# Mensuration Quiz 13

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# Mensuration Quiz 13

### Introduction

Mensuration is one of important topic in Quantitative Aptitude Section. In Mensuration Quiz 13 article candidates can find a question with an answer. By solving this question candidates can improve and maintain, speed, and accuracy in the exams. Mensuration Quiz 13 questions a very useful for different exams such as IBPS PO, Clerk, SSC CGL, SBI PO, NIACL Assistant, NICL AO, IBPS SO, RRB, Railways, Civil Services etc.

### Q1

The diameter of a wheel is 1.4m. If this wheel rotates 500 rotations, how long it can travel?
A. 1050 B. 1100 C. 1400 D. 2200

D
Diameter = 1.4 m.
Circumference of the wheel = πd = $\frac {22}{7}$ × 1.4
= 4.4
We have distance traveled in 1 rotation = 4.4 meter
Distance covered in 500 rotations = 4.4 × 500 = 2200 meters

### Q2

If radius of cylinder and sphere are same and volume of sphere and cylinder are same. what is the ratio between the radius and height of the cylinder?
A. R = H B. R= (3/4)H C. R = 2H D. R = 4H

B
We have Volume and radius of cylinder and sphere are same
Let R be the radius of both the figures.
Then Volume of sphere = $\Rightarrow \frac {4 \pi {(R)}^{3}}{3}$
And Volume of cylinder = $π{R}^{2}h$
Hence equating both the volumes
$\Rightarrow \frac {4 \pi {(R)}^{3}}{3} = π{R}^{2}h$
$\Rightarrow \frac {4R}{3} = h$
$\Rightarrow R = \frac {3h}{4}$

### Q3

Find the cost of carpeting a room 13 m long and 9 m broad with a carpet 75 cm wide at the rate of Rs. 12.40 per meter.
A. 1994.5 B. 1934.4 C. 2080.6 D. 1928.6

B
Length of the room = 13m
Breadth of the room = 9m
Area of the room = length × breadth
Area = 13 × 9 = 117 ${m}^{2}$
Also, we need to cover the entire area with a carpet which is 0.75 m wide.
Hence, length of carpet required = $\frac {Area of room}{width of carpet}$
= $\frac {117}{0.75}$
= 156 m
Cost of 1 m carpet = Rs 12.4
Cost of 156m carpet = Rs 12.4 × 156
= Rs 1934.4

### Q4

An equilateral triangle of side 3 inch each is given. How many equilateral triangles of side 1 inch can be formed from it?
A. 9 B. 15 C. 20 D. 25

A
Area of triangle = $(\frac {\sqrt {3}}{4}){a}^{2}$
Let number of 1 inch triangle = n
∴ $n × (\frac {\sqrt {3}}{4}) × {1}^{2} = (\frac {\sqrt {3}}{4}) × {3}^{2}$
$\Rightarrow$ n = 9
Hence Option (A)

### Q5

The diameter of a wheel is 2.8 m. If this wheel rotates 100 rotations, how long it can travel?
A. 88m B. 888m C. 808m D. 880m

D
Diameter = 2.8 m.
Circumference of the wheel = πd = $\frac {22}{7}$ × 2.8
= 8.8
We have distance traveled in 1 rotation = 8.8 meter
Distance covered in 100 rotations = 8.8 × 100 = 880 meter