Answer: Option D. -> 1188 cm2
Radius and height of a right circular cylinder is 7 cm and 20 cm respectively
Total surface area of right circular cylinder :
$$\eqalign{
& = 2\pi rh + 2\pi r^2 \cr
& = 2\pi r\left( {h + r} \right) \cr
& = 2 \times \frac{{22}}{7} \times 7\left( {20 + 7} \right) \cr
& = 2 \times 22 \times 27 \cr
& = 1188\,sq.\,cm \cr} $$

Answer: Option B. -> Rs. 1650
Length of the fence =$$4\pi R$$
Where, $$R = \frac{{21}}{2}m$$
$$\eqalign{
& 4\pi R \cr
& = \left( {4 \times \frac{{22}}{7} \times \frac{{21}}{2}} \right)m \cr
& = 132\,m \cr} $$
Cost of fencing :
$$\eqalign{
& = {\text{Rs}}{\text{.}}\left( {132 \times \frac{{25}}{2}} \right) \cr
& = {\text{Rs}}{\text{.1650}} \cr} $$

Answer: Option B. -> 20 m × 5 m
We have :
$$\eqalign{
& 2b + l = 30 \cr
& \Rightarrow l = 30 - 2b \cr} $$
$$\eqalign{
& {\text{Area}} = {\text{100 }}{m^2} \cr
& \Rightarrow l \times b = 100 \cr
& \Rightarrow b\left( {30 - 2b} \right) = 100 \cr
& \Rightarrow {b^2} - 15b + 50 = 0 \cr
& \Rightarrow \left( {b - 10} \right)\left( {b - 5} \right) = 0 \cr
& \Rightarrow b = 10{\text{ or }}b = 5 \cr} $$
When, b = 10, $$l$$ = 10 and when b = 5, $$l$$ = 20
Since the garden is rectangular, so its dimension is 20 m × 5 m

Answer: Option A. -> 4.5%
Let the actual length and width of the rectangle be $$l$$ and b respectively.
Then, measured length :
$$ = 100\% {\text{ of }}l = \frac{{11l}}{{10}}$$
Measured width :
$$ = 95\% {\text{ of }}b = \frac{{19b}}{{20}}$$
Actual area = $$lb$$
Measured area :
$$\eqalign{
& = \left( {\frac{{11l}}{{10}} \times \frac{{19b}}{{20}}} \right) \cr
& = \frac{{209lb}}{{200}} \cr} $$
Error in measurement :
$$\eqalign{
& = \left( {\frac{{209lb}}{{200}} - lb} \right) \cr
& = \frac{{9lb}}{{200}} \cr} $$
∴ Error % :$$\eqalign{
& = \left( {\frac{{9lb}}{{200}} \times \frac{1}{{lb}} \times 100} \right)\% \cr
& = 4.5\% \cr} $$

Answer: Option B. -> 60 cm
Length of rectangle = 25 cm
Breadth of rectangle = 15 cm
Area of rectangle :
$$\eqalign{
& = \left( {25 \times 15} \right)c{m^2} \cr
& = 375\,cm \cr} $$
Area of square :
$$\eqalign{
& = \left( {\frac{3}{5} \times 375} \right)c{m^2} \cr
& = 225\,c{m^2} \cr} $$
Side of square :
$$\eqalign{
& = \sqrt {225} \,cm \cr
& = 15\,cm \cr} $$
Perimeter of square :
$$\eqalign{
& = \left( {4 \times 15} \right)cm \cr
& = 60\,cm \cr} $$

Answer: Option B. -> 20 cm
Let the side of the square be x cm
Then, its area = x2 cm2
Area of the rectangle = 3x2 cm2
∴ 40 × $$\frac{3}{2}$$ × x = 3x2
⇔ x = 20 cm

Answer: Option B. -> $$4\frac{8}{{13}}\,cm$$
Area of the triangle :
$$\eqalign{
& = \left( {\frac{1}{2} \times 12 \times 5} \right)c{m^2} \cr
& = 30\,c{m^2} \cr} $$
Hypotenuse :
$$\eqalign{
& = \sqrt {{{12}^2} + {5^2}} \,cm \cr
& = \sqrt {169} \,cm \cr
& = 13\,cm \cr} $$
Let the perpendicular distance of the hypotenuse from the opposite vertex be x cm
Then,
$$\eqalign{
& \Rightarrow \frac{1}{2} \times 13 \times x = 30 \cr
& \Rightarrow x = \frac{{60}}{{13}} \cr
& \Rightarrow x = 4\frac{8}{{13}}\,cm \cr} $$

Answer: Option B. -> 2 : 1
$$\eqalign{
& \Rightarrow \frac{{\pi R_1^2}}{{\pi R_2^2}} = \frac{4}{1} \cr
& \Rightarrow \frac{{R_1^2}}{{R_2^2}} = \frac{4}{1} \cr
& \Rightarrow \frac{{{R_1}}}{{{R_2}}} = \frac{2}{1}{\text{ Or }}2:1 \cr} $$

Answer: Option B. -> 4 m
Let the circumference of front wheel be x metres
Then, Circumference of rear wheel = (x - 1) metres
$$\eqalign{
& \therefore \frac{{600}}{x} - \frac{{600}}{{\left( {x + 1} \right)}} = 30 \cr
& \Rightarrow \frac{1}{{x\left( {x + 1} \right)}} = \frac{1}{{20}} \cr
& \Rightarrow x\left( {x - 1} \right) = 20 \cr
& \Rightarrow \left( {{x^2} + x - 20} \right) = 0 \cr
& \Rightarrow \left( {x + 5} \right)\left( {x - 4} \right) = 0 \cr
& \Rightarrow x = 4\,m \cr} $$

Answer: Option E. -> None of these
$$\eqalign{
& 2\pi R = 2\left( {l + b} \right) \cr
& \Rightarrow 2\pi R = 2(26 + 18)cm \cr
& \Rightarrow R = \left( {\frac{{88}}{{2 \times 22}} \times 7} \right)cm \cr
& \Rightarrow R = 14\,cm \cr} $$
∴ Area of the circle :
$$\eqalign{
& = \pi {R^2} \cr
& = \left( {\frac{{22}}{7} \times 14 \times 14} \right)c{m^2} \cr
& = 616\,c{m^2} \cr} $$