Practice MCQ Questions and Answer on Area

11.

The circumference of the front wheel of a cart is 40 ft long and that of the back wheel is 48 ft long. What is the distance travelled by the cart, when the front wheel has done five more revolutions than the rear wheel ?

• (A) 850 ft
• (B) 950 ft
• (C) 1200 ft
• (D) 1450 ft

Show Answer

 Let the rear wheel make x revolutions Then, the front wheel makes (x + 5) revolutions (x + 5) × 40 = 48x ⇒ 8x = 200 ⇒ x = 25 Distance travelled by the cart : = (48 × 25) ft = 1200 ft

12.

How many squares with side $$\frac{1}{2}$$ inch long are needed to cover a rectangle that is 4 feet long and 6 feet wide ?

• (A) 24
• (B) 96
• (C) 3456
• (D) 13824

Show Answer

 length of rectangle = 4ft = (4 ×12) inch = 48 inch length of rectangle = 6ft = (6 ×12) inch = 72 inch ∴ Number of squares : $$\eqalign{ & = \frac{{48 \times 72}}{{\frac{1}{2} \times \frac{1}{2}}} \cr & = 13824 \cr} $$

13.

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, the n the area of the park (in square metre) is :

• (A) 15360 square metre
• (B) 153600 square metre
• (C) 30720 square metre
• (D) 307200 square metre

Show Answer

 $${\text{Perimeter}} = $$   $${\text{Distance covered in 8 minutes}}$$$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{12000}}{{60}} \times 8} \right)m \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1600{\text{ }}\,m \cr} $$ Let the length = 3x metres and breadth = 2x metres Then, 2 (3x + 2x) = 1600 or, x = 160 Length = 480 m and breadth = 320 m ∴ Area = (480 × 320) m2            = 153600 square metre

14.

If the radius of a circle is increased by 75%, then its circumference will increase by :

• (A) 25%
• (B) 50%
• (C) 75%
• (D) 100%

Show Answer

 Let original radius be R cm Then, original circumference = $$2\pi r$$ cm New radius : $$\eqalign{ & = \left( {175\% {\text{ of }}R} \right)cm \cr & = \left( {\frac{{175}}{{100}} \times R} \right)cm \cr & = \frac{{7R}}{4}\,cm \cr} $$ New circumference : $$\eqalign{ & = \left( {2\pi \times \frac{{7R}}{4}} \right)cm \cr & = \frac{{7\pi R}}{2}\,cm \cr} $$ Increase in circumference : $$\eqalign{ & = \left( {\frac{{7\pi R}}{2} - 2\pi R} \right)cm \cr & = \frac{{3\pi R}}{2}\,cm \cr} $$ Increase % : $$\eqalign{ & = \left( {\frac{{3\pi R}}{2} \times \frac{1}{{2\pi R}} \times 100} \right)\% \cr & = 75\% \cr} $$

15.

Total area of 64 small squares of a chessboard is 400 sq. cm. There is 3 cm wide border around the chess board. What is the length of the side of the chess board ?

• (A) 17 cm
• (B) 20 cm
• (C) 23 cm
• (D) 26 cm

Show Answer

 Area of each square :$$\eqalign{ & = \left( {\frac{{400}}{{64}}} \right)c{m^2} \cr & = 6.25\,c{m^2} \cr} $$ Side of each small square :$$\eqalign{ & = \sqrt {6.25} \,cm \cr & = 2.5\,cm \cr} $$ Since there are 8 squares along each side of the chess board, we have : Side = [(8 × 2.5) + 6] cm         = 26 cm

16.

A rectangular carpet has an area of 120 m² and a perimeter of 46 metres. The length of its diagonal is :

• (A) 23 metres
• (B) 13 metres
• (C) 17 metres
• (D) 21 metres

Show Answer

 Let the length of carpet be l metres and breadth the b metres ∴ Diagonal = $$\sqrt {{l^2} + {b^2}} $$ According to the question, $$\eqalign{ & lb = 120{\text{ and }} {\text{2}}\left( {l + b} \right) = 46 \cr & \Rightarrow \left( {l + b} \right) = 23 \cr} $$ On squaring both sides : $$\eqalign{ & \Rightarrow {\left( {l + b} \right)^2} = {23^2} \cr & \Rightarrow {l^2} + {b^2} + 2lb = 529 \cr & \Rightarrow {l^2} + {b^2} + 2 \times 120 = 529 \cr & \Rightarrow {l^2} + {b^2} = 529 - 240 \cr & \Rightarrow {l^2} + {b^2} = 289 \cr & \therefore \sqrt {{l^2} + {b^2}} = \sqrt {289} = 17 \cr} $$ Diagonal of the carpet = 17 metres

17.

