Solutions

(a) Round off 0.9997 correct to one decimal place.

The digit in the second decimal place is 9, so we round up the first decimal place.

0.9997 → 1.0

(b) The sides of a rectangular plot are 3 m by 5 m correct to the nearest metre. Calculate the least possible perimeter.

Lower bound of 3 m = 2.5 m

Lower bound of 5 m = 4.5 m

Least perimeter = 2(2.5 + 4.5) = 2(7) = 14 m

Given m = 2 × 3^a × 7^b and n = 3 × 7² × 11.

(a) Given that m > 126, find the smallest values of a and b which make m a multiple of 126.

Prime factorisation of 126:

126 = 2 × 3² × 7

For m to be a multiple of 126, we need a ≥ 2 and b ≥ 1.

If a = 2 and b = 1, then m = 2 × 9 × 7 = 126, which is NOT greater than 126.

So increase one exponent to make m the next smallest multiple:

a = 3, b = 1: m = 2 × 27 × 7 = 378

Check: 378 ÷ 126 = 3 ✓ and 378 > 126 ✓

a = 3, b = 1

(b) Explain why 33n is a perfect square.

33n = 33 × (3 × 7² × 11)

33 = 3 × 11

33n = (3 × 11) × 3 × 7² × 11 = 3² × 7² × 11² = (3 × 7 × 11)²

Every prime factor has an even exponent, so 33n is a perfect square.

Given ⁵⁄₆ × ⁴⁸⁶⁄₆₂₅ = ^x4⁄_5^y, find x and y.

Simplify the left-hand side:

^{5 × 486}⁄_{6 × 625} = ²⁴³⁰⁄₃₇₅₀

Divide numerator and denominator by 30:

= ⁸¹⁄₁₂₅

Express 81 and 125 as powers:

81 = 3⁴ and 125 = 5³

So ⁸¹⁄₁₂₅ = ³⁴⁄_5³

Comparing with ^x4⁄_5^y: x = 3, y = 3

Express 1.3 (recurring) as a rational number.

Let x = 1.333...

10x = 13.333...

Subtract: 10x − x = 13.333... − 1.333...

9x = 12

x = ¹²⁄₉ = ⁴⁄₃

E varies directly as the square of W.

E = kW²

From the table: E = 40 when W = 2.

40 = k(2²) = 4k ⟹ k = 10

When W = 5:

f = 10(5²) = 10(25) = 250

Majara bought a laptop for M6 900 including 15% tax. Calculate the cost before tax.

Let P be the cost before tax.

P × 1.15 = 6 900

P = ^{6 900}⁄_1.15 = 6 000

(a) Show that x + y = 80.

Angles inside triangle BCD add up to 180°:

∠DBC + ∠BCD + ∠BDC = 180°

y + (y + 10) + (2x + 10) = 180

2x + 2y + 20 = 180

2x + 2y = 160

Divide by 2: x + y = 80 (Proven)

(b) Find the value of x.

Since CB is extended to A, angles on the straight line at B sum to 180°:

3x + y = 180 (given)

We also have x + y = 80.

Subtract: (3x + y) − (x + y) = 180 − 80

2x = 100 ⟹ x = 50

T-shirt sizes: XS = 16, S = 36, M = 20, L = 30, XL = 18. Total = 120.

(a) State the modal size.

The highest frequency is 36, which corresponds to size S.

(b) Sector angle for size M.

Sector angle = ²⁰⁄₁₂₀ × 360° = 60°

(c) Probability that a randomly chosen T-shirt is XL.

P(XL) = ¹⁸⁄₁₂₀ = ³⁄₂₀

Calculate the height of the cuboid not in contact with water.

Total volume = 300 cm³, length = 10 cm, width = 4 cm.

H = ³⁰⁰⁄_{10 × 4} = ³⁰⁰⁄₄₀ = 7.5 cm

Water depth = 2 cm.

Height not in contact with water = 7.5 − 2 = 5.5 cm

(a) Calculate the length UW.

By Pythagoras' theorem:

UW = √(UV² + VW²) = √(18² + 24²) = √(324 + 576) = √900 = 30 m

(b) Write the value of tan UW.

tan W = ^opposite⁄_adjacent = ^UV⁄_VW = ¹⁸⁄₂₄ = ³⁄₄

Reading graph C on the grid:

(a) The roots of the graph.

The curve crosses the x-axis at x = −3 and x = 1.

(b) The equation of the axis of symmetry.

The axis of symmetry lies halfway between the roots:

x = ^{(−3) + 1}⁄₂ = ⁻²⁄₂ = −1

In a club of 40 members, 30 prefer a picnic and 25 prefer a party. Find the greatest number who prefer neither.

To maximise the number who prefer neither, we need to minimise the union of the two sets.

The smallest possible union is the size of the larger set, if the smaller is entirely contained within it.

Minimum union = max(30, 25) = 30

Greatest number who prefer neither = 40 − 30 = 10

Given p = ^{√(a2(n2 − b))}⁄₂, express a in terms of n, b and p.

Multiply both sides by 2:

2p = √(a²(n² − b))

Square both sides:

4p² = a²(n² − b)

Divide both sides by (n² − b):

a² = ^4p2⁄_{(n² − b)}

Take the square root:

a = ^2p⁄_{√(n² − b)}

(a) x² + kx − 36 has one root as 4. Find k.

