Area - Aptitude MCQ Questions and Solutions with Explanations

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

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 % :
$$\eqalign{ & = \left( {\frac{{41}}{{50}} \times lb \times \frac{1}{{lb}} \times 100} \right)\% \cr & = 82\% \cr} $$

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% :
$$\eqalign{ & = \left( {\frac{{3{x^2}}}{{10{x^2}}} \times 100} \right)\% \cr & = 30\% \cr} $$

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

$${\text{Hence, }}{h_1}:{h_2} = ay:bx$$

$$\eqalign{ & \pi R = 33 \cr & \Rightarrow R = \left( {\frac{{33 \times 7}}{{22}}} \right) \cr & \Rightarrow R = \left( {\frac{{21}}{2}} \right)m \cr} $$
∴ Required distance = 2R = 21 m

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

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

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

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

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 % :
$$\eqalign{ & = \left( {\frac{{lb}}{{100}} \times \frac{1}{{1b}} \times 100} \right)\% \cr & = 1\% \cr} $$