Question Library

For y = ax2 - bx + c for minimum y \(\frac{dy}{dx}\) = 2x - 2 = \(\frac{dy}{dx}\) = 0 ∴ 2x - 2 = 0 x = 1

BD = \(\sqrt{x^2 + x^2}\) = \(\sqrt{12^2 + 12^2}\) = \(\sqrt{144 + 144}\) = 2(144) = 12\(\sqrt{2cm}\)

The total surface area of a cylinder = \(2\pi r (r + h)\) = \(2 \pi (4) (4 + 5)\) = \(8 \pi (9) = 72 \pi cm^{2}\)

% of boxes containing at least 5 defective bolts each = \(\frac{7 + 17 + 10 + 8 + 6}{50}\) = \(\frac{48}{50}\) x \(\frac{100}{1}\) = 96%

ADB = 180o - 152o = 28o 28o + y + 30o = 180o y = 180o - 58o = 122o

Volume of cylinder = \(\pi\)r\(^2\)h = \(\pi\)a\(^2\)y Volume of cone = \(\frac{1}{3}\) \(\pi\)r\(^2\)h = \(\frac{1}{3}\) \(\pi\)a\(^2\)x Total Volume = \(\pi\)a\(^2\)y + \(\frac{1}{3}\) \(\pi\)a\(^2\)x = \(\frac{3\pi a^2y + \pi a^2x}{3}\) = \(\frac{\pi a^2}{3}(x + 3y)\)

(i) \(2\frac{2}{3} - (2\frac{1}{2} - 1\frac{4}{5})\) = \(2\frac{2}{3} - (\frac{5}{2} - \frac{9}{5})\) = \(\frac{8}{3} - (\frac{25 - 18}{10})\) = \(\frac{8}{3} - \frac{7}{10}\) = \(\frac{80 - 21}{30}\) = \(\frac{59}{30}\) (ii) \(\frac{3.25 - 1.64}{2.47 - 2.01}\) = \(\frac{1.61}{0.46}\) = \(\frac{7}{2}\) = \(3\frac{1}{2}\)

(a) No Log \(\sqrt[3]{1.376}\) \(0.1386 \div 3 \) = \(0.0462\) \(\sqrt[4]{0.007}\) \(\bar{3}.8451 \div 4\) = \(\bar{1}.4613\) \(\frac{\sqrt[3]{1.376}}{\sqrt[4]{0.007}}\) = \(0.5849\) Antilog = 3.845 \(\therefore \frac{\sqrt[3]{1.376}}{\sqrt[4]{0.007}} \approxeq 3.85\) (to 3 s.f) (b) \(\frac{\log 81}{\log \frac{1}{3}}\) = \(\frac{\log 3^{4}}{\log 3^{-1}}\) = \(\frac{4 \log 3}{-1 \log 3}\) = \(-4\)

(a) (i) Nobody takes Economics and Chemistry \(n(E \cap C) = \null \). (ii) 4 pupils take Economics and Government; \(n(E \cap G) = 4\) (b) (i) Suppose \(n(C \cap G) = x\) \(22 = 8 + 4 + 8 - 4 - x + x + 7 - x\) \(22 = 23 - x \implies x = 1\) Only 1 person does Chemistry and Government. (ii) Number taking only Government = \(8 - (4 + x)\) \(8 - (4 + 1) = 3\)

(a) Sheets expected from 5 packets = \(480 \times 5 = 2400\) Actual sheet contained in the packets = \(435 + 420 + 405 + 415 + 440 = 2115\) Error in supply = 2400 - 2115 = 285 sheets. \(\therefore\) Percentage error = \(\frac{285}{2400} \times 100% = 11.875%\) \(\approxeq 12%\) (b) If 1 packet = 480 sheets cost N35.00 \(\therefore \text{2400 sheets cost } \frac{2400}{480} \times N35.00 = N175.00\) Amount of buyer was cheated = \(\frac{Error \times N175}{\text{Expected no of sheets}}\) \(\frac{N285 \times 175}{2400} = N20.78\) The buyer was cheated by N21 to the nearest naira.

