Answer: Option D. -> 75%
Let the original circumference be x
Then, new circumference = 50% of x = $$\frac{x}{2}$$
Let original radius = r and new radius = R
$$\eqalign{
& 2\pi r = x \cr
& \Rightarrow r = \frac{{x \times 7}}{{2 \times 22}} \cr
& \Rightarrow r = \frac{{7x}}{{44}} \cr} $$
$$\eqalign{
& 2\pi R = \frac{x}{2} \cr
& \Rightarrow R = \frac{x}{2} \times \frac{7}{{2 \times 22}} \cr
& \Rightarrow R = \frac{{7x}}{{88}} \cr} $$
Original area :
$$\eqalign{
& = \pi {r^2} \cr
& = \left( {\frac{{22}}{7} \times \frac{{7x}}{{44}} \times \frac{{7x}}{{44}}} \right) \cr
& = \frac{{7{x^2}}}{{88}} \cr} $$
New area :
$$\eqalign{
& = \pi {R^2} \cr
& = \left( {\frac{{22}}{7} \times \frac{{7x}}{{88}} \times \frac{{7x}}{{88}}} \right) \cr
& = \frac{{7{x^2}}}{{352}} \cr} $$
Decrease in area :
$$\eqalign{
& = \left( {\frac{{7{x^2}}}{{88}} - \frac{{7{x^2}}}{{352}}} \right) \cr
& = \frac{{21{x^2}}}{{352}} \cr} $$
∴ Decrease % :
$$\eqalign{
& = \left( {\frac{{21{x^2}}}{{352}} \times \frac{{88}}{{7{x^2}}} \times 100} \right)\% \cr
& = 75\% \cr} $$

Answer: Option D. -> 82%
Let length = $$l$$ metres and breadth = b metres
Then, original area = (lb) m2
New length :
$$\eqalign{
& = \left( {140\% {\text{ of }}l} \right)m \cr
& = \left( {\frac{{140}}{{100}} \times l} \right)m \cr
& = \frac{{7l}}{5}m \cr} $$
New breadth :
$$\eqalign{
& = \left( {130\% {\text{ of }}b} \right)m \cr
& = \left( {\frac{{130}}{{100}} \times b} \right)m \cr
& = \frac{{13l}}{{10}}m \cr} $$
New area :
$$\eqalign{
& = \left( {\frac{{7l}}{5} \times \frac{{13b}}{{10}}} \right){m^2} \cr
& = \left( {\frac{{91lb}}{{50}}} \right){m^2} \cr} $$
Increase :
$$\eqalign{
& = \left( {\frac{{91lb}}{{50}} - lb} \right) \cr
& = \frac{{41}}{{50}}lb \cr} $$
∴ Increase % :$$\eqalign{
& = \left( {\frac{{41}}{{50}} \times lb \times \frac{1}{{lb}} \times 100} \right)\% \cr
& = 82\% \cr} $$

Answer: Option D. -> 32 metres
Let the sides of the rectangle be x metres and (x + 48) metres
Then,
$$\eqalign{
& 2\left( {x + x + 48} \right) = 160 \cr
& \Rightarrow 4x + 96 = 160 \cr
& \Rightarrow 4x = 64 \cr
& \Rightarrow x = 16 \cr} $$
So, sides of the rectangle are 16 metres and 64 metres
Area of the rectangle :
= (16 × 64) m2
= 1024 m2
Area of the square = 1024 m2
∴ Side of the square :
= $$\sqrt {1024} $$  m
= 32 metres

Answer: Option D. -> decrease by 30%
Let the length, breadth and height of the room be 3x, 2x and x respectively.
Area of 4 walls :
$$\eqalign{
& = 2\left( {l + b} \right) \times h \cr
& = 2\left( {3x + 2x} \right) \times x \cr
& = 10{x^2} \cr} $$
New length = 6x
New breadth = x
New height = $$\frac{x}{2}$$
New area of four walls :
$$\eqalign{
& = \left[ {2\left( {6x + x} \right)\frac{x}{2}} \right] \cr
& = 7{x^2} \cr} $$
Decrease in area :
$$\eqalign{
& = \left( {10{x^2} - 7{x^2}} \right) \cr
& = 3{x^2} \cr} $$
∴ Decrease% :
$$\eqalign{
& = \left( {\frac{{3{x^2}}}{{10{x^2}}} \times 100} \right)\% \cr
& = 30\% \cr} $$

