Explanation

(a)

(b) Mode = 35.5

(c) Median = \(L_{1} + \frac{(\frac{N}{2} - \sum f_{p}) \times c}{f_{m}}\)

where \(L_{1}\) = lower class boundary of median class = 29.5

\(N = \sum f = 40\) ; \(\sum f_{p}\) = cumulative frequency before median class = 15.

\(f_{m}\) = frequency of median class = 12, c = class interval = 10.

Median = \(29.5 + \frac{(\frac{40}{2} - 15) \times 10}{12}\)

= \(29.5 + \frac{(20 - 15) \times 10}{12}\)

= \(29.5 + 4.17 = 33.67\)

(a)

(b) Mode = 35.5

(c) Median = \(L_{1} + \frac{(\frac{N}{2} - \sum f_{p}) \times c}{f_{m}}\)

where \(L_{1}\) = lower class boundary of median class = 29.5

\(N = \sum f = 40\) ; \(\sum f_{p}\) = cumulative frequency before median class = 15.

\(f_{m}\) = frequency of median class = 12, c = class interval = 10.

Median = \(29.5 + \frac{(\frac{40}{2} - 15) \times 10}{12}\)

= \(29.5 + \frac{(20 - 15) \times 10}{12}\)

= \(29.5 + 4.17 = 33.67\)

Correct Answer: C. k = 1, p = -2

Explanation

Un = kn\(^2\) + pn,

U\(_1\) = -1,

U\(_5\) = 15,

when n = 1

U1 = k(1)\(^2\)+ p(1) = -1

k + p = -1 --------eqn1

when n = 5

U\(_5\) = k(5)\(^2\)+ p(5) = 15

25k + 5p = 15 --------eqn2

multiply eqn1 by 5 and eqn2 by 1

5k + 5p = -5 -------eqn3

25k + 5p = 15 -------eqn4

eqn4 - eqn3

20k = 20

k = 1

sub for k in eqn1

1 + p = -1

p = -1 -1 = -2

Correct Answer: C. \(2(1 - 3m)\)

Explanation

\(\frac{2 - 18m^2}{1 + 3m}\) = \(\frac{2(1 - 9)m^2}{1 + 3m}\)

= \(\frac{2(1 + 3m)(1 - 3m)}{1 + 3m}\)

= \(2(1 - 3m)\)

Correct Answer: A. 25

Explanation

\(f : x \to x^{2} ; g : x \to x + 3\)

\(g(2) = 2 + 3 = 5\)

\(f o g(2) = f(5) = 5^{2} = 25\)

Correct Answer: A. 18 min

Explanation

(a) \(\frac{\frac{3}{2} \log 27 - 3 \log 5\sqrt{5}}{\log 0.6}\)

= \(\frac{\frac{3}{2} \log 3^{3} - 3 \log 5^{\frac{3}{2}}}{\log (\frac{3}{5})}\)

= \(\frac{\frac{2}{3} \times 3 \log ^3 - 3 \times \frac{3}{2} \log 5}{\log 3 - log 5}\)

= \(\frac{\frac{9}{2}(\log 3 - \log 5)}{\log 3 - \log 5}\)

= \(\frac{9}{2} = 4\frac{1}{2}\)

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

(ii) Transformation of A followed by B is BA.

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

= \(\begin{pmatrix} -1 & 7 \\ -1 & 4 \end{pmatrix}\)

Image of (-2, 2) is given by

\(\begin{pmatrix} -1 & 7 \\ -1 & 4 \end{pmatrix} \begin{pmatrix} -2 \\ 2 \end{pmatrix}\)

= \(\begin{pmatrix} 16 \\ 10 \end{pmatrix}\)

Correct Answer: B. 132cm²

Explanation

A prism has 3 rectangular faces and 2 triangular faces and 2 rectangular faces = 10(3 + 4 + 5) = 120

Area of triangular faces = \(\sqrt{s(s - a) (s - b) (s - c)}\)

where s = \(\frac{a + b + c}{2}\)

= \(\frac{3 + 4 + 5}{2}\)

= \(\frac{12}{2}\)

= 6

Area of \(\bigtriangleup\) = \(\sqrt{6(6 - 30(6 - 4)(6 - 5)}\)

= \(\sqrt{6 \times 3 \times 2 \times 1}\) = 6

Area of triangle faces = 2 x 6 = 12cm²

Total surface area = Area of rectangular face + Area of \(\bigtriangleup\) = 120 + 12

= 132cm²

Explanation

< PAB = < ABE = 53° (alternate angles)

< CBE = 180° - < DBC = 180° - 161° = 19°

< ABC = < ABE + < CBE

= 53° + 19° = 72°

(a) In \(\Delta ABC\),

\(AC^{2} = AB^{2} + BC^{2} - 2(AB)(BC) \cos < ABC\)

\(AC^{2} = 15^{2} + 18^{2} - 2(15)(18) \cos 72\)

= \(225 + 324 - 540 \cos 72\)

= \(549 - 166.869\)

\(AC^{2} = 382.131\)

\(AC = \sqrt{382.131} = 19.548m\)

\(\approxeq 19.5m\)

(b) \(\frac{\sin A}{18} = \frac{\sin 72}{19.548}\)

\(\sin A = \frac{18 \times \sin 72}{19.548}\)

\(\sin A = 0.8757\)

\(A = \sin^{-1} (0.8757) = 61.13°\)

The bearing of C from A = 61.13° + 53° = 114.13° \(\approxeq\) 114°.

(c) In \(\Delta ATB\),

\(\frac{15}{BT} = \tan 58\)

\(BT = \frac{15}{\tan 58}\)

\(9.373 m \approxeq 9.37m\)

Correct Answer: B. 30°

Explanation

4m - 15^o = m + 75^o

(Vertically opposite angles are equal)

4m - m = 75 + 15

3m = 90

m = \(\frac{90}{3}\)

m = 30^o

Explanation

(a) < QRT = < RQS (alternate angles)

< RPS = < RQS (angles in the same segment)

\(\therefore\) < RPS = < QRT

(ii) < QRT = < PQS (corresponding angles)

< PQS = < PRS (angles in the same segment)

\(\therefore\) < QTR = < PRS.

(b) \(< ABC = \hat{B} = 180° - 132° = 48°\)

\(< ACB = \hat{C} = 180° - 96° = 84°\)

\(< BAC = \hat{A} = 180° - (48° + 84°) = 48°\)

\(< BAC = < ABC = 48°\)

\(\therefore \Delta ABC = isosceles\).