The circumference of a circle is 100 cm. The side of a square inscribed in the circle is :

• (A) $$50\sqrt{2}c{m}^2$$
• (B) $$\frac{100}{\pi }c{m}^2$$
• (C) $$\frac{50\sqrt{2}}{\pi }c{m}^2$$
• (D) $$\frac{100\sqrt{2}}{\pi }c{m}^2$$

Show Answer

 $$\eqalign{ & 2\pi R = 100 \cr & R = \frac{{100}}{{2\pi }} = \frac{{50}}{\pi } \cr & R = \frac{1}{2} \times {\text{diagonal}} \cr & \Rightarrow {\text{Diagonal}} = 2R \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2 \times 50}}{\pi } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{100}}{\pi } \cr} $$ $$\eqalign{ & \therefore {\text{Area of the square}} = \frac{1}{2} \times {\left( {{\text{diagonal}}} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{a^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \times {\left( {\frac{{100}}{\pi }} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{\sqrt 2 }} \times \frac{{100}}{\pi } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50\sqrt 2 }}{\pi }cm \cr} $$

18.

The length of a room is double its breadth. The cost of colouring the ceiling at Rs. 25 per sq. meter is Rs. 5000 and the cost of painting the four walls at Rs. 240 per sq. metre is Rs. 64800. Find the height of the room :

• (A) 3.5 m
• (B) 4 m
• (C) 4.5 m
• (D) 5 m

Show Answer

 Let the breadth and height of the room be b metres and h metres respectively. Then, length of the room = (2b) metres Area of the ceiling : $$\eqalign{ & = \left( {2b \times b} \right)m \cr & = \left( {2{b^2}} \right){m^2} \cr} $$ $$\eqalign{ & 2{b^2} = \frac{{5000}}{{25}} \cr & 2{b^2} = 200 \cr & {b^2} = 100 \cr & b = 10 \cr} $$ So, length = 20 m, breadth = 10 m Area of 4 walls : $$\eqalign{ & = \left[ {2\left( {20 + 10} \right) \times h} \right]{m^2} \cr & = \left( {60h} \right){m^2} \cr} $$ $$\eqalign{ & \therefore 60h = \frac{{64800}}{{240}} \cr & \Rightarrow 60h = 270 \cr & \Rightarrow h = \frac{{270}}{{60}} \cr & \Rightarrow h = 4.5\,m \cr} $$

19.

Twenty-nine times the area of a square is one square metre less than six times the area of the second square and nine times its side exceeds the perimeter of other square by 1 metre. The difference in the sides of these squares is :

• (A) 5 m
• (B) $$\frac{54}{11}$$ m
• (C) 6 m
• (D) 11 m

Show Answer

 Let the sides of the two squares be x metres and y metres respectively Then, $$ \Rightarrow 29{x^2} = 6{y^2} - 1.....(i)$$ And, $$\eqalign{ & \Rightarrow 9x - 4y = 1 \cr & \Rightarrow 4y = 9x - 1 \cr & \Rightarrow y = \frac{{9x - 1}}{4}.....(ii) \cr} $$ From (i) and (ii), we get : $$\eqalign{ & \Rightarrow 29{x^2} = 6{\left( {\frac{{9x - 1}}{4}} \right)^2} - 1 \cr & \Rightarrow 29{x^2} = 6\left( {\frac{{81{x^2} + 1 - 18x}}{{16}}} \right) - 1 \cr & \Rightarrow 243{x^2} + 3 - 54x - 8 = 232{x^2} \cr & \Rightarrow 11{x^2} - 54x - 5 = 0 \cr & \Rightarrow \left( {x - 5} \right)\left( {11x + 1} \right) = 0 \cr & \Rightarrow x = 5\,\,m \cr & \therefore y = \frac{{9x - 1}}{4} = \frac{{9 \times 5\,\,m - 1}}{4} = 11\,\,m \cr & \text{Required difference} \cr & = \left(11-5\right) m \cr & = 6\,m } $$

20.

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%

Show Answer

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