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Home > Practice > Arithmetic Aptitude > Area > Miscellaneous

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:

A. 15360 m²

B. 153600 m²

C. 30720 m²

D. 307200 m²

Discuss

Solution:

Perimeter = Distance covered in 8 min.
Perimeter =

(\frac{12000}{60} \times 8) 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

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:

A. 2%

B. 2.02%

C. 4%

D. 4.04%

Discuss

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

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

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?

A. 16 cm

B. 18 cm

C. 24 cm

D. Data inadequate

Discuss

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}

4. The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:

A. 40%

B. 42%

C. 44%

D. 46%

Discuss

Solution:

Let original length = x metres and original breadth = y metres

\begin{aligned} Original are = (x y) m^{2} \\ New length = (\frac{120}{100} x) m = (\frac{6}{5} x) m \\ New breadth = (\frac{120}{100} y) m = (\frac{6}{5} y) m \\ New area = (\frac{6}{5} x \times \frac{6}{5} y) m^{2} = (\frac{36}{25} x y) m^{2} \\ The difference between the original area = x y \\ and new area \frac{36}{25} x y is \\ = (\frac{36}{25}) x y - x y \\ = x y (\frac{36}{25} - 1) \\ = x y (\frac{11}{25}) o r (\frac{11}{25}) x y \\ ∴ Increase % \\ = (\frac{11}{25} x y \times \frac{1}{x y} \times 100) % \\ = 44 % \end{aligned}

5. 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?

A. 2.91 m

B. 3 m

C. 5.82 m

D. None of these

Discuss

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 m

6. The diagonal of the floor of a rectangular closet is $7 \frac{1}{2}$ feet. The shorter side of the closet is $4 \frac{1}{2}$ feet. What is the area of the closet in square feet?

A. $5 \frac{1}{4}$ sq. ft.

B. $13 \frac{1}{2}$ sq. ft.

C. 27 sq. ft.

D. 37 sq. ft.

Discuss

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}

7. 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:

A. 10%

B. 10.08%

C. 20%

D. 28%

Discuss

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}

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

A. 20%

B. 24%

C. 30%

D. 33%

Discuss

Solution:

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

AC =

\sqrt{2}

x = (1.41x) m
Saving on 2x metres = (0.59x) m

\begin{aligned} Saving % = (\frac{0.59 x}{2 x} \times 100) % \\ = 30 % (approx) \end{aligned}

9. The diagonal of a rectangle is $\sqrt{41}$ cm and its area is 20 sq. cm. The perimeter of the rectangle must be:

A. 9 cm

B. 18 cm

C. 20 cm

D. 41 cm

Discuss

Solution:

\begin{aligned} \sqrt{l^{2} + b^{2}} = \sqrt{41} \\ Also, l b = 20 \\ {(l + b)}^{2} = (l^{2} + b^{2}) + 2 / b = 41 + 40 = 81 \\ \Rightarrow (l + b) = 9 \\ ∴ Perimeter = 2 (l + b) = 18 c m \end{aligned}

10. 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?

A. 814

B. 820

C. 840

D. 844

Discuss

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}

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