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20 + X + 50 + 40 + 2X + 60 = 260 3X + 170 = 260 3X = 260 - 170 3x = 90 x = 30 Absent for at least 4 days 4 5 6 40 2x 60 i.e 40 + 2x + 60 = 40 + (2 x 30) + 60 = 40 + 60 + 60 = 160

Given, Length of minor arc = 5.5m Angle subtended by minor arc = 360^\circ - 210^\circ = 150^\circ \(\therefore 5.5 = \frac{150}{360} \times 2 \times \frac{22}{7} \times r \) \(\frac{55r}{21} = 5.5\) \(r = \frac{5.5 \times 21}{55}\) r = 2.1m Length of major arc = \(\frac{210}{360} \times 2 \times \frac{22}{7} \times 2.1\) = \(7.7m \approxeq 8m\) (to the nearest metre)

\(BD^2 = 4^2 + 4^2\) \(BD = \sqrt{16 + 16} = \sqrt{32}\) \(BD = 4\sqrt{2} cm\) \((2\sqrt{3})^2 = (2\sqrt{2})^2 + h^2\) \(h^2 = 12 - 8 = 4\) \(h = \sqrt{4} = 2 cm\) Volume of pyramid = \(\frac{a^2 h}{3}\) = \(\frac{4^2 \times 2}{3}\) = \(\frac{32}{3} = 10\frac{2}{3} cm^3\)

Curved surface area or a cone \(=\pi rl\) from the information \(l^2 = 4^2 + 3^2 = 16+9\\ l = \sqrt{25} = 5; ∴ C.S.A\hspace{1mm} = \frac{22}{7}\times 3 \times 5\\ Since \frac{22}{7}=\pi ∴ C.S.A\hspace{1mm} =\hspace{1mm}15\pi\)

< UQR = 29^\circ (alternate angles) < PQR = < PQU + < UQR = 36^\circ + 29^\circ = 65^\circ

\(5\frac{1}{4}\div \left(1\frac{2}{3}- \frac{1}{2}\right)\\ \frac{21}{4}\div \left(1\frac{4-3}{6}\right)\\ \frac{21}{4}\div \left(1\frac{1}{6}\right)\\ \frac{21}{4} \times \frac{6}{7}= 4\frac{1}{2}\)

\(0.5x + 2.6 = 5x + 0.35\\ 0.5x - 5x = 0.35-2.6\\ -4.5x = -2.25\\ x = \frac{-2.25}{-4.5}\\ 0.5\)

Sum of exterior angle of any polygon is 360o (2x+5)o + 2xo + (x-20)o + xo + (3x+10)o + (x + 15)o = 360o; 10x = 350 x = 35

\(M5_{ten} = 1001011_{two}\\ =1 \times 2^6 + 0\times 2^5 + 0\times 2^4 + 1\times 2^3 + 0\times 2^2 + 1\times 2^1 \\ =64+8+2+1=75_{ten}\\ ∴ m = 7 \)

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

Draw a straight horizontal line across the y-axis at y = 3\(\frac{1}{3}\). let the line cut through the curve. the points where the horizontal line meets the curve, trace it down to the x-axis to obtain the values of x when y = 3\(\frac{1}{3}\) take only the positive value of x.

The longest side of a right angle triangle is the hypotenuse. \((x + 2)^{2} = x^{2} + (x + 1)^{2}\) \(x^2 + 4x + 4 = x^2 + x^2 + 2x + 1\) \(2x^{2} - x^{2} + 2x - 4x + 1 - 4 = 0\) \(x^{2} - 2x - 3 = 0\) \(x^{2} - 3x + x - 3 = 0 \implies x(x - 3) + 1(x - 3) = 0\) \((x - 3)(x + 1) = 0 \implies \text{x = 3 or -1}\) \(x > 0 \implies x = 3\) The longest side = 3 + 2 = 5.

\(log2 = 0.3010\hspace{1mm}given\\ log2^y = 1.8062\\ ∴ ylog2 = 1.8062\\ y=\frac{1.8062}{logy}=\frac{1.8062}{0.3010}=6\)

ZY\(^2\) = 12\(^2\) + 6\(^2\) ZY\(^2\) = 144 + 36 = 180 ZY = \(\sqrt{180}\) = \(13.416\) = 13.4 km

20 members in the ratio 3:1 Number of women = \(\frac{1}{4} \times 20\) = 5 Let the number of women to be added = x Total number of members in the committee = 20 + x \(\frac{5 + x}{20 + x} = \frac{2}{5}\) \(5(5 + x) = 2(20 + x)\) \(25 + 5x = 40 + 2x \implies 5x - 2x = 40 - 25\) \(3x = 15 \implies x = 5\)