Solutions

(a) ²⁄₅ ÷ 2

Change division to multiplication by the reciprocal:

²⁄₅ × ¹⁄₂

Cancel the common factor of 2:

= ¹⁄₅

(b) [0.5(10 + 6)] ÷ 4

Brackets first: 10 + 6 = 16

Multiply by 0.5 (same as halving): 0.5 × 16 = 8

Divide by 4: 8 ÷ 4 = 2

Round each number to 1 s.f.:

2.511 → 3

8.71 → 9

7.822 → 8

Substitute into the expression 2.511 × √8.71 + 7.822:

3 × √9 + 8

= 3 × 3 + 8

= 9 + 8

= 17

(a) 300 200 015 in words:

Three hundred million, two hundred thousand and fifteen.

(b) Expand the components:

300 200 015 = 300 000 000 + 200 000 + 15

3 × 10⁸ = 300 000 000, so b = 200 000

In standard form: 200 000 = 2 × 10⁵

Any integer from 6500 up to 7499 rounds to 7000 when rounded to 1 s.f.

The highest integer in this range is 7499.

Distance between centres X and Y is 1 + 1 = 2 units.

Distance between centres X and Z is 1 + 1 = 2 units.

Using Pythagoras on the right-angled triangle YXZ:

YZ² = XY² + XZ² = 2² + 2² = 8

YZ = √8

The segment YZ passes through P and Q, with YP = 1 and QZ = 1.

PQ = YZ − YP − QZ = √8 − 1 − 1 = √8 − 2

So b = 2

(a) 2t + t³

Common factor t:

= t(2 + t²)

(b) x + t − 1 − tx

Group terms: (x − 1) − t(x − 1)

Common factor (x − 1):

= (x − 1)(1 − t)

(a) 0.5 billion = 500 000 000

= 5 × 10⁸

(b) Total mass = 2 kg = 2000 g

Mass of one grain = 2000 ÷ (5 × 10⁸)

= 400 × 10⁻⁸

= 4 × 10⁻⁶ g

(a) Students playing volleyball only = 18

n = 18 ÷ 40

= 0.45 (or ⁹⁄₂₀)

(b) Total probability = 1:

0.2 + m + 0.45 + 0.1 = 1

m = 0.25

Number of students = 0.25 × 40 = 10

(a) Factorise as a difference of squares:

[(3x − y) − (x − 5)][(3x − y) + (x − 5)] = 0

= (2x − y + 5)(4x − y − 5) = 0

Since 2x − y + 5 = 0, the product is zero, so

4x − y − 5 = 0 satisfies the equation. ∎

(b) Solve the system:

1. 2x − y = −5
2. 4x − y = 5

Subtract (1) from (2): 2x = 10

x = 5

Total perimeter = 2l + 2r + πr = 8

2l = 8 − 2r − πr

l = (8 − 2r − πr) ÷ 2

l = 4 − r − ^πr⁄₂ ∎

(a) P is the y-intercept of y = 2x − 3, so P = (0, −3).

PQ = 11 units up, so Q = (0, −3 + 11) = (0, 8)

Parallel lines share gradient m = 2

y = 2x + 8

(b) R is the x-intercept (y = 0): 0 = 2x − 3 → R = (1.5, 0)

Midpoint of QR = ((0 + 1.5)/2, (8 + 0)/2)

= (0.75, 4) or (³⁄₄, 4)

From equation 2: 2a = 3b − 15

Substitute into equation 1:

(3b − 15) + 3b = 3

6b = 18 → b = 3

Substitute back:

2a = 3(3) − 15 = −6 → a = −3

Draw a parallel line through C, extending left.

Interior angle ∠CDE = 55° − 30° = 25°

Reflex angle = 360° − 25°

= 335°

(a) Split the T into two rectangles:

Top: 11 × 3 = 33 cm²

Stem: 3 × (10 − 3) = 21 cm²

Total = 33 + 21 = 54 cm²

(b) Volume = Cross-section × length

= 54 × 20

= 1080 cm³

Apply the 90° anticlockwise rotation about (2, 0):

(x, y) → (2 − (y − 0), 0 + (x − 2))

= (2 − y, x − 2)

Original vertices → image vertices:

• (3, 2) → (0, 1)
• (3, 4) → (−2, 1)
• (6, 1) → (1, 4)

(a) By Pythagoras:

OP = √(8² + 15²)

= √(64 + 225)

= √289

= 17 units

(b) tan θ = opposite ÷ adjacent

= 15 ÷ 8

= ¹⁵⁄₈ (or 1.875)

Multiply every term by the LCM (6):

2(x + 2) + 3(x + B) = 5x + 13

2x + 4 + 3x + 3B = 5x + 13

5x + 4 + 3B = 5x + 13

4 + 3B = 13

3B = 9 → B = 3

(a) Multiply every entry of A by 2:

2A = ( 4   6 ; −2   −10 )

(b) Add corresponding entries:

A + B = ( 3   −1 ; 0   0 )

(a) Range = max − min

= 2 − (−1)

= 3 °C

(b) Mean = Σ(fx) ÷ Σf

= [ (−1 × 6) + (0 × 3) + (1 × 7) + (2 × 4) ] ÷ (6 + 3 + 7 + 4)

= (−6 + 0 + 7 + 8) ÷ 20

= 9 ÷ 20

= 0.45 °C

Convert all to decimals:

0.40 : 0.35 : 0.36

Multiply each by 100 to clear:

40 : 35 : 36

Total frequency = 5 + 7 + 3 = 15

Angle = (7 ÷ 15) × 360°

= 7 × 24°

= 168°

Profit = 754 − 650 = M104

Percentage profit = (104 ÷ 650) × 100%

= 16%