Area - Aptitude MCQ Questions and Solutions with Explanations

Perimeter = Distance covered in 8 min.
Perimeter = $$\left( {\frac{{12000}}{{60}} \times 8} \right){\text{m}}$$
Perimeter = 1600 m
Let length = 3x metres and breadth = 2x metres.
Then, 2(3x + 2x) = 1600 or x = 160
Therefore Length = 480 m and Breadth = 320 m
Therefore Area = (480 x 320) m² = 153600 m²

100 cm is read as 102 cm.
∴ A₁ = (100 x 100) cm² and A₂ (102 x 102) cm²
(A₂ - A₁) = [(102)² - (100)²]
              = (102 + 100) x (102 - 100)
              = 404 cm²
∴ Percentage error
$$\eqalign{ & = \left( {\frac{{404}}{{100 \times 100}} \times 100} \right)\% \cr & = 4.04\% \cr} $$

$$\eqalign{ & \frac{{2\left( {l + b} \right)}}{b} = \frac{5}{1} \cr & \Rightarrow 2l + 2b = 5b \cr & \Rightarrow 3b = 2l \cr & b = \frac{2}{3}l \cr & {\text{Then,}} \cr & {\text{Area = }}216\,c{m^2} \cr & \Rightarrow l \times b = 216 \cr & \Rightarrow l \times \frac{2}{3}l = 216 \cr & \Rightarrow {l^2} = 324 \cr & l = 18\,cm \cr} $$

Let original length = x metres and original breadth = y metres
$$\eqalign{ & {\text{Original}}\,{\text{are}} = \left( {xy} \right){m^2} \cr & {\text{New}}\,{\text{length}} = \left( {\frac{{120}}{{100}}x} \right)m = \left( {\frac{6}{5}x} \right)m \cr & {\text{New}}\,{\text{breadth}} = \left( {\frac{{120}}{{100}}y} \right)m = \left( {\frac{6}{5}y} \right)m \cr & {\text{New}}\,{\text{area}} = \left( {\frac{6}{5}x \times \frac{6}{5}y} \right){m^2} = \left( {\frac{{36}}{{25}}xy} \right){m^2} \cr & {\text{The}}\,{\text{difference}}\,{\text{between}}\,{\text{the}}\,{\text{original}}\,{\text{area = }}xy \cr & {\text{and}}\,{\text{new}}\,{\text{area}}\,\frac{{36}}{{25}}xy\,{\text{is}} \cr & = \left( {\frac{{36}}{{25}}} \right)xy - xy \cr & = xy\left( {\frac{{36}}{{25}} - 1} \right) \cr & = xy\left( {\frac{{11}}{{25}}} \right)\,or\,\left( {\frac{{11}}{{25}}} \right)xy \cr & \therefore {\text{Increase}}\,\% \cr & = \left( {\frac{{11}}{{25}}xy \times \frac{1}{{xy}} \times 100} \right)\% \cr & = 44\% \cr} $$

Area of the park = (60 x 40) m² = 2400 m²
Area of the lawn = 2109 m²
∴ Area of the crossroads = (2400 - 2109) m² = 291 m²
Let the width of the road be x metres. Then,
60x + 40x - x² = 291
⇒ x² - 100x + 291 = 0
⇒ (x - 97)(x - 3) = 0
⇒ x = 3 m

$$\eqalign{ & {\text{Outer}}\,{\text{Side}} \cr & = \sqrt {{{\left( {\frac{{15}}{2}} \right)}^2} - {{\left( {\frac{9}{2}} \right)}^2}ft} \cr & = \sqrt {\frac{{225}}{4} - \frac{{81}}{4}ft} \cr & = \sqrt {\frac{{144}}{4}ft} \cr & = 6ft \cr & \therefore {\text{Area}}\,{\text{of}}\,{\text{closet}} = \left( {6 \times 4.5} \right)sq.\,ft = 27\,sq.\,ft. \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{original}}\,{\text{length}} = x\,{\text{and}} \cr & {\text{original}}\,{\text{breadth}} = y \cr & {\text{Decrease}}\,{\text{in}}\,{\text{area}} \cr & = xy - \left( {\frac{{80}}{{100}}x \times \frac{{90}}{{100}}y} \right) \cr & = {xy - \frac{{18}}{{25}}xy} \cr & = \frac{7}{{25}}xy \cr & \therefore {\text{Decrease}}\,\% = \cr & \left( {\frac{7}{{25}}xy \times \frac{1}{{xy}} \times 100} \right)\% \cr & = 28\% \cr} $$

Let the side of the square(ABCD) be x metres.
Then, AB + BC = 2x metres

AC =$$\sqrt 2 $$ x = (1.41x) m
Saving on 2x metres = (0.59x) m
$$\eqalign{ & {\text{Saving}}\,\% = \left( {\frac{{0.59x}}{{2x}} \times 100} \right)\% \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 30\% \,\left( {{\text{approx}}} \right) \cr} $$

$$\eqalign{ & \sqrt {{l^2} + {b^2}} = \sqrt {41} \cr & {\text{Also}},\,lb = 20 \cr & {\left( {l + b} \right)^2} = \left( {{l^2} + {b^2}} \right) + 2/b = 41 + 40 = 81 \cr & \Rightarrow \left( {l + b} \right) = 9 \cr & \therefore {\text{Perimeter}} = 2\left( {l + b} \right) = 18cm \cr} $$

$$\eqalign{ & {\text{Length}}\,{\text{of}}\,{\text{largest}}\,{\text{tile}} = \cr & {\text{H}}{\text{.C}}{\text{.F}}{\text{.}}\,{\text{of}}\,1517\,cm\,{\text{and}}\,902\,cm = 41\,cm \cr & {\text{Area}}\,{\text{of}}\,{\text{each}}\,{\text{tile}} = \left( {41 \times 41} \right)c{m^2} \cr & \therefore {\text{Required}}\,{\text{number}}\,{\text{of}}\,{\text{tiles}} \cr & = {\frac{{1517 \times 902}}{{41 \times 41}}} \cr & = 814 \cr} $$