Substitute x = 4:

16 + 4k − 36 = 0

4k − 20 = 0

k = 5

(b) Factorise fully 2x² + 9x − 35.

Find two numbers that multiply to 2 × (−35) = −70 and add to 9:

These are +14 and −5.

Rewrite: 2x² + 14x − 5x − 35

Group: 2x(x + 7) − 5(x + 7)

= (2x − 5)(x + 7)

If ^{a − b}⁄_b = ¹⁄₂, find ^a⁄_b.

^{a − b}⁄_b = ^a⁄_b − 1

^a⁄_b − 1 = ¹⁄₂

^a⁄_b = ¹⁄₂ + 1 = ³⁄₂

A quadratic equation with roots −3 and ¹⁄₂, integer coefficients.

Roots: x = −3 and x = ¹⁄₂

Factors: (x + 3) and (2x − 1)

Multiply: (x + 3)(2x − 1) = 0

Expand: 2x² − x + 6x − 3 = 0

2x² + 5x − 3 = 0

Given −7 ≤ ²⁄₃x − 5 < 8.

(a) Solve the inequality.

Add 5 to all parts:

−2 ≤ ²⁄₃x < 13

Multiply all parts by ³⁄₂:

−3 ≤ x < 19.5

(b) Find the largest possible integer value of x.

Since x < 19.5, the largest integer is 19.

Semicircle with diameter ²⁶⁄_π cm. Two chords of 5 cm and 6.6 cm.

Radius = ²⁶⁄_2π = ¹³⁄_π cm

Half-circumference (arc of semicircle) = π × radius = π × ¹³⁄_π = 13 cm

Shaded perimeter = arc + chord 5 + chord 6.6

= 13 + 5 + 6.6

= 23.6 cm

Seeta has 5 mint and 4 chocolate sweets. He chooses two at random, one after the other, and eats them. Find the probability that they are different flavours.

Total = 9 sweets.

P(different) = P(first mint, second chocolate) + P(first chocolate, second mint)

= (⁵⁄₉ × ⁴⁄₈) + (⁴⁄₉ × ⁵⁄₈)

= ²⁰⁄₇₂ + ²⁰⁄₇₂

= ⁴⁰⁄₇₂ = ⁵⁄₉

(a) Describe a single transformation that maps A onto B.

Reading the coordinates:

Triangle A has vertices approximately (2, 4), (5, 2) and (7, 4).

Triangle B has vertices approximately (2, −2), (5, 0) and (7, −2).

The x-coordinates stay the same, while the y-coordinates are negated and shifted.

The image is flipped vertically about the horizontal line midway between them:

Midpoint of A's base (y = 4) and B's base (y = −2) is y = 1.

Check: reflect (5, 2) in y = 1 → (5, 0) ✓; reflect (2, 4) in y = 1 → (2, −2) ✓

Reflection in the line y = 1.

(b) Draw the image of B after a stretch with the y-axis invariant and stretch factor −¹⁄₂.

The stretch multiplies every x-coordinate by −¹⁄₂, while y-coordinates stay the same.

So (2, −2) → (−1, −2), (5, 0) → (−2.5, 0), (7, −2) → (−3.5, −2).

Draw the new triangle on the grid using these coordinates.

(a) Calculate the speed at 1630 hours.

From the graph:

1400 → 1500: distance goes 0 → 60 km in 1 hour. Speed = 60 km/h.

1500 → 1700: distance goes 60 → 100 km in 2 hours. Speed = ⁴⁰⁄₂ = 20 km/h.

At 1630 hours, the car is in the second segment, so speed = 20 km/h.

(b) Represent the distance-time graph on a speed-time graph.

Draw two horizontal line segments on the speed-time axes:

From time 0 to 1 hour: horizontal line at 60 km/h.

From time 1 to 3 hours: horizontal line at 20 km/h.

(a) Find the size of angle OAC in terms of x.

DOA is a straight line (diameter through O).

∠DOC = 2x°, so ∠AOC = 180° − 2x° (angles on a straight line).

OA = OC (radii), so triangle OAC is isosceles with ∠OAC = ∠OCA.

∠OAC = ^{180° − ∠AOC}⁄₂ = ^{180° − (180° − 2x°)}⁄₂ = ^2x°⁄₂ = x°

(b) Find the size of angle ABC in terms of x.

Angle ABC is the inscribed angle subtending arc AC.

The central angle over the same arc is the reflex angle ∠AOC = 360° − (180° − 2x°) = 180° + 2x°.

Inscribed angle = half the central angle over the same arc:

∠ABC = ^{180° − 2x°}⁄₂ = 90° − x°

Check with isosceles triangle: AB = BC, so ∠BAC = ∠BCA = ^{180° − ∠ABC}⁄₂ = ^{180° − (90° − x°)}⁄₂ = ^{90° + x°}⁄₂ = 45° + ^x°⁄₂.

(c) Find the size of angle BCE in terms of x.

TCE is a tangent at C. By the alternate segment theorem, the angle between the tangent and chord CE equals the inscribed angle in the alternate segment:

∠BCE = ∠BAC

From part (b), ∠BAC = 45° + ^x°⁄₂

So ∠BCE = 45° + ^x⁄₂°