C.

$75\sqrt{3}$

Let AB be the tower of height 75 m and C be position of the car.

In 

$$\begin{gathered} ∆\mathrm{ABC}, \\ \cot 30°= \frac{\mathrm{AC}}{\mathrm{AB}} \\ \Rightarrow \mathrm{AC}=\mathrm{AB} \cot 30° \\ \Rightarrow \mathrm{AC}=75\mathrm{m} \mathrm{x} \sqrt{3} \\ \Rightarrow \mathrm{AC}=75\sqrt{3}\mathrm{m} \\ \mathrm{Thus}, \mathrm{the} \mathrm{distance} \mathrm{of} \mathrm{the} \mathrm{car} \mathrm{from} \mathrm{the} \mathrm{base} \mathrm{of} \mathrm{the} \mathrm{tower} \mathrm{is} 75\sqrt{3 }\mathrm{m}. \end{gathered}$$ 

A.

$\frac{1}{2}$

S = { 1, 2, 3, 4, 5, 6 }

let event E be defined as 'getting an even number'.

n(E) = { 2, 4, 6 }

 $$\begin{gathered} \therefore \mathrm{P}\hspace{0.17em}\left(\mathrm{E}\right) = \frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Number} \mathrm{of} \mathrm{possible} \mathrm{outcomes}} =\frac{3}{6} \\ = \frac{1}{2} \end{gathered}$$

C.

$\frac{4}{45}$

S = { 1, 2, 3,........90 }

n(s) = 90

The prime number less than 23 are 2, 3, 5, 7, 11, 13, 17 and 19

Let event E be defined as 'getting a prime number less than 23'.

n(E) = 8

$$\begin{gathered} \therefore \mathrm{p}\left(\mathrm{E}\right) =\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Number} \mathrm{of} \mathrm{possible} \mathrm{outcomes}} = \frac{8}{90} \\ =\frac{4}{45} \end{gathered}$$

B.

44

Let r be the radius of the circle,

From the given information, we have

$$\begin{gathered} 2\mathrm{\pi r} - \mathrm{r} =37 \mathrm{cm} \\ \Rightarrow \mathrm{r} ( 2\mathrm{\pi } - 1 ) = 37 \mathrm{cm} \\ \Rightarrow \mathrm{r} \left( 2 \mathrm{x} \frac{22}{7} - 1\right) = 37 \mathrm{cm} \\ \Rightarrow \mathrm{r} \mathrm{x} \frac{37}{7} = 37 \mathrm{cm} \\ \Rightarrow \mathrm{r} = 7 \mathrm{cm} \\ \therefore \mathrm{Circumference} \mathrm{of} \mathrm{the} \mathrm{circle} = 2\mathrm{\pi r} \\ = 2 \mathrm{x} \frac{22}{7} \mathrm{x} 7 \mathrm{cm} = 44 \mathrm{cm} \end{gathered}$$

C.

-2

 $$\begin{gathered} Comman difference = \frac{1 - 6q}{3q} - \frac{1}{3q} = \frac{1-6q-1}{3q} \\ = \frac{-6q}{3q} \\ = -2 \end{gathered}$$