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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.

Speed = 30 kmh\(^{-1}\) = \(\frac{30 \times 1000}{3600}\) = \(\frac{25}{3} ms^{-1}\) Time = \(\frac{Distance}{Speed}\) = \(\frac{200}{\frac{25}{3}}\) = \(\frac{200 \times 3}{25}\) = 24 seconds

Total capacity of the container = 45 liters = 45 x 1000 = 45000 cm\(^3\) Time to fill the container = \(\frac{45000}{2}\) = 22500 seconds = \(\frac{22500}{3600}\) = 6 hr 15 mins

Price of television = N25,000 Paid N10, 000 Remainder = N(25,000 - 10,000) = N15,000 Eight equal installments = \(\frac{15,000}{8}\) = N1,875 per installment.

\(I = \frac{PRT}{100}\) \(R = \frac{100I}{PT}\) \(R = \frac{100 \times 60}{2000 \times \frac{3}{4}}\) \(R = \frac{6000}{1500}\) = 4%

\(x \propto \frac{1}{y}\) \(x = \frac{k}{y}\) \(\frac{2}{3} = \frac{k}{9}\) \(3k = 18 \implies k = 6\) \(x = \frac{6}{y}\) When y = \(\frac{3}{4}\), x = \(\frac{6}{\frac{3}{4}}\) = \(\frac{6 \times 4}{3}\) = 8

P varies directly as q means p = kq some constant k q varies inversely as r means q = \(\frac{\text{m}}{\text{r}}\) for some constant m Substituting the expression for q into the equation for p: p =\(\frac{\text{km}}{\text{r}}\) This shows that p varies inversely as r.

\(x = -\frac{1}{2}\), y = 4 3x\(^2\)y + xy\(^2\) 3\(\left[-\frac{1}{2}\right]^2 \times 4 + \left(\frac{-1}{2}\right)(4)^2\) 3\(\times \frac{1}{4} \times 4 -\frac{1}{2} \times 16\) 3 - 8 = -5

\(k = m\sqrt{\frac{t - p}{r}}\) \(\frac{k}{m} = \sqrt{\frac{t - p}{r}}\) \((\frac{k}{m})^2 = \frac{t - p}{r}\) \(rk^2 = m^2 (t - p)\) \(\therefore m^2 t = rk^2 + m^2 p\) \(t = \frac{rk^2 + m^2 p}{m^2}\)

Equation with roots -8 and 5: (x + 8)(x - 5) = 0 \(x^2 - 5x + 8x - 40 = 0\) \(x^2 + 3x - 40 = 0\)

3 + 5x - 2x\(^2\) = 0 3 + 6x - x - 2x\(^2\) = 0 3(1 + 2x) - x(1 + 2x) = 0 (3 - x)(1+2x) = 0 3 - x = 0 or 1 + 2x = 0 x = 3 or x = \(\frac{-1}{2}\)