Question Library

(a) \(\frac{2}{5}(x - 2) - \frac{1}{6}(x + 5) \leq 0\) Multiplying through by the LCM of 5 and 6 (i.e 30) \(12(x - 2) - 5(x + 5) \leq 0\) \(12x - 24 - 5x - 25 \leq 0\) \(7x - 49 \leq 0 \implies 7x \leq 49\) \(x \leq 7\). (b) (i) \(\frac{x^{2} - y^{2}}{x^{2} + xy}\) P = \(\frac{(x - y)(x + y)}{x (x + y)}\) P = \(\frac{x - y}{x}\) (ii) When x = -4, y = -6 \(P = \frac{-4 - (-6)}{-4}\) \(P = \frac{2}{-4}\) \(P = - \frac{1}{2}\)

Item Amount Spent (N) Food 4,500 House Rent 3,000 Provisions 2,500 Electricity 2,000 Transportation 5,000 Others 3,000 Total 20,000 Items in degrees : Food : \(\frac{4500}{20000} \times 360^\circ = 81^\circ\) House rent : \(\frac{3000}{20000} \times 360^\circ = 54^\circ\) Provisions : \(\frac{2500}{20000} \times 360^\circ = 45^\circ\) Electricity : \(\frac{2000}{20000} \times 360^\circ = 36^\circ\) Transportation : \(\frac{5000}{20000} \times 360^\circ = 90^\circ\) Others : \(\frac{3000}{20000} \times 360^\circ = 54^\circ\) (a) Pie chart : (b) Percentage of his income spent on transportation : \(\frac{5000}{20000} \times 100% = 25%\)

The sum of exterior angles = 360^\circ \(60^\circ + 2(90 - m)^\circ + 2(30 + 2m)^\circ = 360^\circ\) \(60^\circ + 180^\circ - 2m^\circ + 60^\circ + 4m^\circ = 360^\circ\) \(300^\circ + 2m = 360^\circ\) \(\implies 2m = 360^\circ - 300^\circ = 60^\circ\) \(m = 30^\circ\) (b) In \(\Delta QRS, \cos Q = \frac{\sqrt{3}}{2}\) \(\cos Q = 0.8660 \implies Q = 30^\circ\) \(\therefore < SQR = 30^\circ^\) \(< PQS = 180^\circ - 30^\circ \) (angles on a straight line = 180^\circ) \(< PQS = 150^\circ\)

(a) Perimeter of a rectangle = 2(l + b) \(2l + 2b = 280\) But l = 2.5b, \(2(2.5b) + 2b = 280 \implies 7b = 280\) \(b = 40 m\) \(\therefore l = 2.5(40) \) = \(100 m\) (b) Volume of pyramid = \(\frac{1}{3}l \times b \times h\) = \(\frac{1}{3}(4.5) \times 2.5 \times 4 \) = \(15 m^{3}\).

(a) \(2^{x + y} = 16 \implies 2^{x + y} = 2^{4}\) \(\implies x + y = 4 ..... (1)\) \(4^{x - y} = \frac{1}{32} \implies (2^{2})^{(x - y)} = 2^{-5}\) \(\implies 2x - 2y = -5 .... (2)\) From (1), x = 4 - y. \(\therefore 2(4 - y) - 2y = -5\) \(8 - 2y - 2y = -5 \implies -4y = - 5 - 8 = -13\) \(y = \frac{13}{4}\) \(x = 4 - \frac{13}{4} \) = \(\frac{3}{4}\) (b) \(P \propto \frac{Q^{2}}{R}\) \(\implies P = \frac{kQ^{2}}{R}\) \(k = \frac{PR}{Q^{2}}\) \(k = \frac{36 \times 4}{9} = 16\) \(\implies P = \frac{16Q^{2}}{R}\) \(Q^{2} = \frac{PR}{k}\) \(Q^{2} = \frac{200 \times 2}{16}\) \(Q^{2} = 25\) \(Q = \pm 5\).

