Question Library

(a) \(\frac{0.6 \times 32 \times 0.004}{1.2 \times 0.008 \times 0.16}\) = \(\frac{6 \times 10^{-1} \times 32 \times 4 \times 10^{-3}}{12 \times 10^{-1} \times 8 \times 10^{-3} \times 16 \times 10^{-2}}\) = \(\frac{4}{2 \times 4 \times 10^{-2}}\) = \(\frac{1}{2} \times \frac{1}{10^{-2}\) = \(0.5 \times 10^{2} \) = \(5 \times 10^{1}\) (b) In the diagram above, \(< GHB = 180 - 7x^\circ\) (angles on a straight line) \(< HBN = 180 - 7x^\circ\) (alternate angles) \(< EBN = 3x^\circ\) (corresponding angles) Then \(3x^\circ + (180 - 7x)^\circ = 120^\circ\) \(\implies 3x^\circ - 7x^\circ + 180^\circ = 120^\circ\) \(\implies 180^\circ - 120^\circ = 7x^\circ - 3x^\circ\) \(60^\circ = 4x^\circ\) \(x = 15^\circ\) Hence, \(< GHB = 180^\circ - (7 \times 15^\circ) = 180^\circ - 105^\circ\) = \(75^\circ\).

(a) \(3\sqrt{75} = 3 \times \sqrt{25 \times 3} = 3(5\sqrt{3}) = 15\sqrt{3}\) \(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\) \(\sqrt{108} = \sqrt{36 \times 3} = 6\sqrt{3}\) \(\therefore 3\sqrt{75} - \sqrt{12} + \sqrt{108} \equiv 15\sqrt{3} - 2\sqrt{3} + 6\sqrt{3}\) = \((15 - 2 + 6)\sqrt{3}\) = \(19\sqrt{3}\). (b) \(124_{n} = 232_{five}\) \(124_{n} = (1 \times n^{2}) + (2 \times n^{1}) + (4 \times n^{0}) \) = \(n^{2} + 2n + 4\) \(232_{five} = (2 \times 5^{2}) + (3 \times 5^{1}) + (2 \times 5^{0})\) = \(50 + 15 + 2\) = \(67\) \(\implies n^{2} + 2n + 4 = 67\) \(n^{2} + 2n + 4 - 67 = 0\) \(n^{2} + 2n - 63 = 0\) \(n^{2} - 7n + 9n - 63 = 0\) \(n(n - 7) + 9(n - 7) = 0\) \((n + 9)(n - 7) = 0\) \(\text{n = -9 or 7}\) Hence, \(n = 7\) \(124_{seven} = 232_{five}\)

(a) \(\frac{1}{x} + \frac{1}{y} = 5 ..... (1)\) \(\frac{1}{y} - \frac{1}{x} = 1 ........ (2)\) Substitute p and q for \(\frac{1}{x}\) and \(\frac{1}{y}\) respectively. \(p + q = 5 ......... (1a)\) \(q - p = 1 ....... (2a)\) From (2a), q = 1 + p. Putting into (1a), we have: \(p + (1 + p) = 5\) \(2p + 1 = 5 \implies 2p = 5 - 1 = 4\) \(p = 2\) \(q = 1 + p = 1 + 2 = 3\) \(p = \frac{1}{x} = 2\) \(\implies x = \frac{1}{2}\) \(q = \frac{1}{y} = 3\) \(\implies y = \frac{1}{3}\) Hence, \(x = \frac{1}{2} ; y = \frac{1}{3}\). (b) Total distance travelled = 48 km Total time taken = 45 minutes = \(\frac{45}{60} = 0.75 hr\) Let the man travel a distance of x km on a good surface. Then he travel a distance of (48 - x) km on bad surface. Let the time taken to travel on good surface be t hours, then the time taken to travel on bad surface = (0.75 - t) hrs. Using \(speed = \frac{distance}{time}\) in each case On good surface : \(72 = \frac{x}{t}\) \(\implies x = 72t ... (1)\) On bad surface : \(48 = \frac{48 - x}{0.75 - t}\) \(48(0.75 - t) = 48 - x\) \(\implies 36 - 48t = 48 - x ...... (2)\) Substitute 72t = x in (2) : \(36 - 48t = 48 - 72t \implies 72t - 48t = 48 - 36\) \(24t = 12 \implies t = \frac{12}{24} = 0.5 hours\) From (1), x = 72t \(x = 72 \times 0.5 = 36 km\) Hence, the man travels a distance of 36 km on good road surface.

