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\(\frac{4.314 \times 0.000056}{0.0067}\) \(\frac{0.000242}{0.0067}\) = 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%

Let F be the set of people who can speak French and E be the set of people who can speak English. Then, n(F) = 400 n(E) = 350 n(F ∪ E) = 500 We have to find n(F ∩ E). Now, n(F ∪ E) = n(F) + n(E) – n(F ∩ E) ⇒ 500 = 400 + 350 – n(F ∩ E) ⇒ n(F ∩ E) = 750 – 500 = 250. ∴ 250 people can speak both languages.

A = { 1, 2, 3, 4, 5, 6} B = { 2, 4, 6, 8 } A - B = { 1, 3, 5 } B - A = { 8 } ∴ (A - B) ⋃ (B - A) = {1, 3, 5, 8}

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.

\(\begin{bmatrix}2& -1&3\\4&1&2\\1&-3&1\\\end{bmatrix}\) = 2[(1 x 1) - (2 x -3)] - (-1)[(4 x 1) - (2 x 1)] + 3[(4 x -3) - (1 x 1)] = 2(1 - (-6)) + 1(4 - 2) + 3(-12 - 1) = 2(1 + 6) + 1(2) + 3(-13) = 2(7) + 1(2) + 3(-13) = 14 + 2 - 39 ∴ |D| = -23

Let A = \(\begin {bmatrix} a & b\\ c & d \end {bmatrix}\) i.e \(\begin{bmatrix} a & b\\c & d \end{bmatrix}\) \(\begin{bmatrix} 0 & 1\\2 & -1 \end{bmatrix}\) = \(\begin{bmatrix} 2 & -1\\1 & 0 \end{bmatrix}\) \(\implies \begin {bmatrix} a(0) + b(2) & a(1) + b(-1)\\ c(0) + d(2) & c(1) + d(-1)\end {bmatrix}\) = \(\begin {bmatrix} 2 & -1\\ 1 & 0 \end {bmatrix}\) \(\implies \begin {bmatrix} 2b & a - b\\2d & c - d\end {bmatrix}\) = \(\begin {bmatrix} 2 & -1\\ 1 & 0 \end {bmatrix}\) By comparing 2b = 2 a - b = -1 2d = 1 and c - d = 0 ∴ b = \(^2/_2\) = 1 a - b = -1 ⇒ a - 1 = -1 ∴ a = 0 ∴ d = \(^1/_2\) ⇒ c = d ∴ c = \(^1/_2\) ∴The matrice A = \(\begin {bmatrix} 0 & 1\\^1/_2 & ^1/_2 \end {bmatrix}\)

Let T be the time taken and D is the distance For the case given in the problem, the distance covered is directly proportional to the time taken i.e D \alpha T ⇒ D = ST where S is the constant speed D = 225 km and T = 4.5 hrs ∴ 225 = S x 4.5 S = \(\frac {225}{4.5}\) = 50km/hr When D = 150km D = ST 150 = 50 \(\times\) T T = \(\frac {150}{50}\) = 3 hrs So it will take 3 hrs

Let the third number = \(x\) Then the first number = 100% \(x + 35%x = 135\)%\(x = \frac {135x}{100} = 1.35x\) (Note: 100% \(x = x\)) The second number = 180% \(x = \frac {180x}{100} = 1.80x\) ∴ The ratio of the first number to the second number = \(1.35x : 1.80x = 3 : 4\)

12 men working 8 hours a day can complete the work in 4 days 12 men working 1 hour a day will complete the same work in (8 x 4) days [working less hours a day means they have to work for more days] 1 man working 1 hour a day will complete same work in (8 x 4 x 12) days [fewer number of men working means the remaining ones will have to work for more days ] So, it will take 1 man working 1 hour a day to complete a piece of work in (8 x 4 x 12) days = 384 days Now, 1 man working 16 hours a day will complete the piece of work in (384/16) days = 24 days [working more hours a day means less days to complete the work] ∴ 4 men working 16 hours a day will complete the piece of work in (24/4) days = 6 days [more men working means fewer days to complete the work] ALTERNATIVELY 12 x 8 x 4 = 4 x 16 x d ⇒ d = \(\frac{12 \times 8 \times 4}{4 \times 16}\) ⇒ d = \(\frac{384}{64}\) ⇒ d = 6 ∴ 4 men working 16 hours a day will complete the piece of work in 6 days