Practice MCQ Questions and Answer on Area

21.

The ratio of the outer and the inner perimeters of a circular path is 23 : 22. If the path is 5 metres wide, the diameter of the inner circle is :

• (A) 55 m
• (B) 110 m
• (C) 220 m
• (D) 230 m

Show Answer

 $$\eqalign{ & \frac{{2\pi {R_1}}}{{2\pi {R_2}}} = \frac{{23}}{{22}} \cr & \Rightarrow \frac{{{R_1}}}{{{R_2}}} = \frac{{23}}{{22}} \cr & \Rightarrow {R_1} = \frac{{23{R_2}}}{{22}} \cr & {\text{Also, }}{R_1} - {R_2} = 5\,m \cr & \Rightarrow \frac{{23{R_2}}}{{22}} - {R_2} = 5 \cr & \Rightarrow {R_2} = 110 \cr} $$ ∴ Diameter of inner circle : = (2 × 110) m = 220 m

22.

A coaching institute wants to execute tiling work for one of its teaching halls 60 m long and 40 m wide with a square tile of 0.4 m side. If each tile costs Rs. 5, the total cost of tiles would be :

• (A) Rs. 60000
• (B) Rs. 65000
• (C) Rs. 70000
• (D) Rs. 75000

Show Answer

 Number of tiles required : $$\eqalign{ & = \frac{{{\text{Area of hall}}}}{{{\text{Area of each tile}}}} \cr & = \left( {\frac{{60 \times 40}}{{0.4 \times 0.4}}} \right) \cr & = 15000 \cr} $$ ∴ Total cost of tiles : = Rs. (15000 × 5) = Rs. 75000

23.

If the length and breadth of a rectangular field are increased, the area increases by 50%. If the increase in length was 20 %, by what percentage was the breadth increased ?

• (A) 20%
• (B) 25%
• (C) 30%
• (D) Data inadequate

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 Let the original length and breadth of the rectangle be l and b respectively Then, original area = lb New length = 120% of l = $$\frac{6l}{5}$$ New area = 150% of lb = $$\frac{3lb}{2}$$ New breadth :$$\eqalign{ & = \left( {\frac{{3lb}}{2} \times \frac{5}{{6l}}} \right) \cr & = \frac{{5b}}{4} \cr} $$ Increase in breadth :$$\eqalign{ & = \left( {\frac{{5b}}{4} - b} \right) \cr & = \frac{b}{4} \cr} $$ ∴ Increase % : $$\eqalign{ & = \left( {\frac{b}{4} \times \frac{1}{b} \times 100} \right)\% \cr & = 25\% \cr} $$

24.

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

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

25.

If each side of a square is increased by 10%, its area will be increased by :

• (A) 10%
• (B) 21%
• (C) 44%
• (D) 100%

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 Let the original length of sides be x Then, new length : $$\eqalign{ & = \left( {110\% {\text{ of }}x} \right) \cr & = \frac{{11x}}{{10}} \cr} $$ Original area $${x^2}$$ New area : $$\eqalign{ & = {\left( {\frac{{11x}}{{10}}} \right)^2} \cr & = \frac{{121{x^2}}}{{100}} \cr} $$ Increase in area : $$\eqalign{ & = \left( {\frac{{121{x^2}}}{{100}} - {x^2}} \right) \cr & = \frac{{21{x^2}}}{{100}} \cr} $$ ∴ Increase % : $$\eqalign{ & = \left( {\frac{{21{x^2}}}{{100}} \times \frac{1}{{{x^2}}} \times 100} \right)\% \cr & = 21\% \cr} $$

26.

The dimensions of a room are 12.5 metres by 9 metres by 7 metres. There are 2 doors and 4 windows in the room; each door measures 2.5 metres by 1.2 metres and each window 1.5 metres by 1 metre. Find the cost of painting the walls at Rs. 3.50 per square metre.