(a) Area of the shaded segment = area of sector XOY - area of triangle OXY. \(\implies 504 = \frac{\theta}{360^\circ} \times \pi r^{2} - \frac{1}{2} r^{2} \sin \theta\) \(\implies 504 = \frac{90}{360} \times \frac{22}{7} \times r^{2} - \frac{1}{2} \times r^{2} \sin 90\) \(504 = r^{2}(\frac{22}{28} - \frac{1}{2})\) \(504 = \frac{2r^{2}}{7} \implies r^{2} = \frac{504 \times 7}{2} = 1764\) \(r = \sqrt{1764} = 42 cm\). (b) In the diagram, \(a_{1} = a_{2}\) (base angles of an isosceles triangle) \(\therefore a_{1} + a_{2} = 2a_{1}\) \(\therefore 2a_{1} + 109^\circ = 180^\circ \implies 2a_{1} = 180^\circ - 109^\circ = 71^\circ\) \(a_{1} = a_{2} = \frac{71^\circ}{2} = 35.5^\circ\) Hence, \(< RQS = 66^\circ + 35.5^\circ\) = \(101.5^\circ\)

Let A and B denote the events that the man wins contracts X and Y respectively. Then P(A) = 0.5 P(A') = 1 - 0.5 = 0.5 P(B') = 0.3 P(B) = 1 - 0.3 = 0.7 (a) The probability that the man wins both contracts = \(0.5 \times 0.7 = 0.35\). (b) The probability that the man wins exactly one of the contracts is \(P(A) \times P(B') + P(B) \times P(A')\) = \(0.5 \times 0.3 + 0.7 \times 0.5\) = \(0.15 + 0.35\) = \(0.50\) (c) Neither of the contracts (i.e not X, not Y) = \(P(A') \times P(B')\) = \(0.5 \times 0.3\) = \(0.15\)

(a) \(\frac{3}{2p - \frac{1}{2}} = \frac{\frac{1}{3}}{\frac{1}{4}p + 1}\) \(\implies 3(\frac{1}{4}p + 1) = \frac{1}{3}(2p - \frac{1}{2})\) \(\frac{3}{4}p + 3 = \frac{2}{3}p - \frac{1}{6}\) \(\frac{3}{4}p - \frac{2}{3}p = - 3 - \frac{1}{6}\) \(\frac{1}{12}p = - 3\frac{1}{6}\) \(p = \frac{-19}{6} \div \frac{1}{12}\) \(p = \frac{-19}{6} \times 12 = -38\) (b) \(5% \times ¢760.00 = \frac{5}{100} \times ¢760 = ¢38.00\) So the TV set was actually sold for ¢(760 - 38) = ¢722.00 Using, \(% profit = \frac{\text{SP - CP}}{CP} \times 100%\) \(20% = \frac{760 - x}{x} \times 100%\) \(20% x = (760 - x) \times 100%\) \(x = 5(760 - x) \implies x = 3800 - 5x\) \(6x = 3800 \implies x = ¢633.33\) Thus, the actual profit = actual SP - CP = ¢(722 - 633.33) = ¢88.67 Hence, actual %age profit = \(\frac{88.67}{633.33} \times 100%\) = 14.0006% = 14% (2 sig. figs)

(a) x 0^\circ 30^\circ 60^\circ 90^\circ 120^\circ 150^\circ 180^\circ 210^\circ 240^\circ 270^\circ y 1.0 2.0 2.7 3.0 2.7 2.0 1.0 0.0 -0.7 -1.0 (b) (c) \(\sin x = \frac{1}{4}\) (Given) Multiply through by 2 ; \(2 \sin x = 2 \times \frac{1}{4} = \frac{1}{2}\) \(2 \sin x = \frac{1}{2} \implies 2 \sin x - \frac{1}{2} = 0\) Add \(1\frac{1}{2}\) to both sides ; \(2 \sin x - \frac{1}{2} + 1\frac{1}{2} = 0 + 1\frac{1}{2}\) \(\implies 2 \sin x + 1 = 1\frac{1}{2}\) Draw the line \(y = 1\frac{1}{2}\) on the same axis as \(y = 2 \sin x + 1\). The line \(y = 1\frac{1}{2}\) cuts the curve at points P and Q where x = 15^\circ and x = 168^\circ. Hence, the values of x for which \(\sin x = \frac{1}{4}\)are 15^\circ and 168^\circ.

