Correct Answer: B. -6

Correct Answer: A. 3.0 seconds

Explanation

\(s = ut + \frac{1}{2}gt^{2}\)

\(u = 0; s = 45m; g = 10ms^{-2}\)

\(45 = 0 + \frac{1}{2}(10)t^{2} = 5t^{2}\)

\(t^{2} = \frac{45}{5} = 9\)

\(t = 3.0s\)

Correct Answer: C. - 3

Explanation

Let the given points be:

P(-3, -14) = (x\(_1\), y\(_1\))

Q(t, -5) = (x\(_2\), y\(_2\))

PQ = 9 units (given)

Using the distance formula,

d = √ [ \((x_2 - x_1)^2 + (y_2 - y_1)^2\)]

PQ = √ [ \((t - (-3))^2 + (-5 + 14)^2\)]

\(\implies\)√[ \((t + 3)^2 + 81\)] = 9

Squaring on both sides,

⇒ (t + 3)\(^2\) + 81 = 81

⇒ (t + 3)\(^2\) = 0

⇒ t + 3 = 0

∴ t = -3

Correct Answer: B. \(\frac{2}{3}\)

Explanation

Probability of obtaining a 5 is P(5)=\(\frac{1}{6}\)

Probability of obtaining a 2 is P(2)=\(\frac{1}{6}\)

Probability of obatining either 2 or 5 = P(2?5) = \(\frac{2}{6}\)

Probability of obtaining neither 5 or 2 = \(1 - \frac{2}{6}=\frac{4}{6}=\frac{2}{3}\)

Explanation

(a) \(P \propto Q\) and \(P \propto \frac{1}{R^{2}}\).

\(P \propto \frac{Q}{R^{2}} \implies P = \frac{kQ}{R^{2}}\)

When P = 1, Q = 8, R = 2

\(1 = \frac{8k}{2^{2}} \implies 8k = 4\)

\(k = \frac{4}{8} = 0.5\)

\(P = \frac{Q}{2R^{2}}\)

When P = 3, R = 5, Q = ?

\(3 = \frac{Q}{2(5^{2})}\)

\(3 = \frac{Q}{50} \implies Q = 150\).

(b)

Angular difference : 60° - 20° = 40°

\(\frac{\theta}{360} \times 2\pi r \)

\(r = R\cos \theta\)

\(\frac{40}{360} \times 2 \times 3.142 \times 6400 \cos 20 = \frac{1511687.278}{360}\)

= \(4199.13 km\)

\(Speed = \frac{Distance}{Time}\)

= \(\frac{4199.13}{6}\)

= \(699.85 km/hr\)

\(\approxeq 700 km/hr\) (to 3 significant figures)

(ii) \(D = \frac{\theta}{360} \times 2\pi r\)

\(420 = \frac{\theta}{360} \times 2 \times 3.142 \times 6400\)

\(\theta = \frac{420 \times 360}{2 \times 3.142 \times 6400}\)

\(\theta = \frac{151200}{40217.6} = 3.76°\)

Latitude of C : 20° + 3.76° = 23.76°

Note that due North implies same longitude but different latitude.

Correct Answer: C. 0< x< 1/2

Correct Answer: B. 513

Correct Answer: D. \(\frac{15}{4}\)

Explanation

\(\left(\frac{16}{81}\right)^{-\frac{3}{4}}\times \sqrt{\frac{100}{81}}\\

\frac{1}{\left(\sqrt[4]{\frac{16}{81}}\right)^3}\times \frac{10}{9}=\frac{1}{\left(\frac{2}{3}\right)^3}\times\frac{10}{9}\\

=\frac{27}{8}\times \frac{10}{9}=\frac{15}{4}\)

Correct Answer: A. 36°

Explanation

The sum of angles in a triangle = 180º

(180 - 108)º = 72º

Isosceles triangle has both two equal sides and two equal angles.

∠OSN = ∠SON = \(\frac{72}{2}\)

= 36º

Correct Answer: A. 3√2+4