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Converting each number to base 10 \(110_{two} = 1 \times 2^2 + 1 \times 2^1 + 0 \times 2^0\) = 1 \times 4 + 1 \times 2 + 0 \times 1 = 4 + 2 + 0 = \(6_{ten}\) \(31_{eight} = 3 \times 8^1 + 1 \times 8^0\) = 3 \times 8 + 1 \times 1 = 24 + 1 = \(25_{ten}\) \(42_{five} = 4 \times 5^1 + 2 \times 5^0\) = 4 \times 5 + 2 \times 1 = 20 + 2 = \(22_{ten}\) Hence, \(31_{eight} > 42_{five} > 110_{two}\) In ascending order, \(110_{two}, 42_{five}, 31_{eight}\)

\(2^{5x} \div 2^x = \sqrt[5]{2^{10}}\) applying the laws of indices \(2^{5x - x} = 2^{10(1/5)}\) \(2^{4x} = 2^{10(1/5)}\) \(2^{4x} = 2^2\) Equating the powers then 4x = 2 therefore, x = \(\frac{2}{4}\) = \(\frac{1}{2}\)

\(log_3 x + log_3 (x - 8) = 2\) \(log_3 x(x - 8) = log_39\) since 2 = \(log_39\) \(log_3\) cancels out ⇒ x(x - 8) = 9 ⇒ \(x^2 - 8x = 9\) ⇒ \(x^2 - 8x - 9 = 0\) ⇒ \(x^2 - 9x + x - 9 = 0\) ⇒ x(x - 9) + 1(x - 9) = 0 ⇒ (x - 9)(x + 1) = 0 ⇒ x = 9 or x = -1 Since we can't have a log of negative numbers, ∴ x = 9.

\(3\sqrt{12} + 10\sqrt{3} - \frac{6}{\sqrt{3}}\) = \(3\sqrt{4\times3} + 10\sqrt{3} - \frac{6}{\sqrt{3}}\) = \(6\sqrt{3} + 10\sqrt{3} - \frac{6}{\sqrt{3}}\) treating like a fraction, then = \(\frac{18 + 30 - 6}{\sqrt{3}}\) = \(\frac{42}{\sqrt{3}}\) rationalizing = \(14\sqrt{3}\)

\(\frac{4.314 \times 0.000056}{0.0067}\) \(\frac{0.000242}{0.0067}\) = 0.036 ( to 3 decimal places)

%error=5%, measured height = 5.98m let the actual height = y error=x - 5.98 (since 'y' is more than 5.98) %error = \(\frac{error}{actual height}\times 100%\) 5% = \(\frac{y - 5.98}{y}\times 100%\) \(\frac{5}{100} = \frac{y - 5.98}{y}\) 5y = 100(y - 5.98) 5y = 100y - 598 5y - 100y = - 598 -95y = - 598 y = \(\frac{-598}{-95}\) y = 6.29m( to 2 d.p).

Number of students that scored at most 5 marks = 12 + 1 + 9 + 6 + 3 = 31 % of student = \(\frac{31}{53} \times 100%\) = 58.5%

Let x = original price A decrease of 22% = (100-22)% = 78% of the original price. 78% of x = 27.30 = \(\frac{78x}{100} = 27.30\) 78x = 2730 then, x = \(\frac{2730}{78}\) x = 35 Therefore, the original price of the shoe was $35.00.

Error = 16.8 - 15 = 1.8cm % error = \(\frac{error}{ Actual value}\times 100%\) = \(\frac{1.8}{ 15}\times100%\) = \(0.12\times100%\) = 12.00%

Using the venn diagram above \mu = 30 n(W) = 15 n(M) = 13 \(n(W ∪ M)^1 = 6\) Let x = number of students that study both woodwork and metalwork i.e. n(W ∩ M) = x Number of students that study only woodwork,\(n(W ∩ M^1)\) = \(15 - x\) Number of students that study only metalwork, \(n(W^1 ∩ M)\) = \(13 - x \) Bringing all together, \(n(W ∩ M^1)\) +\( n(W^1 ∩ M)\) + \(n(W ∩ M)\) + \(n(W ∪ M)^1\) = \(\mu\) ∴ (15 - x) + (13 - x) + x + 6 = 30 ⇒ 34 - x = 30 ⇒ 34 - 30 = x ∴ x = 4 \(n(W ∩ M^1)\) = \(15 - 4 = 11\) ∴ The number of students that study woodwork but not metalwork is 11.

