Question Library

Volume of a cone = \(\frac{1}{3} \pi r^2h\) h2 = 52 = 32 = 25 - 9 = 16 h = \(\sqrt{16}\) h = 4cm v = \(\frac{1}{3} \times \frac{22}{7} \times 3^2 \times 4\) \(\frac{1}{3} \times \frac{22}{7} \times 9 \times 4\) = \(\frac{22 \times 3 \times 4}{7}\) = 37.7cm3 = 38cm3

(x + 2)(x\(^2\) - 3x + 2) + 2(x + 2)(x - 1) = (x + 2)m (x + 2)[(x\(^2\) - 3x + 2) + 2(x - 1)] = (x + 2)M divide both side by (x + 2) (x\(^2\) - 3x + 2) + 2(x - 1) = M x\(^2\) - 3x + 2 + 2x - 2 = M x\(^2\) - 3x + 2 + 2x - 2 = M x\(^2\) - 3x + 2x = M x\(^2\) - x = M M = x\(^2\) - x

L2 = 962 + 282 = 9216 + 784 = 10000 L = \(\sqrt{10000}\) = 100cm curved surface area = \(\pi r l\) = \(\frac{22}{7} \times 28 \times 100\) = 8800cm2 area of cone = area of sector area of sector = 8800cm2

by Pythagoras x2 = 62 + 82 36 + 64 = 100 x = \(\sqrt{100}\) x = 10

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

\(\frac{45^o}{360}\) x 100% = 12\(\frac{1}{2}\)%

(x \(\cap\) y2) = [a, b] n(x \(\cap\) y2) = 2

(a) \(7x + 4 < \frac{1}{2}(4x + 3)\) \(7x + 4 < 2x + \frac{3}{2}\) \(7x - 2x < \frac{3}{2} - 4\) \(5x < -\frac{5}{2}\) \(\frac{5x}{5} < \frac{-\frac{5}{2}}{5}\) \(x < -\frac{1}{2}\). (b) When Salem gets N2 ------> Sunday gets 50 kobo (Recall N1 = 100 kobo) Salem gets 200 kobo ------> Sunday gets 50 kobo. Sunday gets N4 -----> Shaka gets N2. ====> Sunday gets 400 kobo ------> Shaka gets 200 kobo. When Sunday gets 400 kobo ------> Salem gets 1600 kobo. \(\therefore\) Salem : Sunday : Shaka = 1600 : 400 : 200 = 8 : 2 : 1 \(\therefore \text{Shaka's share} = \frac{1}{11} \times N1,100 = N100\)

(a) Present ages of a father and his son are in the ratio 10 : 3. Let the sum of their ages be x. The son's age = 15 years = \(\frac{3}{13} \times x = 15\) \(x = \frac{15 \times 13}{3} = 65\) \(\therefore\) The Father's present age = 65 - 15 = 50 years. In z years time, the ratio of their ages = 2 : 1 \(\frac{50 + z}{15 + z} = \frac{2}{1}\) \(\implies 50 + z = 2(15 + z)\) \(50 + z = 30 + 2z \implies 50 - 30 = 2z - z\) \(20 = z\) Therefore, in 20 years' time, the ratio of their age will be 2 : 1. (b) \(\frac{x + y + z}{3} = 6 \implies x + y + z = 18 .... (1)\) \(\frac{x + y + z + l + u + v + w}{7} = 9 \implies x + y + z + l + u + v + w = 63 .... (2)\) Putting (1) into (2), we have \(18 + l + u + v + w = 63 \) \(\therefore l + u + v + w = 63 - 18 = 45\) Mean of l, u, v and w = \(\frac{45}{4} = 11.25\).

