C.

$\frac{1}{5}$

$$\begin{gathered} \mathrm{The} \mathrm{multiple} \mathrm{of} 4 \mathrm{between} 1 \mathrm{and} 15 \mathrm{are} 4, 8, 12. \mathrm{the} \mathrm{probability} \mathrm{of} \mathrm{getting} \mathrm{a} \\ \mathrm{multiple} \mathrm{of} 4 = \frac{3}{15} = \frac{1}{5} \end{gathered}$$

D.

$90°$

$$\begin{gathered} \mathrm{TA}=\mathrm{TP}\Rightarrow \angle \mathrm{TPA} = \angle \mathrm{TPA} \\ \mathrm{TB}= \mathrm{TP}\Rightarrow \angle \mathrm{TBP}=\angle \mathrm{TPB} \\ \therefore \angle \mathrm{TAP}+\angle \mathrm{TBP}=\angle \mathrm{TPA}+\angle \mathrm{TPB} = \angle \mathrm{APB} \\ \angle \mathrm{TAP}+\angle \mathrm{TBP}+\angle \mathrm{APB} = 180° [\because \mathrm{Sum} \mathrm{of} ...180°] \\ \therefore \angle \mathrm{APB}+\angle \mathrm{APB}=180° \\ \therefore 2\angle \mathrm{APB}=180° \\ \therefore \angle \mathrm{APB}=90° \end{gathered}$$

B.

3

$$\begin{gathered} \mathrm{K}, 2\mathrm{K}-1, 2\mathrm{K}+1 \mathrm{are} \mathrm{in} \mathrm{Arithmetic} \mathrm{Progression} \\ 2\mathrm{x}(2\mathrm{k}-1) = \mathrm{k}+2\mathrm{k}+1 \\ 4\mathrm{k}-2=3\mathrm{k}+1 \\ \mathrm{k}=3 \end{gathered}$$

A.

$\frac{7}{8}$

There are in all 2³ = 8 combinations or outcomes for the gender of the 3 children

The 8 combinations are as follows

BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG

Thus the probability of having at least one boy in a family is $\frac{7}{8}$

B.

10$\sqrt{2}$

$$\begin{gathered} \mathrm{Given}: \angle \mathrm{AOB} \mathrm{is} \mathrm{given} \mathrm{as} 90° \\ ∆\mathrm{AOB} \mathrm{is} \mathrm{an} \mathrm{isosceles} \mathrm{trianglesince} \mathrm{OA}=\mathrm{OB} \\ \mathrm{Therefore} \angle \mathrm{OAB}= \angle \mathrm{OBA}= 45° \\ \mathrm{Thus} \angle \mathrm{AOP}=45° \mathrm{and} \angle \mathrm{BOP}= 45° \\ \mathrm{Hence} ∆\mathrm{AOP} \mathrm{and} ∆\mathrm{BOP} \mathrm{also} \mathrm{are} \mathrm{isosceles} \mathrm{triangles} \\ \mathrm{Thus} \mathrm{let} \mathrm{AP}=\mathrm{PB}=\mathrm{OP}=\mathrm{x} \\ \mathrm{Using} \mathrm{pythagoras} \mathrm{theorem} \\ {\mathrm{x}}^2 + {\mathrm{x}}^2 = {10}^2 \\ \mathrm{Thus} 2{\mathrm{x}}^2 = 100 \\ \mathrm{x}=5\sqrt{2} \\ \mathrm{Hence} \mathrm{length} \mathrm{of} \mathrm{chord} \mathrm{AB} = 2\mathrm{x} = 10\sqrt{2} \end{gathered}$$