Problems on Areas Question and Answers : Quantitative Aptitude

Mensuration : Quantitative Aptitude
  • The sides of a rectangle are in the ratio of 6 : 5 and its area is 1331 sq.m. Find the perimeter

    of rectangle.

    A.) 120 m B.) 121 Cm
    C.) 121 m D.) None of these
    Answer: Option 'C'

    Let 6x and 5x be sides of the rectangle

    6x × 5x = 1331

    11x2 = 1331

    x2 = 1331/11 = 121 => x = 11

    Length = 6x = 6 × 11 = 66

    Breadth = 5x => 5 × 11 = 55

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



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

    A.) 17 Cm B.) 17 m
    C.) 17 m2 D.) 16 m
    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 = √152+82

    √225+64 => √289 = 17 m



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

    A.) 46 Cm B.) 48 Cm
    C.) 46 m D.) None of these
    Answer: Option 'A'

    Second side = √172-82

    = √289-64

    = 15 Cm

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



  • 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:

    A.) 350 B.) 450
    C.) 450 D.) 495
    Answer: Option 'C'

    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



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

    A.) 1 : 3 B.) 3 : 1
    C.) 1 : 4 D.) 4 : 1
    Answer: Option 'D'

    Lenth of the diagonals be 2x and x units.

    areas are 1/2 × (2x)2 and (1/2 × x2)

    Required ratio = 1/2 × 4 x2 : 1/2 x2 = 4 : 1



  • 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.

    A.) Rs. 25660 B.) Rs. 25560
    C.) Rs. 25650 D.) Rs. 26550
    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



  • 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?

    A.) 20 metres B.) 25 metres
    C.) 27 metres D.) 18 metres
    Answer: Option 'A'

    lb = 460 m2...(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



  • 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.) 128 sq. ft. B.) 126 sq. ft.
    C.) 136 sq. ft. D.) 116 sq. ft.
    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.



  • 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?

    A.) 98 B.) 88
    C.) 99 D.) 89
    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



  • 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?

    A.) 395 B.) 385
    C.) 415 D.) 405
    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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