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log (y + 8) + log (y - 8) = 2log 3 + 2log 5 ⇒ log (y + 8)(y - 8) = 2log 3 + 2log 5 (log ab = log a + log b) ⇒ log (y\(^2\) -8y + 8y - 64) = log 3\(^2\) + log 5\(^2\) ⇒ log (y\(^2\) - 64) = log 3\(^2\) + log 5\(^2\) ⇒ log (y\(^2\) - 64) = log 9 + log 25 ⇒ log (y\(^2\) - 64) = log (9 \times 25) ⇒ log(y\(^2\) - 64) = log 225 ⇒ y\(^2\) - 64 = 225 ⇒ y\(^2\) = 225 + 64 ⇒ y\(^2\) = 289 ⇒ y = \pm\sqrt289 ∴ y = \pm17

\(log_2 (6 - x) = 3 - log_2 x\) ⇒ \(log_2 (6 - x) = 3 log_2 2 - log_2 x\) (since \(log_2\) 2 = 1) ⇒ \(log_2 (6 - x) = log_2 2^3 - log_2 x\) \((a log\) c = \(log\) c\(^a)\) ⇒ \(log_2 (6 - x) = log_2 8 - log_2 x\) ⇒\(log_2 (6 - x) = log_2 \frac{8}{x}\) (\(log\) a - \(log\) b = \(log \frac{a}{b})\) ⇒ \(6 - x = \frac{8}{x}\) ⇒ \(x (6 - x) = 8\) ⇒ \(6x - x^2 = 8\) ⇒ \(x^2 - 6x + 8 = 0\) ⇒ \(x^2 - 4x - 2x + 8 = 0\) ⇒ \(x (x - 4) - 2(x - 4) = 0\) ⇒ \((x - 4)(x - 2) = 0\) ⇒ \(x - 4 = 0 or x - 2 = 0\) ∴ x = 4 or 2

\(\frac {3 - \sqrt 3}{2 + \sqrt 3} = a + b\sqrt 3\) Rationalize = \(\frac {3 - \sqrt 3}{2 + \sqrt 3} \times \frac {2 - \sqrt 3}{2 - \sqrt 3}\) = \(\frac {(3 - \sqrt 3)}{(2 + \sqrt 3)} \frac {(2 - \sqrt 3)}{(2 - \sqrt 3)}\) = \(\frac {6 - 3 \sqrt 3 - 2 \sqrt 3 + (\sqrt 3)^2}{4 - 2 \sqrt 3 + 2 \sqrt 3 - (\sqrt 3)^2}\) = \(\frac {6 - 5 \sqrt 3 + 3}{4 - 3}\) = \(\frac {9 - 5 \sqrt 3}{1} = 9 - 5 \sqrt 3\) = 9 + (-5) \(\sqrt 3\) \(\therefore a = 9, b = - 5\)

Original population = 1,230 New population = 1,040 Decrease in population = 1,230 – 1,040 = 190 Percentage decrease in population = decrease in population x 100% original population = \(\frac {190}{1,230}\) x 100 = 15.45%

Let the total number of items in the man's shop = \(y\) Number of Brand A's items in the man's shop = \(\frac {1}{9} y\) Remaining items = 1 - \(\frac {1}{9} y = \frac {8}{9} y\) Number of Brand B's items in The man's shop = \(\frac{5}{8} of \frac{8}{9}y = \frac{5}{9}y\) Total of Brand A and Brand B's items = \(\frac{1}{9}y + \frac{5}{9}y = \frac{2}{3}y\) Number of Brand C's items in the man's shop = 1 - \(\frac{2}{3}y = \frac{1}{3}y\) \(\implies\frac{1}{3}y\) = 81 (Given) \(\implies y\) = 81 x 3 = 243 ∴ The total number of items in the man's shop = 243 ∴ Number of Brand B's items in the man's shop = \(\frac{5}{9}\) x 243 = 135 ∴ The number of more Brand B items than Brand C = 135 - 81 =54

16.54 x 10\(^{-5}\) - 6.76 x 10\(^{-8}\) + 0.23 x 10\(^{-6}\) ⇒ 1.654 x 10\(^{-4}\) - 6.76 x 10\(^{-8}\) + 2.3 x 10\(^{-7}\) ⇒ 1.654 x 10\(^{-4}\) - 0.000676 x 10\(^{-4}\) + 0.0023 x 10\(^{-4}\) ⇒ (1.654 - 0.000676 + 0.0023) x 10\(^{-4}\) ∴ 1.655624 x 10\(^{-4}\) ≃ 1.66 x 10\(^{-4}\)

% error = \(\frac{Error}{Actual Value}\) x 100% Student's Value = 18mL Actual Value = 18.4mL Error = 18.4 - 18 = 0.4 ∴ % error = \(\frac{0.4}{18.4}\) x 100% = 2.17%

\(\frac{5}{8} - \frac{3}{4} \div \frac{5}{12} x \frac{1}{4}\) ⇒ \(\frac{5}{8} - (\frac{3}{4} \div \frac{5}{12}) \times \frac{1}{4}\) ⇒ \(\frac{5}{8} - (\frac{3}{4} \times \frac{12}{5}) \times \frac{1}{4}\) ⇒ \(\frac{5}{8} - (\frac{9}{5} \times \frac{1}{4})\) ⇒ \(\frac{5}{8} - \frac{9}{20}\) ∴ \(\frac{7}{40}\)

The two trailing zeros in the number are not significant, but the other five are, making it a five-figure number.

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}

\(\begin{bmatrix}2& -1&3\\4&1&2\\1&-3&1\\\end{bmatrix}\) = 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 = \(1.35x : 1.80x = 3 : 4\)