Correct Answer: D. 29^o

Explanation

PSQ = 54^o

?PQS = 180^o - (97 + 54)^o

= 29^o

Correct Answer: D. -24

Correct Answer: A. 15

Explanation

G \(\alpha\) H²

G = KH²

4 = K(3)²

4 = 9k; K = \(\frac{4}{9}\)

100 = \(\frac{4}{9}H^2\)

4H² = 900

H² = \(\frac{900}{4}\)

H² = 225

H = \(\sqrt{225}\)

H = 15

Explanation

\(\begin{vmatrix} x - 3 & -4 & 3 \\ 5 & 2 & 2 \\ 2 & -4 & 6 - x \end{vmatrix} = -24\)

\((x - 3)[2(6 - x) + 8] + 4[5(6 - x) - 4] + 3[- 20 - 4] = -24\)

\((x - 3)[12 - 2x + 8] + 4[30 - 5x - 4] + 3[-24] = -24\)

\((x - 3)(20 - 2x) + 4(26 - 5x) - 72 = -24\)

\((x - 3)(20 - 2x) + 4(26 - 5x) = - 24 + 72 = 48\)

\(20x - 2x^{2} - 60 + 6x + 104 - 20x = 48\)

\(-2x^{2} + 6x - 4 = 0\)

\(2x^{2} - 6x + 4 = 0\)

Solving the equation, we get x = 1 or 2.

Correct Answer: B. \(\frac{11}{20}\)

Explanation

at most 2 marks = 5 + 2 + 4 students = 11 students

probability(at most 2 marks) = \(\frac{11}{20}\)

Correct Answer: C. 1 cm

Correct Answer: D. \(\frac{n+1-r}{r}\)

Explanation

\(^{n}C_{r} = \frac{n!}{(n-r)! r!}\)

\(^{n}C_{r - 1} = \frac{n!}{(n - (r - 1))! (r - 1)!}\)

\(^{n}C_{r} ÷ ^{n}C_{r - 1} = \frac{n!}{(n - r)! r!} ÷ \frac{n!}{(n-(r-1))!(r-1)!}\)

= \(\frac{n!}{(n-r)! r!} \times \frac{(n-(r-1)! (r-1)!}{n!}\)

= \(\frac{(n + 1 - r)! (r - 1)!}{(n - r)! r!}\)

= \(\frac{(n+1-r)(n-r)! (r-1)!}{(n-r)! r (r - 1)!}\)

= \(\frac{n + 1 - r}{r}\)

Explanation

(a) px + qy = z

\(p \begin{pmatrix} 2 \\ 3 \end{pmatrix} + q \begin{pmatrix} 5 \\ -2 \end{pmatrix} = \begin{pmatrix} -4 \\ 13 \end{pmatrix}\)

\(\begin{pmatrix} 2p + 5q \\ 3p - 2q \end{pmatrix} = \begin{pmatrix} -4 \\ 13 \end{pmatrix}\)

Thus,

\(2p + 5q = -4 .....(1)\)

\(3p - 2q = 13 ......(2)\)

Solving the equations, we have p = 3 and q = -2.

(b)(i)

(ii) See attached graph for the quadrilateral W(2, 3), X(4, -1), Y(-3, -4) and Z(-3,2).

(iii) See attached for the image for the quadrilateral WXYZ under an anticlockwise rotation of 90° is \(W_{1}(-3, 2), X_{1}(1, 4), Y_{1}(4, -3)\) and \(Z_{1}(-2, -3)\).

Correct Answer: A. -5

Explanation

Let \(f(x) = y\), then we have

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

Let \(f^{1}(x) = x; x = 2y-1\), replacing y with x,

\(f^{1}(x) = 2x - 1 \implies f^{1}(-2) = 2(-2) -1= -5\)

Explanation

(a) The range; 20 - 15 = 5.

(b) They drew the frequency distribution table as

\(\sum\) f = 11\(\sum\) fx = 191 \(\sum\)f\(^2\) = 3337

(c) Median age as 17.

(d)(i) Mean = \(\frac{\sum fx}{\sum f}\) = \(\frac{191}{11}\) = 17.3636

(d)(ii). The standard deviation was \(\sqrt{\frac{2227}{11} - (\frac{191}{11})^2}\)

= \(\sqrt{1.8677}\) = 1.37(correct to two decimal places)