mensuration area volumes Model Questions & Answers, Practice Test for ssc cgl tier 1 2023
ssc cgl tier 1 2023 SYLLABUS WISE SUBJECTS MCQs
Number System
Ratio Proportion & Partnership
Averages
Percentages
Profit & Loss
Time & Work
Time & Distance
Simple Interest & Compound Interest
Mensuration: Area & Volumes
Algebraic Expressions
Linear Equations
Squareroots & Cuberoots
A square is drawn such that its vertices are lying on a circle of radius 201 mm. What is the ratio of area of circle to that of square ?
Answer: (b)
$D_S = 2R$
$D_S^2 = (2R)^2$
${\text"Area of circle"}/{\text"Area of square"} = {πR^2}/{1/2 D_S^2} = π/2 = {22}/{14} = {11}/7$
The diameter of the Moon is approximately one-fourth of the diameter of the Earth. What is the ratio (approximate) of their volumes?
Answer: (c)
The diameter of Moon is approximately one-fourth of the diameter of Earth.
Let radius of Moon = r,
then radius of Earth = 4r
Required ratio =
${\text"Volume of Moon"}/{\text"Volume of Earth"} = {4/3 π r^3}/{4/3 π (4r)^3} = r^3/{64r^3} = 1/{64}$ = 1 : 64
Consider the following
The length of a side of a cube is 1 cm. Which of the following can be the distance between any two vertices?
I. 1 cm
II. 2 cm
III. $√3$ cm
Select the correct answer using the codes given below.
Answer:(d)
The distance between vertices B and C is 1 cm.
The distance between A and B is $√{1^2 + 2^2} = √2$ cm
The distance between diagonal B and D is
$√{1^2 + 1^2 + 1^2} = √3$ cm
A person rides a bicycle round a circular path of radius 50 m. The radius of the wheel of the bicycle is 50 cm. The cycle comes to the starting point for the first time in 1 h. What is the number of revolutions of the wheel in 15 min?
Answer: (c)
∵ Circumference of circular path = 2π × 50 m = 10000π cm
and circumference of wheel = 2π × 50 = 100π cm
∴ Distance covered in 60 min = 10000π cm
Distance covered in 1 min = ${10000}/{60}$ π
Distance covered in 15 min
= ${10000}/{60}$ π × 15 = 2500π cm
∴ Number of revolutions = ${2500π}/{100π}$ = 25
A cylindrical tube open at both ends is made of metal. The internal diameter of the tube is 6 cm and length of the tube is 10 cm. If the thickness of the metal used is 1 cm, then the outer curved surface area of the tube is
Answer: (d)
Internal diameter of the tube = 6 cm
∴ Internal radius (r) = 3 cm
Height of the tube (h) = 10 cm
Thickness of the metal = 1 cm
∴ Outer radius (R) = Thickness of the metal + Internal radius = 1 + 3 = 4 cm
∴ Outer curved surface area
= 2πrh + 2π $(R^2 - r^2)$
= 2π(3)(10) + 2π (16 - 9)
= 60π + 14π = 74 π sq cm
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