Area - Aptitude MCQ Questions and Solutions with Explanations

Let r₁ and r₂ be the radii of the in-circle and circum-circle of a square respectively and let each side of the square be a.
Then,
$$\eqalign{ & {r_1} = \frac{a}{2} \cr & {r_2} = \frac{1}{2} \times {\text{diagonal of the sequence}} \cr & {r_2} = \frac{1}{2} \times \sqrt 2 a \cr & {r_2} = \frac{{\sqrt 2 a}}{2}cm \cr} $$
∴ Required ratio :
$$\eqalign{ & = \frac{{\pi \times {{\left( {\frac{a}{2}} \right)}^2}}}{{\pi \times {{\left( {\frac{{\sqrt 2 a}}{2}} \right)}^2}}} \cr & = 1:2 \cr} $$

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 % :
$$\eqalign{ & = \left( {\frac{{21{x^2}}}{{352}} \times \frac{{88}}{{7{x^2}}} \times 100} \right)\% \cr & = 75\% \cr} $$

Let the length of carpet be l metres and breadth the b metres
∴ Diagonal = $$\sqrt {{l^2} + {b^2}} $$
According to the question,
$$\eqalign{ & lb = 120{\text{ and }} {\text{2}}\left( {l + b} \right) = 46 \cr & \Rightarrow \left( {l + b} \right) = 23 \cr} $$
On squaring both sides :
$$\eqalign{ & \Rightarrow {\left( {l + b} \right)^2} = {23^2} \cr & \Rightarrow {l^2} + {b^2} + 2lb = 529 \cr & \Rightarrow {l^2} + {b^2} + 2 \times 120 = 529 \cr & \Rightarrow {l^2} + {b^2} = 529 - 240 \cr & \Rightarrow {l^2} + {b^2} = 289 \cr & \therefore \sqrt {{l^2} + {b^2}} = \sqrt {289} = 17 \cr} $$
Diagonal of the carpet = 17 metres

$$\eqalign{ & \frac{l}{{2\left( {l + b} \right)}} = \frac{1}{3} \cr & \Rightarrow 3l = 2l + 2b \cr & \Rightarrow l = 2b \cr & \Rightarrow \frac{1}{b} = \frac{2}{1} = 2:1 \cr} $$

Let original length = $$l$$ metres and original breadth = b metres
Original area : $$ = \left( {lb} \right){m^2}$$
New length :
$$\eqalign{ & = \left( {\frac{{150l}}{{100}}} \right)m \cr & = \left( {\frac{{3l}}{2}} \right)m \cr & \text{New breadth :} \cr & = \left( {\frac{{150b}}{{100}}} \right)m \cr & = \left( {\frac{{3b}}{2}} \right)m \cr & \text{New area :} \cr & = \left( {\frac{{3l}}{2} \times \frac{{3b}}{2}} \right){m^2} \cr & = \left( {\frac{{9lb}}{4}} \right){m^2} \cr & \text{Increase} = 1 - \frac{9lb}{4} = \frac{5lb}{4} \cr & \therefore \text{ Increase % :} \cr & = \left( {\frac{{5lb}}{4} \times \frac{1}{{lb}} \times 100} \right)\% \cr & = 125\% \cr} $$

Area of the path :
$$\eqalign{ & = \left[ {\left( {26 \times 16} \right) - \left( {24 \times 14} \right)} \right]{m^2} \cr & = \left( {416 - 336} \right){m^2} \cr & = 80\,{m^2} \cr} $$
∴ Number of tiles required to cover the path :
$$\eqalign{ & = \frac{{{\text{Area of path}}}}{{{\text{Area of each tile}}}} \cr & = \left( {\frac{{80 \times 100 \times 100}}{{20 \times 20}}} \right) \cr & = 2000 \cr} $$

Length of largest tile = H.C.F. of 624 and 480 cm = 48 cm
Area of each tile = (48 × 48) cm²
∴ Required number of tiles :
$$\eqalign{ & = \left( {\frac{{624 \times 480}}{{48 \times 48}}} \right) \cr & = 130 \cr} $$

$${\text{Perimeter}} = $$   $${\text{Distance covered in 8 minutes}}$$
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{12000}}{{60}} \times 8} \right)m \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1600{\text{ }}\,m \cr} $$
Let the length = 3x metres and breadth = 2x metres
Then,
2 (3x + 2x) = 1600
or, x = 160
Length = 480 m and breadth = 320 m
∴ Area = (480 × 320) m²
           = 153600 square metre

Original breadth = 3 m
Original length :
$$\eqalign{ & = \left( {1.44 \times 3} \right)m \cr & = 4.32\,m \cr} $$
New breadth :
$$\eqalign{ & = \left( {125\% {\text{ of 3}}} \right)m \cr & = \left( {\frac{{125}}{{100}} \times 3} \right)m \cr & = 3.75\,m \cr} $$
New length :
$$\eqalign{ & = \left( {140\% {\text{ of 4}}{\text{.32}}} \right)m \cr & = \left( {\frac{{140}}{{100}} \times 4.32} \right)m \cr & = 6.048\,m \cr} $$
Original area :
$$\eqalign{ & = \left( {4.32 \times 3} \right){m^2} \cr & = 12.96\,{m^2} \cr} $$
New area :
$$\eqalign{ & = \left( {6.048 \times 3.75} \right){m^2} \cr & = 22.68\,{m^2} \cr} $$
Increase in area :
$$\eqalign{ & = \left( {22.68 - 12.96} \right){m^2} \cr & = 9.72\,{m^2} \cr} $$
∴ Increase in cost :
$$\eqalign{ & = {\text{Rs}}{\text{. }}\left( {9.72 \times 45} \right) \cr & = {\text{Rs}}{\text{.}}\,{\text{437}}{\text{.40}} \cr} $$

(A) Diagonal of the rectangle :
$$\eqalign{ & = \sqrt {{{\left( {51} \right)}^2} + {{\left( {49} \right)}^2}} m \cr & = \sqrt {2601 + 2401} \,m \cr & = \sqrt {5002} \,m \cr} $$
Diagonal of the square :
$$\eqalign{ & = 50\sqrt 2 \,m \cr & = \sqrt {5000} \,m \cr} $$

(B) Diagonal of a square intersect at right angles but those of a rectangle do not.

(C) Perimeter of rectangle :
= 2(51 + 49) m
= 200 m
Perimeter of square :
= (4 × 50) m
= 200 m

(D) Area of rectangle :
= (50 × 50) m²
= 2500 m²