(a) \(\otimes\) 1 5 9 10 1 1 5 9 10 5 5 3 1 6 9 9 1 4 2 10 10 6 2 1 (i) \((9 \otimes 5) \otimes (10 \otimes 10) = 1 \otimes 1 = 1\) (ii) (1) \(10 \otimes m = 2\) By comparison, \(10 \otimes 9 = 2\). (2) \(n \otimes n = 4\) From the table, \(9 \otimes 9 = 4\) Hence, n = 9. (b) Let the fraction = \(\frac{m}{n}\) \(\frac{m}{n} = \frac{3}{4}..... (1)\) \(\implies n = \frac{4m}{3} ....... (2)\) \(2m = n + 34 ...... (3)\) Put (2) in (3), \(2m = \frac{4m}{3} + 34\) \(2m - \frac{4m}{3} = 34 \implies \frac{2m}{3} = 34\) \(m = \frac{34 \times 3}{2} = 51\) \(n = \frac{4m}{3} = \frac{4 \times 51}{3} = 68\) Hence, the fraction = \(\frac{51}{68}\).

(a) (i) \((16 - 2x) + (6 + x) + (19 - 3x) + 5x + 7x + 8x + 4x + 4 = 125\) \(45 + 20x = 125 \implies 20x = 125 - 45 = 80\) \(x = \frac{80}{20} = 4\) (ii) \(n(P \cup Q \cap R') = (16 - 2x) + 5x + (6 + x)\) = \(22 + 4x\) = \(22 + 4(4) = 22 + 16 = 38\) (b) In the diagram above, < XYW = 90^\circ (angle in a semi-circle; XW is a diameter) < XWY + 90^\circ + 50^\circ = 180^\circ (sum of angles in a triangle) < XWY = 180^\circ - 140^\circ = 40^\circ (i) < WYZ = 40^\circ (alternate angles; \(\overline{XW} || \overline{YZ}\)) (ii) < WOZ = \(2 \times 40^\circ \) (angle at centre = twice angle at circumference) = 80^\circ \(\therefore < WEO = 180^\circ - (40^\circ + 80^\circ)\) = \(180^\circ - 120^\circ \) = \(60^\circ\) \(\therefore < YEZ = < WEO = 60^\circ\) (vertically opposite angles).

(a) \((x - 2)(x - 3) = 12\) \(x^{2} - 3x - 2x + 6 = 12 \implies x^{2} - 5x + 6 - 12 = 0\) \(x^{2} - 5x - 6 = 0\) \(x^{2} - 6x + x - 6 = 0 \implies x(x - 6) + 1(x - 6) = 0\) \( (x - 6)(x + 1) = 0 \implies \text{x = 6 or -1}\) (b) The shaded portion comprises two shaded segments labelled A and B. Area of segment A = \(\frac{60^\circ}{360^\circ} \times \pi r^{2} - \frac{1}{2} r^{2} \sin 60^\circ\) \(\frac{1}{6} \times \frac{22}{7} \times 7 \times 7 - \frac{1}{2} \times 7 \times 7 \times 0.8660\) = \(25.667 - 21.217\) = \(4.45 cm^{2}\) Also, the area of the segment B = \(4.45 cm^{2}\). Hence, the area of the shaded portion = \(4.45 cm^{2} + 4.45 cm^{2}\) = \(8.9 cm^{2}\) \(\approxeq 9 cm^{2}\) (to the nearest whole number).

