Question Library

(a) From the table, \(\sum f = x + 155\) P(obtaining a 3) = \(\frac{\text{no of times a 3 was obtained}}{\sum f}\) \(\frac{x}{x + 155} = 0.225\) \(x = 0.255x + (155 \times 0.225)\) \(x - 0.225x = 34.875 \) \(0.775x = 34.875\) \(x = \frac{39.875}{0.775}\) \(x = 45\) \(\sum f = 155 + x = 155 + 45 = 200\) Hence, the die was thrown 200 times. (b) Set of even numbers = {2, 4, 6} Set of prime numbers = {2, 4, 5} Therefore, \(N \cup M = (2, 3, 4, 5, 6)\) \(n(N \cup M) = 30 + 45 + 28 + 40 + 32 = 175\) \(P(N \cup M) = \frac{175}{200} = 0.875\)

(a) In the diagram above, area of \(\Delta PSR = \frac{1}{2} \times |PR| \times 6\) \(36 = \frac{1}{2} \times |PR| \times 6\) \(36 = 3 \times |PR|\) \(|PR| = 12 cm\) But |PQ| = |TS| = 8cm (opp. sides of a parallelogram) Hence, |QR| = |PR| - |PQ| = 12 - 8 = 4cm. (b) In the diagram above, \(\beta = 60^\circ\) (alternate angles) \(\tan \beta = \frac{10.65}{|QB|}\) \(|QB| = \frac{10.65}{\tan 60}\) = \(\frac{10.65}{1.732} = 6.149m\) |PO| = |QB| = 6.149m (opp. sides of a rectangle) In \(\Delta POT, \tan 45 = \frac{|TO|}{6.149}\) \(|TO| = \tan 45 \times 6.149 = 1 \times 6.149 = 6.149m\) |OB| = |PQ| = 10.65m (opp sides of a rectangle) \(h = |OB| - |OT| \) = \(10.65 - 6.149\) = \(4.501 m \approxeq 4.50m\) (3 sig. figs) Hence, the height of the flagpole = 4.50m

(a) (b)(i) From the Venn diagram, x + x + 7 + 8 = n(B) 2x + 15 = 47 2x = 47 - 15 = 32 x = 16 (ii) Number of people who travelled by at least 2 means = 16 + 7 + 8 + 3 = 34

(a) \(\frac{x - 2}{4} = \frac{x + 2}{2x}\) \(2x(x - 2) = 4(x + 2)\) \(2x^{2} - 4x = 4x + 8\) \(2x^{2} - 4x - 4x - 8 = 2x^{2} - 8x - 8 = 0\) Divide through by 2, we have \(x^{2} - 4x - 4 = 0\) \(x^{2} - 4x = 4\) Taking the square of \(\frac{b}{2}\) and add to both sides, \(x^{2} - 4x + (-2)^{2} = 4 + (-2)^{2}\) \((x - 2)^{2} = 8\) \(x - 2 = \pm {\sqrt{8}}\) \(x = 2 \pm \sqrt{8}\) \(x = 2 \pm 2.828\) \(x = 4.828\) or \(x = -0.828\). Hence, x = 4.83 or -0.83 (2 decimal place). (b)(i) In the diagram above, < QRS = 180^\circ - 52^\circ = 128^\circ (opp. angles of a cyclic quadrilateral). \(x_{1} = x_{2}\) (base angles of isosceles triangle) \(x_{1} + x_{2} + 128^\circ = 180^\circ\) (sum of angles of a triangle) \(2x_{2} = 180^\circ - 128^\circ = 52^\circ\) \(x_{2} = 26^\circ\) Hence, < SQT = 26^\circ. (ii) < PQS = 90^\circ (angles in a semi-circle) Hence, < PQT = 90^\circ - 26^\circ = 64^\circ.

(a) In the diagram above, \(\theta = y\) (vertically opposite angles) \(\alpha = 180^\circ - 110^\circ = 70^\circ\) (angles on a straight line) Thus, \(\beta = \theta + \alpha\) (sum of interior opp. angles) \(\beta = y + 70^\circ ..... (1)\) \(x + r + \beta = 180^\circ ..... (2)\) But 2x = r = y Hence, \(x + 2x + (2x + 70) = 5x + 70 = 180\) \(5x = 180 - 70 = 110\) \(x = 22^\circ\) (b) Let Nx represent the donation collected by the girls. Then, the donation collected by the boys = N(x + 600). Also, let y represent the average collection, \(x_{g}\), of the girls. Then the average collection, \(x_{b}\), of the boys = N(y + 100). Thus, \(x_{g} = \frac{x}{12} = y\) \(x = 12y ... (1)\) \(x_{b} = \frac{x + 600}{10} = y + 100\) \(x + 600 = 10(y + 100) .... (2)\) Substitute 12y for x in (2) : \(12y + 600 = 10y + 1000\) \(12y - 10y = 1000 - 600 \) \(2y = 400 \implies y = 200\) From (1), \(x = 12y\) \(\therefore x = 12 \times 200 = N2,400\). Thus, the girls collected N2,400 The boys collected N2,400 + N600 = N3000 Total = N(2,400 + 3,000) = N5,400

(a) Weight (kg) Frequency 62 1 63 5 64 12 65 14 66 10 67 6 68 2 (b) Using an assumed mean of 65kg \(x\) \(f\) \(fx\) \(d = x - A\) \(fd\) \(fd^{2}\) 62 1 62 -3 -3 9 63 5 315 -2 -10 20 64 12 768 -1 -12 12 65 14 910 0 0 0 66 10 660 1 10 10 67 6 402 2 12 24 68 2 136 3 6 18 \(\sum\) = 50 3253 3 93 (i) Mean \(\bar{x} = \frac{\sum fx}{\sum f}\) = \(\frac{3253}{50}\) = 65.06 kg (ii) \(SD = \sqrt{\frac{\sum fd^{2}}{\sum f} - (\frac{\sum fd^{2}}{\sum f})^{2}}\) \(SD = \sqrt{\frac{93}{50} - (\frac{3}{50})^{2}}\) = \(\sqrt{\frac{93}{50} - \frac{9}{2500}}\) = \(\sqrt{\frac{4650 - 9}{2500}}\) = \(\sqrt{\frac{4641}{2500}}\) = \(1.3625 \approxeq 1.36\) (2 d.p.)

