Answer: Option B. -> 25%
Let the original length and breadth of the rectangle be l and b respectively
Then, original area = lb
New length = 120% of l = $$\frac{6l}{5}$$
New area = 150% of lb = $$\frac{3lb}{2}$$
New breadth :$$\eqalign{
& = \left( {\frac{{3lb}}{2} \times \frac{5}{{6l}}} \right) \cr
& = \frac{{5b}}{4} \cr} $$
Increase in breadth :$$\eqalign{
& = \left( {\frac{{5b}}{4} - b} \right) \cr
& = \frac{b}{4} \cr} $$
∴ Increase % :
$$\eqalign{
& = \left( {\frac{b}{4} \times \frac{1}{b} \times 100} \right)\% \cr
& = 25\% \cr} $$

Answer: Option C. -> 800%
Let the original radius be R
New radius = (100 + 200)% of R = 300% of R = 3R
Original area = $$\pi $$R2
New area = $$\pi $$ × (3R)2 = 9$$\pi $$R2
Increase in area :
$$\eqalign{
& = \left( {9\pi {{\text{R}}^2} - \pi {{\text{R}}^2}} \right) \cr
& = 8\pi {{\text{R}}^2} \cr} $$
∴ Increase %:
$$\eqalign{
& = \left( {\frac{{8\pi {{\text{R}}^2}}}{{\pi {{\text{R}}^2}}} \times 100} \right)\% \cr
& = 800\% \cr} $$

Answer: Option D. -> 8π sq. units
Area of square = 4 sq.units
Side of square = $$\sqrt 4 $$ = 2 units
Diagonal of square = $$2\sqrt 2 $$ units
Radius of the circle = $$2\sqrt 2 $$ units
Radius of the circle :
$$\eqalign{
& = \pi {r^2} \cr
& = \pi \times {\left( {2\sqrt 2 } \right)^2} \cr
& = 8\pi \,\text{sq. units} \cr} $$

Answer: Option D. -> $$20\sqrt 2 $$
Let the radius of circle be r cm and side of square be a cm
Then circumference of circle = $$2\pi r$$ and
Perimeter of square = 4a
According to the question,
$$\eqalign{
& 2\pi r = 4a \times \frac{{110}}{{100}} \cr
& \Rightarrow 2\pi r = \frac{{44a}}{{10}} \cr
& \Rightarrow r = \frac{{44a}}{{2\pi \times 10}} \cr
& \Rightarrow r = \frac{{11a}}{{5\pi }} \cr
& \Rightarrow a = \frac{{5\pi r}}{{11 }} . . . . .(i) \cr
& {\text{Also, }}\pi {r^2} - {a^2} \Rightarrow 216 \cr} $$
$$ \Rightarrow \pi {r^2} - \frac{{25\pi {r^2}}}{{121}} = 216$$     $$\left[ {{\text{from equation (i)}}} \right]$$
$$\eqalign{
& \Rightarrow \frac{{121\pi {r^2} - 25\pi {r^2}}}{{121}} = 216 \cr
& \Rightarrow {r^2}\left[ {121\pi - 25{\pi ^2}} \right] = 26136 \cr} $$
$$ \Rightarrow {r^2}\left[ {121 \times \frac{{22}}{7} - 25 \times \frac{{22}}{7} \times \frac{{22}}{7}} \right]$$      $$ = 26136$$
$$\eqalign{
& \Rightarrow {r^2}\left[ {\frac{{2662}}{7} - \frac{{12100}}{{49}}} \right] = 26136 \cr
& \Rightarrow {r^2}\left[ {\frac{{6534}}{{49}}} \right] = 26136 \cr
& \Rightarrow {r^2} = \frac{{26136 \times 49}}{{6534}} \cr
& \Rightarrow {r^2} = 196 \cr
& \Rightarrow r = 14\,cm \cr
& \therefore a = \frac{{5\pi r}}{{11}} = 5 \times \frac{{22}}{7} \times \frac{{14}}{{11}} = 20\,cm \cr} $$
Hence, diagonal of square $$ = \sqrt 2 a = 20\sqrt 2 \,cm$$

Answer: Option C. -> 132 cm
Given length of the piece of wire = 84 cm
Length of the piece of wire = Circumference of circle
$$\eqalign{
& = 2\pi r \cr
& = 2 \times \frac{{22}}{7} \times 84 \cr
& = 528\,cm \cr} $$
Length of each side of square = a
∴ Perimeter of square = 4a = 528 cm
∴ Side of square = $$\frac{{528}}{4}$$ = 132 cm

Answer: Option B. -> 9 ft
Let the width of the table be x feet.
Then, length of the table = (x + 4) ft
$$\eqalign{
& \therefore x\left( {x + 4} \right) = 45 \cr
& \Rightarrow {x^2} + 4x - 45 = 0 \cr
& \Rightarrow {x^2} + 9x - 5x - 45 = 0 \cr
& \Rightarrow x\left( {x + 9} \right) - 5\left( {x + 9} \right) = 0 \cr
& \Rightarrow \left( {x + 9} \right)\left( {x - 5} \right) = 0 \cr
& \Rightarrow x = 5 \cr} $$
Hence, length of the table = (5 + 4) = 9 feet

Answer: Option B. -> 154 cm2
Radius of the required circle :$$\eqalign{
& = \left( {\frac{1}{2} \times 30} \right)cm \cr
& = 15\,cm \cr} $$
Area of the circle :$$\eqalign{
& = \left( {\frac{{22}}{7} \times 7 \times 7} \right)\,c{m^2} \cr
& = 154\,c{m^2} \cr} $$

Answer: Option B. -> 154 cm2
Radius of circle = 7cm
Given area of rectangle :
= Area of circle
= $${\frac{22}{7} \times 7 \times 7}$$
= 154 cm2

Answer: Option A. -> 75
Given length and width of a square base plate of brass is x cm and 1 mm
Volume of the plate of square base = Area of base × height
$$\eqalign{
& = {x^2} \times \frac{1}{{10}} \cr
& = \frac{{{x^2}}}{{10}}\,cu.cm. \cr} $$
According to the question,$$\eqalign{
& \Rightarrow \frac{{{x^2}}}{{10}} \times 8.4 = 4725 \cr
& \Rightarrow {x^2} = \frac{{4725 \times 10}}{{8.4}} \cr
& \Rightarrow {x^2} = 5625 \cr
& \Rightarrow x = \sqrt {5625} = 75\,cm \cr} $$

Answer: Option C. -> 200 m2
Let the breadth of the rectangle be x metres
Then, length of the rectangle = 2x metres
⇒ 2(2x + x) = 60
⇒ 6x = 60
⇒ x = 10
So, length = 20 m, breadth = 10 m
∴ Area = (20 × 10) m2 = 200 m2