Solutions

(a) 0.71 − 1.2

Align decimals to subtract:

1.20 − 0.71 = 0.49

Since you are subtracting a larger number from a smaller one, the result is negative.

= −0.49

(b) 2⁴⁄₇ ÷ ⁹⁄₁₄

1. Convert the mixed number into an improper fraction:

2⁴⁄₇ = ^{(2 × 7) + 4}⁄₇ = ¹⁸⁄₇

2. Convert division to multiplication by the reciprocal:

¹⁸⁄₇ × ¹⁴⁄₉

3. Simplify by cross-cancelling: ¹⁸⁄₉ = 2 and ¹⁴⁄₇ = 2

4. Calculate: 2 × 2 = 4

Work out ${\left(\frac{1\times {10}^2}{9}\right)}^{-\frac{3}{2}}$

1. Evaluate inside brackets:

${\left(\frac{100}{9}\right)}^{-\frac{3}{2}}$

2. Invert the fraction to remove the negative sign from the power:

${\left(\frac{9}{100}\right)}^{\frac{3}{2}}$

3. Take the square root of the fraction (denominator of the power):

${\left(\sqrt{\frac{9}{100}}\right)}^3={\left(\frac{3}{10}\right)}^3$

4. Cube both values:

$\frac{3^3}{{10}^3}=\frac{27}{1000}$

Estimate 2.511 × √8.71 + 7.822 by rounding each number to 1 significant figure:

2.511 → 3 (rounded to 1 s.f.)

8.71 → 9 (rounded to 1 s.f.)

7.822 → 8 (rounded to 1 s.f.)

Substitute and compute using BODMAS:

3 × √9 + 8 = 3 × 3 + 8 = 9 + 8 = 17

Given $X=\left(\begin{array}{cc}4& \mathrm{-3}\end{array}\right)$ and $Y=\left(\begin{array}{cc}1& \mathrm{-1}\\ 4& \mathrm{-4}\end{array}\right)$

(a) Predict the order of matrix XY:

Matrix X has dimensions 1 × 2 and Matrix Y has dimensions 2 × 2. The inner numbers match for multiplication, and the outer numbers give the final order: 1 × 2.

(b) Explain why the inverse of Y does not exist:

Find the determinant of matrix Y:

det(Y) = (1 × −4) − (−1 × 4) = −4 − (−4) = −4 + 4 = 0

The determinant of matrix Y is equal to 0, making it a singular matrix which cannot have an inverse.

(a) Write 300 200 015 in words:

Three hundred million, two hundred thousand and fifteen.

(b) Given that 300 200 015 = 3 × 10⁸ + b + 15, express b in standard form:

Expand the number by place value:

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

Comparing this to the equation shows that b = 200 000.

Converting to standard form yields 2 × 10⁵.

Find the value of b when the distance PQ is √8 − b:

1. Draw straight lines connecting centres XY and XZ. Since the circles are identical and have a radius of 1 unit, the lengths are XY = 2 units and XZ = 2 units.

2. Since XY ⊥ XZ, triangle △YXZ is right-angled. Use Pythagoras' theorem to find line YZ:

$Y{Z}^2=X{Y}^2+X{Z}^2=2^2+2^2=4+4=8⟹YZ=\sqrt{8}$

3. The line segment YZ contains the radii YP = 1 unit and QZ = 1 unit.

4. Compute distance PQ = YZ − YP − QZ = √8 − 1 − 1 = √8 − 2.

5. Compare √8 − 2 to √8 − b.

b = 2

Arrange a, a^−a, a⁰, a^1/a in order of size, starting with the smallest (given a > 1 is an integer):

Substitute an integer value greater than 1, like a = 2:

a^−a = 2⁻² = ¹⁄₄ = 0.25

a⁰ = 2⁰ = 1

a^1/a = 2^1/2 = √2 ≈ 1.41

a = 2

Comparing values: 0.25 < 1 < 1.41 < 2

a^−a, a⁰, a^1/a, a

(a) Factorise completely: 2t + t³

Pull out the highest common factor, which is t: t(2 + t²)

(b) Factorise completely: x + t − 1 − tx

Group by terms: (x − 1) + (t − tx) = (x − 1) − t(x − 1)

Factor out the common binomial element (x − 1).

