Question Library

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

The graph should start at its lowest point(-1) at -\(\frac{\pi}{2}\), rise to cross the origin (0, 0), and continue upward to reach its peak (1) at \(\frac{\pi}{2}\). It then turns downward, crossing the x- axis(0) at \(\pi\) and ends back at its lowest point (-1) at \(\frac{3\pi}{2}\)

The two dice produce 36 equally likely outcomes in total. The sums divisible by 3 are exactly 3, 6, 9, and 12. The complete list of favorable pairs \((d_1, d_2)\) is: Sum = 3 (2 ways): (1,2), (2,1) Sum = 6 (5 ways): (1,5), (2,4), (3,3), (4,2), (5,1) Sum = 9 (4 ways): (3,6), (4,5), (5,4), (6,3) Sum = 12 (1 way): (6,6) Total favourable outcomes: 2 + 5 + 4 + 1 = 12. Therefore, the probability is \(\frac{12}{36} = \frac{1}{3}.\)

\(\begin{array}{c|c} \text{Age in years} & 10 & 20 & 30 & 40 & 50 \\\hline \text{frequency} & 3 & 4 & 5 & 6 & 2\end{array}\) \(\sum f \) = number of people involved in the pollution \(\sum f\) = 3 + 4 + 5 + 6 + 2 = 20

To calculate the circumference of a circle, use the formula C = \pid, where "C" is the circumference, "d" is the diameter, and \pi is 3.14 Or The ratio of the circumference to the diameter of a circle is the mathematical constant \pi (pi), which is approximately 3.14.

- If two solids are similar, the ratio of their volumes is equal to the cube of the ratio of their corresponding sides. Given: Heights of cylinder --> 21 : 7 or 3 : 1 = \(\frac{3}{1}\) = 3 times height enlargement Volume / Capacity of larger cylinder = 3\(^3\) \(\times\) 250g = 27 \(\times\) 250g = 6,750 or 6.75kg

(a) \(p \propto r^{2}\) \(\implies p = kr^{2}\) \(3.2 = k(4^{2}) \implies 3.2 = 16k\) \(k = \frac{3.2}{16} = 0.2\) \(\therefore p = 0.2r^{2}\) When r = 6.5, \(p = 0.2 (6.5^{2}) \) = \(0.2(42.25)\) \(p = 8.45\) (b) \(\frac{x}{2} + \frac{y}{4} = 1 ... (1)\) \(\frac{x}{3} - \frac{y}{4} = -\frac{1}{6} .... (2)\) Adding equation (1) and (2), we have \(\frac{x}{2} + \frac{x}{3} = 1 - \frac{1}{6}\) \(\frac{5x}{6} = \frac{5}{6} \implies x = 1\) Put x = 1 in (1), \(\frac{1}{2} + \frac{y}{4} = 1 \implies \frac{y}{4} = 1 - \frac{1}{2}\) \(\frac{y}{4} = \frac{1}{2} \implies y = 2\) \((x, y) = (1, 2)\).

(a) \(\sqrt{50} - 3\sqrt{2}(2\sqrt{2} - 5) - 5\sqrt{32}\) \(\sqrt{25 \times 2} - (3 \times 2)\sqrt{2 \times 2} + (3 \times 5)\sqrt{2} - 5\sqrt{16 \times 2}\) = \(5\sqrt{2} - 6(2) + 15\sqrt{2} - 5(4\sqrt{2})\) = \(5\sqrt{2} - 12 + 15\sqrt{2} - 20\sqrt{2}\) = \((5 + 15 - 20)\sqrt{2} - 12\) = -12. (b) \(\frac{1}{2} \log_{10} (\frac{25}{4}) - 2 \log_{10} (\frac{4}{5}) + \log_{10} (\frac{320}{125})\) = \(\log_{10} (\frac{25}{4})^{\frac{1}{2}} - \log_{10} (\frac{4}{5})^{2} + \log_{10} (\frac{320}{125})\) = \(\log_{10} (\frac{5}{2}) - \log_{10} (\frac{16}{25}) + \log_{10} (\frac{320}{125})\) = \(\log_{10} (\frac{\frac{5}{2} \times \frac{320}{125}}{\frac{16}{25}}\) = \(\log_{10} (\frac{\frac{160}{25}}{\frac{16}{25}}\) = \(\log_{10} 10 = 1\)

(a) \(625^{\frac{3}{8}} \times 5^{\frac{1}{2}} \div 25\) = \((5^{4})^{\frac{3}{8}} \times 5^{\frac{1}{2}} \div 5^{2}\) = \(5^{\frac{3}{2}} \times 5^{\frac{1}{2}} \div 5^{2}\) = \(5^{\frac{3}{2} + \frac{1}{2} - 2}\) = \(5^{0} = 1\) (b)(i) \(\tan (\theta + 25)^\circ = 5.145\) \(\theta + 25 = \tan^{-1} (5.145)\) \(\implies \theta + 25 = 79^\circ\) \(\theta = 79^\circ - 25^\circ = 54^\circ\) (ii) \(5 \cos \theta - 1 = 0\) \(\implies 5 \cos \theta = 1\) \(\cos \theta = \frac{1}{5} = 0.2\) \(\theta = \cos^{-1} (0.2) = 78.463^\circ\) \(\approxeq 78.5^\circ\) (to one d.p)

