## Problems on Areas Question and Answers : Quantitative Aptitude

- 1. The sides of a rectangle are in the ratio of 6 : 5 and its area is 1331 sq.m. Find the perimeter of rectangle.

Answer: Option 'C'

Let 6x and 5x be sides of the rectangle

6x × 5x = 1331

11x^{2} = 1331

x^{2} = 1331/11 = 121 => x = 11

Length = 6x = 6 × 11 = 66

Breadth = 5x => 5 × 11 = 55

Perimeter = 2 (l+b) => 2 (66+55) = 121 m

- 2. A rectangle measures 8 Cm on length and its diagonal measures 17 Cm. What is the perimeter of the rectangle?

Answer: Option 'C'

Second side = √17^{2}-8^{2}

= √289-64

= 15 cm

Perimeter = 2 (l+b) = 2(8+5) Cm = 2(23) = 46 Cm

- 3. A rectangular courtyard 3.78 m lang and 5.25 m broad is to be paved exactly with square tiles, all of the same size. The minimum number of such tiles is:

Answer: Option 'B'

l = 378 Cm and b = 525 Cm

Maximum length of a square tile

= HCF of (378,525) = 21 Cm

Number of tiles = (378×525)/(21×21) = (18×25) = 450

- 4. The ratio of the areas of two squares, one having double its diagonal then the other is:

Answer: Option 'D'

Lenth of the diagonals be 2x and x units.

areas are 1/2 × (2x)^{2} and (1/2 × x^{2})

Required ratio = 1/2 × 4 x^{2} : 1/2 x^{2} = 4 : 1

- 5. A rectangular mat has an area of 120 sq.metres and perimeter of 46 m. The length of its diagonal is:

^{2}

Answer: Option 'B'

rectangular area = l × b = 120 and perimeter = 2(l+b) = 46

l+b = 23

(l-b)^{2} - 4lb = (23)^{2} - 4 × 120 = (529-480) = 49 = l-b = 7

l+b = 23, l-b = 7, we get l = 15, b = 8

Diagonal = √15^{2}+8^{2}

√225+64 => √289 = 17 m

- 6. The length of a room is 6 m and width is 4.75 m. What is the cost of paying the floor by slabs at the rate of Rs. 900 per sq. metre.

Answer: Option 'C'

Area = 6 × 4.75 sq. metre.

Cost for 1 sq. metre. = Rs. 900

Hence total cost = 6 × 4.75 × 900 = 6 × 4275 = Rs. 25650

- 7. The area of a rectangle plot is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the plot?

Answer: Option 'A'

Answer: Option 'A'

lb = 460 m^{2}...(Equation 1)

Let the breadth = b

Then length, l = b × (100+15)/100 = 115b/100 ...(Equation 2)

From Equation 1 and Equation 2,

115b/100 × b = 460

b2 = 46000/115 = 400

⇒ b = √400 = 20 m

- 8. A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, find out the area of the parking space in square feet? a

Answer: Option 'A'

Let l = 9 ft.

Then l + 2b = 37

=> 2b = 37 - l = 37 - 9 = 28

=> b = 28/2 = 14 ft

Area = lb = 9 × 14 = 126 sq. ft.

- 9. A rectangular field has 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?

Answer: Option 'B'

Given that area of the field = 680 sq. feet

=> lb = 680 sq. feet

Length(l) = 20 feet

=> 20 × b = 680

=> b = 680/20 = 34 feet

Required length of the fencing = l + 2b = 20 + (2 × 34) = 88 feet

- 10. A large field of 900 hectares is divided into two parts. The difference of the areas of the two parts is one-fifth of the average of the two areas. What is the area of the smaller part in hectares?

Answer: Option 'C'

Let the areas of the parts be x hectares and (900 - x) hectares.

Difference of the areas of the two parts = x - (900 - x) = 2x - 900

one-fifth of the average of the two areas = 1/5[x+(900−x)]/2

= 1/5 × (900/2) = 450/5 = 90

Given that difference of the areas of the two parts = one-fifth of the average of the two areas

=> 2x - 900 = 90

=> 2x = 990

⇒ x = 990/2 = 495

Hence, area of smaller part = (900 - x) = (900 – 495) = 405 hectares.

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