# Mensuration – Quiz 6

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# Mensuration – Quiz 6

### Introduction

Mensuration is one of important topic in Quantitative Aptitude Section. In Mensuration – Quiz 6 article candidates can find questions with answer. By solving this questions candidates can improve and maintain, speed, and accuracy in the exams. Mensuration – Quiz 6 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

The length of the longest rod that can be placed in a room which is 12 m long, 9 m broad and 8 m high is
A. 27 m B. 19 m C. 17 m D. 13 m

C
Length of the longest rod = Diagonal of the room
= $\sqrt{{12}^{2} + {9}^{2} + {8}^{2}}$
$\sqrt{289}$ = 17 m

### Q2

The cross-section of a canal is in the form of a trapezium. If canal top is 10 m wide, the bottom is 6 m wide and area of the cross section is 72 m^2, then depth of the canal is
A. 10 m B. 7 m C. 6 m D. 9 m

D
Let h the height of the trapezium
Hence area of cross-section of the channel in the form of trapezium
$\frac{1}{2}$(10 + 6) * h = 72
i.e, 8h = 72
h = 9 m

### Q3

A steel wire bent in the form of a square of area 121 ${cm}^{2}$. If the same wire is bent in the form of a circle, then the area of the circle is
A. 130 B. 136 C. 154 D. None of these

C
Side of a square = $\sqrt{121}$ = 11 cm
i.e, Perimeter of circle = 4 * 11 = 44 cm
2πr = 44
where r is radius of the circle
i.e, 2 * $\frac{22}{7}$ * r = 44
r = 7 cm
i.e, Area of the circle = π${r}^{2}$ = $\frac{22}{7} * {7}^{2} = 154{cm}^{2}$

### Q4

The diameter of a garden roller is 1.4 m and it is 2 m long. How much area will it cover in 5 revolutions ?
A. 40 B. 44 C. 48 D. 36

B
Garden roller is in the form of a cylinder whose radius is 0.7 m and height is 2 m.
i.e, Area covered in 5 revolutions
= 5 * 2πrh
= 5 * 2 * $\frac{22}{7}$ * 0.7 * 2
= 44 ${m}^{3}$

### Q5

A bicycle wheel makes 5000 revolutions in moving 11 km. Find diameter of the wheel
A. 55 cm B. 60 cm C. 65 cm D. 70 cm

D
Circumference of the wheel = 2πr = $\frac{11000 metres}{5000}$
πr = $\frac{11}{10}$
r = $\frac{11}{10} * \frac{7}{22} = \frac{7}{20}$metre
$\frac{7}{20} * 100$ = 35 cm
Diameter of the wheel = 70 cm.