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The ratio of men to women = 3:1 Number of committee members = 20 number of men in the committee = \(\frac{3}{4} \times 20 = 15\) Number of women in the committee = \(\frac{1}{4} \times 20 = 5\) New ratio = 3:2, and let the number of women to be added to make the ratio 3:2 be x 3:2 = 15:5+x \(\frac{3}{2} = \frac{15}{5 + x }\) 15 + 3x = 30 3x = 30 - 15 = 15 x = \(\frac{15}{3}\) = 5 Therefore, the number of women to be added to make the ratio of men to women 3:2 = 5.

Amount = principal + Interest I = \(\frac{ P\times R \times T}{100}\) P = #200,000, R = 5%, T = 3 yrs I = = \(\frac{ 200,000\times 5 \times 3}{100}\) = 2000 x 15 = #30,000 A = 200,000 + 30,000 = #230,000.

The boy bought 5 oranges for #24.00. Therefore, the cost price per orange is: c.p = \(\frac{24}{5}\) = #4.800 The boy sold 4 oranges for #30.00. Therefore, the selling price per orange is: s.p = \(\frac{30}{4}\) = # 7.50 Profit per orange is calculated as: Profit per orange = SP per orange − CP per orange = =7.50 − 4.80 = 2.70 If the boy sold 4 oranges, the total profit from selling 4 oranges is: Total Profit = 4 \times 2.70 = #10.80

y ∝ \(\frac{1}{x}\) y = \(\frac{k}{x}\) y = 4, x = \(\frac{1}{2}\), find x when y = 10 4 = \(\frac{k}{\frac{1}{2}}\) k = 2 Therefore, x = \(\frac{2}{10} = \frac{1}{5}\)

U = K\(\sqrt{V}\) 24 = K\(\sqrt{9}\) K = 8 U = 8\(\sqrt{V}\) 16 = 8\(\sqrt{V}\) 2 = \(\sqrt{V}\) V = 4

B = \(\frac{K}{C^{\frac{1}{3}}}\) 2 = \(\frac{K}{27^{\frac{1}{3}}}\) K = 2 x 3 = 6

x = \(\frac{K}{y}\) 9 = \(\frac{K}{4}\) K = 9 x 4 = 36 Therefore, the law connecting x and y is: x = \(\frac{36}{y}\)

The shaded part is after 1, and on the x-axis, showing that x is greater than 1 i.e x > 1

3x + 4y < 12, x \ge 0, y \ge 0 represent the graph above.

x\(^3\) - 19x - 30 test x = 5 5\(^3\) −19(5) − 30 = 125 − 95 − 30 = 0 (x - 5) is a factor. x\(^3\) - 19x - 30 divided by (x - 5) = x\(^2\) + 5x + 6 factorizing x\(^2\) + 5x + 6, we have (x+2)(x+3) (x+2)(x+3)(x - 5) = 0 x = -2, -3, and 5.

S\(_\infty\) = \(\frac{a}{1- r}\) since 1 > r S\(_\infty\) = 100, r = \(\frac{-1}{2}\), a = ? 100 = \(\frac{a}{1- \frac{-1}{2}}\) = \(\frac{a}{\frac{3}{2}}\) ( - - = +) a = 100 x \(\frac{3}{2}\) = 150 Therefore, the first term (a) = 150

T\(_7\) = ar\(^{n-1}\) = ar\(^{6}\) a = - 10, r = - 5 T\(_7\) = -10(-5)\(^{6}\) T\(_7\) = - 10 x 15625 = - 156250

p * q = 2p + pq + q (p*2) - (p*1) = 40 firstly, p = p and q = 2, secondly p = p and q = 1 then, 2p + 2p + 2 - (2p + p + 1) = 40 4p + 2 - 3p - 1 = 40 p + 1 = 40 p = 40 - 1 = 39

Given the operation defined by \( x * y = x^y \), \[x * 2 = x \implies x^2 = x\] Rearranging gives: \[x^2 - x = 0 \implies x(x - 1) = 0\] Thus, the possible values of \( x \) are: \[\boxed{0 \text{ and } 1}\]

From the diagram above, Calculate the distance between any two vertices as follows. PQ = \(\sqrt{(6 - 3)^2 + (2-2)^2}\) = \(\sqrt{9}\) = 3 QR = \(\sqrt{(6 - 4)^2 + (2-5)^2}\) = \(\sqrt{13}\) PR = \(\sqrt{(4 - 3)^2 + (5 - 1)^2}\) = \(\sqrt{10}\) \(\therefore\), perimeter = 3 + \(\sqrt{13}\) + \(\sqrt{10}\)