Question 1 1 / -0
Two cuboids, each of breadth 3 cm, length 6 cm and height 6 cm, are joined to form a cube. What is the total surface area of the resultant cube?
Solution
When two cuboids are joined to form a cube, then side of the cube becomes 6 cm each.
Total surface area of the resultant cube = 6
(Side)
2 = 6
(6)
2 cm
2 = 216 cm
2
Question 2 1 / -0
If the volume of a cube is (12)3 m3 , then the total surface area of the cube is equal to
Solution
Volume of a cube = a
3 , where 'a' is the edge of the cube.
Given, volume of the cube = (12)
3 m
3 a
3 = (12)
3 m
3 Taking cube root, we get
a = 12 m
Now, total surface area of the cube = 6a
2 = 6(12)
2 = 6
12
12
= 6
144
= 864 m
2
Question 3 1 / -0
If the radius of a hemisphere is 3 cm, then its total surface area is
Solution
Given:
Radius of the hemisphere (r) = 3 cm
Total surface area of the hemisphere = 3
r
2 = 3
(3)
2 cm
2 = 3
3
3 cm
2 = 27
cm
2
Question 4 1 / -0
The length and the breadth of a cuboidal box are 2 m and 0.5 m, respectively. If the lateral surface area of the box is 7.5 m2 , its height is
Solution
Lateral surface area of the cuboid = 2h(
+ b)
h = Height of the cuboid
b = Breadth of the cuboid
= Length of the cuboid
Given:
= 2 m, b = 0.5 m, h = ?
Lateral surface area of the cuboid = 7.5 m
2 (Given)
Therefore, 7.5 = 2h(2 + 0.5)
= h
h =
=
= 1.5
Thus, the height of the box is 1.5 m.
Hence, answer option 2 is correct.
Question 5 1 / -0
The height of a right circular cone, whose base radius is 5 cm and volume is 25
cm
3 , is
Solution
Volume of a cone =
where r = radius of the cone
h = height of the cone
According to the question: 25
=
25
=
h = 3 cm
Question 6 1 / -0
If V, p and q are the volume, base radius and height of a right circular cylinder, respectively, then which of the following relationships is true?
Solution
Volume of the cylinder =
r
2 h …. (1)
Given, volume = V
Radius = p
Height = q
Substituting the values in (1), we get
V =
p
2 q
Question 7 1 / -0
A hemisphere of diameter d is surmounted by a cone of radius
(as shown in the figure) unit. If the height of the figure is twice the height of the hemisphere, which of the following expressions gives the volume of the figure?
Solution
Radius of the hemisphere (r) =
Height of the given figure = 2(height of the hemisphere) = 2
= h
1 = d
So, height of the cone = h = d - r =
Thus, volume of the given figure = volume of the hemisphere + volume of the cone
=
r
3 +
r
2 h
=
Question 8 1 / -0
A cone of height 15 cm and diameter 7 cm is mounted on the base of a hemisphere of same radius. Find the volume of the combination so formed.
Solution
Height of the cone (h) = 15 cm
Diameter of the cone = 7 cm
Radius of the cone =
cm
Radius of the hemisphere =
cm
Volume of the combination = Volume of the cone + Volume of the hemisphere
=
r
2 h +
r
3 =
r
2 [h + 2r]
=
cm
3 = 282.33 cm
3
Question 9 1 / -0
A rectangular cuboid block of dimensions 18 cm x 12 cm x 9 cm is cut into an exact number of equal cubes. Find the possible number of cubes.
Solution
Dimensions of the block = 18 cm x 12 cm x 9 cm
Question 10 1 / -0
A sphere of copper, whose diameter is 18 cm, is melted and converted into a wire of diameter 4 mm. The length of the wire (in metres) is
Solution
Volume of sphere = Volume of wire (cylinder)
× (9)
3 =
×
× h
h = 24300 cm = 243 m
Question 11 1 / -0
A hollow iron pipe is 42 cm long and its inner radius is 7 cm. If the thickness of the pipe is 1.5 cm and iron weights 10 gm/cm3 , find the mass of the pipe (in kg).
Solution
External radius = 7 + 1.5 = 8.5 cm
Internal radius = 7 cm.
