Area

1.The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:

1. 15360
2. 153600
3. 30720
4. 307200

Answer:Option 1

Explanation:

Solution:

Perimeter = Distance covered in 8 min. =

(\frac{12000}{60} \times 8) m = 1600 m

$\left( {\frac{{12000}}{{60}} \times 8} \right)m = 1600\,m$ 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.

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2.An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is:

1. 2%
2. 2.02%
3. 4%
4. 4.04%

Answer:Option 1

Explanation:

Solution:

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 =

(\frac{404}{100 \times 100} \times 100) % = 4.04 %

$\left( {\frac{{404}}{{100 \times 100}} \times 100} \right)\% = 4.04\%$

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3.The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle?

1. 16 cm
2. 18 cm
3. 24 cm
4. Data inadequate

Answer:Option 1

Explanation:

Solution:

\begin{aligned} \frac{2 (l + b)}{b} = \frac{5}{1} \\ \Rightarrow 2 l + 2 b = 5 b \\ \Rightarrow 3 b = 2 l \\ b = \frac{2}{3} l \\ Then, Area = 216 c m^{2} \\ \Rightarrow l \times b = 216 \\ \Rightarrow l \times \frac{2}{3} l = 216 \\ \Rightarrow l^{2} = 324 \\ l = 18 c m \end{aligned}

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

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4.A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?

1. 2.91 m
2. 3 m
3. 5.82 m
4. None of these

Answer:Option 1

Explanation:

Solution:

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.

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5.The diagonal of the floor of a rectangular closet is 7 ¹/₂ feet. The shorter side of the closet is 4 ¹/₂ feet. What is the area of the closet in square feet?

1. 5 ¹/₄
2. 13 ¹/₂
3. 27
4. 37

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Outer Side \\ = \sqrt{{(\frac{15}{2})}^{2} - {(\frac{9}{2})}^{2}} f t \\ = \sqrt{\frac{225}{4} - \frac{81}{4} f t} \\ = \sqrt{\frac{144}{4} f t} \\ = 6 f t \\ ∴ Area of closet = (6 \times 4.5) s q . f t = 27 s q . f t . \end{aligned}

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

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6.A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is:

1. 10%
2. 10.08%
3. 20%
4. 28%

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Let original length = x and \\ original breadth = y \\ Decrease in area \\ = x y - (\frac{80}{100} x \times \frac{90}{100} y) \\ = (x y - \frac{18}{25} x y) \\ = \frac{7}{25} x y \\ ∴ Decrease % = \\ (\frac{7}{25} x y \times \frac{1}{x y} \times 100) % = 28 % \end{aligned}

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

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7.A man walked diagonally across a square lot. Approximately, what was the percent saved by not walking along the edges?

1. 20
2. 24
3. 30
4. 33

Answer:Option 1

Explanation:

Solution:

Let the side of the square(ABCD) be x metres.

AC = √2x = (1.41x) m.
Saving on 2x metres = (0.59x) m.

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8.What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?

1. 814
2. 820
3. 840
4. 844

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Length of largest tile = \\ H .C .F . of 1517 c m and 902 c m = 41 c m \\ Area of each tile = (41 \times 41) c m^{2} \\ ∴ Required number of tiles \\ = (\frac{1517 \times 902}{41 \times 41}) \\ = 814 \end{aligned}

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

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9.The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:

1. 1520 m²
2. 2420 m²
3. 2480 m²
4. 2520 m²

Answer:Option 1

Explanation:

Solution:

We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103.
Solving the two equations, we get: l = 63 and b = 40.
∴ Area = (l x b) = (63 x 40) m² = 2520 m².

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10.The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area?

1. 25% increase
2. 50% increase
3. 50% decrease
4. 75% decrease

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Let original length = x and \\ Original breadth = y \\ Original area = x y \\ New length = \frac{x}{2} \\ New breadth = 3 y \\ New area = (\frac{x}{2} \times 3 y) = \frac{3}{2} x y \\ ∴ Increase % = (\frac{1}{2} x y \times \frac{1}{x y} \times 100) \\ = 50 % \end{aligned}

$\eqalign{ & {\text{Let}}\,{\text{original}}\,{\text{length}} = x\,{\text{and}}\, \cr & {\text{Original}}\,{\text{breadth}} = y \cr & {\text{Original}}\,{\text{area}} = xy \cr & {\text{New}}\,{\text{length}} = \frac{x}{2} \cr & {\text{New}}\,{\text{breadth}} = 3y \cr & {\text{New}}\,{\text{area}} = \left( {\frac{x}{2} \times 3y} \right) = \frac{3}{2}xy \cr & \therefore {\text{Increase}}\,\% = \left( {\frac{1}{2}xy \times \frac{1}{{xy}} \times 100} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 50\% \cr}$

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