(x − 1)(1 − t)

Simplify $\frac{7-7m}{m^2-3m+2}$

1. Factorise the numerator: 7(1 − m) = −7(m − 1)

2. Factorise the quadratic denominator: m² − 3m + 2 = (m − 1)(m − 2)

3. Construct fraction and cancel common terms:

$\frac{\mathrm{-7}(m-1)}{(m-1)(m-2)}=\frac{\mathrm{-7}}{m-2}=\frac{7}{2-m}$

(a) Write the number of sugar grains (0.5 billion) in standard form:

0.5 billion = 500 000 000 = 5 × 10⁸

(b) Find the mass of each sugar grain in grams from a 2 kg packet:

Convert the total weight to grams: 2 kg = 2000 g.

$\text{Mass per grain}=\frac{\text{Total Mass}}{\text{Total Grains}}=\frac{2000}{5\times {10}^8}=400\times {10}^{-8}=4\times {10}^{-6}$

= 4 × 10⁻⁶ g

Find the value of 2 sin CAB given tan DCA = ⁻⁵⁄₁₂:

1. Angles on a straight line sum to 180°, so ∠ACB = 180° − ∠DCA. Since tan(180° − θ) = −tan θ, we know tan ∠ACB = ⁵⁄₁₂.

2. In right-angled triangle △ABC, tan ∠ACB = ^Opposite⁄_Adjacent = ^AB⁄_BC = ⁵⁄₁₂. Set side AB = 5 and BC = 12.

3. Use Pythagoras' theorem to find hypotenuse AC:

AC = √(5² + 12²) = √(25 + 144) = √169 = 13

4. Calculate sin ∠CAB = ^Opposite⁄_Hypotenuse = ^BC⁄_AC = ¹²⁄₁₃.

5. Evaluate expression: 2 × ¹²⁄₁₃ = ²⁴⁄₁₃.

Given y varies directly as (x² + 1):

(a) Find the constant of variation:

Write variation equation: y = k(x² + 1). Substitute the table values x = 2, y = 15:

$15=k(2^2+1)⟹15=5k⟹k=3$

(b) Find the value of a:

Use formula with calculated constant: y = 3(x² + 1). Substitute the table values x = 3, y = a:

$a=3(3^2+1)=3(9+1)=3\left(10\right)=30$

(a) Determine the probability n that a student will take part in volleyball only:

n = ^{Students in V only}⁄_{Total Students} = ¹⁸⁄₄₀ = 0.45

(b) Find the number of students that will take part in both football and volleyball:

Total sum of probabilities in a Venn diagram equals 1:

$0.2+m+n+0.1=1⟹0.2+m+0.45+0.1=1⟹m+0.75=1⟹m=0.25$

$\text{Number of students}=0.25\times 40=10$

(a) If 2x − y + 5 = 0 satisfies (3x − y)² − (x − 5)² = 0, show that 4x − y − 5 = 0 also satisfies it:

Factorise the equation as a difference of two squares (A² − B² = (A − B)(A + B)):

$\left[\right(3x-y)-(x-5\left)\right]\left[\right(3x-y)+(x-5\left)\right]=0$

$(2x-y+5)(4x-y-5)=0$

Since the problem states that 2x − y + 5 = 0 is a valid solution, it makes the first half of the product zero. This leaves the alternative factor, 4x − y − 5 = 0, as a mathematically required root that satisfies the equation.

(b) Hence, determine the value of x:

Solve the system of equations derived from both valid factors:

1. 2x − y = −5
2. 4x − y = 5

Subtract equation (1) from equation (2) to eliminate y:

(4x − 2x) = 5 − (−5) ⟹ 2x = 10 ⟹ x = 5.

(a) Calculate the heights of other bars:

Frequency Density (FD) = ^Frequency⁄_{Class Width}

For group 15 < x ≤ 25: Width = 10. FD = ²⁸⁄₁₀ = 2.8. Given graph bar height is 7, the multiplier scale factor is ⁷⁄_2.8 = 2.5.

For group 25 < x ≤ 30: Width = 5. FD = ⁴⁰⁄₅ = 8. Bar height = 8 × 2.5 = 20.

For group 30 < x ≤ 50: Width = 20. FD = ¹²⁄₂₀ = 0.6. Bar height = 0.6 × 2.5 = 1.5.

