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(a) (b)(i) \(C \cap A' = {15}\) (ii) \(A' \cap (B \cup C) = {2, 6, 12, 15}\)

(a) Area of the major sector = \(\frac{\theta}{360^\circ} \times \pi r^{2}\) = \(\frac{360 - 120}{360} \times \frac{22}{7} \times 7 \times 7\) = \((\frac{2}{3} \times 154) cm^{2}\) Volume = \(\text{Area of the major sector} \times \text{height (cross section)}\) = \(\frac{2}{3} \times 154 \times 15 = 1540 cm^{3}\) (b) Curved surface area = \(\frac{240}{360} \times 2\pi rh\) = \(\frac{240}{360} \times 2 \times \frac{22}{7} \times 7 \times 15\) = \(440 cm^{2}\) Area of rectangular surface = \(2 \times l \times b\) = \(2 \times 7 \times 15 = 210 cm^{2}\) Area of top and bottom of cylinder (major sector) = \(2 \times \frac{240}{360} \times \frac{22}{7} \times 7 \times 7\) = \(205.33 cm^{2}\) \(\therefore \text{Total surface area} = 440 + 205.33 + 210 = 855.33 cm^{2}\)

Hint: Use BODMAS, in other words, do multiplication of the second and the last first before subtracting value obtained from the first.

\((\sqrt[3]{64a^{3}})^{-1} = (\sqrt[3]{(4a)^{3}})^{-1}\) = \((4a)^{-1} \) = \(\frac{1}{4a}\)

\(\frac{p^{2} - q^{2}}{2pq} = \frac{(p + q)(p - q)}{2pq}\) = \(\frac{(1 + \sqrt{2} - (1 - \sqrt{2}))(1 + \sqrt{2} + 1 - \sqrt{2})}{2(1 + \sqrt{2})(1 - \sqrt{2})}\) = \(\frac{(2\sqrt{2})(2)}{-2}\) = \(-2\sqrt{2}\)

Total cost = N(250,000 + 70,000) = N320,000 Selling price = N400,000 (given) Gain = SP - CP = N(400,000 - 320,000) = N80,000 Gain % = gain/CP x 100 = (80,000/320,000) x 100 Gain % = 25%

\(x = \frac{y}{2} \) \(\left(\frac{x^{3}}{y^{3}}+\frac{1}{2}\right) \div \left(\frac{1}{2} - \frac{x^{2}}{y^{2}}\right)\) \(\frac{x^3}{y^3} + \frac{1}{2} = (\frac{y}{2})^{3} \div y^{3} + \frac{1}{2}\) = \(\frac{y^{3}}{8} \times \frac{1}{y^3} + \frac{1}{2}\) = \(\frac{1}{8} + \frac{1}{2}\) = \(\frac{5}{8}\) \(\frac{1}{2} - \frac{x^2}{y^2} = \frac{1}{2} - (\frac{y}{2})^{2} \div y^2)\) = \(\frac{1}{2} - \frac{y^2}{4} \times \frac{1}{y^2}\) = \(\frac{1}{2} - \frac{1}{4}\) = \(\frac{1}{4}\) \(\therefore \left(\frac{x^{3}}{y^{3}}+\frac{1}{2}\right) \div \left(\frac{1}{2} - \frac{x^{2}}{y^{2}}\right) = \frac{5}{8} \div \frac{1}{4}\) = \(\frac{5}{2}\)

Principal, P = Amount, A - Interest, I. A = P + I I = (P.T.R)/100 = (P x 5 x 2)/100 = 10P/100 = P/10 But A = P + I, => 5500 = P + (P/10) => 55000 = 10P + P => 55000 = 11P Thus P = 55000/11 = N5,000

\(\frac{(0.14)^{2} \times 0.275}{7(0.02)} = \frac{(0.14)^{2} \times 0.275}{0.14}\) = \(0.14 \times 0.275\) = \(0.0385 \approxeq 0.039\) Note: Embedded zeroes are counted just the same as non-zero digits and For the number of decimal places stated, count that number of digits to the right of the decimal and underline it. The next number to its right is called the 'rounder decider'. If the 'rounder decider' is 5 or more, then round the previous digit up by 1

\(\frac{a^{3x} - 26a^{2x} + 156a^{x} - 216}{a^{2x} - 24a^{x} + 108}\) Let \(a^{x} = z\) \(\therefore = \frac{z^{3} - 26z^{2} + 156z - 216}{z^{2} - 24y + 108} ... (i)\) Dividing (i) above, we get \(z - 2\) = \(a^{x} - 2\)

\(y = px^{2} + q ... (i)\) \(y = 2x^{2} - 1 ... (ii)\) At x = 2, (i): \(y = p(2^{2}) + q = 4p + q\) (ii): \(y = 2(2^{2}) - 1 = 7\) \(\therefore \text{The coordinates of the point of intersection = (2, 7)}\) (i): \(7 = 4p + q \implies p = \frac{7 - q}{4}\)

By definition a*b = a + b + 1. Let the inverse of the element 2 be x, Therefore 2*x = -1 i.e. 2 + x + 1 = -1 3 + x = -1 x = -1 - 3 x = -4

Given: 4x2 - 9y2 + 20x + 25 Collect like terms: 4x2 + 20x + 25 - 9y2 (2x + 5)(2x + 5) - 9y2 (2x + 5)2 - (3y)2 (2x - 3y +5)(2x + 3y + 5)

(r/L) = (\theta/360^\circ) Given \theta = 300, and L = 7.2cm, => r = (300 x 7.2)/360 r = 6cm

From the diagram above,

Source: Myschool Past Questions Archive (Record #2122)