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Mensuration Test 10
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Mensuration Test 10
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  • Question 1/20
    1 / -0

    A rectangular box has dimensions 1.6 m × 1 m × 0.6 m. How many cubical boxes each of side 20 cm can be fit inside the rectangular box? 

    Solutions

    Explanation: 

    20 cm = 0.2 m

    So number of boxes that can fit = 1.6 × 1 × 0.6/0.2 × 0.2 × 0.2

     

  • Question 2/20
    1 / -0

    What is the volume of a cylinder whose curved surface area is 704 cm2 and height is 14 cm?

    Solutions

    Explanation: 

    2ᴨrh = 704, h = 14

    Solve both, so r = 8

    Volume = ᴨr2h

     

  • Question 3/20
    1 / -0

    If the volume of wire remains the same but its radius decreases to 1/3rd of previous, then the new length of the wire will be how many times of the previous length?

    Solutions

    Explanation: 

    Wire will be taken as cylinder.

    R = r/3

    ᴨr2h = ᴨR2H

    r2h = (r/3)2H

    so H = 9h

     

  • Question 4/20
    1 / -0

    A cone with diameter of its base as 12 cm is formed by melting a spherical ball of diameter 6 cm. What is the height of the cone? 

    Solutions

    Explanation: 

    Radius of cone = 12/2 = 6, radius of ball = 6/2 = 3

    Volumes will be equal, so

    (1/3)ᴨ r2h = (4/3)ᴨ R3

    62h = 4* 33

    So h = 3

     

  • Question 5/20
    1 / -0

    The total cost of painting the walls of a room is Rs 475. Find the cost of painting the walls of another room whose length, breadth and height each are double than the dimensions of the previous room.

    Solutions

    Explanation: 

    Area of first room = 2(l+b)*h

    After all dimensions doubled, new area = 2(2l+2b)*2h = 4[2(l+b)*h ] = 4 times previous area, so cost of painting = 4*475

     

  • Question 6/20
    1 / -0

    A circular wire of diameter 84 cm is bent into a rectangle with sides ratio 6 : 5. What are the respective sides of the rectangle? 

    Solutions

    Explanation: 

    Length of wire = 2ᴨr = 2(22/7)*42 = 264 cm

    Perimeter of rectangle = 2(6x+5x) = 264

    Solve, x = 12

    So dimensions – 12*6, 12*5

     

  • Question 7/20
    1 / -0

    The ratio of the outer and the inner perimeters of a circular path is 9 : 8. The path is 3 metres wide. What is the diameter of the outer circle? 

    Solutions

    Explanation: 

    2ᴨr/2ᴨR = 9/8

    So r/R = 9/8, so r = (9/8)R

    r-R = 3, so (9/8)R – R = 3

    R = 24, so r = (9/8)*24 = 27

     

  • Question 8/20
    1 / -0

    Four circular cardboard pieces, each of diameter 28 cm are placed in such a way that each piece touches two other pieces. Find the area of the space enclosed by the four pieces. ​​​​​​

    Solutions

    Explanation:

    Required area = area of square shown in figure – 4*areas of 1/4th parts of each circle, so

    Required area = 28*28 – 4*(1/4)*(22/7)*14*14

     

  • Question 9/20
    1 / -0

    A solid sphere of radius 20 cm is melted to form 12 spherical balls. Find the radius of each of the ball.

    Solutions

    Explanation: 

    Volume of each of the 8 ball = [(4/3)*(22/7)*203]/8

    So (4/3)*(22/7)*r3 = [(4/3)*(22/7)*203]/8

    Solve, r = 10

     

  • Question 10/20
    1 / -0

    The perimeter of a rectangular plot is 210 m. Find the cost of gardening 2 m broad boundary around rectangular plot whose perimeter The cost of gardening is Rs 14 per m2.