(b) Calculate an estimate of the number of people whose age is under 28 years:

Take all 28 people from the first group, plus a proportional fraction of the middle class up to age 28:

$\text{People}=28+(\frac{28-25}{30-25}\times 40)=28+(\frac{3}{5}\times 40)=28+24=52$

Show, in terms of r, that the width l = 4 − r − ^πr⁄₂:

The total perimeter is made up of the base rectangle side (2r), the two vertical side lengths (2l), and the open curved semicircle arc (πr):

$2l+2r+\pi r=8$

$2l=8-2r-\pi r$

Divide the entire expression by 2 to isolate l:

$l=4-r-\frac{\pi r}{2}$

(a) Write the equation of the line passing through Q and parallel to y = 2x − 3:

Find the coordinates of point P (the y-intercept of the line): P = (0, −3). Point Q is on the y-axis, located 11 units above P: Q = (0, −3 + 11) = (0, 8). Parallel lines have identical gradients (m = 2).

y = 2x + 8

(b) Find the coordinates of the midpoint of line QR:

Find coordinate R (the x-intercept where y = 0):

0 = 2x − 3 ⟹ x = 1.5 ⟹ R = (1.5, 0).

$\text{Midpoint}=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})=(\frac{0+1.5}{2},\frac{8+0}{2})=(0.75,4)$

(a) Find the size of angle JNM:

Angles subtended at the circumference by the same circular arc segment are equal (∠JNM = ∠JLM).

65°

(b) Find the size of angle KLJ:

The angle subtended by a diameter inside a semi-circle is always right-angled (∠KLM = 90°).

∠KLJ = ∠KLM − ∠JLM = 90° − 65° = 25°

(a) State the length of AX:

The diagonals of a parallelogram bisect each other: AX = ^{Total line AC}⁄₂ = ¹⁶⁄₂ = 8 cm.

(b) Find the ratio of AX : XY:

Since XC = AX = 8 cm and YC = 6 cm, calculate segment length XY = XC − YC = 8 − 6 = 2 cm. The ratio is 8 : 2.

4 : 1

(c) Find the area of △CFY:

The linear scale factor (k) between similar triangles △ABX and △CFY is determined by corresponding side components along the shared diagonal: k = ^YC⁄_AX = ⁶⁄₈ = ³⁄₄. The area scale factor is k² = (³⁄₄)² = ⁹⁄₁₆.

Area = 32 × ⁹⁄₁₆ = 2 × 9 = 18

18 cm²

Find g(x) when f(x) = 3x + 2 and fg(x) = 3x − 13:

Substitute function g(x) into the expression for f(x):

$3\left(g\right(x\left)\right)+2=3x-13⟹3g\left(x\right)=3x-15⟹g\left(x\right)=x-5$

Solve the inequality $\frac{x-5}{-2}\ge 2$:

Multiply both sides by −2 and invert the direction of the inequality sign:

$x-5\le -4⟹x\le -4+5⟹x\le 1$

Find the value of k:

Simplify the fraction on the right-hand side first by dividing by 6: ^{15x + 24}⁄₁₈ = ^{5x + 8}⁄₆.

Match standard structural denominators on the left side to target 6: let m = 2 and k = 9.

$\frac{x+2}{2}+\frac{3x+3}{9}=\frac{x+2}{2}+\frac{x+1}{3}=\frac{3(x+2)+2(x+1)}{6}=\frac{3x+6+2x+2}{6}=\frac{5x+8}{6}$

This matches perfectly.

k = 9

Write the three inequalities which satisfy the shaded region R:

Identify the boundaries from the boundary lines:

1. Region is below the line y = 2x + 4 ⟹ y ≤ 2x + 4
2. Region is below the line y = −2x + 12 ⟹ y ≤ −2x + 12
3. Region is above the line y = −0.5x + 4 ⟹ y ≥ −0.5x + 4

Describe fully a single transformation that maps A onto its final image:

1. Apply the translation vector step:

$\left(\begin{array}{c}5\\ 2\end{array}\right)+\left(\begin{array}{c}\mathrm{-3}\\ 3\end{array}\right)=\left(\begin{array}{c}2\\ 5\end{array}\right)$

2. Apply an anticlockwise 90° rotation transformation rule about the origin, which maps any coordinate (x, y) → (−y, x):

(2, 5) → (−5, 2).

3. Compare original position A(5, 2) to final coordinates (−5, 2). The y-value is unchanged, while the x-value has been mirrored across the central vertical axis.

Reflection in the y-axis (or the vertical line x = 0).