• (A) Rs. 150.50
• (B) Rs. 1011.50
• (C) Rs. 1101.50
• (D) Cannot be determined

Show Answer

 Area of 4 walls : $$\eqalign{ & = 2\left( {l + b} \right) \times h \cr & = \left[ {2\left( {12.5 + 9} \right) \times 7} \right]{m^2} \cr & = 301\,{m^2} \cr} $$ Area of 2 doors and 4 windows : $$\eqalign{ & = \left[ {2\left( {2.5 \times 1.2} \right) + 4\left( {1.5 \times 1} \right)} \right]{m^2} \cr & = 12\,{m^2} \cr} $$ ∴ Area to be painted : $$\eqalign{ & = \left( {301 - 12} \right){m^2} \cr & = 289\,{m^2} \cr} $$ Cost of painting : $$\eqalign{ & = {\text{Rs}}{\text{.}}\left( {289 \times 3.50} \right) \cr & = {\text{Rs}}{\text{. }}1011.50 \cr} $$

27.

A wire can be bent in the form of a circle of radius 56 cm. If it is bent in the form of a square, then its area will be :

• (A) 3520 cm2
• (B) 6400 cm2
• (C) 7744 cm2
• (D) 8800 cm2

Show Answer

 Length of wire : $$\eqalign{ & = 2\pi \times R \cr & = \left( {2 \times \frac{{22}}{7} \times 56} \right)cm \cr & = 352\,cm \cr} $$ Side of the square : $$\eqalign{ & = \frac{{352}}{4}\,cm \cr & = 88\,cm \cr} $$ Area of the square : $$\eqalign{ & = \left( {88 \times 88} \right)c{m^2} \cr & = 7744\,c{m^2} \cr} $$

28.

A can go round a circular path 8 times in 40 minutes. If the diameter of the circle is increased to 10 times the original diameter, then the time required by A to go round the new path once, travelling at the same speed as before, is :

• (A) 20 min
• (B) 25 min
• (C) 50 min
• (D) 100 min

Show Answer

 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 : $$\eqalign{ & = \left( {\frac{{40}}{{8\pi d}} \times 10\pi d} \right)\min \cr & = 50\,\min \cr} $$

29.

The perimeter of a square is equal to twice the perimeter of a rectangle of length 8 cm and breadth 7 cm. What is the circumference of a semi-circle whose diameter is equal to the side of the square ? (rounded off to two decimal places)

• (A) 23.57 cm
• (B) 38.57 cm
• (C) 42.46 cm
• (D) 47.47 cm

Show Answer

 Perimeter of rectangle : $$\eqalign{ & = \left[ {2\left( {8 + 7} \right)cm} \right] \cr & = 30\,cm \cr} $$ Perimeter of square : $$\eqalign{ & = \left( {2 \times 30} \right)cm \cr & = 60\,cm \cr} $$ Side of the square : $$\eqalign{ & = \left( {\frac{{60}}{4}} \right)cm \cr & = 15\,cm \cr} $$ Radius of the semi-circle $$ = \left( {\frac{{15}}{2}} \right)cm$$ ∴ Circumference : $$\eqalign{ & = \left( {\frac{{22}}{7} \times \frac{{15}}{2}} \right)cm \cr & = \left( {\frac{{165}}{7}} \right)cm \cr & = 23.57\,cm \cr} $$

30.

The base of an isosceles is 14 cm and its perimeter is 36 cm. Find its area.

• (A) 42$$\sqrt{2}$$ sq. cm
• (B) 42 sq. cm
• (C) 84 sq. cm
• (D) 48 sq. cm

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 Let each equal side of isosceles triangle be x cm Perimeter of an isosceles triangle = 36 cm $$\eqalign{ & \therefore x + x + 14 = 36 \cr & \Rightarrow 2x = 36 - 14 \cr & \Rightarrow x = \frac{{22}}{2} \cr & \Rightarrow x = 11\,cm \cr} $$ BD = DC = 7cm From ΔABD By using Pythagoras theorem : $$\eqalign{ & AD = \sqrt {A{B^2} - B{D^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {{{11}^2} - {7^2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {121 - 49} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \sqrt {72} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 3 \times 2\sqrt 2 \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 6\sqrt 2 \,cm \cr} $$ ∴ Area of ΔABC $$\eqalign{ & = \frac{1}{2} \times BC \times AD \cr & = \frac{1}{2} \times 14 \times 6\sqrt 2 \cr & = 42\sqrt 2 {\text{ sq}}{\text{. cm}} \cr} $$