Area - Aptitude MCQ Questions and Solutions with Explanations

Let the radius of the pool be R ft.
Radius of the pool including the wall = (R + 4)ft.
Area of the concrete wall :
$$\eqalign{ & = \pi \left[ {{{\left( {R + 4} \right)}^2} - {R^2}} \right]sq.ft \cr & = \pi \left[ {\left( {R + 4 + R} \right)\left( {R + 4 - R} \right)} \right]sq.ft \cr & = 8\pi \left( {R + 2} \right)sq.ft \cr & 8\pi \left( {R + 2} \right) = \frac{{11}}{{25}}\pi {R^2} \cr & \Rightarrow 11{R^2} = 200\left( {R + 2} \right) \cr & \Rightarrow 11{R^2} - 200R - 400 = 0 \cr & \Rightarrow 11{R^2} - 220R + 20R - 400 = 0 \cr & \Rightarrow 11R\left( {R - 20} \right) + 20\left( {R - 20} \right) = 0 \cr & \Rightarrow \left( {R - 20} \right)\left( {11R + 20} \right) = 0 \cr & \Rightarrow R = 20 \cr} $$
∴ Radius of the pool = 20 ft.

Diameter of the circle :
$$\eqalign{ & = AC = \sqrt {{{\left( {AB} \right)}^2} + {{\left( {BC} \right)}^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {{8^2} + {6^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {100} \,cm \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 10\,cm \cr} $$
Radius = 5 cm
Required area = (Area of the circle) - (Area of the rectangle)

$$\eqalign{ & = \left[ {\left( {\frac{{22}}{7} \times 5 \times 5} \right) - \left( {8 \times 6} \right)} \right]c{m^2} \cr & = \left( {\frac{{550}}{7} - 48} \right)c{m^2} \cr & = \left( {\frac{{214}}{7}} \right)c{m^2} \cr & = 30.57\,c{m^2} \approx 30.6\,c{m^2} \cr} $$

Let original diameter be d metre
Then, its circumference = $$\left( {\pi d} \right)$$ metres
Time taken to cover $$\left( {8\pi d} \right)$$ m = 40 min
New diameter = $$\left( {10d} \right)$$ m
Then, its circumference = $$\left( {\pi \times 10d} \right)$$ m
∴ Time taken to go round it once :
$$\eqalign{ & = \left( {\frac{{40}}{{8\pi d}} \times 10\pi d} \right)\min \cr & = 50\,\min \cr} $$

Let the length of rectangle be $$l$$ metre and the breadth of the rectangle be $$b$$ metre
Then area of the rectangle = $$l$$ × $$b$$
$$lb$$ = 150 m² . . . . . (i)
According to the question,
$$\eqalign{ & \left( {l - 5} \right) \times \left( {b + 2} \right) = 150 - 30 \cr & \Rightarrow \left( {l - 5} \right) \times \left( {b + 2} \right) = 120 \cr & \Rightarrow lb - 5b + 2l - 10 = 120 \cr & \Rightarrow 150 - 5b + 2l - 10 = 120 \cr & \Rightarrow 5b - 2l = 20 \cr & \Rightarrow \frac{{5 \times 150}}{l} - 2{l^2} = 20l \cr & \Rightarrow 2{l^2} + 20l - 750 = 0 \cr & \Rightarrow {l^2} + 10l - 375 = 0 \cr & \Rightarrow l\left( {l + 25} \right) - 15\left( {l + 25} \right) = 0 \cr & \Rightarrow \left( {l + 15} \right)\left( {l + 25} \right) = 0 \cr} $$
On solving both equations we get,
$$l$$ = 15 m and $$b$$ = 10 m
Side of square = length of rectangle (given)
So, the perimeter of the square :
= 4 × $$l$$
= 4 × 15
= 60 m

Each edges of cube = a cm
∴ Total surface area of cube = 6a²
⇒ 6a² = 864
⇒ a² = $$\frac{864}{6}$$ = 144
⇒ a = $$\sqrt {144} $$ = 12 cm
∴ Volume of cube :
= a³ cm³
= (12 × 12 × 12) cm³
= 1728 cm³

Let the length and breadth of the field be l and b km respectively
Then,
2 (l + b) = 6
⇒ l + b = 3
And lb = 2
(l - b)² = (l + b)² - 4lb
⇒ (l - b)² = 3² - 4 × 2
⇒ (l - b)² = 1
⇒ (l - b) = 1 km

Let the length of the rectangle be x metres.
Then, breath of the rectangle = $$\left( {\frac{{168}}{x}} \right)m$$
$$\eqalign{ & \therefore \sqrt {{x^2} + {{\left( {\frac{{168}}{x}} \right)}^2}} = 25 \cr & \Rightarrow \sqrt {{x^2} + \frac{{28224}}{{{x^2}}}} = 25 \cr & \Rightarrow {x^2} + \frac{{28224}}{{{x^2}}} = 625 \cr & \Rightarrow {x^4} - 625{x^2} + 28224 = 0 \cr & \Rightarrow {x^4} - 576{x^2} - 49{x^2} + 28224 = 0 \cr & \Rightarrow {x^2}\left( {{x^2} - 576} \right) - 49\left( {{x^2} - 576} \right) = 0 \cr & \Rightarrow \left( {{x^2} - 576} \right)\left( {{x^2} - 49} \right) = 0 \cr & \Rightarrow {x^2} = 576\,\,or\,\,{x^2} = 49 \cr & \Rightarrow x = 24\,\,\,or\,\,\,x = 7 \cr} $$
Hence, length = 24 m and breadth = 7 m

Let the width of the path be x
Then,
$$\left[ {\left( {38 \times 32} \right) - \left\{ {\left( {38 - 2x} \right)\left( {32 - 2x} \right)} \right\}} \right]$$       $$ = 600$$
$$\eqalign{ & \Rightarrow \left[ {1216 - \left( {1216 - 140x + 4{x^2}} \right)} \right] \cr & \Rightarrow 4{x^2} - 140x + 600 = 0 \cr & \Rightarrow {x^2} - 35x + 150 = 0 \cr & \Rightarrow {x^2} - 30x - 5x + 150 = 0 \cr & \Rightarrow \left( {x - 30} \right)\left( {x - 5} \right) = 0 \cr & \Rightarrow x = 5\,m\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\therefore x \ne 30} \right] \cr} $$

Area left after laying black tiles :
$$\eqalign{ & = \left[ {\left( {20 - 4} \right) \times \left( {10 - 4} \right)} \right]{\text{sq}}{\text{.ft.}} \cr & = 96\,{\text{sq}}{\text{.ft.}} \cr} $$
Area under white tiles :
$$\eqalign{ & = \left( {\frac{1}{3} \times 96} \right){\text{sq}}{\text{.ft.}} \cr & = 32\,{\text{sq}}{\text{.ft.}} \cr} $$
Area under blue tiles :
$$\eqalign{ & = \left( {96 - 32} \right){\text{sq}}{\text{.ft.}} \cr & = 64\,{\text{sq}}{\text{.ft.}} \cr} $$
Number of blue tiles :
$$\eqalign{ & = \frac{{64}}{{\left( {2 \times 2} \right)}} \cr & = 16 \cr} $$

Area = 1 hect. = 10000 sq. m
Side = $$\sqrt {10000} $$   m = 100 m
Side of the other square = 101 m
Difference in their areas :
= [(101)² - (100)²] m²
= [(101 + 100) (101 - 100)] m²
= 201 m²