Volume of iron =
× ((R × R) - (r × r))h = 22/7 × ((8.5 × 8.5) - (7 × 7)) × 42 = 3069 cm
3 Mass of iron pipe = (3069 cm
3 × 10 gm/cm
3 )
= 30690 gm
= 30.69 kg
Question 12 1 / -0
Water is pumped out through a cylindrical pipe of internal diameter 28 cm. How much water is pumped out in one hour if the flow velocity of water is 4 cm/s?
Solution
Volume of water pumped out in a second =
(r = internal radius of the pipe, h = distance covered by water in one second)
=
cm
3 = 2464 cm
3 Volume of water pumped out in an hour = 2464 × 3600 cm
3 = 88,70,400 cm
3
Question 13 1 / -0
The water level in a cylindrical beaker, having a base radius 10 cm, is at a height of 14 cm from the base. If a metallic sphere of radius 3 cm is put into this beaker, what will be the rise in the water level?
Solution
Let the water rise to a level of h.
Then, volume of sphere = volume of cylinder of height h and radius 10 cm – volume of water initially
4
9 = 100 h – 14
100
36 = 100 h – 1400
h =
h = 14.36 cm
Rise in water level = (14.36 – 14) cm = 0.36 cm
Question 14 1 / -0
A tank 12 m long and 8 m wide contains water up to a height of 2.5 m. The total area of the wet surface is
Solution
Area of the wet surface = [2(lb + bh + lh) - lb] = 2(bh + lh) + lb = [2 (8 x 2.5 + 12 x 2.5) + 12 x 8] m2 = 196 m2
Question 15 1 / -0
If the dimensions of a cuboid are increased by 15%, 10% and 5%, respectively. Find the approximate percentage increase in its volume.
Solution
Let the length of the cuboid = x units, breadth = y units and height = z units
Then, volume = xyz cubic units
After increment in the dimensions,
Volume =
cubic units
Total increase in volume = 0.32825xyz cubic units
Percentage increase = 0.32825xyz × 100/xyz = 32.825 = 33% (approximately)
Question 16 1 / -0
A wooden box measures externally 4 m x 3 m x 2 m. The thickness of all four sides is 5 cm. If the capacity of the box is 18.096 m3 , find the thickness of the bottom.
Solution
Let the thickness of the bottom = x m
Question 17 1 / -0
A cylinder is formed by folding a rectangular tin sheet along its length, and the dimensions of the sheet are 44 cm by 20 cm. What is the curved surface area of the cylinder?
Solution
As the rectangular sheet has been folded along its length, circumference of base of the cylinder = 2
× r = 44 cm and height of the cylinder = 20 cm.
Now, curved surface area of the cylinder = 2
× r × h = 44 × 20 = 880 cm
2
Question 18 1 / -0
The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume is
of the volume of the cone, the height at which the section is made is
Solution
Let height and radius of big cone be H and R, respectively.
Let height and radius of small cone be h and r, respectively.
We know,
Volume of big cone, V =
Volume of small cone,
Dividing, we get
(Since volume of smaller cone is
of big cone)
R
2 H = 8r
2 h
30R
2 = 8r
2 h [
H = 30 cm]
... (2)
From (1) and (2),
... (3)
Substituting equation (3) in (1), we get
h =
h = 30
h = 15 cm
So, option (3) is correct.
Question 19 1 / -0
A sphere of diameter 6 cm is dropped into a right circular cylindrical vessel, partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, then by how much level will water rise in the cylindrical vessel?
Solution
Radius of sphere = 6/2 = 3 cm
Radius of cylinder = 12/2 = 6 cm
Volume of sphere = (4/3)
r
3 = 4/3
3
3
3 = 36
cm
3 r
2 H -
r
2 h = 36
H - h =
= 1 cm
Question 20 1 / -0
A circus tent is in the form of a cone over a cylinder. The diameter of the base is 6m, the height of cylindrical part is 15 m and the total height of the tent is 19 m. The canvas required for the tent is
Solution
Height of cylinder (H) = 15 m
Height of cone 'h' = 19 - 15 = 4 m
Radius =
= 3 m
Slant height of cone =
= 5 m
Total canvas required = CSA of cone + CSA of cylinder
=
r
+ 2
rH
=
r(5 + 2
15)
=
= 330 sq. m
Question 21 1 / -0
The external dimensions of an open box are 25 cm (length), 20 cm (breadth) and 12 cm (height). Its thickness is 2 cm. If 1 cm3 of metal used in the box weighs 0.5 gm, find the weight of empty box.
