Correct Answer: B. 20

Explanation

The line: \(3y + 6x\) = 48

Divide through by 3

⇒ y + 2\(x\) = 16

⇒ y = -2\(x\) + 16

∴ The gradient of the line = -2

The points: A(-2, k) and B (4, 8)

m =\(\frac{y2 - y1}{x2 - x1} = \frac{8 - k}{4 - (-2)}\)

⇒ m =\(\frac{8 - k}{4 + 2} = {8 - k}{6}\)

Since the line passes through the points

∴ -2 = \(\frac{8 - k}{6}\)

⇒ \(\frac{-2}{1} = \frac{8 - k}{6}\)

⇒ 8 - k = -12

⇒ k = 8 + 12

∴ k = 20

Correct Answer: B. 25°

Explanation

(a) Area of the shaded segment = area of sector XOY - area of triangle OXY.

\(\implies 504 = \frac{\theta}{360°} \times \pi r^{2} - \frac{1}{2} r^{2} \sin \theta\)

\(\implies 504 = \frac{90}{360} \times \frac{22}{7} \times r^{2} - \frac{1}{2} \times r^{2} \sin 90\)

\(504 = r^{2}(\frac{22}{28} - \frac{1}{2})\)

\(504 = \frac{2r^{2}}{7} \implies r^{2} = \frac{504 \times 7}{2} = 1764\)

\(r = \sqrt{1764} = 42 cm\).

(b) In the diagram, \(a_{1} = a_{2}\) (base angles of an isosceles triangle)

\(\therefore a_{1} + a_{2} = 2a_{1}\)

\(\therefore 2a_{1} + 109° = 180° \implies 2a_{1} = 180° - 109° = 71°\)

\(a_{1} = a_{2} = \frac{71°}{2} = 35.5°\)

Hence, \(< RQS = 66° + 35.5°\)

= \(101.5°\)

Correct Answer: C. 20

Explanation

\(p(boys) = \frac{8}{14} = \frac{4}{7}\)

\(p(girls) = \frac{6}{14} = \frac{3}{7}\)

(a) p(both are of the same sex) = p(both are boys or both are girls)

= \(\frac{8}{14} \times \frac{7}{13} + \frac{6}{14} \times \frac{5}{13}\)

= \(\frac{86}{182}\)

= \(\frac{43}{91}\)

(b) p(they are of different sex) = p(first is a boy and second is a girl or first is a girl and second is a boy)

= \(\frac{8}{14} \times \frac{6}{13} + \frac{6}{14} \times \frac{8}{13}\)

= \(\frac{96}{182}\)

= \(\frac{48}{91}\)

Correct Answer: C. 4,2

Explanation

(a) Time taken to reach the ground

Using S = ut + 1/2gt\(^2\)

(u= 0 for a body falling from rest)

50 = 0 x t + 1/2 x 10t\(^2\)

5t\(^2\) = 50;

t\(^2\) =50/5 = 10

t = √10 =3.16

Distance when it strikes the ground This journey is independent of gravity

d =vt =30 x 3.16 = 94.8m

(b) Using Lami's rule,

\(\frac{20}{sin 34}\) = \(\frac{T_1}{sin 90}\) = \(\frac{T_2}{sin 56}\)

= T\(_1\) \(\frac{20.sin 90}{sin 34}\) = T\(_2\) \(\frac{20.sin 56}{sin 34}\)

T\(_1\) = 35.79N and T\(_2\) = 29.65N

Explanation

(a) \(A(x, y) \to (2x - y , -5x + 3y)\)

Matrix \(A = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}\)

(b) \(A = \begin{pmatrix} 2 & -1 \\ -5 & 2 \end{pmatrix} ; B = \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\)

(i) \(A^{2} - B^{2} = \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ -5 & 2 \end{pmatrix} - \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix} \begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\)

= \(\begin{pmatrix} 9 & -5 \\ -25 & 14 \end{pamtrix} - \begin{pmatrix} 14 & 5 \\ 25 & 9 \end{pmatrix}\)

= \(\begin{pmatrix} -5 & -10 \\ -50 & 5 \end{pmatrix}\)

= \(-5 \begin{pmatrix} 1 & 2 \\ 10 & -1 \end{pmatrix}\)

(ii) \(C = B^{2} A\)

= \(\begin{pmatrix} 14 & 5 \\ 25 & 9 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ -5 & 3 \end{pmatrix}\)

= \(\begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}\)

(iii) \(\begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 10 \\ 18 \end{pmatrix}\)

\(3x + y = 10 ... (1)\)

\(5x + 2y = 18 ... (2)\)

Multiply (1) by 2 :

\(6x + 2y = 20 ... (3)\)

\((3) - (2) : (6x + 2y) - (5x + 2y) = (20 - 18)\)

\(x = 2\)

Put x = 2 in (1) :

\(3x + y = 10 \implies 3(2) + y = 10\)

\(6 + y = 10 \implies y = 10 - 6 = 4\)

\(M(2, 4)\).

(c) C is the inverse of matrix A and A is the inverse of matrix C.

Correct Answer: B. (2, -7)

Explanation

Given \(M(4, -1)\) and \(N(x, y)\) with midpoint \(P(3, -4)\).

\(\implies \frac{4 + x}{2} = 3; \frac{-1 + y}{2} = -4\)

\(\therefore x = 2; y = -7\)

N = (2, -7)

Explanation

XWZ = 90° - (45° + 23°) = 22°

WYX = 180° - (45° + 95°) = 40°

< XZW = 180° - (89° + 22°) = 69°

(a)(i) \(\frac{XY}{\sin 95} = \frac{180}{\sin 40°}\)

\(XY = \frac{180 \sin 95}{\sin 40} = 278.96m\)

\(\approxeq 279m\)

(ii) \(\frac{ZW}{\sin 89°} = \frac{180}{\sin 69°}\)

\(ZW = \frac{180 \sin 89°}{\sin 69°} = 192.76m\)

\(\approxeq 193m\)

(b) < ZNX = <ZXN (base angles of an isosceles triangle)

= 44°^.

\(\therefore\) Bearing of Z from N = 270° + 44° = 314°.