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-We see that as it passes through centre of curvature it reflects back its path -Incident to the mirror at 90º - and with the normal at 0º

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

Lens formula: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\) where f, d\(_o\) and d\(_i\) are the focal length, object distance and image distance respectively. The distance from the lens to the focal point is called the focal length. For converging lenses, the focal length is always positive, while diverging lenses always have negative focal lengths. ⇒ \(\frac{-1}{30} = \frac{1}{d_o} + \frac{1}{20}\) \(\frac{1}{d_o} = \frac{-1}{30} - \frac{1}{20}\) \(\frac{1}{d_o} = \frac{-2 + -3}{60}\) → \frac{-5}{60}\) = \frac{-1}{12}\) \(d_o = - 12 cm\) The object distance -12cm = 12cm on the other side of the lens.

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

\(\begin{array}{c|c} 3 & \text{27 rem 0} \\ \hline 3 & \text{ 9 rem 0} \\ \hline 3 & \text{ 3 rem 0} \\ \hline 3 & \text{ 1 rem 1}\\ \hline & 0 \end{array}\) 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