Answer: Option B. -> 12 m
We have :
$$lb$$ = 60 and $$\sqrt {{l^2} + {b^2}} + l = 5b$$
Now,
$$\eqalign{
& {l^2} + {b^2} = {\left( {5b - l} \right)^2} \cr
& \Rightarrow 24{b^2} - 10lb = 0 \cr
& \Rightarrow 24{b^2} - 600 = 0 \cr
& \Rightarrow {b^2} = 25 \cr
& \Rightarrow b = 5 \cr
& \therefore l = \left( {\frac{{60}}{b}} \right) \cr
& \,\,\,\,\,\,\,\, = \left( {\frac{{60}}{5}} \right)m \cr
& \,\,\,\,\,\,\,\, = 12\,m \cr} $$
So, length of the carpet = 12 m

Answer: Option C. -> Rs. 100
Length of the fence :
= 2(60 + 40) m
= 200 m
Cost of fencing :
= Rs. (200 × 50)
= Rs. 10000
Area of the road :
= [(64 × 44) - (60 × 40)] m2
= (2816 - 2400) m2
= 416 m2
Let the cost of tiling the road be Rs. x per sq.m
∴ 416x + 10000 = 51600
⇒ 416x = 41600
⇒ x = Rs. 100

Answer: Option C. -> 120 m × 36 m
Let the side of each square formed by fencing parallel to breadth be x metres and that of each square formed by fencing parallel to length be y metres.
Then,
$$\eqalign{
& 3{x^2} + 3{y^2} = 4320 \cr
& \Rightarrow {x^2} + {y^2} = 1440.....(i) \cr} $$
And,
$$\eqalign{
& x\left( {3x + y} \right) = 4320 \cr
& \Rightarrow 3{x^2} + xy = 3\left( {{x^2} + {y^2}} \right) \cr
& \Rightarrow xy = 3{y^2} \cr
& \Rightarrow x = 3y.....(ii) \cr} $$
From (i) and (ii), we have :
$$\eqalign{
& {\left( {3y} \right)^2} + {y^2} = 1440 \cr
& \Rightarrow 10{y^2} = 1440 \cr
& \Rightarrow {y^2} = 144 \cr
& \Rightarrow y = 12 \cr} $$
$${\text{So, }}x = 36$$
Length of rectangular plot :
$$\eqalign{
& = 3x + y \cr
& = \left( {3 \times 36 + 12} \right)m \cr
& = 120\,m \cr} $$
Breadth of rectangular plot = x = 36 m

Answer: Option B. -> 21%
Let the original length of sides be x
Then, new length :
$$\eqalign{
& = \left( {110\% {\text{ of }}x} \right) \cr
& = \frac{{11x}}{{10}} \cr} $$
Original area $${x^2}$$
New area :
$$\eqalign{
& = {\left( {\frac{{11x}}{{10}}} \right)^2} \cr
& = \frac{{121{x^2}}}{{100}} \cr} $$
Increase in area :
$$\eqalign{
& = \left( {\frac{{121{x^2}}}{{100}} - {x^2}} \right) \cr
& = \frac{{21{x^2}}}{{100}} \cr} $$
∴ Increase % :
$$\eqalign{
& = \left( {\frac{{21{x^2}}}{{100}} \times \frac{1}{{{x^2}}} \times 100} \right)\% \cr
& = 21\% \cr} $$

Answer: Option D. -> 72 cm
Let a = 3x cm, b = 4x and c = 5x
Then, s = 6x
$$\eqalign{
& A = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} \cr
& \,\,\,\,\,\,\, = \sqrt {6x \times 3x \times 2x \times x} \cr
& \,\,\,\,\,\,\, = \left( {6{x^2}} \right)c{m^2} \cr
& \therefore 6{x^2} = 216 \cr
& \Rightarrow {x^2} = 36 \cr
& \Rightarrow x = 6 \cr} $$
So, a = 18 cm, b = 24 cm and c = 30 cm
Perimeter :
= (18 + 24 + 30) cm
= 72 cm

Answer: Option C. -> $$\sqrt 3 $$ : 2
Let the side of the square be a cm
Then, the length of its diagonal = $$\sqrt 2 $$ a cm
Area of equilateral triangle with side :
$$\eqalign{
& = \sqrt 2 a \cr
& = \frac{{\sqrt 3 }}{4} \times {\left( {\sqrt 2 a} \right)^2} \cr
& = \frac{{\sqrt 3 {a^2}}}{2} \cr} $$
∴ Required ratio :
$$\eqalign{
& = \frac{{\sqrt 3 {a^2}}}{2}:{a^2} \cr
& = \sqrt 3 :2 \cr} $$

Answer: Option D. -> 250%
$$\eqalign{
& {\text{Required % }} = \left[ {\frac{{\pi \times {{\left( 5 \right)}^2}}}{{2\pi \times 5}} \times 100} \right]\% \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 250\% \cr} $$

Answer: Option C. -> 1200 ft
Let the rear wheel make x revolutions
Then, the front wheel makes (x + 5) revolutions
(x + 5) × 40 = 48x
⇒ 8x = 200
⇒ x = 25
Distance travelled by the cart :
= (48 × 25) ft
= 1200 ft

Answer: Option D. -> 14 cm
Let the length of the side of the square be x cm
Then, radius of each semi-circle = $$\left( {\frac{x}{2}} \right)$$ cm
Total area :
$$\eqalign{
& = \left[ {{x^2} + \frac{\pi }{2}{{\left( {\frac{x}{2}} \right)}^2} + \frac{\pi }{2}{{\left( {\frac{x}{2}} \right)}^2}} \right]c{m^2} \cr
& = \left( {{x^2} + \frac{\pi }{4} \times {x^2}} \right)c{m^2} \cr} $$
$$\eqalign{
& \therefore {x^2} + \frac{\pi }{4} \times {x^2} = 350 \cr
& \Rightarrow {x^2} + \frac{{22{x^2}}}{{28}} = 350 \cr
& \Rightarrow {x^2} + \frac{{11{x^2}}}{{14}} = 350 \cr
& \Rightarrow \frac{{25{x^2}}}{{14}} = 350 \cr
& \Rightarrow {x^2} = \left( {\frac{{350 \times 14}}{{25}}} \right) \cr
& \Rightarrow {x^2} = 196 \cr
& \Rightarrow x = 14\,cm \cr} $$

Answer: Option C. -> 7744 cm2
Length of wire :
$$\eqalign{
& = 2\pi \times R \cr
& = \left( {2 \times \frac{{22}}{7} \times 56} \right)cm \cr
& = 352\,cm \cr} $$
Side of the square :
$$\eqalign{
& = \frac{{352}}{4}\,cm \cr
& = 88\,cm \cr} $$
Area of the square :
$$\eqalign{
& = \left( {88 \times 88} \right)c{m^2} \cr
& = 7744\,c{m^2} \cr} $$