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(a) \(P = x : x \leq 2 = {..., -1, 0, 1, 2}\) \(Q = x : -7 < x < 15 = {-6, -5, -4, -3, -2, -1, 0, ..., 13, 14}\) \(R = x : -2 \leq x < 19 = {-2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18}\) (i) \(P \cap Q = {-6, -5, -4, -3, -2, -1, 0, 1, 2}\) (ii) \(R' = {..., -5, -4, -3, 19, 20, ...}\) \(Q \cup R' = {..., -1, 0, 1, 2, 3, ...}\) \(\therefore P \cap (Q \cup R') = {..., -1, 0, 1, 2}\) (b)(i) Class intervals Midpoint (x) Freq (f) \(fx\) 20 - 29 24.5 3 73.5 30 - 39 34.5 5 172.5 40 - 49 44.5 10 445 50 - 59 54.5 13 708.5 60 - 69 64.5 6 387 70 - 79 74.5 3 223.5 \(\sum \) 40 2010 (ii) \(Mean = \frac{\sum fx}{\sum f} = \frac{2010}{40}\) = \(50.25\)

(a) Let the number be N. \(\frac{N}{3} + 8 = 18 - \frac{N}{2}\) \(\frac{N}{3} + \frac{N}{2} = 18 - 8 = 10\) \(\frac{5N}{6} = 10 \implies 5N = 60\) \(N = 12\) (b)

(a) x -3 -2 -1 0 1 2 3 4 \(y = 3x^{2} - 5x - 7\) 35 15 1 -7 -9 -5 5 21 (b) (c) \(y + 3x + 2 = 0 \implies y = -3x - 2\) when x = -2, y = 6 - 2 = 4. when x = 3, y = -9 - 2 = -11. (d) \(3x^{2} - 2x - 5 = 0 = 3x^{2} - 5x - 7 = -3x - 2\) The roots of the equation \(3x^{2} - 2x - 5 = 0\) is when the line \(y = -3x - 2\) cuts the equation i.e. x = 1.0 or 1.7.

(a) \(S = \frac{n}{2} [2a + (n - 1)d]\) \(2S = n[2a + (n - 1) d]\) \(\frac{2S}{n} = 2a + (n - 1)d\) \(\frac{2S}{n} - 2a = (n - 1) d\) \(d = \frac{\frac{2S}{n} - 2a}{(n - 1)}\) = \(\frac{2S - 2an}{n(n - 1)}\) (b) (i) Given: circle ABP To prove < AOB = 2 < APB Construction : Join PO produced to Q. Proof : /OA/ = /OB/ (radii) \(\therefore x_{1} = x_{2} \) (base angles of isosceles triangle) \(\therefore < AOQ = x_{1} + x_{2}\) (exterior angle of triangle AOP) \(\therefore < AOQ = 2x_{2} (x_{1} = x_{2})\) Similarly, < AOB = < BOQ - < AOQ. = \(2y_{2} - 2x_{2} = 2(y_{2} - x_{2}) = 2 \times < APB\) \(\therefore < AOB = 2 < APB.\) (ii) \(z = 2x \) (angle subtended at the centre) \(z = 2(126^\circ) = 252^\circ\) \(\therefore x = 360^\circ - 252^\circ = 108^\circ\) \(\therefore y = \frac{1}{2} \times 108^\circ = 54^\circ\) \(x = 108^\circ ; y = 54^\circ ; z = 252^\circ\)

(a) Difference in longitudes = 124^\circ + 56^\circ = 180^\circ Distance PQ = \(\frac{180^\circ}{360^\circ} \times 2\pi r\) where \(r = R \cos \theta\) \(/PQ/ = \frac{180}{360} \times 2 \times 3.142 \times R \cos 68.7^\circ\) \(/PQ/ = 3.142 \times 6400 \times 0.3633\) = \(7305.53 km\) \(\approxeq 7300 km\) (3 sig. figs) (b) Red balls = 4, white = 3 ; green = 5 Total = 4 + 3 + 5 = 12 balls. (i) P(green) = \(\frac{5}{12}\) P(not green) = \(1 - \frac{5}{12} = \frac{7}{12}\) (ii) P(picking red and white) = P( red then white) or P(white then red) = \(\frac{4}{12} \times \frac{3}{11} + \frac{3}{12} \times \frac{4}{11}\) = \(\frac{2}{11}\)