(a)(i) \(p > q .... (1)\) \(p + q = 3(p - q) ...... (2)\) From (2), \(p + q = 3p - 3q\) \(p - 3p = - 3q - q \implies - 2p = - 4q\) \(p = 2q\) (ii) \(\frac{p^{2} + q^{2}}{pq}\) = \(\frac{(2q)^{2} + q^{2}}{2q \times q}\) \(\frac{4q^{2} + q^{2}}{2q^{2}}\) = \(\frac{5q^{2}}{2q^{2}}\) = \(\frac{5}{2}\) (b) Selling price = \(\frac{100 \times 66}{25}\) = \(N264.00\) Using \(\frac{SP - CP}{CP} = % gain\) \(\frac{264 - CP}{CP} = \frac{32}{100}\) \(100(264 - CP) = 32CP\) \(26400 = 32CP + 100CP = 132CP\) \(CP = \frac{26400}{132}\) = \(N200\) When he sells 20 for N50, Selling price = \(\frac{100 \times 50}{20} = N250.00\) Hence, he made gain. \(% gain = \frac{250 - 200}{200} \times 100%\) \(\frac{1}{4} \times 100%\) = \(25% gain\)

(a) (i) (3 - x), 6, (7 - 5x). First term = a = (3 - x) common ratio = \(\frac{6}{3 - x} = \frac{7 - 5x}{6}\) \(36 = (3 - x)(7 - 5x)\) \(36 = 21 - 15x - 7x + 5x^{2}\) \(36 = 21 - 22x + 5x^{2}\) \(5x^{2} - 22x + 21 - 36 = 0\) \(5x^{2} - 22x - 15 = 0\) \(5x^{2} - 25x + 3x - 15 = 0 \implies 5x(x - 5) + 3(x - 5) = 0\) \((x - 5)(5x + 3) = 0\) \(x = -\frac{3}{5} \text{ or = } 5\) (ii) Constant ratio = \(\frac{6}{3 - x} = \frac{7 - 5x}{6}\) Using x = 5, \(r = \frac{7 - 5(5)}{6} = \frac{-18}{6}\) = \(-3\) (b) Considering \(\Delta\) AOC in the above diagram. \(b^{2} = 4^{2} + 3^{2}\) \(b^{2} = 16 + 9 = 25\) \(b = \sqrt{25} = 5 cm\) Considering \(\Delta\) ACD, Using cosine rule, \(\cos D = \frac{a^{2} + b^{2} - c^{2}}{2ab} = \frac{a^{2} + c^{2} - b^{2}}{2ac}\) = \(\frac{6^{2} + 7^{2} - 5^{2}}{2 (6)(7)}\) = \(\frac{36 + 49 - 25}{84}\) \(\cos D = \frac{60}{84}\) \(\cos D = 0.714\) \(D = \cos^{-1} (0.714)\) \(D = 44.4^\circ \approxeq 44^\circ\) (to the nearest degree)

Marks frequency cum freq lower bound upper bound 1 - 10 2 2 0.5 10.5 11 - 20 3 5 10.5 20.5 21 - 30 5 10 20.5 30.5 31 - 40 13 23 30.5 40.5 41 - 50 19 42 40.5 50.5 51 - 60 31 73 50.5 60.5 61 - 70 13 86 60.5 70.5 71 - 80 9 95 70.5 80.5 81 - 90 4 99 80.5 90.5 91 - 100 1 100 90.5 100.5 (a) see the diagram above. (b) (i) The 60th percentile is where the line from the cumulative frequency of 60 touches the curve i.e. 60th percentile = 58.5 (ii) If the pass mark was fixed at 35%, then the frequency will be 13 with the class mark 31 - 40.

Midpoint of a line PQ where P has coordinates (\(x_1\), \(y_1\)) and Q has coordinates (\(x_2\), \(y_2\)) is given as \((\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})\). \(\therefore\) If Q has coordinates (r, s), then \(\frac{-2 + r}{2} = 2\) and \(\frac{1 + s}{2} = 3\) \(-2 + r = 4 \implies r = 6\) \(1 + s = 6 \implies s = 5\) Q = (6, 5)

The expression(P n Q) u R is shaded by identifying two specific areas: 1. (P n Q): the overlapping region between circles P and R 2. u R: the entire area of the circle R

\(\frac{3\sqrt{5} \times 4\sqrt{6}}{2 \sqrt{3} \times 3\sqrt{2}}\) = \(\frac{\sqrt{5} \times 2\sqrt{6}}{\sqrt{2} \times \sqrt{3}}\) = \(\frac{\sqrt{5} \times 2 \sqrt{6}}{\sqrt{6}}\) = 2\(\sqrt{5}\)