Correct Answer: B. 40√3m /3

Explanation

\(\tan 30 = \frac{x}{40}\)

\(x = 40 \tan 30 = \frac{40}{\sqrt{3}}\)

= \(\frac{40\sqrt{3}}{3} m\)

Explanation

(a)

Given : Circle ABC, centre O.

To prove: < AOB = < ACB

Construction : Join CO produced to P.

Proof: With the lettering in the figure, OA = OB (radii)

\(x_{1} = x_{2}\) (base angles of isosceles triangle)

\(\therefore < AOP = x_{1} + x_{2}\) (exterior angle of triangle AOC)

\(\therefore < AOP = 2x_{2} (x_{1} = x_{2})\) (base angles of an isosceles triangle)

Similarly, \(< BOP = 2y_{2}\)

\(\therefore < AOB = 2x_{2} + 2y_{2} \)

= \(2(x_{2} + y_{2})\)

\(\implies < AOB = 2 \times < ACB\) (proved)

(b) From the figure, \(< ACB = \frac{130°}{2} = 65°\) (angle subtended at the centre of the circle)

\(< ACO = 26°\) (Base angles of isosceles triangle ACO)

\(\therefore < BCO = 65° - 26° = 39°\)

\(< OBC = 39°\) (base angles of isosceles triangle OBC)

\(\therefore < COB = 180° - (39° + 39°) = 102°\)

Correct Answer: D. 54°

Explanation

Ondo = \(\frac{126}{840} \times 360°\)

= \(54°\)

Explanation

n(R) = number of red balls = 12;

n(W) = number of white balls = 16;

n(B) = number of blue balls = 8

Total number of balls = 36

(a) P(three are red) = \(\frac{12}{36} \times \frac{11}{35} \times \frac{10}{34}\)

= \(\frac{11}{357}\)

(b) P(first is blue and other two red) = \(\frac{8}{36} \times \frac{12}{35} \times \frac{11}{34}\)

= \(\frac{44}{1785}\)

(c) P(two are white and one is blue) = \(\frac{16}{36} \times \frac{15}{35} \times \frac{8}{34}\)

= \(\frac{16}{357}\)

Correct Answer: A. -2

Explanation

when n = 2

(-1)\(^{n-2}\) 2\(^{n+1}\) = 2

When n = 3

(-1)\(^{n-2}\) 2\(^{n+1}\) = -4

Sum = 2 - 4 = -2

Explanation

(a) \(\log_{10} (\frac{30}{16}) - 2 \log_{10} (\frac{5}{9}) + \log_{10} (\frac{400}{243})\)

= \(\log_{10} (\frac{30}{16}) - \log_{10} (\frac{5}{9})^{2} + \log_{10} (\frac{400}{243})\)

= \(\log_{10} (\frac{30}{16} \times \frac{400}{243} \div \frac{25}{81})\)

= \(\log_{10} (\frac{30}{16} \times \frac{400}{243} \times \frac{81}{25})\)

= \(\log_{10} (10) = 1\)

(b) \(\sqrt{\frac{P}{Q}} = \sqrt{\frac{3.6 \times 10^{-3}}{2.25 \times 10^{6}}\)

= \(\sqrt{\frac{36 \times 10^{-4}}{2.25 \times 10^{6}}\)

= \(\frac{6 \times 10^{-2}}{1.5 \times 10^{3}}\)

= \(4 \times 10^{-2 - 3} = 4 \times 10^{-5}\)

Correct Answer: D. 1/2

Explanation

\(\frac{3}{4} \div 1\frac{1}{4} \times (1\frac{1}{2} - \frac{2}{3})\)

\(\frac{3}{4} \div \frac{5}{4} \times (\frac{9 - 4}{6})\)

= \(\frac{3}{4} \times \frac{4}{5} \times \frac{5}{6}\)

= \(\frac{1}{2}\)

Explanation

In \(\Delta\) BPY, \(\tan 35° = \frac{h}{x}\)

\(h = x \tan 35 = 0.7 x ... (1)\)

In \(\Delta\) BPX, \(\frac{h}{x + 4000} = \tan 15\)

\(\frac{h}{x + 4000} = 0.3\)

\(h = 0.3 (x + 4000) = 0.3x + 1200 ... (2)\)

Equating (1) and (2) as the values of h, we have

\(0.7x = 0.3x + 1200 \implies 0.7x - 0.3x = 1200\)

\(0.4x = 1200 \times x = \frac{1200}{0.4} = 3000m\)

\(\therefore h = 0.7x = 0.7 (3000) = 2100m\)

\(\text{The total height = } 5200m + 2100m = 7300m\)

Correct Answer: A. 40°

Explanation

(a) 5, 9, 13, ... is an A.P with the first term = 5 and common difference = 9 - 5 = 13 - 9 = 4.

\(T_{n} = a + (n - 1)d\) (terms of an A.P)

\(T_{25} = 5 + (25 - 1) \times 4\)

= \(5 + 24 \times 4\)

= \( 5 + 96 = 101\)

(b) \(\frac{8}{9}, \frac{-4}{3}, 2, ...\) is a G.P with first term = \(\frac{8}{9}\)

\(r = \frac{T_{2}}{T_{1}} = -\frac{4}{3} \div \frac{8}{9} = \frac{-4}{3} \times \frac{9}{8} = -\frac{3}{2}\)

\(T_{n} = ar^{n - 1}\) (terms of a G.P)

\(T_{5} = (\frac{8}{9})(-\frac{3}{2})^{5 - 1}\)

= \(\frac{8}{9} \times \frac{81}{16} = \frac{9}{2}\)

(c) \(T_{n} = ar^{n - 1}\) (terms of a G.P)

\(T_{3} = ar^{2} = 48 ... (1)\)

\(T_{6} = ar^{5} = 14\frac{2}{9} ... (2)\)

\((2) \div (1) : r^{3} = \frac{128}{9} \div 48 = \frac{128}{9} \times \frac{1}{48}\)

\(r^{3} = \frac{8}{27}\)

\(r = \sqrt[3]{\frac{8}{27}} = \frac{2}{3}\)

From (1), \(ar^{2} = 48 \implies a \times (\frac{2}{3})^{2} = 48\)

\(4a = 48 \times 9 \implies a = \frac{48 \times 9}{4} = 108\)

\(T_{2} = 108 \times \frac{2}{3} = 72\)

\(T_{4} = 48 \times \frac{2}{3} = 32\)

\(\therefore\) The first four terms of the sequence are \(108, 72, 48, 32\).