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Since < URP = 25^\circ, then < XPU = 4 x 25^\circ = 100^\circ \(\therefore\) < TPQ = 180^\circ - 100^\circ = 80^\circ \(\therefore\) < PQT = 180^\circ - (80^\circ + 25^\circ) = 75^\circ < SQR = 75^\circ - 25^\circ = 50^\circ (exterior angle = 2 opp interior angles) \(\therefore\) < USQ = 180^\circ - 50^\circ = 130^\circ

Let x be the exterior angle; interior = 3x; but x + 3x = 180o 4x = 180o; x = 45 = \(\frac{360}{45}\) = 8

From the figure, < PQR = 68^\circ \(\therefore\) < QRS = < QSR = \(\frac{180 - 68}{2}\) (base angles of an isos. triangle) = 56^\circ \(\therefore\) < PRS = 90^\circ - 56^\circ = 34^\circ (angles in a semi-circle)

Area = 1/2 x 16 x 16sin60o = 64√3cm2

Let the radius of the arc PQ = r and the radius of the arc XY = R. Length of arc PQ = \(\frac{\theta}{360} \times 2\pi r = 1\) Length of arc XY = \(\frac{\theta}{360} \times 2\pi R = 3\) Ratio of the arc = \(\frac{r}{R} = \frac{360 \times 2\pi \theta}{2\pi \theta \times 360 \times 3}\) = \(\frac{1}{3}\) Ratio of their area = \((\frac{1}{3})^2 = \frac{1}{9}\) = 1 : 9

Area of \(\Delta\)DCF = 24cm2 Area of Quad= 2 x 24 = 48cm2

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

3 - \(\frac{1}{3}\) - (\(\frac{5}{4}\) x \(\frac{2}{3}\)) + 1\(\frac{2}{5}\) = \(\frac{10}{3}\) - \(\frac{5}{6}\) + \(\frac{7}{5}\) = \(\frac{100 - 25 + 42}{30}\) = \(\frac{117}{30}\) = 3.9 \(\approx\) 4

\(\begin{array}{c|c} 7056 \\ \text{- 4577} \\\hline 2257 \end{array}\) By trial and error method Let the base to 8 i.e. Let n = 8 and it is easily verified that the subtraction holds. The subtraction does not hold when other values of n are tried n = 8

(\(\frac{1}{0.05} \div \frac{1}{5.005}\)) - (0.05 x 2.05) (\(\frac{1}{0.05}\) x 5.005) - (0.05 x 2.05) \(\frac{5.005}{0.05}\) - 0.1025 100.1 - 0.1025 = 99.9975 99.998 to 3 decimal place.

\(\frac{62}{3}\) = 20.6666..... = 20.7 to 3 sig. fig.

P produces 20,000 bags P reduces production by 5% i.e. reduction = \(\frac{5}{100}\) x \(\frac{20000}{1}\) = 1000 Q produces 15000 bags Q increases production by 5% i.e. increase = \(\frac{5}{100}\) x \(\frac{15000}{1}\) = 750 Effective loss by the two = 1000 - 750 = 250 bags

% of water in the mixture = \(\frac{\text{Total Amount of water}}{\text{Total quantity of spirit}}\) x \(\frac{100}{1}\) \(\frac{3(\frac{20}{100}) + 5 (\frac{15}{100})}{3 + 5}\) x \(\frac{100}{1}\) = \(\frac{\frac{6}{10} + \frac{75}{100}}{8}\) x \(\frac{100}{1}\) = \(\frac{0.6 + 0.75}{8}\) x \(\frac{100}{1}\) = \(\frac{1.35}{8}\) x \(\frac{100}{1}\) = \(\frac{33.75}{2}\) = 16.875 = 16\(\frac{7}{8}\)

2log \(\frac{2}{5}\) - log\(\frac{72}{125}\) + log 9 [\(\frac{2}{5}\))2 x 9] = log \(\frac{4}{25}\) x \(\frac{9}{1}\) x \(\frac{125}{72}\) = log \(\frac{72}{125}\) = log \(\frac{5}{2}\) = log \(\frac{10}{4}\) = log 10 - log 4 = log10 10 - log10 22 = 1 - 2 log2

\(\frac{1}{1 + \sqrt{5}}\) - \(\frac{1}{1 - \sqrt{5}}\) = \(\frac{1 - \sqrt{5} - 1 - \sqrt{5}}{(1 + \sqrt{5}) (1 - \sqrt{5}}\) = \(\frac{-2\sqrt{5}}{1 - 5}\) = \(\frac{-2\sqrt{5}}{- 4}\) = \(\frac{1}{2}\sqrt{5}\)