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Mensuration Test 2
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Mensuration Test 2
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  • Question 1/10
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    The area of a rectangular field is 144 m². If the length had been 6 metres more, the area would have been 54 m² more. The original length of the field is

    Solutions

    Let the length and breadth of the original rectangular field be x m and y m respectively.

    Area of the original field

    = x × y = 144 m²

    ∴  … (i)

    If the length had been 6 m more, then area will be

    (x + 6) y = 144 + 54

    ⇒ (x + 6) y = 198 … (ii)

    Putting the value of x from eq (i) in eq (ii), we get

    ⇒ 144 + 6y = 198

    ⇒ 6y = 54 ⇒ y = 9 m

    Putting the value of y in eq (i) we get x = 16 m

  • Question 2/10
    1 / -0

    The length and breadth of a playground are 36 m and 21 m respectively. Poles are required to be fixed all along the boundary at a distance 3 m apart. The number of poles required will be

    Solutions

    Given, playground is rectangular.

    Length = 36 m, Breadth = 21 m

    Now, perimeter of playground

    = 2( 21 + 36) = 114

    Now, poles are fixed along the boundary at a distance 3m.

    ∴ Required no. of poles = ¹¹⁴⁄₃ = 38

  • Question 3/10
    1 / -0

    The cost of carpeting a room 18 m long with a carpet 75 cm wide at Rs 4.50 per metre is Rs 810. The breadth of the room is:

    Solutions

    Length of the carpet

    Area of the room = Area of the carpet

    ∴ Breadth of the room

    = 7.5 m.

  • Question 4/10
    1 / -0

    If the perimeter and diagonal of a rectangle are 14 and 5 cms respectively, find its area.

    Solutions

    In a rectangle,

    ⇒ 

    49 = 25 + 2 × area

    ​​​​​​​

  • Question 5/10
    1 / -0

    A square carpet with an area 169 m² must have 2 metres cut-off one of its edges in order to be a perfect fit for a rectangular room. What is the area of rectangular room?

    Solutions

    Side of square carpet

    After cutting of one side,

    Measure of one side

    = 13 – 2 = 11 m

    and other side = 13 m (remain same)

    ∴ Area of rectangular room

    = 13 × 11 = 143 m²

  • Question 6/10
    1 / -0

    A picture 30 inch × 20 inch has a frame 2½ inch wide. The area of the picture is approximately how many times the area of the frame?

    Solutions

    Length of frame

    = 30 + 2.5 × 2 = 35 inch

    Breadth of frame

    = 20 + 2.5 × 2 = 25 inch

    Now, area of picture

    = 30 × 20 = 600 sq. inch

    Area of frame

    = (35 × 2.5) + (25 × 2.5) = 150

  • Question 7/10
    1 / -0

    The floor of a rectangular room is 15 m long and 12 m wide. The room is surrounded by a verandah of width 2 m on all its sides. The area of the verandah is:

    Solutions

    Area of the outer rectangle = 19 × 16 = 304 m²

    Area of the inner rectangle

    = 15 × 12 = 180 m²

    Required area

    = (304 – 180) = 124 m²

  • Question 8/10
    1 / -0

    A typist uses a paper 12 inch by 5 inch length wise and leaves a margin of 1 inch at the top and the bottom and a margin of ½ inch on either side. What fractional part of the paper is available to him for typing?

    Solutions

    Area of paper

    = 12 × 5 = 60 sq. inch

    Area of typing part

    = (12 – 2) × (5 – 1)

    = (10 × 4) sq. inch

    ∴ Required fraction

  • Question 9/10
    1 / -0

    A rectangular lawn 70 m × 30 m has two roads each 5 metres wide, running in the middle of it, one parallel to the length and the other parallel to the breadth. Find the cost of gravelling the road at the rate of Rs 4 per square metre.

    Solutions

    Total area of road

    = Area of road which parallel to length + Area of road which parallel to breadth – overlapped road

    = 70 × 5 + 30 × 5 – 5 × 5

    = 350 + 150 – 25

    = 500 – 25 = 475 m²

    ∴ Cost of gravelling the road

    = 475 × 4 = Rs 1900

  • Question 10/10
    1 / -0

    The length and breadth of the floor of the room are 20 feet and 10 feet respectively. Square tiles of 2 feet length of different colours are to be laid on the floor. Black tiles are laid in the first row on all sides. If white tiles are laid in the one-third of the remaining and blue tiles in the rest, how many blue tiles will be there?

    Solutions

    Area left after laying black tiles

    = [(20 – 4) × (10 – 4)] sq. ft.

    = 96 sq. ft.

    Area under white tiles

     sq. ft

    = 32 sq. ft.

    Area under blue tiles

    = (96 – 32) sq. ft = 64 sq. ft.

    Number of blue tiles

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