(a) By cosine rule, \(/PQ/^{2} = 20^{2} + 15^{2} - 2 \times 20 \times 15 \times \cos 145^\circ\) = \(400 + 225 - 600 \cos 145^\circ\) = \(625 + 600 \times 0.8192\) = \(491.52 + 625\) \(/PQ/^{2} = 1116.52 \implies /PQ/ = \sqrt{1116.52} = 33.41\) \(\approxeq 33km\) ( to the nearest whole number) (b) By sine rule, \(\frac{\sin < QPR}{15} = \frac{\sin 145}{33.41}\) \(\sin < QPR = \frac{15 \sin 145}{33.41}\) \(\sin < QPR = \frac{15 \times 0.5736}{33.41} = 0.2575\) \(< QPR = 14.92^\circ\) \(\therefore\) Bearing of Q from P = 25^\circ + 14.92^\circ = 39.92^\circ. \(\approxeq 40^\circ\) (to nearest whole number)

\(\begin{array}{c|c} \text{class intervals} & Fre(F) & \text{class-marks(x)} & Fx & (x - x)& (x - x)^2 & F(x - x)^2 \\ \hline 1 - 3 & 5 & 2 & 10 & -3 & 9 & 45\\ 4 - 6 & 8 & 5 & 40 & 0 & 0 & 0 \\ 7 - 9 & 5 & 8 & 40 & 3 & 9 & 45 \\ \hline & 18 & & 90 & & & 90 \end{array}\) x = \(\frac{\sum fx}{\sum f}\) = \(\frac{90}{18}\) = 5 S.D = \(\frac{\sum f(x - x)^2}{\sum f}\) = \(\frac{90}{18}\) = \(\sqrt{5}\)

Total number of students (y) using the subject French ( 90 / 360 )º x Y = 30 Y = 30 x 4 Y = 120 History- 90º + French- 90º + Economics- 150º + CRK = 360º CRK = 360º - 330º = 30º Number that offered CRK: ( 30 / 360 )º x 120 = 10

You can convert it to base 10 and square, then re-convert it after the operation. OR You can multiply it straight applying the rules of binary multiplication.

Ratio = 6 + 7 + 8 = 21 21 = N10.50 1 = 1050K/21 = 50K 6 = 6 x 50K = N3.00 7 = 7 x 50K = N3.50 8 = 8 x 50K = N4.00 the Largest share = N4.00

No expanded working has been attached for this question yet. The verified answer is Option A.

Log2a = Log84 Log2a = Log882/3 → 2/3Log88 → 2/3 x 1 Log2a = 2/3 Recall; If Logax = y ∴ ay = x Log2a = 2/3 22/3 = a

S.P = N600.00 (100 + 50)% = N600 150% = N600 1% = \(\frac{600}{150}\) 100% = \(\frac{600}{150} \times 100%\) = N400

A.P 11, 4, -3 1st term = 11 A.P = a, a + d, a + 2d ...... a + (n - 1)d If a = 11 a + d = 4 d = 4 - 11 = -7 nth term = a + (n-1)d          = 11 + (n-1)(-7)          = 11 - 7n + 7          = 18 - 7n

16/9, x, 1, y => a, ar, ar2, ar3 ar2 = 1 => 16r2/9 = 1 => 16r2 9 => r2 = 9/16 => r = 3/4 ar2 = y = ar2 x r = 1 x 3/4 = 3/4 x xy = 4/3 x 3/4 = 1