Let < POQ = \(\theta\) \(\implies\) SR = \(4\theta\). \(< PSQ = \frac{\theta}{2} ; < SPR = 2\theta\) In \(\Delta PSK\), \(2\theta + \frac{\theta}{2} = 90^\circ\) \(\frac{5\theta}{2} = 90^\circ \implies 5\theta = 180^\circ\) \(\theta = \frac{180^\circ}{5} = 36^\circ\) \(\therefore < SOR = 4\theta = 4 \times 36^\circ = 144^\circ\) Length of arc = \(\frac{\theta}{360^\circ} \times 2\pi r\) \(\frac{144}{360} \times 2 \times \frac{22}{7} \times 7 \) = \(17.6 cm\).

x 0 1 2 3 4 5 7 2 7 7 2 3 9 fx 0 7 14 6 12 45 (a)(i) Median = \(\frac{1}{2} (N + 1)\) = \(\frac{1}{2}(30 + 1) = 15.5\) = 15th and 16th position = \(\frac{2 + 2}{2} = 2\) (ii) Mean = \(\frac{\sum fx}{\sum f}\) = \(\frac{84}{30}\) = \(2.8 \approxeq 3\) (nearest whole no). (b)

(a) \(\frac{\frac{1}{3}c^{2} - \frac{2}{3}cd}{\frac{1}{2}d^{2} - \frac{1}{4}cd}\) = \(\frac{\frac{1}{3}c(c - 2d)}{\frac{1}{4}d(2d - c)}\) = \(\frac{\frac{1}{3}c (c - 2d)}{-\frac{1}{4}d (c - 2d)}\) = \(-\frac{4c}{3d}\) (b)(i) In \(\Delta XYP\), \(\frac{7}{XP} = \sin 44^\circ\) \(XP = \frac{7}{\sin 44}\) = \(10.08 cm\) < XPM + 44^\circ + 46^\circ = 180^\circ < XPM = 90^\circ. In \(\Delta XPM\), \(XM^{2} = XP^{2} + PM^{2}\) \(XM^{2} = (10.08)^{2} + 9^{2}\) = \(101.6064 + 81 = 182.6064\) \(XM = \sqrt{182.6064}\) = \(15.513 cm \approxeq 13.5 cm\) In \(\Delta XYP\), \(\frac{7}{YP} = \tan 44\) \(YP = \frac{7}{\tan 44}\) \(YP = 7.249 cm \approxeq 7.2 cm\) In \(\Delta MPF\), \(\frac{PF}{9} = \cos 46\) \(PF = 9 \cos 46 = 6.2519 cm\) \(YF = YP + PF = 7.249 + 6.2519\) = \(13.5009 cm \approxeq 13.5 cm\) (ii) \(\frac{10.1}{9} = \tan \theta\) \(\tan \theta = 1.122\) \(\theta = \tan^{-1} (1.122)\) = 48.296^\circ

(a) Items Percentage(%) Degree in pie chart Pension 5 \(\frac{5}{100} \times 360^\circ = 18^\circ\) Income tax 25 \(\frac{25}{100} \times 360^\circ = 90^\circ\) Food 40 \(\frac{40}{100} \times 360^\circ = 144^\circ\) Transport 10 \(\frac{10}{100} \times 360^\circ = 36^\circ\) Rent 12.5 \(\frac{12.5}{100} \times 360^\circ = 45^\circ\) Others 7.5 \(\frac{7.5}{100} \times 360^\circ = 27^\circ\) 100 360^\circ A pie- chart showing the monthly contributions and expenditure pattern of an employee in 1999. (b)(i) \(360^\circ = N10,800.00\) \(\text{Pension} - 18^\circ = \frac{18}{360} \times N10,800\) = \(N540\) (ii) \(360^\circ = N10,800.00\) \(\text{Income tax} - 90^\circ = \frac{90}{360} \times N10,800\) = \(N2700\) (c) The percentage sum of pension and income tax = (5 + 25)% = 30% Take home pay = 100% - 30% = 70% = \(\frac{70}{100} \times N10,800\) = \(N7560\)

(a)(i) U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} A = {1, 2, 5, 7} A' = U - A = {3, 4, 6, 8, 9, 10} (ii) \(A \cap B = {1, 2, 5, 7} \cap {1, 3, 6, 7} = {1, 7}\) \((A \cap B)' = {2, 3, 4, 5, 6, 8, 9, 10}\) (iii) \(A \cup B = {1, 2, 5, 7} \cup {1, 3, 6, 7}\) = \({1, 2, 3, 5, 6, 7}\) \((A \cup B)' = {4, 8, 9,10}\) (iv) {1, 3, 6}, {1, 3, 7}, {1, 6, 7} and {3, 6, 7}. (b) The nth term of the sequence is given as \(U_{n} = \frac{n + 1}{n(n + 2)}, n = 1, 2,3\) For the 15th term, \(U_{15} = \frac{16}{15 \times 17}\) (c)(i) \(T_{n} = a + (n - 1)d\) \(T_{n} = 3 + (n - 1)4 = 3 + 4n - 4\) = \(4n - 1\) (ii) \(4n - 1 > 100\) \(4n > 100 + 1 \implies 4n > 101\) \(n > 25.25\) The least term greater than 100 = 26th term.

(a) Class Boundaries Tally Frequency 59.5 - 64.5 || 2 64.5 - 69.5 ||| 3 69.5 - 74.5 |||| | 6 74.5 - 79.5 |||| |||| | 11 79.5 - 84.5 |||| ||| 8 84.5 - 89.5 |||| || 7 89.5 - 94.5 || 2 94.5 - 99.5 | 1 (b) An histogram to represent the distribution of 40 students in an examination. (c) Modal mark : \(\frac{74.5 + 79.5}{2} = \frac{154}{2} = 77\) (d) Number of students who obtained marks greater than 79 : 8 + 7 + 2 + 1 = 18. Total number of students = 40 \(\therefore\) P(mark greater than 79) = \(\frac{18}{40} = \frac{9}{20}\)