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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 : \(\frac{5000}{20000} \times 100% = 25%\)

(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) 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 \(r = \frac{1}{2}\). \(T_{2} = ar = 8 \implies \frac{a}{2} = 8\) \(a = 16\) \(\therefore \text{The first 6 terms} = 16, 8, 4, 2, 1, \frac{1}{2}\)

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

Number of itineraries = 7P4 = \(\frac{7!}{(7-4)!}\) =\(\frac{7!}{3!}\) =\(\frac{7 \times 6 \times 5 \times 4 \times 3!}{3!}\) = 840

K + 2K + 3K + 75 + 105 = 360 (at a point) 6K + 180 = 360 6K = 360 - 180 6K = 180 k = 180/6 = 30º

\(^{10}C_6 = \frac{10!}{(10-6)!6!}=\frac{10!}{4!6!}\\ =\frac{(10\times 9\times 8\times 7 \times 6!)}{4\times 3\times 2\times 1\times 6!}\\ =210\)

f(x) = 2x\(^3\) - x\(^2\) - 4x – 4 f’(x) = 6x\(^2\) - 2x – 4 As f’(x) = 0 Implies 6x\(^2\) - 2x – 4 = 0 3x – x – 2 = 0 (By dividing by 2) (3x + 2)(x - 1) = 0 3x + 2 = 0 implies x = \(\frac{-2}{3}\) Or x - 1 = 0 implies x = 1 f’(x) = 6x\(^2\) - 2x – 4 f’’(x) = 12x – 2 At max point f’’(x) < 0 ∴f’’(x) = 12x – 2 at x = 1 = 12(1) – 2 = 12 – 2 = 10 ∴ Max at x = \(\frac{-2}{3}\)

x⊕y = xy + x + y = -3/4 (6) + (-3/4) + 6 = -9/2 - 3/4 + 6 = (-18-3+3+24) / 4 = 3/4

x = -2, x = -1 and x = 1 then the factors; x+2, x+1 and x-1 Product of the factors; (x+2)(x+1)(x-1) = y = (x + 2)(x2 - x + x - 1) = y = (x+2)(x2-1) x3 - x + 2x2 - 2 = y x3 + 2x2 - x - 2 = y

2t2 + t - 15 = (2t-5)(t+3)

W ∝ L2 W = KL2 K = W/L2 K = 6/42 K = 6/16 = 3/8 W = 3/8 L2 W = 3/8(√17)2 W = 3/8 x 17 W = 51/8 = 63/8

a*e = a + e + 1 = a implies e+ 1 = 0 ∴ e = -1 7 * e = -1 ∴ a + 7 + 1 = -1 a + 8 = -1 a+8 = -1 a = -1-8 a = -9

\(\sqrt{12}\) -\(\sqrt{147}\) + y\(\sqrt{3}\) = 0 \(\sqrt{4\times 3}\) - \(\sqrt{49\times 3}\) + y\(\sqrt{3}\) = 0 2\(\sqrt{3}\) - 7\(\sqrt{3}\) + y\(\sqrt{3}\) = 0 y\(\sqrt{3}\) = 7\(\sqrt{3}\) - 2\(\sqrt{3}\) y = \(\frac{5\sqrt{3}}{\sqrt{3}}\) y = 5