Solutions

(a) Work out ²⁄₅ − ¹⁄₆.

LCM of 5 and 6 is 30.

²⁄₅ = ¹²⁄₃₀

¹⁄₆ = ⁵⁄₃₀

¹²⁄₃₀ − ⁵⁄₃₀ = ⁷⁄₃₀

(b) Work out 1.25 ÷ 0.5.

Multiply both by 10: 12.5 ÷ 5

= 2.5

Evaluate ¹⁄₁₀ × 6 ÷ 30%.

Convert 30% to a fraction: 30% = ³⁄₁₀

Rewrite: ⁶⁄₁₀ ÷ ³⁄₁₀

Multiply by the reciprocal:

⁶⁄₁₀ × ¹⁰⁄₃ = ⁶⁄₃ = 2

Express 605 as a product of its prime factors.

605 ÷ 5 = 121

121 ÷ 11 = 11

So 605 = 5 × 11 × 11 = 5 × 11²

Write 33¹⁄₃% as a fraction in its simplest form.

Convert the mixed number to an improper fraction:

33¹⁄₃ = ¹⁰⁰⁄₃

Divide by 100 to remove the percentage sign:

¹⁰⁰⁄₃ ÷ 100 = ¹⁰⁰⁄₃ × ¹⁄₁₀₀ = ¹⁄₃

Given v² = u² + 2ax.

(a) Find v when u = 8, a = 4 and x = 10.

v² = 8² + 2(4)(10)

v² = 64 + 80 = 144

v = √144 = 12

(b) Make a the subject of the formula.

v² − u² = 2ax

a = ^{v2 − u2}⁄_2x

Describe the locus represented by line l.

The arcs shown at Q indicate that line l divides angle PQR into two equal angles.

The locus is the angle bisector of angle PQR — the set of all points equidistant from lines QP and QR.

Find the area of the shaded region in the form a + bπ.

Triangle OAB has base AB = 22 cm and height OC = 6 cm (radius perpendicular to tangent).

Area of triangle OAB = ¹⁄₂ × 22 × 6 = 66 cm²

The sector has central angle ∠AOC + ∠BOC = 60° + 60° = 120°.

Area of sector = ¹²⁰⁄₃₆₀ × π × 6² = ¹⁄₃ × 36π = 12π cm²

Shaded area = 66 − 12π

(a) Work out 3.2 × 10⁸ − 1.5 × 10⁷ in standard form.

Rewrite 1.5 × 10⁷ as 0.15 × 10⁸.

3.2 × 10⁸ − 0.15 × 10⁸ = (3.2 − 0.15) × 10⁸

= 3.05 × 10⁸

(b) Work out (1⁷⁄₉)^−1/2.

Convert 1⁷⁄₉ to an improper fraction: ¹⁶⁄₉

(¹⁶⁄₉)^−1/2 = (⁹⁄₁₆)^1/2

= ^√9⁄_√16 = ³⁄₄

A car travels 75 km in 1 hour 15 minutes. Calculate the speed in km/h.

Time = 1 hour 15 minutes = 1¹⁵⁄₆₀ = 1.25 hours

Speed = ^Distance⁄_Time = ⁷⁵⁄_1.25 = 60 km/h

Solve 3x − 1 = 11.

3x = 11 + 1

3x = 12

x = 4

Solve the simultaneous equations:

2x = ¹⁄₃y … (1)

y = 9x − 3 … (2)

From (1): y = 6x

Substitute into (2):

6x = 9x − 3

3 = 3x

x = 1

y = 6(1) = 6

(a) Calculate the length of BC.

By Pythagoras: BC² = AB² + AC²

BC² = 8² + 15² = 64 + 225 = 289

BC = √289 = 17 km

(b) Calculate the bearing of C from B.

Angle at B in triangle ABC = 180° − 90° − 28° = 62°

The north line at B is vertically up. BA points due south (bearing 180°).

Angle from BA to BC = 62°, measured clockwise toward C.

Bearing of C from B = 180° − 62° = 118°

Complete the pie chart table.

Total animals = 12 + 18 + 4 + 2 = 36

Sector angle per animal = ^360°⁄₃₆ = 10°

Horses: 4 × 10° = 40°

Donkeys: 2 × 10° = 20°

Check: 120° + 180° + 40° + 20° = 360° ✓

(a) Find the length of line segment AB where A(1, 3) and B(2, 4).

AB = √((2 − 1)² + (4 − 3)²)

= √(1² + 1²)

= √2

(b) Find the x-coordinate of C(a, 8).

Gradient of AB = ^{4 − 3}⁄_{2 − 1} = 1

Since the points are collinear, the gradient of BC is also 1:

^{8 − 4}⁄_{a − 2} = 1

⁴⁄_{a − 2} = 1

a − 2 = 4

a = 6

Calculate the size of one of the four equal angles.

Interior angle sum of a hexagon = (6 − 2) × 180° = 720°

Two angles are right angles: 2 × 90° = 180°

Remaining sum = 720° − 180° = 540°

The four equal angles each = ^540°⁄₄ = 135°

(a) Find the size of a°.

AB is tangent at A, so ∠OAB = 90°.

a = ∠OAB − ∠OAC = 90° − 50° = 40°

(b) Find the size of b°.

In triangle OAC: OA = OC (radii), so it's isosceles.

∠OCA = ∠OAC = 50°

∠AOC = 180° − 50° − 50° = 80°

b = 80°