Answer: Option C. -> $$ay : bx$$
$$\eqalign{
& \frac{a}{b} = \frac{{\frac{1}{2}x \times {h_1}}}{{\frac{1}{2}y \times {h_2}}}bx{h_1} = ay{h_2} \cr
& \Leftrightarrow \frac{{{h_1}}}{{{h_2}}} = \frac{{ay}}{{bx}} \cr} $$
\[\left[ \begin{gathered}
{\text{Ratio of areas}} = \frac{a}{b}{\text{ }} \hfill \\
{\text{Ratio of base}} = x:y \hfill \\
\end{gathered} \right]\]
$${\text{Hence, }}{h_1}:{h_2} = ay:bx$$

Answer: Option C. -> 21 m
$$\eqalign{
& \pi R = 33 \cr
& \Rightarrow R = \left( {\frac{{33 \times 7}}{{22}}} \right) \cr
& \Rightarrow R = \left( {\frac{{21}}{2}} \right)m \cr} $$
∴ Required distance = 2R = 21 m

Answer: Option B. -> 1% decrease
Let the original length and breadth of the rectangle be $$l$$ and b respectively.
Then,
Original area = $$lb$$
New length :
$$ = 110\% {\text{ of }}l = \frac{{11}}{{10}}l$$
New breadth :
$$ = 90\% {\text{ of }}b = \frac{{9}}{{10}}b$$
New area :
$$\eqalign{
& = \left( {\frac{{11}}{{10}}l \times \frac{{9}}{{10}}}b \right) \cr
& = \frac{{99}}{{100}}lb \cr} $$
Decrease in area :
$$\eqalign{
& = \left( {lb - \frac{{99}}{{100}}}lb \right) \cr
& = \frac{{lb}}{{100}} \cr} $$
∴ Decrease % :
$$\eqalign{
& = \left( {\frac{{lb}}{{100}} \times \frac{1}{{1b}} \times 100} \right)\% \cr
& = 1\% \cr} $$

Answer: Option B. -> (32 - 8π) sq. units
Radius of each circle = 2 units
Area of the shaded region :
$$=$$ Area of the rectangle $$-$$ Area of two circles
$$\eqalign{
& = \left[ {\left( {8 \times 4} \right) - 2 \times \pi {{\left( 2 \right)}^2}} \right]\text{sq. units} \cr
& = \left( {32 - 8\pi } \right)\text{sq. units} \cr} $$

Answer: Option D. -> $$1\frac{{\pi {r^2}}}{2}$$
We have :
Required area :
$$\eqalign{
& = \frac{{\pi {r^2}{\theta _1}}}{{360}} + \frac{{\pi {r^2}{\theta _2}}}{{360}} + \frac{{\pi {r^2}{\theta _3}}}{{360}} \cr
& = \frac{{\pi {r^2}}}{{360}}\left( {{\theta _1} + {\theta _2} + {\theta _3}} \right) \cr
& = \frac{{\pi {r^2} \times 180}}{{360}} \left[ {\because {\theta _1} + {\theta _2} + {\theta _3} = 180^ \circ} \right] \cr
& = \frac{{\pi {r^2}}}{2}Or = \frac{1}{2}\pi {r^2} \cr} $$

Answer: Option B. -> 160 m
Length of the field :
$$\eqalign{
& = \left( {3.25 \times 100} \right)m \cr
& = 325\,m \cr} $$
∴ Breadth of the field :$$\eqalign{
& = \left( {\frac{{52000}}{{325}}} \right)m \cr
& = 160\,m \cr} $$