Practice | Hashtable

content Computer Science Arithmetic Aptitude Data Interpretation Logical Reasoning Verbal Reasoning Non-Verbal Reasoning

About About Privacy Policy

Hashtable

Home
Practice
PYQ
Mock Test
Register Free

Home Practice PYQ Mock Test Register Free

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

81. The number of circular pipes with an inside diameter of 1 inch which will carry the same amount of water as a pipe with an inside diameter of 6 inches is :

A. $6 π$

B. $12$

C. $36$

D. $36 π$

Discuss

Solution:

Let the length of each pipe be

l

inches
Then, volume of water in thinner pipe :

\begin{aligned} = [π \times {(\frac{1}{2})}^{2} \times 1] cu.inch \\ = (\frac{π l}{4}) cu.inch \end{aligned}

Volume of water in thinker pipe :

\begin{aligned} = (π \times 3^{2} \times l) cu.inch \\ = (9 π l) cu.inch \end{aligned}

∴ Required number of pipes :

\begin{aligned} = \frac{9 π l}{(\frac{π l}{4})} \\ = 36 \end{aligned}

82. A right triangle with sides 3 cm, 4 cm and 5 cm is rotated about the side of 3 cm to form a cone. The volume of the cone so formed is :

A. 12π cm³

B. 15π cm³

C. 16π cm³

D. 20π cm³

Discuss

Solution:

Clearly, we have r = 3 cm and h = 4 cm
∴ Volume :

\begin{aligned} = \frac{1}{3} π r^{2} h \\ = (\frac{1}{3} \times π \times 3^{2} \times 4) π r^{3} \\ = 12 π c m^{3} \end{aligned}

83. The radius of the base and height of a metallic solid cylinder are r cm and 6 cm respectively. It is melted and recast into a solid cone of the same radius of base. The height of the cone is :

A. 9 cm

B. 18 cm

C. 27 cm

D. 54 cm

Discuss

Solution:

Let the height of the cone be h cm
Then,

\begin{aligned} π \times r^{2} \times 6 = \frac{1}{3} \times π \times r^{2} \times h \\ \Rightarrow h = 18 c m \end{aligned}

84. For a sphere of radius 10 cm, What percent of the numerical value of its volume would be the numerical value of the surface area ?

A. 24%

B. 26.5%

C. 30%

D. 45%

Discuss

Solution:

Volume of the sphere :

= [\frac{4}{3} π {(10)}^{3}] c m^{3}

Surface area of the sphere :

= [4 π {(10)}^{2}] c m^{2}

∴ Required percentage :

\begin{aligned} = [\frac{4 π {(10)}^{2}}{\frac{4}{3} π {(10)}^{3}} \times 100] % \\ = 30 % \end{aligned}

85. How many lead shots each 3 mm in diameter can be made from a cuboid of dimensions 9 cm × 11 cm × 12 cm ?

A. 7200

B. 8400

C. 72000

D. 84000

Discuss

Solution:

Volume of each lead shot :

\begin{aligned} = [\frac{4}{3} π \times {(\frac{0.3}{2})}^{3}] c m^{3} \\ = (\frac{4}{3} \times \frac{22}{7} \times \frac{27}{8000}) c m^{3} \\ = \frac{99}{7000} c m^{3} \end{aligned}

∴ Number of lead shots :

\begin{aligned} = (9 \times 11 \times 12 \times \frac{7000}{99}) \\ = 84000 \end{aligned}

86. A metallic sphere of radius 5 cm is melted to make a cone with base of the same radius. What is the height of the cone ?

A. 5 cm

B. 10 cm

C. 15 cm

D. 20 cm

Discuss

Solution:

Let the height of the cone be h cm
Then,

\begin{aligned} \frac{4}{3} π \times {(5)}^{3} = \frac{1}{3} π \times {(5)}^{2} \times h \\ \Rightarrow h = 20 c m \end{aligned}

87. A hemisphere of lead of radius 6 cm is cast into a right circular cone of height 75 cm. The radius of the base of the cone is :

A. 1.4 cm

B. 2 cm

C. 2.4 cm

D. 4.2 cm

Discuss

Solution:

Let the radius of the cone be R cm
Then,

\begin{aligned} \frac{1}{3} π \times R^{2} \times 75 = \frac{2}{3} π \times 6 \times 6 \times 6 \\ \Rightarrow R^{2} = (\frac{2 \times 6 \times 6 \times 6}{75}) \\ \Rightarrow R^{2} = \frac{144}{25} \\ \Rightarrow R^{2} = \frac{{(12)}^{2}}{{(5)}^{2}} \\ \Rightarrow R = \frac{12}{5} \\ \Rightarrow R = 2.4 c m \end{aligned}

88. Base of a right prism is a rectangle, the ratio of whose length and breadth is 3 : 2. If the height of the prism is 12 cm and total surface area is 288 sq.cm the volume of the prism is :

A. 291 cm³

B. 288 cm³

C. 290 cm³

D. 286 cm³

Discuss

Solution:

Let the length of base be 3a cm and breadth be 2a cm
Total surface area of prism :
= [Perimeter of base × height] + [2 × Area of base]
= [2 (3a + 2a) × 12 + 2 × 3a × 2a] sq.cm
= (120a + 12a²) sq.cm
According to the question,
120a + 12a² = 288
⇒ a² + 10a = 24
⇒ a² + 10a - 24 = 0
⇒ a² + 12a - 2a - 24 = 0
⇒ a (a + 12) - 2 (a + 12) = 0
⇒ (a - 2)(a + 12) = 0
⇒ a = 2 because a

\neq

-12
∴ Volume of prism :
= Area of base × Height
= (3a × 2a × 12)cu.cm
= 72a² cu.cm
= (72 × 2 × 2)cu.cm
= 288 cu.cm

89. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients ?

A. 38L

B. 40L

C. 39.5L

D. 38.5L

Discuss

Solution:

Diameter of bowl = 7 cm
∴ Radius of bowl =

\frac{2}{7}

cm
Height = 4 cm
∴ Volume of cylindrical bowl :

\begin{aligned} = π r^{2} h \\ = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 4 \\ = 154 c u . c m \end{aligned}

Hence, volume of soup for 250 patients :

\begin{aligned} = 154 \times 250 \\ = 38500 c m^{3} \\ = 38.5 L \end{aligned}

90. The breadth of a room is twice its height and half its length. The volume of the room is 512 cu.m. The length of the room is :

A. 16 m

B. 18 m

C. 20 m

D. 32 m

Discuss

Solution:

Let the height of the room be x metres
Then, breadth = 2x metres and length = 4x metres
∴ Volume of the room :
= (4x × 2x × x) m³
= (8x³) m³
8x³ = 512
⇒ x³ = 64
⇒ x = 4
∴ Length of the room is :
= 4x
= (4 × 4)
= 16 m

1 2 3 4 5 6 7 8 9 10

Sign up for our newsletter

Are you passionate about opportunities in the realm of Computer Science within the government sector? Don't miss out on crucial updates, job openings, and career prospects tailored specifically straight to your inbox.

Email address

By subscribing, you agree with our Terms of Service and Privacy Policy.

Resources

Practice PYQ Mock Test

Company

About us Contact us Privacy policy Terms of service

Hashtable