    Solutions

    Explanation: 

    Given 2(l+b) = 210

    2 m broad boundary means increase in l and b by 4 m

    So area of the boundary will be [(l+4)(b+4) – lb] = 4(l+b) + 16 = 2*[2(l+b)] + 16 = 2*210 + 16 = 436 m2

    So cost of gardening = 436*14

     

  • Question 11/20
    1 / -0

    Perimeter of a square and an equilateral triangle is equal. If the diagonal of the square is 10√2 cm, then find the area of equilateral triangle?

    Solutions

    Explanation :

    Diagonal of a square = a√2 = 10√2

    so a = 10, perimeter of square = 4*10 = 40 = 3x (x is the length of each side of triangle)

    x = 40/3, so are of equilateral triangle = √3/4*40/3*40/3 = (400√3)/9 cm^2

     

  • Question 12/20
    1 / -0

    Length of a rectangular field is increased by 10 meters and breadth is decreased by 4 meters, area of the field remains unchanged. If the length decreased by 5 meters and breadth is increased by 7 meters, again the area remains unchanged. Find the length and breadth of the rectangular field.

    Solutions

    Explanation :

    Length = l and breadth = b,

    (l +10)*(b-4) = lb and (l-5)*(b+7) = lb

    Solve both equation to get l and b

     

  • Question 13/20
    1 / -0

    If the length of the rectangle is increased by 20%, by what percent should the width be reduced to maintain the same area?

    Solutions

    Explanation :

    let length = 100 and breadth = 100

    now new length = 120 and let breadth = b

    so, 100*100 = 120*b

    b = 250/3, so % decrease = 100 – 250/3 = 50/3 = 16.67%

     

  • Question 14/20
    1 / -0

    A cone whose height is half of its radius is melted to from a hemi-sphere. Find the ratio of the radius of the hemi-sphere to that of cone.

    Solutions

    Explanation :

    volume will remains constant. So,

    V = 1/3*22/7*r^2*r/2 (volume of cone) and V = 2/3*22/7*R^3 (volume of hemi-sphere)

    So, R/r = 1:4

     

  • Question 15/20
    1 / -0

    Find the number of spherical balls of radius 1 cm that can be made from a cylinder of height 8 cm and diameter 14 cm?

    Solutions

    Explanation :

    (22/7)*7*7*8 = x*(4/3)*(22/7)*1^3 (x = number of spherical balls)

     

  • Question 16/20
    1 / -0

    A rectangle whose sides are in the ratio 6:5 is formed by bending a circular wire of radius 21cm. Find the difference between the length and breadth of the rectangle?

    Solutions

    Explanation :

    circumference of the wire = 2*(22/7)*21 = 22*6

    perimeter of rectangle = 2*11x = 22*6, so x= 6

    difference = 36 -30 = 6cm

     

  • Question 17/20
    1 / -0

    A right circular cone is exactly fitted inside a cube in such a way that the edges of the base of the cone are touching the edges of one face of the cube and the vertex is on the opposite face of the cube. If the volume of the cube is 512 cubic cm. find the approximate volume of the cone?

    Solutions

    Explanation :

    when cone is completely fitted inside the cube, then diameter of cone = side of cube and height of cone = height of cube

    so, volume = (1/3)*(22/7)*4*4*8 = 134 (approx)

     

  • Question 18/20
    1 / -0

    if the radius of a cylinder is doubled and height is halved, what is the ratio between the new volume and the previous volume?

    Solutions

    Explanation :

    New volume = (22/7)*4r^2*h/2 and old volume = (22/7)*r^2*h

    so ratio = 2:1

     

  • Question 19/20
    1 / -0

    A cone of radius 12 cm and height 5 cm is mounted on a cylinder of radius 12 cm and height 19 cm. Find the total surface area of the figure thus formed?

    Solutions

    Explanation :

    total surface area = curved surface area of cone + curved surface area of cylinder + base area

     = (22/7)*12*13 + (22/7)*12*19 + (22/7)*12*12 = 2376 cm^2

     

  • Question 20/20
    1 / -0

    A rectangular garden is 30 meter long and 20 meter broad. It has 6 meter wide pavements all around it both on its inside and outside. Find the total area of pavements?

    Solutions

    Explanation :

    Required area = 42*32 – 18*8 = 1200

     

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