(a), (b) (c) < ZMY = 120^\circ.

(a) \(\frac{3p + 4q}{3p - 4q} = 2\) \(3p + 4q = 2(3p - 4q)\) \(3p + 4q = 6p - 8q\) \(3p - 6p = - 8q - 4q\) \(-3p = - 12q\) \(p = 4q\) \(\frac{p}{q} = \frac{4}{1}\) \(\therefore p : q = 4 : 1\). (b)(i) Perimeter of cross- section = 4 + x + 4 + 2 + y + 2 i.e 34 = 12 + x + y x + y = 22 ..... (1) From the diagram, |PQ| = |UR| (opp. sides of a rectangle) i.e. x = 2 + d + 2 x - 4 = d ..... (2) From (1), y = 22 - x is the circumference of the semi-circle. \(22 - x = \frac{2\pi r}{2} = \pi r\) \(r = \frac{d}{2} \) \(22 - x = \frac{22}{7} \times \frac{(x - 4)}{2}\) \(154 - 7x = 11x - 44\) \(154 + 44 = 11x + 7x \) \(198 = 18x\) \(x = \frac{198}{18} = 11 m\) Hence, |PQ| = 11 m. (ii) Area of cross section = Area of rectangle PQRU - area of semi-circle Area of rectangle = \(11 \times 4 = 44 m^{2}\) Area of semi-circle = \(\frac{\pi r^{2}}{2} \) = \(\frac{22}{7} \times (\frac{(11 - 4)}{2})^{2} \times \frac{1}{2}\) = \(19.25 m^{2}\) Area of cross section = \(44 - 19.25 = 24.75 m^{2}\).

(a) x 0^\circ 30^\circ 60^\circ 90^\circ 120^\circ 150^\circ 180^\circ 210^\circ 240^\circ 270^\circ 300^\circ 330^\circ 360^\circ y 2.0 3.2 3.6 3.0 1.6 -0.2 -2.0 -3.2 -3.6 -3.0 -1.6 0.2 2.0 (b) (c)(i) The equation, \(3\sin x + 2\cos x = 0\) has solution where the curve cuts the x- axis, i.e. at A(x = 147^\circ) and B(x = 325.5^\circ). (ii)First, rearrange \(2 + 2\cos x + 3\sin x = 0\) to have the main graph content \(3\sin x + 2\cos x\) on one side of the equation; i.e. \(3\sin x + 2\cos x = -2\). Add the line \(y = -2\) to the graph. The line intersects the curve at the points (x = 180^\circ) and (x = 292.5^\circ). Hence, these are the solutions.

(a) First, change the given linear equation to the gradient- intercept form: \(2x + y = 6 \implies y = 6 - 2x\) Thus, \(m_{1} = -2, c = 6\). Thus, for a line to be parallel, the gradient \(m_{2} = -2\) (condition for parallelism). Using \(y - y_{1} = m(x - x_{1})\), we have \(y - (-3) = -2(x - 2)\) \(y + 3 = - 2x + 4\) \(2x + y = 4 - 3 = 1\) Thus, the line passing through (2, -3) parallel to 2x + y = 6 is 2x + y = 1. (b) (i) \(\Delta\) 2 3 5 7 2 0 3 1 7 3 3 7 7 7 5 1 7 3 7 7 7 7 7 7 (ii) (a) \(2 \Delta (5 \Delta 7) = 2 \Delta 7 = 7\) (b) \(2 \Delta n = 5 \Delta 7\) \(2 \Delta n = 7\) By comparison, n = 7.

If logx 64 = 3, then \(64 = x^3\) \(4^3 = x^3\) Since the indices are equal, x = 4 Hence, x log\(_2\)8 = 4(3.log\(_2\)2) = where log\(_2\)2 = 1(1) = 12 X 1 = 12

If m = 4, n = 9, r = 16, then \(\frac{m}{n}\) - 1\(\frac{7}{9}\) + \(\frac{n}{r}\) = \(\frac{4}{9}\) - \(\frac{16}{9}\) + \(\frac{9}{16}\) = \(\frac{64 - 256 + 81}{144}\) = \(\frac{-111}{144}\) = - \(\frac{37}{48}\)

In a school with students' population 810, the sectoral angle for a class of 72 students is = \(\frac{72}{810}\) x 360º = 32º

Percentage birthrate = \(\frac{62}{6250} \times 100%\) = \(\frac{6200}{6250}\) = 0.992% \(\approx\) 1.0%.

1¼ ÷ [ 2 ÷ ¼] of 28 Apply BODMAS rules \( \frac{5}{4}\) ÷ [ 2 ÷ ¼ × 28 ] \( \frac{5}{4} ÷ \frac{2}{7} \) \( \frac{5}{4} × \frac{7}{2} \) \( \frac{35}{8}\) = \(4 \frac{3}{8}\)