Questions on Area - Quantitative Aptitude

[Q] Find the percentage of error in the calculated area of the square if an error 2% in excess is made while measuring the side of  it. 
(a) 3.04%
(b) 2.04%
(c) 4.04%
(d) 5.04%

Solution

Answer   (c)
Explanation
---- 

Error = 2%
Let the correct value of the side of the square = 100

Then, the measured value = 100×(100+2)100=102 (∵ error 2% in excess)

Correct area of the square = 100 × 100 = 10000

Calculated area of the square = 102 × 102 = 10404

Error = 10404 - 10000 = 404

Percentage Error = ErrorActual Value×100=40410000×100=4.04%

[Q] 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. The area of the lawn is 2109 sq. m. Find the width of the road?
(a) 4
(b) 5
(c) 3
(d) 6

Solution

Answer   (c)
Explanation
---- 
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.

[Q] When a towel is bleached, loses 20% of its length and 10% of its breadth. Find the percentage decrease in area?
(a) 18&
(b) 28%
(c) 38%
(d) 48%

Solution

Answer   (b)
Explanation
---- 
Let original length and breadth = 100.
Then, original area = 100 × 100 = 10000

After losing 20% of length⇒New length = Original length×(100−20)100=100×80100=80After losing 10% of breadth⇒New breadth= Original breadth×(100−10)100=100×90100=90

New area = 80  × 90 = 7200



Decrease in area = Original Area - New Area = 10000 - 7200 = 2800

Percentage decrease in area = Decrease in AreaOriginal Area×100=280010000×100=28%

[Q] Find 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) 841
(b) 414
(c) 541
(d) 814

Solution

Answer   (d)
Explanation
---- 

Hence, HCF of 1517 and 902 = 41



Hence, side length of largest square tile we can take = 41 cm

Area of each square tile = 41 × 41 cm2

Number of tiles required = 1517×90241×41=37×22=407×2=814

[Q] A rectangular plot measuring 90 metres by 50 metres is to be enclosed by wire fencing such that poles of the fence are 5 metres apart. Find the number of poles.
(a) 56
(b) 45
(c) 36
(d) 65

Solution

Answer   (a)
Explanation
---- 

perimeter of the wire fencing =  = 2(90 + 50) = 280 metres
Two poles are 5 metres apart.

Hence number of poles required = ²⁸⁰⁄₅ = 56