Question Library

Let the number of goats be X C.P of X goats = N4000X S.P of X goats = (75/100) x 4000X = N3000X 3000X = 180,000 => X = 180000/3000 = 60. X = 60 goats

The time (t) to do the work is inversely proportional to the number of workers (n). \(\implies t \propto \frac{1}{n}\) \(t = \frac{k}{n}\) \(5 = \frac{k}{45} \implies k = 45 \times 5 = 225\) \(\therefore t = \frac{225}{n}\) For 25 men, \(t = \frac{225}{25} = 9\) \(\therefore\) 25 men will do the work in 9 days.

\(\frac{x + 2}{4} - \frac{2x - 3}{3}\) < 4 \(\frac{3(x + 2) - 4(2x - 3)}{12}\) < 4 3x + 6 - 8x + 12 < 48 18 - 5x < 48 \(\implies\) - 5x < 48 - 18 -5x < 30 \(\implies \)x > - 6

\(\frac{x}{r + a} = \frac{a}{r}\) \(\implies rx = ar + a^{2}\) \(rx - ra = a^{2} \implies r = \frac{a^2}{x - a}\)

Let y = f(x). \(y = 3x + 4 \implies x = \frac{y - 4}{3}\) Replace y with x. \(y = \frac{x - 4}{3} \)

\(p \ast q = pq + p + q\) \(2 \ast (3 \ast 4) \) \(3 \ast 4 = 12 + 3 + 4 = 19\) \(2 \ast 19 = 38 + 2 + 19 = 59\)

The series is geometric with common ratio \(\frac{1}{3}\). \(S_{\infty} = \frac{a}{1 - r}\) = \(\frac{1}{1 - \frac{1}{3}} \) = \(\frac{1}{\frac{2}{3}}\) = \(\frac{3}{2}\)

\(x \propto \sqrt{n}\) \(x = k \sqrt{n}\) \(9 = k \sqrt{9} \implies 9 = 3k\) \(k = 3\) \(x = 3 \sqrt{n}\) When n = 17/9, \(x = 3 \times \sqrt{\frac{17}{9}} = \sqrt{17}\)

Volume of a frustrum with top of radius R and bottom r and height h = \(\frac{1}{3} \pi (R^{2} + Rr + r^{2})\) V = \(\frac{1}{3} \pi \times 4 \times (6^2 + 6 \times 4 + 4^2)\) = \(\frac{304}{3} \pi cm^{3}\)

2y = 4x - 8 \(\implies\) y = 2x - 4. When x = 0, y = -4. When y = 0, x = 2. The midpoint between (0, -4) and (2, 0) = \((\frac{0 + 2}{2}, \frac{-4 + 0}{2})\) = \((1, -2)\)

2y - ax + 4 = 0 \(\implies\) 2y = ax - 4. y + (x/4) - 7 = 0 \(\implies\) y = -x/4 + 7 Slope of line 1 = a/2 Slope of line 2 = -1/4 -1/4 x a/2 = -1 (product of the slope of two perpendicular lines = -1) a/8 = 1 \(\implies\) a = 8.

Range = Highest value - Lowest value = (k + 6) - (k - 5) = 11

To get the answer, we simply do \(\frac{5!}{(5 - 3)!}\) = \(\frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}\) = 60

30 - x + x + 40 - x = 80 - 20 70 - x = 60 - x = 60 - 70 - x = - 10 ∴ x = 10 Music only = 30 - x = 30 - 10 20 P(music only) = 20/80 = 1/4 = 0.25

m = \(\frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 0}{0 - (4)} = \frac{2}{4} = \frac{1}{2}\) \(\frac{y_2 - y_1}{x_2 - x_1} \geq m\) \(\frac{y - 0}{x + 4} \geq \frac{1}{2}\) 2y \(\geq\) x + 4, -x + 2y \(\geq\) 4 = px + qy \(\geq\) 4 p = -1, q = 2