Correct Answer: D. 1:2

Explanation

\(\frac{x}{y}\) = \(\frac{1}{4} \div \frac{3}{8}\) = \(\frac{1}{4} \times \frac{8}{3}\) = \(\frac{2}{3}\)

\(\frac{y}{z}\) = \(\frac{1}{3} \div \frac{4}{9}\) = \(\frac{1}{3} \times \frac{9}{4}\) = \(\frac{3}{4}\)

But,

x = \(\frac{2}{5}T_1\), y = \(\frac{3}{5}T_1\)

y = \(\frac{3}{7}T_2\), z = \(\frac{4}{7}T_2\)

Using y = y

\(\frac{3}{5}T_1\) = x = \(\frac{3}{7}T_2\)

\(\frac{T_1}{T_2}\) = \(\frac{3}{7}\) x \(\frac{5}{3}\) = \(\frac{15}{21}\)

\(T_1 = 15\) and \(T_2 = 21\)

Thus , x = \(\frac{2}{5}\) x 15 = 6

y = \(\frac{3}{5}\) x 15 = 9

y = \(\frac{3}{7}\) x 21 = 9

z = \(\frac{4}{7}\) x 21 = 12

Hence; x : z = 6 : 12

= 1 : 2

Correct Answer: D. Rhombus

Correct Answer: A. \(\frac{153}{250}\)

Explanation

0.612 = \(\frac{0.612}{1}\) x \(\frac{1000}{1000}\)

= \(\frac{612}{1000}\)

= \(\frac{153}{250}\)

Correct Answer: C. 3

Explanation

\(\begin{vmatrix} k & k \\ 4 & k \end{vmatrix} + \begin{vmatrix} 2 & 3 \\ -1 & k \end{vmatrix} = 6\)

\(\begin{vmatrix} k & k \\ 4 & k \end{vmatrix} = (k^{2} - 4k)\)

\(\begin{vmatrix} 2 & 3 \\ -1 & k \end{vmatrix} = (2k + 3)\)

\(\therefore (k^{2} - 4k) + (2k + 3) = k^{2} - 2k + 3 = 6\)

\(k^{2} - 2k - 3 = 0\), factorising, we have \(k + 1 = 0\) or \(k - 3 = 0\)

Since k > 0, k = 3.

Correct Answer: B. 9

Explanation

Mean = \(\frac{\sum f x}{\sum f}\)

= \(\frac{350}{10}\)

= 35

= \(\frac{\sum f |d|}{\sum f}\)

= \(\frac{90}{10}\)

= 9

Correct Answer: D. 1\(\frac{4}{5}\)

Explanation

2\(\frac{1}{4} \times 3\frac{1}{2} \div 4 \frac{3}{8}\)

= \(\frac{9}{4} \times \frac{7}{2} \div \frac{35}{8}\)

= \(\frac{9}{4} \times \frac{7}{2} \div \frac{8}{35}\)

= \(\frac{9}{5}\)

= 1 \(\frac{4}{5}\)

Correct Answer: D. 9\(\sqrt{3m}\)

Explanation

tan 30\(^o\) = \(\frac{h}{27}\)

h = 27 tan 30\(^o\)

= 27 x \(\frac{1}{\sqrt{3}}\)

= \(\frac{27}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\)

= \(\frac{27 \sqrt{3}}{3}\)

= 9\(\sqrt{3m}\)

Correct Answer: B. -1

Explanation

\(\begin{pmatrix} 2 & x \\ 3 & 5 \end{pmatrix} = 13\)

\(\begin{vmatrix} 2 & x \\ 3 & 5 \end{vmatrix} = (2 \times 5) - (3 \times x) = 13\)

\(10 - 3x = 13 \implies -3x = 3; x = -1\)

Correct Answer: B. 0.0084N

Explanation

\(F = mass \times acceleration\) but \(accl = \frac{v - u}{t}\)

\(\therefore F = m(\frac{v - u}{t})\)

\(Mass = 28g = 0.028kg\)

\(v = 5.4 ms^{-1}; u = 0; t = 18secs\)

\(\therefore F = 0.028(\frac{5.4 - 0}{18}) = 0.028 \times 0.3 = 0.0084N\)

Explanation

Spearmann's rank correlation coefficient = \(1 - \frac{6\sum D^{2}}{N(N^{2} - 1)}\)

= \(1 - \frac{6(36)}{8(64 - 1)}\)

= \(1 - \frac{216}{504}\)

= \(1 - 0.4286\)

= \(0.5714\)

\(\approxeq 0.57\).