(a) \(4x^{2} = 11x + 21 \implies 4x^{2} - 11x - 21 = 0\) \(a = 4, b = - 11 , c = -21\) \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) \(x = \frac{-(-11) \pm \sqrt{(-11)^{2} - 4(4)(-21)}}{2(4)}\) \(x = \frac{11 \pm \sqrt{121 + 336}}{8}\) \(x = \frac{11 \pm \sqrt{457}}{8}\) \(x = \frac{11 \pm 21.378}{8}\) \(x = \frac{11 + 21.378}{8}\) or \(x = \frac{11 - 21.378}{8}\) \(x = \frac{32.378}{8}\) or \(x = \frac{-10.378}{8}\)\ \(x = \text{4.047 or -1.297}\) \(x \approxeq \text{4.05 or -1.30}\).(2 decimal place). (b) (i) \(I = \frac{PRT}{100}\) \(I = £(1560 - 1500) = £60\) \(60 = \frac{1500 \times R \times 1}{100}\) \(60 = 15R \implies R = 4%\) (ii) Interest for the second year = \(\frac{1560 \times 4 \times 1}{100}\) = \(£62.40\) (c) Initial price of the car \(N300,000.00\) Depreciation after first year (25%) \(\frac{25}{100} \times 300,000 = N75,000\) Value after 1st year N225,000.00 Depreciation after 2nd year(20%) \(\frac{20}{100} \times 225,000 = N45,000\) Value after 2nd year N180,000.00

Class Interval Tally Classmark (x) Freq (f) \(fx\) 21 - 25 IIII || 23 7 161 26 - 30 |||| |||| | 28 11 308 31 - 35 |||| |||| 33 9 297 36 - 40 |||| |||| 38 9 342 41 - 45 |||| | 43 6 258 46 - 50 |||| ||| 48 8 384 \(\sum\) 50 1750 (b) (c) Mode : \(L_{1} + (\frac{f_{0} - f_{1}}{2f_{0} - f_{1} - f_{2}})t\) Where \(L_{1}\) = lower class boundary of modal class = 25.5 \(f_{0}\) = frequency of modal class = 11 \(f_{1}\) = frequency of pre-modal class = 7 \(f_{2}\) = frequency of post modal class = 9 \(t\) = interval mark = 5. Mode : \(25.5 + (\frac{11 - 7}{22 - 7 - 9})\times 5 = 25.5 + 3.3\) = 28.8. (d) Mean : \(\frac{\sum fx}{\sum f} = \frac{1750}{50}\) = 35.

(a) (i) \(\cos C = \frac{a^{2} + b^{2} - c^{2}}{2ab}\) \(\cos C = \frac{12^{2} + 10^{2} - 17^{2}}{2(12)(10)}\) \(\cos C = \frac{-45}{240} = -0.1875\) \(C = \cos^{-1} (-0.1875)\) = \(100.81^\circ\) (ii) Area of \(\Delta ABC = \frac{1}{2} ab \sin C\) = \(\frac{1}{2} \times 12 \times 10 \times \sin 100.81\) = \(60 \times 0.9822\) = \(58.936 m^{2}\) (b) \(\tan 63 = \frac{38}{x}\) \(x = \frac{38}{\tan 63}\) \(x = 19.362 m\) \(\approxeq 19 m\)

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

(a) 105^\circ + 26^\circ + x = 180^\circ 131^\circ + x = 180^\circ x = 180^\circ - 131^\circ = 49^\circ. 131^\circ + x + 131^\circ + 360 - y = 360^\circ 131^\circ + 49^\circ + 131^\circ + 360 - y = 360^\circ 671^\circ - y = 360^\circ y = 671^\circ - 360^\circ = 311^\circ. (b)(i) \(\stackrel\frown{ABC} = \frac{1}{2} \stackrel\frown{AOC} reflex\) = \(\frac{1}{2} \times 190^\circ = 95^\circ\) (ii) \(\stackrel\frown{ADC} = \frac{1}{2} \stackrel\frown{AOC}\) = \(\frac{1}{2} (360^\circ - 190^\circ)\) = \(\frac{1}{2} \times 170^\circ\) = 85^\circ (c) (i) \(\stackrel\frown{ABC} = \frac{360 - 190}{360} \times 2\pi r\) = \(\frac{170}{360} \times 2 \times 3.142 \times 7\) = \(20.77 cm\) \(\approxeq 20.8 cm\) (3 significant figures) (ii) \(AD = 2 \times 7\cos 35\) = \(14 \times 0.8192\) = \(11.468 cm \approxeq 11.5 cm\) (3 significant figures)

