# Mensuration Quiz 10

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# Mensuration Quiz 10

### Introduction

Mensuration is one of important topic in Quantitative Aptitude Section. In Mensuration Quiz 10 article candidates can find questions with answer. By solving this questions candidates can improve and maintain, speed, and accuracy in the exams. Mensuration Quiz 10 questions are 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

A solid metallic cylinder of base radius 5 cm and height 7 cm is melted to form cones, each of height 1 cm and base radius 1 mm. Find the number of cones?
A. 53500 B. 49500 C. 51500 D. 52500

D
Number of cones = $\frac {Volume of Cylinder}{Volume of one cone}$
= $\frac {π \times 5 \times 5 \times 7}{(\frac {1}{3 \pi} \times \frac {1}{10} \times \frac {1}{10} \times 1)}$
= 52500

### Q2

A hollow cylindrical tube open at both ends is made of plastic 4 cm thick. If the external diameter be 54 cm and the length of the tube be 490 cm, find the volume of plastic.
A. 320000 ${cm}^{3}$ B. 340000 ${cm}^{3}$ C. 306300 ${cm}^{3}$ D. 308000 ${cm}^{3}$

D
Volume of Plastic = $πh ({R}^{2} - {r}^{2}) = \frac {22}{7} \times 490({27}^{2} - {23}^{2}) = 308000{cm}^{3}$

### Q3

Find the area of a white sheet required to prepare a cone with a height of 21cm and the radius of 20cm.
A. 3080 ${cm}^{2}$ B. 2300 ${cm}^{2}$ C. 3460 ${cm}^{2}$ D. 3600 ${cm}^{2}$

A
r = 20 ; h = 21
l = $\sqrt {(400 + 441)}$ = 29 cm
Total surface area = $πrl + π{r}^{2} = πr (l + r) = \frac {22}{7} \times 20(49) = 3080 {cm}^{2}$

### Q4

The radius of a circle is 4 m. What is the radius of another circle whose area is 16 times of that first?
A. 16 m B. 64 m C. 256 m D. 400 m

A
Ratio of areas = ${(ratio of radii)}^{2}$
$\frac {16}{1} = {(ratio of radii)}^{2}$
ratio of radii = $\frac {4}{1}$
The required radius = 16 m

### Q5

A circular path runs round a circular garden. If the difference between the circumference of the outer circle and inner circle is 88m. Find the width of Path?
A. 14 m B. 15 m C. 18 m D. 13 m

A
Width of the Road = R – r
2πR – 2πr = 88
R – r = 14 m

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