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\(\frac{0.00275 \times 0.0064}{0.025 \times 0.08}\) Removing the decimals = \(\frac{275 \times 64}{2500 \times 800}\) = \(\frac{88}{10^4}\) \(88 x 10^{-4} = 88 x 10^{1} x 10^{-4} = 8.8 x 10^{-3}\)

use "T" to represent the total profit. The first receives \(\frac{1}{3}\) T remaining, 1 - \(\frac{1}{3}\) = \(\frac{2}{3}\)T The seconds receives the remaining, which is \(\frac{2}{3}\) also \(\frac{2}{3}\) x \(\frac{2}{3}\) = \(\frac{4}{9}\) The third receives the left over, which is \(\frac{2}{3}\)T - \(\frac{4}{9}\)T = (\(\frac{6 - 4}{9}\))T = \(\frac{2}{9}\)T The third receives \(\frac{2}{9}\)T which is equivalent to N12000 If \(\frac{2}{9}\)T = N12, 000 T = \(\frac{12 000}{\frac{2}{9}}\) Total share[T] = N54, 000 The first receives \(\frac{1}{3}\) of T → \(\frac{1}{3}\) * N54, 000 = N18,000 The second receives \(\frac{4}{9}\) of T → \(\frac{4}{9}\) * N54, 000 = N24,000 The third receives \(\frac{2}{9}\)T which is equivalent to N12000. Adding the three shares give total profit of N54,000

You can use any whole numbers (eg. 1. 2. 3) to represent all the mangoes in the basket. If the first child takes \(\frac{1}{4}\) it will remain 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\) Next, the second child takes \(\frac{3}{4}\) of the remainder which is \(\frac{3}{4}\) i.e. find \(\frac{3}{4}\) of \(\frac{3}{4}\) = \(\frac{3}{4}\) x \(\frac{3}{4}\) = \(\frac{9}{16}\) the fraction remaining now = \(\frac{3}{4}\) - \(\frac{9}{16}\) = \(\frac{12 - 9}{16}\) = \(\frac{3}{16}\)

Interest I = \(\frac{PRT}{100}\) ∴ R = \(\frac{100 \times 1}{100 \times 5}\) = \(\frac{100 \times 7.50}{500 \times 5}\) = \(\frac{750}{500}\) = \(\frac{3}{2}\) = 1.5%

The diagonal of a rhombus is a line segment that joins any two non-adjacent vertices. A rhombus has two diagonals that bisect each other at right angles. i.e this splits 6cm into 3cm each AND 8cm to 4cm Using Hyp\(^2\) = adj\(^2\) + opp\(^2\) Hyp\(^2\) = 3\(^2\) + 4\(^2\) Hyp\(^2\) = 25 Hyp = 5 ∴ Length (L) is 5cm because a rhombus is a quadrilateral with 4 equal lengths

Length of Arc AB = \(\frac{\theta}{360}\) 2\(\pi\)r = \(\frac{48}{360}\) x 2\(\frac{22}{7}\) x \(\frac{21}{2}\) = \(\frac{4 \times 22 \times \times 3}{30}\) \(\frac{88}{10}\) = 8.8cm Perimeter = 8.8 + 2r = 8.8 + 2(10.5) = 8.8 + 21 = 29.8cm

Volume of a cylinder = \pir\(^2\)h First convert 3m to cm by multiplying by 100 Volume of External cylinder = \pi \times 13\(^2\) \times 300 Volume of Internal cylinder = \pi \times 10\(^2\) \times 300 Hence; Volume of External cylinder - Volume of Internal cylinder Total volume (v) = \pi (169 - 100) \times 300 V = \pi \times 69 \times 300 V = 20700\picm\(^3\)

For a regular polygon of n sides n = \(\frac{360}{\text{Exterior angle}}\) Exterior < = 180^\circ - 140^\circ = 40^\circ n = \(\frac{360}{40}\) = 9 sides

The required locus is angle bisector of the two lines

Mean of 10 numbers = 16 The total sum of numbers = 16 x 10 = 160 Mean of 11 numbers = 18 Total sum of numbers = 11 x 18 = 198 The 11th no. = 198 - 160 = 38

Given that 4, 16, 30, 20, 10, 14 and 26 Adding up = 120 nos \(\geq\) 16 are 16 + 30 + 20 + 26 = 92 The requires sum of angles = \(\frac{92}{120}\) x \(\frac{360^o}{1}\) = 276o

Total pairs: C(4,2) = 6. Favorable (even sum): both odd (1 way: {1,3}) or both even (1 way: {2,4}) → 2 ways. P = \(\frac{2}{6}\) = \(\frac{1}{3}\).

Given from 41 to 56 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56 The nos multiple of 9 are: 45, 54 P(multiple of 9) = \(\frac{2}{16}\) = \(\frac{1}{8}\)

125\(^{\frac{-1}{3}}\) x 49\(^{\frac{-1}{2}}\) x 10\(^0\) = 5\(^{-1}\) x 7\(^{-1}\) x 1 = \(\frac{1}{35}\)

3\(^{2x}\) = 27 3\(^{2x}\) = 3\(^3\) 2x = 3 x = 1.5