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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-2 section-1

Each section contains 100 MCQs on Area .
11
The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:
Discuss
Answer & Solution
Answer: Option D
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
12
The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area?
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & {\text{Let}}\,{\text{original}}\,{\text{length}} = x\,{\text{and}}\, \cr & {\text{Original}}\,{\text{breadth}} = y \cr & {\text{Original}}\,{\text{area}} = xy \cr & {\text{New}}\,{\text{length}} = \frac{x}{2} \cr & {\text{New}}\,{\text{breadth}} = 3y \cr & {\text{New}}\,{\text{area}} = \left( {\frac{x}{2} \times 3y} \right) = \frac{3}{2}xy \cr & \therefore {\text{Increase}}\,\% = {\frac{1}{2}xy \times \frac{1}{{xy}} \times 100} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 50\% \cr} $$
13
The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ 26.50 per metre is Rs. 5300, what is the length of the plot in metres?
Discuss
Answer & Solution
Answer: Option E
Solution:
$$\eqalign{ & {\text{Let}}\,{\text{breadth}} = x\,{\text{metres}} \cr & {\text{Then,}}\,{\text{length}} = \,\left( {x + 20} \right)\,{\text{metres}} \cr & {\text{Perimeter}} = {\frac{{5300}}{{26.50}}} m = 200m \cr & \therefore 2\left[ {\left( {x + 20} \right) + x} \right] = 200 \cr & \Rightarrow 2x + 20 = 100 \cr & \Rightarrow 2x = 80 \cr & \Rightarrow x = 40 \cr & {\text{Hence,}}\,{\text{length}} = x + 20 = 60m \cr} $$
14
A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?
Discuss
Answer & Solution
Answer: Option D
Solution:
We have: l = 20 ft and lb = 680 sq. ft.
So, b = 34 ft.
∴ Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.
15
A tank is 25 m long, 12 m wide and 6 m deep. The cost of plastering its walls and bottom at 75 paise per sq. m, is:
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & {\text{Area}}\,{\text{to}}\,{\text{be}}\,{\text{plastered}} \cr & = \left[ {2\left( {l + b} \right) \times h} \right] + \left( {l \times b} \right) \cr & = \left\{ {\left[ {2\left( {25 + 12} \right) \times 6} \right] + \left( {25 \times 12} \right)} \right\}{m^2} \cr & = \left( {444 + 300} \right){m^2} \cr & = 744{m^2} \cr & \therefore {\text{Cost}}\,{\text{of}}\,{\text{Plastering}} \cr & = Rs.\,\left( {744 \times \frac{{75}}{{100}}} \right) \cr & = Rs.\,558 \cr} $$
16
The length of a room is 5.5 m and width is 3.75 m. Find the cost of paving the floor by slabs at the rate of Rs. 800 per square metre.
Discuss
Answer & Solution
Answer: Option D
Solution:
Area of the floor
= (5.5 × 3.75)m2
= 20.625m2
∴ Cost of paying
= Rs. (800 × 20.625)
= Rs. 16500
17
An artist has completed one-fourth of a rectangular oil painting. When he paint another 100 square centimetres of the painting, he would complete three-quarters of the painting. If the height of the oil painting is 10 cm, determine the length (in cm) of the oil panting.
Discuss
Answer & Solution
Answer: Option C
Solution:
Let the area of the whole painting be x cm2
Then,
$$\eqalign{ & \Rightarrow \frac{1}{4}x + 100 = \frac{3}{4}x \cr & \Rightarrow \frac{1}{2}x = 100 \cr & \Rightarrow x = 200 \cr} $$
∴ Area of painting = 200 cm2
And, Height = 10 cm
Length of painting :
$$\eqalign{ & = \left( {\frac{{200}}{{10}}} \right)cm \cr & = 20\,\,cm \cr} $$
18
The length of a rectangle is increased by 60%. By what percent would the width have to be decreased so as to maintain the same area ?
Discuss
Answer & Solution
Answer: Option A
Solution:
Let original length = x and original breadth = y
Then, original area = xy
New length :
$$\eqalign{ & = \frac{{160x}}{{100}} \cr & = \frac{{8x}}{5} \cr} $$
Let the new breadth = z
Then,
$$\eqalign{ & \Rightarrow \frac{{8x}}{5} \times z = xy \cr & \Rightarrow z = \frac{{5y}}{8} \cr} $$
∴ Decrease in breadth :
$$\eqalign{ & = \left( {\frac{{3y}}{8} \times \frac{1}{y} \times 100} \right)\% \cr & = 37\frac{1}{2}\% \cr} $$
19
The following squares represent the monthly income of two families
Area mcq question image
If the monthly income of family A is Rs. 40000, the monthly income of family B is ?
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\frac{{{\text{Monthly income of family A}}}}{{{\text{Monthly income of family B}}}}$$       $$ = \frac{{{\text{Area of square A}}}}{{{\text{Area of square B}}}}$$
$$\eqalign{ & \Rightarrow \frac{{40000}}{x} = \frac{{{2^2}}}{{{3^2}}} \cr & \Rightarrow x = \left( {\frac{{40000 \times 9}}{4}} \right) \cr & \Rightarrow x = 90000 \cr} $$
20
What will be the length of the diagonal of that square plot whose area is equal to the area of a rectangular plot of length 45 metres and breadth 40 metres ?
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & {\text{Area :}} \cr & = \left( {45 \times 40} \right){m^2} \Leftrightarrow \frac{1}{2} \times {\left( {{\text{Diagonal}}} \right)^2} \cr & = 1800\,\,\,\, \Leftrightarrow \,\,\,\,{\text{Diagonal}} \cr & = 60\,\,m \cr} $$