Correct Answer: B. 20

Explanation

Let the sum of the men's ages be f, so that

\(\frac{f}{n} = 50 .... (1)\)

Also, \(\frac{f - (55 + 63)}{n - 2} = 50 - 1 = 49 .... (2)\)

From (1), \(f = 50n\)

From (2), \(f - 118 = 49(n - 2) = 49n - 98\)

\(f = 49n - 98 + 118 = 49n + 20\)

\(\therefore f = 50n = 49n + 20\)

\(50n - 49n = n = 20\)

Correct Answer: B. 30

Explanation

M = Mercedez; T = Toyota

Number of staff = 0 + (20 - 5) + (15 - 5) + 5

0 + 15 + 10 + 5 = 30

Correct Answer: A) 354

Explanation

(a) \(9^{2x - 1} = \frac{81^{x - 2}}{3^{x}}\)

\((3^{2})^{2x - 1} = \frac{(3^{4})^{x - 2}}{3^{x}}\)

\(3^{4x - 2} = \frac{3^{4x - 8}}{3^{x}}\)

\(3^{4x - 2} = 3^{4x - 8 - x}\)

\(3^{4x - 2} = 3^{3x - 8}\)

\(\therefore 4x - 2 = 3x - 8 \implies 4x - 3x = -8 + 2\)

\(x = -6\)

(b) \(\sqrt{\frac{0.81 \times 10^{-5}}{2.25 \times 10^{7}}}\)

= \(\sqrt{\frac{0.81}{2.25}} \times \sqrt{10^{-5 - 7}}\)

= \(\frac{0.9}{1.5} \times \sqrt{10^{-12}}\)

= \(0.6 \times 10^{-6}\)

= \(6 \times 10^{-7}\)

Correct Answer: A. y = 5x + 2

Correct Answer: B. N54 000.00

Correct Answer: C. -1

Explanation

Let the linear transformation T be represented by the following:

\(T : \begin{pmatrix} x \\ y \end{pmatrix} \to \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 3 & -1 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} -1 & 7 \\ 4 & 11 \end{pmatrix}\).

\(3a + 2b = -1 ... (1)\)

\(-a + 4b = 7 ... (2)\)

\(3c + 2d = 4 ... (3)\)

\(-c + 4d = 11 ... (4)\)

\((2) \times 3 : -3a + 12b = 21... (5)\)

\((1) + (5) : 14b = 20 \implies b = \frac{10}{7}\)

Substitute for b in (1),

\(3a + 2(\frac{10}{7}) = -1 \implies 3a = \frac{-27}{7}\)

\(a = \frac{-9}{7}\)

Solving for c and d,

\((4) \times 3 : -3c + 12d = 33 ... (6)\)

\((3) + (6) : 14d = 37 \implies d = \frac{37}{14}\)

\(-c + 4(\frac{37}{14}) = 11 \implies -c = 11 - \frac{74}{7} = \frac{3}{7}\)

\(c = \frac{-3}{7}\)

\(\therefore T = \begin{pmatrix} \frac{-9}{7} & \frac{10}{7} \\ \frac{-3}{7} & \frac{37}{14} \end{pmatrix}\)

\(T : \begin{pmatrix} x \\ y \end{pmatrix} \to \begin{pmatrix} \frac{-9}{7} & \frac{10}{7} \\ \frac{-3}{7} & \frac{37}{14} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}\)

From \(P : (x, y) \to (x + y, x + 2y)\)

\(P : \begin{pmatrix} x \\ y \end{pmatrix} \to \begin{pmatrix} x + y \\ x + 2y \end{pmatrix}\)

\(P : \begin{pmatrix} x \\ y \end{pmatrix} \to \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}\)

\(\therefore P = \begin{pmatrix} 1 & 1 \\ 1 & 2 \\end{pmatrix}\)

(b) \(TP = \begin{pmatrix} \frac{-9}{7} & \frac{10}{7} \\ \frac{-3}{7} & \frac{37}{14} \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}\)

= \(\begin{pmatrix} \frac{-9}{7} + \frac{10}{7} & \frac{-9}{7} + \frac{20}{7} \\ \frac{-3}{7} + \frac{37}{14} & \frac{-3}{7} + \frac{74}{14} \end{pmatrix}\)

= \(\begin{pmatrix} \frac{1}{7} & \frac{11}{7} \\ \frac{31}{14} & \frac{34}{7} \end{pmatrix}\)

(c) The image of X(4, 3) under TP:

\(TP : \begin{pmatrix} 4 \\ 3 \end{pmatrix} \to \begin{pmatrix} \frac{1}{7} & \frac{11}{7} \\ \frac{31}{17} & \frac{34}{7} \endd{pmatrix} \begin{pmatrix} 4 \\ 3 \end{pmatrix}\)

= \(\begin{pmatrix} \frac{4}{7} + \frac{33}{7} \\ \frac{62}{7} + \frac{102}{7} \end{pmatrix}\)

= \(\begin{pmatrix} \frac{37}{7} \\ \frac{164}{7} \end{pmatrix}\)

The image of X(4, 3) under TP is \((\frac{37}{7}, \frac{164}{7})\).

Correct Answer: B. \(3\)

Explanation

\(\begin{vmatrix} m-2 & m+1 \\ m+4 & m-2 \end{vmatrix} = -27\)

\((m^{2} - 4m + 4) - (m^{2} + 5m + 4) = -27\)

\(-9m = -27 \implies m = 3\)

Correct Answer: B. 22