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\(\log_4 x = -3\) \(x = 4^{-3}\) = \(\frac{1}{64}\)

Let the number = d. \(\log d = \bar{1}.8732\) \(\log \sqrt{d} = \frac{\bar{1}.8732}{2}\) = \(\frac{\bar{2} + 1.8732}{2}\) = \(\bar{1}.9366\) \(\therefore \sqrt{d} = Antilog (\bar{1}.9366)\) = 0.86

(a) Total number of fruits = 16 limes + 14 apples = 30 (i) P(both are good limes) = \(\frac{10}{30} \times \frac{9}{29} = \frac{3}{29}\) (ii) P(both are bad fruits) = \(\frac{12}{30} \times \frac{11}{29} = \frac{22}{145}\) (iii) P(one is a good apple and the other a bad lime) = \((\frac{8}{30} \times \frac{6}{29}) + (\frac{6}{30} \times \frac{8}{29})\) = \(\frac{8}{145} + \frac{8}{145}\) = \(\frac{16}{145}\) Note : The arrangement is not specific. Hence, it can be a good apple first and then a bad lime or a bad lime first and then a good apple. (b) \(\log_{3} (4x + 1) - \log_{3} (3x - 5) = 2\) \(\log_{3} (\frac{4x + 1}{3x - 5}) = 2\) \((\frac{4x + 1}{3x - 5}) = 3^{2}\) \(\frac{4x + 1}{3x - 5} = 9 \implies 4x + 1 = 9(3x - 5)\) \(4x + 1 = 27x - 45 \implies 23x = 46\) \(x = 2\)

\(\frac{10}{\sqrt{32}}\) = \(\frac{10}{\sqrt{16\times 2}}\) = \(\frac{10}{4\sqrt{2}}\) \(\frac{10\times \sqrt{2}}{4\sqrt{2}\times\sqrt{2}}\) = \(\frac{10\sqrt{2}}{4\times 2}\) = \(\frac{5}{4}\sqrt{2}\)

By trial and error method 1728 is a perfect cube or try multiplying 3 same numbers based on the given options, or use the long division method, and group the divisors(same) in 3s. i.e., 12 x 12 x 12 = 1728, making 1728 a perfect cube.

(a) \(\frac{4.56 \times 3.6}{0.12} \equiv \frac{4.56 \times 360}{12}\) = \(4.56 \times 30 \) = \(136.8 \) = \(1.368 \times 10^{2}\) (b) \((73.8)^{2} - (26.2)^{2}\) Using the method of difference of two squares, = \((73.8 + 26.2) \times (73.8 - 26.2)\) = \(100 \times 47.6\) = \(4760\) (c) \(\sqrt{1\frac{19}{81}}\) = \(\sqrt{\frac{100}{81}}\) = \(\frac{10}{9}\)

\(\frac{7}{19}\) as a percentage will be \(\frac{7}{19}\times \frac{100}{1}\% = 36.8\%\)

398753 \(\approxeq\) 399000 (to 3 s.f)

\(2\frac{1}{6} + 2\frac{7}{12}\) = \(\frac{13}{6} + \frac{31}{12}\) = \(\frac{26 + 31}{12}\) = \(\frac{57}{12} = \frac{19}{4}\) \(\frac{19}{4} - 3\frac{1}{4}\) = \(\frac{19}{4} - \frac{13}{4}\) = \(\frac{6}{4}\) = \(1\frac{1}{2}\)

(a) Taxable income = \(N150,000 - N40,000 = N110,000\) Tax payment = \(N110,000 \times 0.25 = N27,500\) Income left = \(N(150,000 - 27,500) = N122,500\) (b) \(% gain = \frac{\text{actual profit}}{\text{cost price}} \times 100%\) = \(\frac{SP - CP}{CP} \times 100%\) \(\therefore \frac{75 - CP}{CP} = \frac{30}{100} \implies 0.3CP = 75 - CP\) \(1.3CP = 75 \implies CP = N57.69\) Total cost price = \(650 \times \frac{750}{13} = N37,500\) 350 copies sold at N75 ==> \(SP = 350 \times N75 = N26,250\) 300 copies sold at 5% discouunt ==> \((75 - \frac{5}{100} \times 75) \times 300\) = \((75 - 3.75) \times 300 = N21,375\) Total selling price = \(N(26,250 + 21,375) = N47, 625\) Profit = \(N(47,625 - 37,500) = N10,125\) \(% = \frac{10,125}{37,500} \times 100% = 27%\)

C.P = N50000 Loss = x% = \(\frac{\text{x}}{100}\) x 50000 = 500x S.P = C.P - Loss = 50,000 - 500x Profit = 40% = \(\frac{40}{100}\) x (50000 - 500x) = 10,000 \(\frac{2}{5}\) x (50000 - 500x) = 10,000 100,000 - 1000x = 50,000 1000x = 50000 x = 50

4321\(_7\) + 1234\(_7\) = 5555\(_7\) Missing number = 12341\(_7\) - 5555\(_7\) = 3453\(_7\)

Given P\(^T\) = -P, b = 1 If P = \(\begin{pmatrix}a & b\\ c & d\end{pmatrix}\), then, P\(^T\) = \(\begin{pmatrix}a & c\\ b & d\end{pmatrix}\) \(\begin{pmatrix}a & c\\ b & d\end{pmatrix}\) = -\(\begin{pmatrix}a & b\\ c & d\end{pmatrix}\) = \(\begin{pmatrix}-a & -1\\ -c & -d\end{pmatrix}\) comparing a = -a ⇒ 0, b = 1, c = -1, d = -d ⇒ 0 \(\begin{pmatrix}a & b\\ c & d\end{pmatrix}\) = \(\begin{pmatrix}0 & 1\\ -1 & 0\end{pmatrix}\)

\(\begin{pmatrix}t-4 & 0 & 0\\ -1 & t+1 & 1\\3 & 4 & t-2\end{pmatrix}\) = 0 t - 4\(\begin{pmatrix}t+1 & 1\\4 & t-2\end{pmatrix}\) + 0\(\begin{pmatrix}-1 & 1\\3 & t-2\end{pmatrix}\) + 0\(\begin{pmatrix}-1 & t+1\\3 & 4\end{pmatrix}\) t - 4[(t + 1)(t - 1) - 4] = 0 (t - 4)[ t\(^2\) - t - 6] = 0 = (t - 4)((t - 3)(t + 2) = 0 t = 4 or -2 or 3

Given that the ages are in the ratio 3: 4: 5. Let the sum of their ages be t. \(\therefore\) The youngest age = \(\frac{3}{12} t\) The eldest age = \(\frac{5}{12} t\) \(\implies \frac{5}{12} t - \frac{3}{12} t = 18\) \(\frac{2t}{12} = 18 \implies t = \frac{18 \times 12}{2}\) t = 108 years. The sum of their ages = 108 years.