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Area Aptitude Test Paper 4

Area Aptitude Test Paper 4

16) The difference between the length and the perimeter of a rectangle is 100 cms. What is the breadth of the rectangle?

  1. 80 cms
  2. 60 cms
  3. 100 cms
  4. Data Inadequate

Answer: D

Explanation:

Let the length of the rectangle be ‘x’ and breadth of the rectangle be ‘y’

According to the question:

2(x + y) – x = 100

2x + 2y – x = 100

x + 2y = 100

From this we cannot find ‘y’ (breadth), so the given data is inadequate.


17) The sides of a rhombus are 10 cm in length, and one diagonal is 16 cm. The area of the rhombus is

  1. 96 cm2
  2. 95 cm2
  3. 94 cm2
  4. 93 cm

Answer: A

Explanation:

When d1 and d2 are the diagonals of rhombus then,
The Side of rhombus = Area Aptitude

Area Aptitude

Squaring both sides

202 = 162 + d22

400 = 256 + d22

d22 = 400 – 256

d2 = √144

d2 = 12

Area of rhombus =1/2( d1 × d2 )

=1/2 ( 16 × 12 )=96 cm2


18) If each side of the square is increased by 50%, what will be the ratio between the new area and the original area of the square?

  1. 5: 4
  2. 9: 4
  3. 4: 5
  4. 4: 9

Answer: B

Explanation:

Area Aptitude


19) The sides of a rectangular field are in the ratio 3: 4 and its area is 7500 m2. What is the cost of fencing it at 25 paise per meter?

  1. Rs. 87.50
  2. Rs. 77.50
  3. Rs. 67.50
  4. Rs. 57.50

Answer: A

Explanation:

Ratio between the sides of rectangle = 3: 4

Let the ratio constant be x then,

Length = 3x and breadth = 4x

Area = L x B

7500=3x × 4x=12x2
Area Aptitude= x2=625

x=25

Length = 3 x 25 = 75 m, and Breadth = 4 x 25 = 100 m

Perimeter = 2(75 + 100) = 2 x 175 = 350 m

Cost of fencing 1 meter = 25 paise

Cost of fencing 350 m = 350 x 25 = 8750 paise

In rupees: Rs. 87.50


20) If the size of a tile is 9″ by 9″, how many tiles are required to cover a 12 ft. wide and 18 ft. long floor?

  1. 32
  2. 384
  3. 216
  4. 24

Answer: B

Explanation:

Area Aptitude

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