Solution
l
e = 25 cm, b
e = 20 cm, h
e = 12 cm
l
i = 25 cm - (2 x 2) cm = 21 cm; b
i = 20 cm - (2 x 2) cm = 16 cm; h
i = 12 cm - (2 x 1) cm = 10 cm
External volume = (25 x 20 x 12) cm
3 = 6000 cm
3 Internal volume = (21 x 16 x 10) cm
3 = 3360 cm
3 Volume of metal = (6000 - 3360) cm
3 = 2640 cm
3 Weight of empty box = 2640 x
kg
Weight = 1.32 kg
Hence, answer option 2 is correct.
Question 22 1 / -0
A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of the cylinder is 24 m. The height of the cylindrical portion is 11 m, while the vertex of the cone is 16 m above the ground. What is the area of the curved surface for the conical portion?
Solution
Radius of cone = 12 m
Height of cone = 5 m
Hence, slant height =
m
= 13 m
Curved surface area of cone =
=
m
2 =
Question 23 1 / -0
Directions: Statement II: If the length of a cuboid is increased by 25%, the width is decreased by 75% and the height is decreased by 20%, then the volume decreases by 75%.
Solution
Statement I: Let the radius of the cone be r and the height be h.
New radius (R) =
=
New height (H) =
Volume (V1) =
New volume (V2) =
Change in volume:
=
=
=
=
= -80%
Therefore, the volume decreases by 80%.
Hence, Statement I is incorrect.
Statement II: Let the length of the cuboid be l, the width be w and the height be h.
New length (L) =
=
=
New width (W) =
=
=
New height (H) =
=
=
Volume (V) = lwh
New volume (V1) = LWH
Change in volume:
=
=
=
= 75%
Therefore, the volume increases by 75%.
Hence, Statement II is incorrect.
Hence, option 1 is the correct answer.
Question 24 1 / -0
The total cost of white-washing the four walls of a room is Rs. 6520. The length, width and height of the room are 20 m, 16 m and 25 m, respectively. Find the approximate values of the following:2 cost of white-washingthe four walls2 is ignored
Solution
(a)
Area of four walls = 2(l + b)h
= 2(20 + 16) × 25
= 1800 m
2 Now, per m
2 cost of white-washing the four walls = Total cost/Total area = Rs.
= Rs. 3.62
(b) Area of doors and windows = 34 m
2 Area of four walls = (1800 - 34) m
2 = 1766 m
2 (Ignoring the area of windows and doors)
Therefore, cost required for white-washing = 1766 x 3.62 = Rs. 6392.92
Question 25 1 / -0
Match the following:
Column - I Column - II (P) The curved surface area of a right circular cone, whose base radius is 13 cm and slant height is 15 cm, is 1. 1734 cm2 (Q) If the volume of a cube is 4913 cm3 , then the total surface area of the cube is equal to 2. 252π cm2 (R) The lateral surface area of a right circular cylinder with base radius 7 cm and height 20 cm is 3. 880 cm2 (S) The volumes of two spheres are in the ratio 64 : 27. If the sum of their radii is 21 cm, the difference between their surface areas will be 4. 195π cm2
Solution
(P) Given: Base radius (r) of the cone = 13 cm
Slant height (
) of the cone = 15 cm
Curved surface area of the cone = πr
= π × 13 × 15 = 195π cm
2 (Q) Volume of a cube = a
3 , where 'a' is the edge of the cube.
Given volume of the cube = 4913 cm
3 a
3 = 4913 cm
a =
a = 17 cm
Now, total surface area of the cube = 6a
2 = 6(17)
2 = 6 × 17 × 17 = 1734 cm
2 (R) Radius of the cylinder = 7 cm
Height of the cylinder = 20 cm
Lateral surface area of a right circular cylinder = 2πrh = 2 × (22/7) × 7 × 20 = 880 cm
2 (S) Volume of sphere 1/Volume of sphere 2 = 64/27
Let the radii of sphere 1 be r
1 and that of sphere 2 be r
2 4πr
1 3 /4πr
2 3 = 64/27
r
1 /r
2 = 4/3
r
1 = (4/3)r
2 (4/3)r
2 + r
2 = 21
7r
2 /3 = 21
r
2 = 9 m
r
1 = 12 m
Difference between the surface areas = 4πr
1 2 - 4πr
2 2 = 4π(12
2 - 9
2 ) = 252π sq. m
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