y = 3 ± 652

x = -2.53 or x = 5.53 (to 2 dec pl)

3 marks

17

2008 practice (June 06) 43005 H214 (a) The equations of four lines are given below

Line P: y = 3x + 5

Line Q: y = 4 – 3x

Line R: y + 3x = 8

Line S: y – 3x = 1

(i) Name the lines that are parallel to the line y = 3x.

(ii) Which line goes through the point (2, 7) ?

(b) Write down the gradient of the line y + 2x = 7

1 mark

1 mark

1 mark

Rewrite in the form y = mx + c

y = -3x + 4

y = -3x + 8

y = 3x + 1

lines P and S have gradient = 3

when x = 2, y = 7 only on line S

rewrite as y = -2x + 7

gradient = -2

18

2008 practice (June 06) 43005 H215 D, E and F are three points on level ground.

TF is the vertical side of a building.

The angle of elevation of the top of the building from D is 160.

The angle of elevation of the top of the building from E is 350.

The distance DE is 30m.

Calculate the height of the building marked h in the diagram.

5 marks

16° 35°

30 cm

h

T

FD E

19


16° 35°

30 cm

h

T

FD E

x

Use the sine rule to find x in this triangle DET

xsin 16

= 30sin 19

145

19

x = 30 x sin 16 ÷ sin 19 x = 25.39904644

Use SOHCAHTOA to find h from x in the RAT, TEF h = 25.399 x sin 35

h = 14.6 cm (to 3 sig figs)

5 marks

20


16 The surface area of a sphere is 1380 cm2.

Calculate the volume of the sphere.

State the units of your answer.

Helpful formulas from page 2

4πr2 = 1380

r2 = 1380 ÷ (4xπ) Use brackets on your calculator

r = √109.8169107 r = 10.47935641

Volume = 4 π r3 ÷ 3

= 4 π x 10.4793 ÷ 3

= 4820 cm3 (to 3 sig figs)

5 marks

21

2008 practice (June 06) 43005 H217 Solve the equation

3y +7

– 1y +6

= 12

5 marks

Multiply by (y + 7)

Multiply by (y + 6)

Multiply by 2 6(y + 6) – 2(y + 7) = (y + 6)(y + 7)

Multiply out

3 – y +7y +6

= y +72

3(y +6) – (y +7) = (y +6)(y +7)2

6y + 36 – 2y - 14 = y2 +13y + 42

Rearrange to quadratic equation

0 = y2 +9y + 20

Factorise to solve (y + 5)(y + 4) = 0 y = -5 or y = -

4

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Total: out of 70 a rough guide

grade D C B A A*

score 16 26 36 46 56

Question 1 percentages Question 2 solve equations Question 3 area of parallelogram

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