Question Library

x\(^2\) + 4 = 0 x\(^2\) = -4 You can't apply a square root to a negative number. So the answer is None of the above.

surface area of formula = \pir\(^2\) If the radius is reduced then let its radius be x. Its area is \pix\(^2\) . x : r = 2 : 5, so \(\frac{x}{r}\) = \(\frac{2}{5}\) → 5x = 2r Hence, x = 0.4r. Hence the area of the new circle is \pi (0.4r)\(^2\) = 0.16\pi r\(^2\) . The ratio of the two areas is 0.16\pir\(^2\) : \pir\(^2\) = 0.16 : 1 = 16 : 100 = 4 : 25.

Area of the rectangular picture = L x B = 8 x 6 = 48 sq.cm. Area of the whole surface (which is gotten by adding \(\frac{1}{2}\) on every side to the original picture size) is 9 x 7 = 63 sq. cm area of the frame is 63 - 48 = 15 sq. cm

\(3\frac{7}{8} + 1\frac{1}{3} = 4\frac{21 + 8}{24}\) = \(4\frac{29}{24}\) \(\equiv 5\frac{5}{24}\) \(1\frac{2}{3} - \frac{1}{8} = \frac{5}{3} - \frac{1}{8}\) = \(\frac{40 - 3}{24}\) = \(\frac{37}{24}\) \(5\frac{5}{24} - \frac{37}{24} = \frac{125}{24} - \frac{37}{24}\) = \(\frac{11}{3}\) = 3\(\frac{2}{3}\)

\((x + 3y + 5)(2x^{2} + 5y + 2)\) = \(2x^{3} + 5xy + 2x + 6yx^{2} + 15y^{2} + 6y + 10x^{2} + 25y + 10\) = \(2x^{3} + 5xy + 2x + 6yx^{2} + 15y^{2} + 31y + 10x^{2} + 10\)

\(\frac{3}{5}\) = 0.60, \(\frac{9}{16}\) = 0.56, \(\frac{34}{59}\) = 0.58, \(\frac{71}{97}\) = 0.73 Hence, \(\frac{9}{16} < \frac{34}{59} < \frac{3}{5} < \frac{71}{97}\)

Sum of n terms of Geometric progression is \(S_{n} = \frac{a(1 - r^n)}{1 - r}\) In the given series, a (the first term) = 1 and r (the common ratio) = x. \(S_{n} = \frac{1(1 - x^{n})}{1 - x}\) a = 1, and as n tends to infinity \(S_{n} = \frac{1}{1 - x}\)

\(N \propto P_{A}\); \(N \propto P_{B}\); \(N \propto \frac{1}{D^{2}}\) \(\therefore N \propto \frac{P_{A} P_{B}}{D^{2}}\) \(N = \frac{k P_{A} P_{B}}{D^{2}}\)

\(\sqrt{170 - 20 \sqrt{30}} = \sqrt{a} - \sqrt{b}\) Squaring both sides, \(170 - 20\sqrt{30} = a + b - 2\sqrt{ab}\) Equating the rational and irrational parts, we have \(a + b = 170 ... (1)\) \(2 \sqrt{ab} = 20 \sqrt{30}\) \(2 \sqrt{ab} = 2 \sqrt{30 \times 100} = 2 \sqrt{3000} \) \(ab = 3000 ... (2)\) From (2), \(b = \frac{3000}{a}\) \(a + \frac{3000}{a} = 170 \implies a^{2} + 3000 = 170a\) \(a^{2} - 170a + 3000 = 0\) \(a^{2} - 20a - 150a + 3000 = 0\) \(a(a - 20) - 150(a - 20) = 0\) \(\text{a = 20 or a = 150}\) \(\therefore b = \frac{3000}{20} = 150 ; b = \frac{3000}{150} = 20\) \(\sqrt{170 - 20\sqrt{30}} = \sqrt{20} - \sqrt{150}\) or \(\sqrt{150} - \sqrt{20}\) = \(2\sqrt{5} - 5\sqrt{6}\) or \(5\sqrt{6} - 2\sqrt{5}\)

\(\begin{array}{c|c} Nos. & log \\ \hline 222.9 & 2.3482\end{array}\) N : b Look for the antilog of .3482 which gives 222.9