ai. To find all the elements in the set M, we have to solve the inequality. 2n - 3 \le 37 2n = 37 + 3 = 40. 2n \le 40 n \le \(\frac{40}{2}\) n \le 20 so, the set M contains all positive numbers n such that n is less than or equal to 20 Therefore, set M elements are {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20} (\alpha) Pr(multiple of 3) = \(\frac{multiples of 3}{ Total number of element in M}\) But the multiples of 3 in set M are = {3, 6, 9, 12, 15, 18} = 6 ஃ Pr(multiples of 3) = \(\frac{6}{20} = \frac{3}{10}\) (\beta) Pr( factors of 10) = \(\frac{№ of factors of 10}{ Total № of set M elements}\) factors of 10 in set M = { 1, 2, 5, and 10} = 4 ஃ Pr( factors of 10) = \(\frac{4}{20} = \frac{1}{5}\) bi. Let the total amount shared = y Gapson's age = \(\frac{3}{2}\times Kontor's age\). Since kontor is 20yrs old then Gapson's age = \(\frac{3}{2}\times20\) = 30years. sharing ratio = 20 : 30 = 2 : 3 total ratio = 2 + 3 = 5 Since Kontor received Le200,000 \(\frac{2}{5} \times y\) = 200,000 \(\frac{2y}{5} = 200,000\) cross multiplying, we have 2y = 5 x 200,000 = 1000,000 y = \(\frac{1000,000}{2}\) = 500,000. ஃ Total amount shared = Le500,000. ii. Gapson's share = Total - kontor's share = 500,000 - 200,000 = Le300,000.

The symbolic form for the statement "If Siah is from Foya, then Foya is in Lofa" is: p ⇒ q This is read as "p implies q". It means that if p is true, then q must also be true. In other words, if Siah is from Foya, then Foya must be in Lofa.

m:n = \(2\frac{1}{3} : 1\frac{1}{5}\) = m : n = \(\frac{7}{3} : \frac{6}{5}\) \(\frac{7}{3} : \frac{6}{5}\) = \(\frac{7}{3} \div \frac{6}{5}\) \(\frac{m}{n}\) = \(\frac{7}{3} \times \frac{5}{6}\) \(\frac{m}{n}\) = \(\frac{35}{18}\) = m = \(\frac{35n}{18}\) n : q = \(1\frac{1}{2} : 1\frac{1}{3}\) = \(\frac{3}{2} : \frac{4}{3}\) \(\frac{n}{q}\) = \(\frac{3}{2} \times\frac{3}{4}\) \(\frac{n}{q}\) = \(\frac{9}{8}\) = q = \(\frac{8n}{9}\) q : m = \(\frac{8n}{9}\) : \(\frac{35n}{18}\) \(\frac{q}{m}\) = \(\frac{8n}{9} \div \frac{35n}{18}\) \(\frac{q}{m}\) = \(\frac{8n}{9}\times\frac{18}{35n}\) =\(\frac{q}{m} = \frac{16}{35}\) = q : m = 16 : 35

a. The total time taken for the entire journey is the sum of time taken for each journey. But Average speed = \(\frac { total distance covered}{ time taken}\) then Time taken = \(\frac { total distance covered}{ time taken}\) for the first journey(t1), d = 112km and speed = 70km/hr \(t_1 = \frac{112}{70} = \frac{8}{5}\)hrs for the second journey(t2), distance = 60km and speed = 50km/hr \(t_2 = \frac{60}{50} = \frac{6}{5}\)hrs Therefore, Total time taken = \( t_1 + t_2 = \frac{8}{5} + \frac{6}{5}\) = 2.8hrs. b. if \(\frac{x}{y}\) = 2 , then x = 2y and if \(\frac{y}{z}\) = 3 then z = \(\frac{y}{3}\), substituting, the values of x and z into \(\frac{ x + y}{y + z }\) = \(\frac{2y + y}{y + \frac{y}{3}} = 3y\div\frac{4y}{3} = 3y\times\frac{3}{4y} = \frac{9}{4}\) Therefore, \(\frac{ x + y}{y + z } = \frac{9}{4}\)

\(M ∝ n^2\sqrt{q}\) \(M = Kn^2\sqrt{q}\) K = \(\frac{M}{n^2\sqrt{q}}\) K = \(\frac{24}{2^2\sqrt4}\) k = \(\frac{24}{8} = 3\) Now, let's find M when n = 5 and q = 9 M = \(Kn^2\sqrt{q}\) M = \( 3\times5^2\sqrt9\) \(M = 3\times25\times3\) Therefore, M = 225.