Area Aptitude Solved Examples - Set 01 - ObjectiveBooks
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# Area Aptitude Solved Examples - Set 01

Question No. 01
A rectangular lawn is 80 meters long and 60 meters wide. The time taken by a man to walk along its diagonal at the speed of 18 km per hour is
(A) 25 sec.
(B) 20 sec.
(C) 18 sec.
(D) 10 sec.
Explanation:
Diagonal, d = √(80² + 40²) = √10000 = 100
Speed, v = 18 km/hr = 18 × (5/18) m/sec = 5 m/sec
∴ Time taken, t = d/v = 100/5 = 20 sec.

Question No. 02
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
Explanation:
Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm
Area of each tile = (41 × 41) cm2
∴ Required number of tiles = (1517 × 902)/ (41 × 41) = 814

Question No. 03
A tank is 25 m long, 12 m wide and 6 m deep. The cost of plastering its walls and bottom at 75 paisa per sq. m is:
(A) Rs. 456
(B) Rs. 458
(C) Rs. 558
(D) Rs. 568
Explanation:
Area to be plastered = = [2(l + b) × h] + (l × b)
= [{2(25 + 12) × 6} + (25 × 12)] m2
= (444 + 300) m2
= 744 m2.
∴ Cost of plastering = Rs. 744 × (75/100) = Rs. 558

Question No. 04
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, then the area of the park (in sq. m) is:
(A) 15,360
(B) 1,53,600
(C) 30,720
(D) 3,07,200
Explanation:
Perimeter = Distance covered in 8 min. = (12,000/60) × 8 m = 1600 m.
Let length = 3x meters and breadth = 2x meters.
Then, 2(3x + 2x) = 1600 or x = 160
∴ Length = 480 m and Breadth = 320 m.
∴ Area = (480 × 320) m2 = 153600 m2.

Question No. 05
The length of a rectangular plot is 20 meters more than its breadth. If the cost of fencing the plot @ 26.50 per meter is Rs. 5300, what is the length of the plot in meters?
(A) 40
(B) 50
(C) 120
(D) None of these
Explanation:
Then, length = (x + 20) meters.
Perimeter = (5300/26.50) m = 200 m.
∴ 2[(x + 20) + x] = 200
=> 2x + 20 = 100
=> 2x = 80
=> x = 40
Hence, length = x + 20 = 60 m.

Question No. 06
An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is:
(A) 2%
(B) 2.02%
(C) 4%
(D) 4.04%
Explanation:
100 cm is read as 102 cm.
∴ A1 = (100 x 100) cm2 and A2 (102 x 102) cm2.
(A2 - A1) = [(102)2 - (100)2]
= (102 + 100) x (102 - 100)
= 404 cm2.
∴ Percentage error = [{404/(100 × 100)} × 100] % = 4.04%

Question No. 07
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?
(A) 34
(B) 40
(C) 68
(D) 88
Explanation:
We have: l = 20 ft and lb = 680 sq. ft.
So, b = 34 ft.
∴ Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.

Question No. 08
The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle?
(A) 16 cm
(B) 18 cm
(C) 24 cm
(D) None of these
Explanation:
According to question; [2(l + b)]/b = 5/1
=> 2l + 2b = 5b
=> 3b = 2l
∴ b = 2l/3
Then, Area = 216 cm2
=> l × b = 216
=> l × (2l/3) = 216
=> l² = 324
=> l = 18 cm.

Question No. 09
What is the total surface area of a right circular cone of height 14 cm and base radius 7 cm?
(A) 344.35 cm2
(B) 462 cm2
(C) 498.35 cm2
(D) None of these
Explanation:
h = 14 cm, r = 7 cm.
So, l = √[(7)2 + (14)2]= √245 = 7√5 cm.
∴ Total surface area = π rl + π r2
= [{(22/7) × 7 × 7√5} + {(22/7) × 7 × 7}] cm2
= [154(√5 + 1)] cm2
= (154 × 3.236) cm2
= 498.35 cm2.

Question No. 10
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. If the area of the lawn is 2109 sq. m, then what is the width of the road?
(A) 2.91 m
(B) 3 m
(C) 5.82 m
(D) None of these
Explanation:
Area of the park = (60 × 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 meters.
Then, 60x + 40x - x2 = 291
=> x2 - 100x + 291 = 0
=> (x - 97)(x - 3) = 0
=> x = 3
∴ The width of the road is 3 meters.

Area of plane Surfaces:
Formula:                Area Aptitude Formulas
Solved Examples:  Solved Examples: Set 01
Practice Test:         Practice Test: 01
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