(a) x 0^\circ 60^\circ 120^\circ 180^\circ 240^\circ 300^\circ 360^\circ \(\sin x\) 0 0.8660 0.8660 0 -0.8660 -0.8660 0 \(\cos x\) 1 0.5 -0.5 -1 -0.5 -0.5 1 \(3\sin x\) 0 2.598 2.598 0 -2.598 -2.598 0 \(2\cos x\) 2 1 -1 -2 -1 -1 2 y 2 3.598 1.598 -2 -3.598 -3.598 2 y 2.00 3.60 1.60 -2.00 -3.60 -3.60 2.00 (b) (c) From graph, \(x_{1} = 120 + \frac{1}{5} \times 15 = 120 + 3 = 123^\circ\) \(x_{2} = 360^\circ - \frac{2}{3} \times 15 = 360 - 10 = 350^\circ\) (d) From graph, \(180^\circ - 15 < x < 300 + \frac{3}{5} \times 15\) \(165^\circ < x < 300 + 9 = 165^\circ < x < 309^\circ\)

(a) 3, x, y, 18. Note : When given the consecutive terms of an A.P to be \(a, b, c, d\), then \(\frac{a + c}{2} = b ; \frac{b + d}{2} = c\). \(\therefore \frac{3 + y}{2} = x \implies 3 + y = 2x .... (1)\) Also, \(\frac{x + 18}{2} = y \implies x + 18 = 2y ..... (2)\) From (2), x = 2y - 18. Putting that in (1), we have \(3 + y = 2(2y - 18) \implies 3 + y = 4y - 36\) \(3 + 36 = 4y - y \implies 3y = 39\) \(y = 13\) \(x = 2y - 18\) \(x = 2(13) - 18 = 26 - 18 = 8\) \((x, y) = (8, 13)\). (b)(i) G.P \(T_{2} = ar ; T_{3} = ar^{2} ; T_{4} = ar^{3}\) \(T_{2} + T_{3} = 6T_{4}\) \(ar + ar^{2} = 6ar^{3}\) \(6ar^{3} - ar^{2} - ar = 0\) \(6ar^{3} - 3ar^{2} + 2ar^{2} - ar = 0\) \(3ar^{2}(2r - 1) + ar(2r - 1) = 0\) \((3ar^{2} + ar)(2r - 1) = 0\) \(ar(3r + 1)(2r - 1) = 0\) \(\implies 3r + 1 = 0 ; 2r - 1 = 0\) \(r = \frac{-1}{3} ; r = \frac{1}{2}\) (ii) Since r is positive, then \(r = \frac{1}{2}\). \(T_{2} = ar = 8 \implies \frac{a}{2} = 8\) \(a = 16\) \(\therefore \text{The first 6 terms} = 16, 8, 4, 2, 1, \frac{1}{2}\)

Arrange in ascending order 2, 5, 5, 5, 6, 6, 6, 6, 8, 8 Position of median = N/2 = 10/2 = 5th and 6th ∴Median = (6+6) / 2 = 12/2 = 6 Mode the item that repeats itself most = 6 ∴Product of modal and median number = 6 x 6 = 36

Number of grapes = 6 Number of banana = 11 Number of oranges = 13 Total = 30 P(Grape) = 6/30, P(Banana) = 11/30 P(Orange) = 13/30 ∴P(Either grape or banana) = 6/30 + 11/30 = 17/30

students who made the trip = 2 + 3 + 4 + 5 + 6 + 8 + 1 = 29