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Home > Practice > Arithmetic Aptitude > Volume and Surface Area > Miscellaneous

71. The ratio of the volume of a hemisphere and a cylinder circumscribing this hemisphere and having a common base is :

A. 1 : 2

B. 2 : 3

C. 3 : 4

D. 4 : 5

Discuss

Solution:

Let the radius of the hemisphere be be r cm
Then, radius of the cylinder = r cm
Height of the cylinder = r cm
∴ Required ratio :

\begin{aligned} = \frac{Volume of hemisphere}{Volume of cylinder} \\ = \frac{\frac{2}{3} π r^{3}}{π r^{2} \times r} \\ = \frac{2}{3} O r 2 : 3 \end{aligned}

72. The volume of a right circular cone which is obtained from a wooden cube of edge 4.2 dm wasting minimum amount of wood is :

A. 19404 dm³

B. 194.04 dm³

C. 19.404 dm³

D. 1940.4 dm³

Discuss

Solution:

The volume of cone should be maximum
∴ Radius of the base of cone :

\begin{aligned} = \frac{Edge of cube}{2} \\ = \frac{4.2}{2} \\ = 2.1 dm. \end{aligned}

Height of cone = Edge of cube = 4.2 dm.
∴ Volume of cone :

\begin{aligned} = \frac{1}{3} π r^{2} h \\ = (\frac{1}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 4.2) cu .dm . \\ = 19.404 cu .dm . \end{aligned}

73. If the diameter of a sphere is 6 m, its hemisphere will have a volume of :

A. $18 π$ m³

B. $36 π$ m³

C. $72 π$ m³

D. None of these

Discuss

Solution:

Volume of hemisphere :

\begin{aligned} = (\frac{2}{3} π \times 3 \times 3 \times 3) m^{3} \\ = (18 π) m^{3} \end{aligned}

74. The length of the longest rod that can be placed in a room of dimensions 10 m × 10 m × 5 m is :

A. 15 $\sqrt{3}$ m

B. 15 m

C. 10 $\sqrt{2}$ m

D. 5 $\sqrt{3}$ m

Discuss

Solution:

Required length :

\begin{aligned} = \sqrt{{(10)}^{2} + {(10)}^{2} + {(5)}^{2}} m \\ = \sqrt{225} m \\ = 15 m \end{aligned}

75. The sum of the radius and the height of a cylinder is 19 m. The total surface area of the cylinder is 1672 m², what is the volume of the cylinder ?

A. 3080 m³

B. 2940 m³

C. 3220 m³

D. 2660 m³

Discuss

Solution:

Let the radius of the cylinder be r and height be h
Then, r + h = 19
Again, total surface area of cylinder =

(2 π r h + 2 π r^{2})

Now,

\begin{aligned} 2 π r (h + r) = 1672 \\ \Rightarrow 2 π r \times 19 = 1672 \\ \Rightarrow 38 π r = 1672 \\ ∴ π r = \frac{1672}{38} = 44 m \\ ∴ r = \frac{44 \times 7}{22} = 14 m \end{aligned}

∴ Height = 19 - 14 = 5 m
Volume of cylinder :

\begin{aligned} = π r^{2} h \\ = \frac{22}{7} \times 14 \times 14 \times 5 \\ = 22 \times 2 \times 14 \times 5 \\ = 3080 m^{3} \end{aligned}

76. Given that 1 cu. cm of marble weights 25 gms, the weight of a marble block 28 cm in width and 5 cm thick is 112 kg. The length of the block is :

A. 26.5 cm

B. 32 cm

C. 36 cm

D. 37.5 cm

Discuss

Solution:

Let length = x cm
Then,

\begin{aligned} x \times 28 \times 5 \times \frac{25}{1000} = 112 \\ \Rightarrow x = (112 \times \frac{1000}{25} \times \frac{1}{28} \times \frac{1}{5}) \\ \Rightarrow x = 32 c m \end{aligned}

77. An open box is made by cutting the congruent squares from the corners of a rectangular sheet of cardboard of dimension 20 cm × 15 cm. If the side of each square is 2 cm, the total outer surface area of the box is :

A. 148 cm²

B. 284 cm²

C. 316 cm²

D. 460 cm²

Discuss

Solution:

Clearly,

l

= (20 - 4) cm = 16 cm
b = (15 - 4) cm = 11 cm and
h = 2 cm
∴ Outer surface area of the box :

\begin{aligned} = [2 (l + b) \times h] + l b \\ = [{2 (16 + 11) \times 2} + 16 \times 11] \\ = (108 + 176) \\ = 284 c m^{2} \end{aligned}

78. V₁, V₂, V₃ and V₄ are the volumes of four cubes of side lengths x cm, 2x cm, 3x cm and 4 cm respectively. Some statements regarding these volumes are given below :
(i) V₁ + V₂ + 2V₃ < V₄
(ii) V₁ + 4V₂ + V₃ < V₄
(iii) 2(V₁ + V₃) + V₂ = V₄
Which of these statements area correct ?

A. 1 and 2

B. 2 and 3

C. 1 and 3

D. 1, 2 and 3

Discuss

Solution:

Clearly, we have :
V₁ = x³,
V₂ = (2x)³ = 8x³
V₃ = (3x)³ = 27x³
V₄ = (4x)³ = 64x³
(i) V₁ + V₂ + 2V₃
= x³ + 8x³ + 2 × 27x³
= 63x³ < V₄

(ii) V₁ + 4V₂ + V₃
= x³ + 4 × 8x³ + 27 x³
= 60x³ < V₄

(iii) 2(V₁ + V₃) + V₂
= 2 (x³ + 27x³) + 8x³
= 64x³ = V₄

79. A circular well with a diameter of 2 metres, is dug to a depth of 14 metres. What is the volume of the earth dug out ?

A. 32 m³

B. 36 m³

C. 40 m³

D. 44 m³

Discuss

Solution:

Volume :

\begin{aligned} = π r^{2} h \\ = (\frac{22}{7} \times 1 \times 1 \times 14) m^{3} \\ = 44 m^{3} \end{aligned}

80. If the radius of the base of a right circular cylinder is halved, keeping the height same, what is the ratio of the volume of the reduced cylinder to that of the original one ?

A. 1 : 2

B. 1 : 4

C. 1 : 8

D. 8 : 1

Discuss

Solution:

Let original radius = R
Then, new radius =

\frac{R}{2}

\begin{aligned} ∴ \frac{Volume of reduced cylinder}{Volume of original cylinder} \\ = \frac{π \times {(\frac{R}{2})}^{2} \times h}{π \times R^{2} \times h} \\ = \frac{1}{4} O r 1 : 4 \end{aligned}

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