(a) Scores (x) Frequency (f) \(fx\) 1 2 2 2 5 10 3 13 39 4 11 44 5 9 45 6 10 60 \(\sum f = 50\) \(\sum fx = 200\) \(Mean (\bar{x}) = \frac{\sum fx}{\sum f} = \frac{200}{50} = 4\) \(d = x - 4\) \(|d|\) f \(f|d|\) -3 3 2 6 -2 2 5 10 -1 1 13 13 0 0 11 0 1 1 9 9 2 2 10 20 \(\sum f|d| = 58\) Hence, Mean Deviation = \(\frac{\sum f|d|}{\sum f} = \frac{58}{50}\) = 1.16 (b) Let E denote the event of getting a score of at least 4. \(n(E) = 11 + 9 + 10 = 30\) \(p(E) = \frac{n(E)}{n(S)} = \frac{30}{50}\) = \(0.6\)

(a) \(5 \cos (x + 8.5)^\circ - 1 = 0\) \(5 \cos (x + 8.5)^\circ = 1\) \(\cos (x + 8.5)^\circ = \frac{1}{5} = 0.2\) \((x + 8.5)^\circ = \cos^{-1}(0.2) = 78.463^\circ\) \(x = 78.463^\circ - 8.5 = 69.963^\circ\) \(\approxeq 70^\circ\) (to the nearest degree). (b)(i) (ii) By the cosine rule, \(|QR|^{2} = 32^{2} + 24^{2} - 2 \times 32 \times 24 \times \cos 45\) \(|QR|^{2} = 1024 + 576 - 1536 \cos 45\) = \(1600 - 1086.1056\) \(|QR|^{2} = 513.8944\) \(|QR| = \sqrt{513.8944} = 22.669 km\) \(\approxeq 22.67 km\) ( 2 decimal place) (iii) By the sine rule, \(\frac{32}{\sin \alpha} = \frac{22.67}{\sin 45}\) \(\sin \alpha = \frac{32 \times \sin 45}{22.67}\) = \(0.9981\) \(\alpha = \sin^{-1} (0.9981) = 86.4787^\circ\) The diagram below shows all the angles at Q; The bearing of R from Q is given by the reflex angle NQR. Thus reflex < NQR = 360^\circ - (86.47^\circ + 30^\circ) = 360^\circ - 116.47^\circ = 243.53^\circ Hence, the bearing of R from Q = 244^\circ (to the nearest degree).

(a) \(f : x \to 2x^{2} - 1 ; g : x \to 3x + 2\) (i) \(f(x - 1) - 7 = 0\) \(f(x - 1) = 2(x - 1)^{2} - 1 = 2(x^{2} - 2x + 1) - 1\) = \(2x^{2} - 4x + 2 - 1\) \(f(x - 1) - 7 = 2x^{2} - 4x + 1 - 7 = 2x^{2} - 4x - 6 = 0\) \(2x^{2} - 6x + 2x - 6 = 0 \implies 2x(x - 3) + 2(x - 3) = 0\) \((2x + 2)(x - 3) = 0 \implies 2x = -2; x = 3\) \(x = -1 ; 3\) (ii) \(\frac{f(-\frac{1}{2}) . g(3)}{f(4) - g(5)}\) \(f(-\frac{1}{2}) = 2(-\frac{1}{2})^{2} - 1 = \frac{1}{2} - 1 = -\frac{1}{2}\) \(g(3) = 3(3) + 2 = 9 + 2 = 11\) \(f(4) = 2(4^{2}) - 1 = 32 - 1 = 31\) \(g(5) = 3(5) + 2 = 15 + 2 = 17\) \(\frac{f(-\frac{1}{2}) . g(3)}{f(4) - g(5)} = \frac{(-\frac{1}{2}) . (11)}{31 - 17}\) = \(\frac{-\frac{11}{2}}{14}\) = \(-\frac{11}{28}\) (b) \(m \ast n = \frac{-n}{m^{2} + 1}; m, n \in R\) \(-3 \ast -10 = \frac{-(-10)}{(-3)^{2} + 1}\) = \(\frac{10}{10} = 1\) \(-10 \ast -3 = \frac{-(-3)}{(-10)^{2} + 1}\) = \(\frac{3}{101}\) Hence, \(\ast\) is not commutative as \(m \ast n \neq n \ast m\).

Perimeter of the sector = \(2r + \frac{\theta}{360^\circ} \times 2\pi r\) = \(2(10.5) + \frac{120}{360} \times 2 \times \frac{22}{7} \times 10.5\) = \(21 + 22\) = 43 cm

Angle in a semi-circle is 90º when OM is subtended : (90 - 25)º = 65º