Given:  P = Rs. 600,  n = 30 months  and   r = 10%

$$\begin{gathered} \therefore \mathrm{I} = \mathrm{Rs}. \left( 600 \times \frac{30 ( 30 + 1 ) }{2 \times 12} \times \frac{10}{100} \right) \\ = \mathrm{Rs}. 2325 \end{gathered}$$

Since sum deposited = p x n = Rs.  600 x 30

                               = Rs. 18000

Thus, the maturity value = Rs. ( 18000 + 2325 ) = Rs. 20325.

Total number of cards = 10

( i ) Prime number card is 2.

$\Rightarrow$ Number of favourable outcomes = 1

$\therefore \mathrm{Required} \mathrm{probability} = \frac{1}{10}$

( ii ) Cards having number divisible by  4  are  4,  8,  12,  16,  20.

$\Rightarrow \; Number\; of\; favourable\; outcomes\; =\; 5$

$\therefore \mathrm{Required} \mathrm{probability} = \frac{5}{10} = \frac{1}{2}$

( iii ) Cards having number that is multiple of  6  are  6,  12,  18.

$\Rightarrow \; Number\; of\; favourable\; outcomes\; =\; 3$

$\therefore \mathrm{Required} \mathrm{probability} = \frac{3}{10}$

( iv ) Odd number card is not there.

$\Rightarrow \; Number\; of\; favourable\; outcomes\; =\; 0$

$\therefore \mathrm{Required} \mathrm{probability} = 0$

$$\begin{gathered} 2 \left[ \begin{array}{cc}\mathrm{x}& 7\\ 9& \mathrm{y} - 5 \end{array}\right] + \left[ \begin{array}{cc}6& - 7\\ 4& 5\end{array} \right] = \left[ \begin{array}{cc}10& 7\\ 22& 15\end{array} \right] \\ \Rightarrow \left[ \begin{array}{cc}2\mathrm{x}& 14\\ 18& 2 \mathrm{y} - 10 \end{array}\right] + \left[ \begin{array}{cc}6& - 7\\ 4& 5\end{array} \right] = \left[ \begin{array}{cc}10& 7\\ 22& 15\end{array} \right] \\ \Rightarrow \left[ \begin{array}{cc}2 \mathrm{x} + 6& 14 - 7\\ 18 + 4& 2 \mathrm{y} - 10 + 5 \end{array}\right] = \left[ \begin{array}{cc}10& 7\\ 22& 15\end{array} \right] \\ \Rightarrow \left[ \begin{array}{cc}2 \mathrm{x} + 6& 7\\ 22& 2 \mathrm{y} - 5 \end{array}\right] = \left[ \begin{array}{cc}10& 7\\ 22& 15\end{array} \right] \end{gathered}$$

$\Rightarrow$ 2 x + 6 = 10    and   2 y - 5 = 15

$\Rightarrow \; 2\; x\; =\; 4 \; \; and \; \; 2\; y\; =\; 20$

$\Rightarrow \; x\; =\; 2 \; \; and \; \; y\; =\; 10$

Let the radius of the cylindrical vessel be  r  and its height be  h.

$\Rightarrow$ Height = h = 25 cm.

( i ) Circumference  of the base = 132 cm.

$$\begin{gathered} \Rightarrow 2 \mathrm{\pi } \mathrm{r} = 132 \\ \Rightarrow 2 \times \frac{22}{7} \times \mathrm{r} = 132 \\ \Rightarrow \mathrm{r} = 21 \mathrm{cm} \end{gathered}$$ 

$$\begin{gathered} ( \mathrm{ii} ) \mathrm{Volume} \mathrm{of} \mathrm{cylinder} = \mathrm{\pi } {\mathrm{r}}^2 \mathrm{h} \\ = \frac{22}{7} \times 21 \times 21 \times 25 \\ = 22 \times 3 \times 21 \times 25 \\ = 34650 {\mathrm{cm}}^3 \end{gathered}$$ 

( k - 3 ),  ( 2 k + 1 )  and  ( 4 k + 3 )  are three consecutive terms of an A.P.

$\Rightarrow$ 2 ( 2 k + 1 ) = ( k - 3 ) + ( 4 k + 3 )

$\Rightarrow \; 4\; k\; +\; 2\; =\; k\; -\; 3\; +\; 4\; k\; +\; 3$

$\Rightarrow \; 4\; k\; +\; 2\; =\; 5\; k$

$\Rightarrow \; k\; =\; 2$