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Let the number be y, subtract y from 2 i.e 2 - y 2 - y = 4 < \(\frac{1}{5}\) y, 2 - y = \(\frac{y}{5}\) - 4 2 - y + 4 = \(\frac{y}{5}\) 6 = \(\frac{y}{5}\) + y 6 = \(\frac{y + 5y}{5}\) 6 = \(\frac{6y}{5}\) multiplying through by 5 6 * 5 = 6y \(\frac{30}{6}\) = y = 5

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

\(\begin{array}{c|c} 3 & \text{27 rem 0} \\ \hline 3 & \text{ 9 rem 0} \\ \hline 3 & \text{ 3 rem 0} \\ \hline 3 & \text{ 1 rem 1}\\ \hline & 0 \end{array}\) Hence the correct answer is 10003

The sum, S of ratio is S = 5 + 3 + 2 = 10. But highest share = \(\frac{5}{10} \times T\), where T is the total number of apples. Thus, \(40 = \frac{5}{10} \times T\), given 40 x 10 = 5T, \(T = \frac{40 \times 10}{5} = 80\) Hence the smallest share = \(\frac{2}{10} \times 80\) = 16 apples

\(\frac{1.25 \times 0.025}{0.05}\) \( = \frac{125 \times 10^{-2} \times 25 \times 10^{-3}}{5 \times 10^{-2}}\) = 125 x 5 x 10-3 = 625 x 0.001 = 0.625 = 0.6 Approx to 1 d.p.

Using \(S.I =\frac{P \times T \times R}{100}\) \(600 =\frac{3000 \times T \times 8}{100}\) \(T =\frac{600 \times 100}{3000 \times 8}\) \(\frac{20}{8}\) = 2\(\frac{1}{2}\) years

\(\frac{3^{-5n}}{9^{1-n}} \times 27^{n + 1}\) \(\frac{3^{-5n}}{3^{2(1-n)}} \times 3^{3(n + 1)}\) \(3^{-5n} \div 3^{2(1-n)} \times 3^{3(n + 1)}\) \(3^{-5n - 2(1-n) + 3(n + 1)}\) \(3^{-5n - 2 + 2n + 3n + 3}\) \(3^{-5n + 5n + 3 - 2}\) \(3^{1}\) = 3

log1041/3 = 1/3 log104 = 1/3 x 0.6021 = 0.2007

\(\frac{\sqrt{5}(\sqrt{147} - \sqrt{12}}{\sqrt{15}}\) \(\frac{\sqrt{5}(\sqrt{49 \times 3} - \sqrt{4 \times 3}}{\sqrt{5 \times 3}}\) \(\frac{\sqrt{5}(7\sqrt{3} - 2\sqrt{3}}{\sqrt{5} \times \sqrt{3}}\) \(\frac{\sqrt{3} (7 - 2}{\sqrt{3}}\) = 5

P = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} Q = {-1, 3, 5, 7, 11, 13, 17, 19, 23} P \(\cap\) Q = {3, 5, 7, 11, 13, 17, 19}

S = \(\sqrt{t^2 - 4t + 4}\) S2 = t2 - 4t + 4 t2 - 4t + 4 - S2 = 0 Using \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) Substituting, we have; Using \(t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(4 - S^2)}}{2(1)}\) \(t = \frac{4 \pm \sqrt{16 - 4(4 - S^2)}}{2}\) \(t = \frac{4 \pm \sqrt{16 - 16 + 4S^2}}{2}\) \(t = \frac{4 \pm \sqrt{4S^2}}{2}\) \(t = \frac{2(2 \pm S)}{2}\) Hence t = 2 + S or t = 2 - S

Let f(x) = x2 - x - k Then by the factor theorem, (x - 4): f(4) = (4)2 - (4) - k = 0 16 - 4 - k = 0 12 - k = 0 k = 12

Let f(p) = 6p3 - p2 - 47p + 30 Then by the remainder theorem, (p - 3): f(3) = remainder R, i.e. f(3) = 6(3)3 - (3)2 - 47(3) + 30 = R 162 - 9 - 141 + 30 = R 192 - 150 = R R = 42

P \(\propto\) mu, p \(\propto \frac{1}{q}\) p = muk ................ (1) p = \(\frac{1}{q}k\).... (2) Combining (1) and (2), we get P = \(\frac{mu}{q}k\) 4 = \(\frac{m \times u}{1}k\) giving k = \(\frac{4}{6} = \frac{2}{3}\) So, P = \(\frac{mu}{q} \times \frac{2}{3} = \frac{2mu}{3q}\) Hence, P = \(\frac{2 \times 6 \times 4}{3 \times \frac{8}{5}}\) P = \(\frac{2 \times 6 \times 4 \times 5}{3 \times 8}\) p = 10

\(r \propto \frac{1}{\sqrt{s}}, r \propto \frac{1}{\sqrt{t}}\) \(r \propto \frac{1}{\sqrt{s}}\) ..... (1) \(r \propto \frac{1}{\sqrt{t}}\) ..... (2) Combining (1) and (2), we get \(r = \frac{k}{\sqrt{s} \times \sqrt{t}} = \frac{k}{\sqrt{st}}\) This gives \(\sqrt{st} = \frac{k}{r}\) By taking the square of both sides, we get st = \(\frac{k^2}{r^2}\) s = \(\frac{k^2}{r^{2}t}\)