(a) Given: Circle ABC with centre O To prove : < AOB = 2 < ACB Construction: Join CO produced to P. Proof: With lettering as in the figure, OA = OB (radii) ; \(x_{1} = x_{2}\) (base angles of an isosceles triangle) \(\therefore < AOP = x_{1} + x_{2}\) (exterior angle of triangle AOC) \(\therefore < AOP = 2x_{2} (x_{1} = x_{2})\) Also, \(< BOP = 2y_{2}\) (similar proof as the earlier done ones) \(\therefore < AOB = 2x_{2} + 2y_{2} = 2(x_{2} + y_{2})\) \(< AOB = 2 \times < ABC\) (proven) (b) From the diagram, < OQR = 32^\circ and < MPQ = 15^\circ. (i) \(\therefore < ORQ = 32^\circ\) (base angles of an isosceles triangle) \(< OQR = 180^\circ - (32^\circ + 32^\circ) = 180^\circ - 64^\circ = 116^\circ\) \(\therefore < QPR = \frac{116^\circ}{2} = 58^\circ\) (angle subtended at the centre) (ii) Join MO and MQ, Since < MPQ = 15^\circ then < MOQ = 2 < MPQ (angle at the centre) \(\therefore < MOQ = 2 \times 15^\circ = 30^\circ\) \(\therefore < OMQ = < MQO \) (base angles of an isosceles triangle) \(\therefore < MQO = \frac{180^\circ - 30^\circ}{2} = 75^\circ\)

Let T and B be the top and base of the control tower respectively; P, the point of pilot's observation and PS, the line of observation. (a) Shortest distance is |PB|. \(\therefore \cos 33^\circ = \frac{8.5}{PB}\) \(PB = \frac{8.5}{\cos 33} = \frac{8.5}{0.8387}\) \(|PB| = 10.135km \approxeq 10.1km\) (b) Height of the control tower = |SB| - |ST| \(\therefore \tan 33^\circ = \frac{|SB|}{8.5 km} \implies |SB| = 8.5 \tan 33^\circ\) \(|SB| = 5.52 km\) \(\tan 30^\circ = \frac{|ST|}{8.5 km} \implies |ST| = 8.5 \tan 30^\circ\) \(|ST| = 4.907 km\) \(\therefore\) The height of the tower = 5.52 km - 4.907 km = 0.613 km.

first work out the expression and then correct the answer to 4 s.f = 241.34..............(A) (3 x 10-\(^3\))\(^2\)............(B) = 3\(^2\)x\(^2\) = \(\frac{1}{10^3}\) x \(\frac{1}{10^3}\) (Note that x\(^2\) = \(\frac{1}{x^3}\)) = 24.34 x 3\(^2\) x \(\frac{1}{10^6}\) = \(\frac{2172.06}{10^6}\) = 0.00217206 = 0.002172(4 s.f)

\(\sqrt{27}\) + \(\frac{3}{\sqrt{3}}\) = \(\sqrt{9 \times 3}\) + \(\frac{3 \times {\sqrt{3}}}{{\sqrt{3}} \times {\sqrt{3}}}\) = 3\(\sqrt{3}\) + \(\sqrt{3}\) = 4\(\sqrt{3}\)

\(\frac{4\frac{3}{4} - 6\frac{1}{4}}{4\frac{1}{5} \text{ of } 1\frac{1}{4}}\) \(\frac{19}{4}\) - \(\frac{25}{4}\)............(A) \(\frac{21}{5}\) x \(\frac{5}{4}\).............(B) Now work out the value of A and the value of B and then find the value \(\frac{A}{B}\) A = \(\frac{19}{4}\) - \(\frac{25}{4}\) = \(\frac{-6}{4}\) B = \(\frac{21}{5}\) x \(\frac{5}{4}\) = \(\frac{105}{20}\) = \(\frac{21}{4}\) But then \(\frac{A}{B}\) = \(\frac{-6}{4}\) \(\div\) \(\frac{21}{4}\) = \(\frac{-6}{4}\) x \(\frac{4}{21}\) = \(\frac{-24}{84}\) = \(\frac{-2}{7}\)