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Area - Aptitude MCQ Questions and Solutions with Explanations

Area MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Area with easy and logical explanations. Area MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Page-5 section-2

Each section contains 100 MCQs on Area .
41
A rectangular field has dimensions 25 m by 15 m. Two mutually perpendicular passages, 2 m wide have been left in its central part and grass has been grown in rest of the field. The area (in sq. metres) under the grass is :
Discuss
Answer & Solution
Answer: Option B
Solution:
Area of the field :
= (25 × 15) m2
= 375 m2
Area of the passages :
= (25 × 2 + 15 × 2 - 2 × 2) m2
= 76 m2
Area under grass :
= (375 - 76) m2
= 299 m2
42
A room is $$12\frac{1}{4}$$ m long and 7 m wide. The maximum length of a square tile to fill the floor of the room with whole number of tiles should be :
Discuss
Answer & Solution
Answer: Option C
Solution:
Length of largest tile :
= H.C.F. of $$12\frac{1}{4}$$ m and 7 m
= H.C.F. of 12.25 m and 7 m
= H.C.F. of 1225 cm and 700 cm = 175 cm
43
If the area of a square increase by 69%, then the side of the square increases by :
Discuss
Answer & Solution
Answer: Option B
Solution:
Le original area = 100 cm2
Then, new area = 169 cm2
⇒ Original side = 10 cm
New side = 13 cm
Increase on 10 cm = 3 cm
Increase % :
$$\eqalign{ & = \left( {\frac{3}{{10}} \times 100} \right)\% \cr & = 30\% \cr} $$
44
The cost of papering the four walls of a room is Rs. 475. Each one of the length, breadth and height of another room is double that of this room. The cost of papering the walls of this new room is :
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & {A_1} = 2\left( {l + b} \right) \times h \cr & {A_2} = 2\left( {2l + 2b} \right) \times 2h \cr & \,\,\,\,\,\,\,\, = 8\left( {l + b} \right) \times h \cr & \,\,\,\,\,\,\,\, = 4{A_1} \cr & {\text{ Required cost}} = {\text{Rs}}{\text{.}}\left( {4 \times 475} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{.1900}} \cr} $$
45
The area of two similar triangles are 12 cm2 and 48 cm2. If the height of the smaller one is 2.1 cm, then the corresponding height of the bigger one is :
Discuss
Answer & Solution
Answer: Option B
Solution:
Note : The areas of two similar triangles are in the ratio of the squares of the corresponding altitudes. Let the length of the required altitude be x cm
Then,
$$\eqalign{ & \frac{{12}}{{48}} = \frac{{{{\left( {2.1} \right)}^2}}}{{{x^2}}} \cr & \Rightarrow {x^2} = \left( {4.41 \times 4} \right) \cr & \Rightarrow x = 2.1 \times 2 \cr & \Rightarrow x = 4.2\,cm \cr} $$
46
A square and an equilateral triangle have equal perimeters. If the diagonal of the square is $$12\sqrt 2 $$ cm, then the area of the triangle is :
Discuss
Answer & Solution
Answer: Option D
Solution:
Let the side of the square be a cm
Then, its diagonal = $$\sqrt 2 $$ a cm
Now, $$\sqrt 2 $$ a = $$12\sqrt 2 $$
⇒ a = 12 cm
Perimeter of the square = 4a = 48 cm
Perimeter of the equilateral triangle = 48 cm
Each side of the triangle = 16 cm
Area of the triangle :
$$\eqalign{ & = \left( {\frac{{\sqrt 3 }}{4} \times 16 \times 16} \right)c{m^2} \cr & = \left( {64\sqrt 3 } \right)c{m^2} \cr} $$
47
Which of the following figures has the longest perimeter ?
Discuss
Answer & Solution
Answer: Option C
Solution:
(a) Perimeter = (4 × 10) cm = 40 cm
(b) Perimeter = 2(12 + 9) cm = 42 cm
(c) Perimeter = $$\left( {2 \times \frac{{22}}{7} \times 7} \right)$$   cm = 44 cm
(d) Perimeter = (4 × 9) cm = 36 cm
48
The perimeter of a circular field and a square field are equal. If the area of the square field is 12100 m2, the area of the circular field will be :
Discuss
Answer & Solution
Answer: Option C
Solution:
Side of the square field = $$\sqrt {12100} $$   m = 110 m
Perimeter of the circle field :
= Perimeter of the square field
= (4 × 110) m
= 440 m
$$\eqalign{ & 2\pi R = 440 \cr & \Rightarrow R = \frac{{440 \times 7}}{{2 \times 22}} = 70\,m \cr} $$
∴ Area of the circular field :
$$\eqalign{ & = \left( {\frac{{22}}{7} \times 70 \times 70} \right){m^2} \cr & = 15400\,{m^2} \cr} $$
49
A toothed wheel of diameter 50 cm is attached to a smaller wheel of diameter 30 cm. How many revolutions will the smaller wheel make the larger one makes 15 revolutions ?
Discuss
Answer & Solution
Answer: Option C
Solution:
Distance covered by smaller wheel in 1 revolution :
$$\eqalign{ & = \left( {2\pi \times 15} \right)cm \cr & = \left( {30\pi } \right)cm \cr} $$
Distance covered by larger wheel in 1 revolution :
$$\eqalign{ & = \left( {2\pi \times 25} \right)cm \cr & = \left( {50\pi } \right)cm \cr} $$
Let,
$$k \times 30\pi = 15 \times 50\pi $$
Then,
$$k = \left( {\frac{{15 \times 50\pi }}{{30\pi }}} \right) = 25$$
∴ Required number of revolution = 25
50
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)
Discuss
Answer & Solution
Answer: Option A
Solution:
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 :
$$\eqalign{ & = \left( {\frac{{22}}{7} \times \frac{{15}}{2}} \right)cm \cr & = \left( {\frac{{165}}{7}} \right)cm \cr & = 23.57\,cm \cr} $$