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-7 \(\leq\) 9 - 8x < 16 - x -7 \(\leq\) 9 - 8x and 9 - 8x < 16 - x -7 - 9 \(\leq\) -8x and -8x + x < 16 - 9 -16 \(\leq\) -8x and -7x < 7 \(\therefore\) x \(\leq\) 2 and -1 < x -1 < x \(\leq\) 2.

\(T_n = 2^{2n - 1}\) \(T_1 = 2^{2(1) - 1} \) = 2 \(T_2 = 2^{2(2) - 1}\) = 8 \(T_3 = 2^{2(3) - 1}\) = 32 \(T_4 = 2^{2(4) - 1}\) = 128 \(T_1 + T_2 + T_3 + T_4 = 2 + 8 + 32 + 128\) = 170

\(\int_{-1} ^{2} (2x^2 + x) \mathrm {d} x\) = \([\frac{2x^{2 + 1}}{3} + \frac{x^{1 + 1}}{2}]_{-1} ^{2}\) = \([\frac{2x^{3}}{3} + \frac{x^{2}}{2}]_{-1} ^{2}\) = \((\frac{2(2)^{3}}{3} + \frac{2^2}{2}) - (\frac{2(-1)^{3}}{3} + \frac{(-1)^{2}}{2})\) = \((\frac{16}{3} + 2) - (\frac{-2}{3} + \frac{1}{2})\) = \(\frac{22}{3} - (-\frac{1}{6})\) = \(\frac{22}{3} + \frac{1}{6}\) = \(\frac{15}{2}\) = \(7\frac{1}{2}\)

\(p \propto \frac{1}{\sqrt{q}}\) \(\implies p = \frac{k}{\sqrt{q}}\) when p = 3, q = 16. \(3 = \frac{k}{\sqrt{16}}\) \(k = 3 \times 4 = 12\) \(\therefore p = \frac{12}{\sqrt{q}}\) when p = 4, \(4 = \frac{12}{\sqrt{q}} \implies \sqrt{q} = \frac{12}{4}\) \(\sqrt{q} = 3 \implies q = 3^2 \) \(q = 9\)

200 mangoes at 4 for N2.50 \(\implies\) Total cost price = \(\frac{200}{4} \times N 2.50\) = N 125.00 Since 30 mangoes got spoilt \(\implies\) Left over = 200 - 30 = 170 mangoes 170 mangoes at 2 for N 2.40 \(\implies\) Total selling point = \(\frac{170}{2} \times N 2.40\) = N 204.00 Profit : N (204.00 - 125.00) = N 79.00 % profit = \(\frac{79}{125} \times 100%\) = 63.2% profit.

S.I = \(\frac{PRT}{100}\) \(\implies\) N 4890 = \(\frac{8550 \times 3 \times x}{100}\) \(x = \frac{4890 \times 100}{8550 \times 3}\) \(x = 19.06%\) \(x \approxeq 19%\)

81\(^{\frac{-3}{4}}\) x 25\(^{\frac{1}{2}}\) x 243\(^{\frac{2}{5}}\) = \((\sqrt[4]{81})^{-3} \times \sqrt{25} \times (\sqrt[5]{243})^2\) = \(\frac{5 \times 3^2}{3^{3}}\) = \(\frac{5}{3}\)

(\frac{(0.5436)^3}{0.017 \times 0.219}\) = \(\frac{0.16063}{0.017 \times 0.219}\) = 43.1 (to 3 s.f)

\(s = (4t + 3)(t - 2)\) \(\frac{\mathrm d s}{\mathrm d t} = (4t + 3)(1) + (t - 2)(4)\) = \(4t + 3 + 4t - 8\) = 8t - 5 \(\frac{\mathrm d s}{\mathrm d t} (t = 5 secs) = 8(5) - 5\) = 40 - 5 = 35 units per sec

Since there are 4 angles given, the polygon is a quadrilateral. Sum of angle in a quadrilateral = 360^\circ \(\therefore\) 2x + 5x + x + 4x = 360^\circ 12x = 360^\circ x = 30^\circ

Actual weight = 0.18g Error = 0.21g - 0.18g = 0.03g % error = \(\frac{0.03}{0.18} \times 100%\) = 16.7%

\((\frac{6}{0.32} \div \frac{2}{0.084})^{-1}\) = \((\frac{600}{32} \div \frac{2000}{84})^{-1}\) = \((\frac{600}{32} \times \frac{84}{2000})^{-1}\) = \((\frac{63}{80})^{-1}\) = \(\frac{80}{63}\) = 1.3 (to 1 decimal place)

2\(^{x + y}\) = 16 ; 4\(^{x - y}\) = \(\frac{1}{32}\). \(\implies 2^{x + y} = 2^4\) \(x + y = 4 ... (1)\) \(2^{2(x - y)} = 2^{-5} \) \(2^{2x - 2y} = 2^{-5}\) \(\implies 2x - 2y = -5 ... (2)\) Solving the equations (1) and (2) simultaneously, we have x = \(\frac{3}{4}\) and y = \(\frac{13}{4}\)

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

3x + y = 21 ... (i); xy = 30 ... (ii) From (ii), \(y = \frac{30}{x}\). Putting the value of y in (i), we have 3x + \(\frac{30}{x}\) = 21 \(\implies\) 3x\(^2\) + 30 = 21x 3x\(^2\) - 21x + 30 = 0 3x\(^2\) - 15x - 6x + 30 = 0 3x(x - 5) - 6(x - 5) = 0 (3x - 6)(x - 5) = 0 3x - 6 = 0 \(\implies\) x = 2. x - 5 = 0 \(\implies\) x = 5. If x = 2, y = \(\frac{30}{2}\) = 15; If x = 5, y = \(\frac{30}{5}\) = 6.