# Mensuration Quiz 12

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# Mensuration Quiz 12

### Introduction

Mensuration is one of important topic in Quantitative Aptitude Section. In Mensuration Quiz 12 article candidates can find questions with answer. By solving this questions candidates can improve and maintain, speed, and accuracy in the exams. Mensuration Quiz 12 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

Water flows into a tank 200 m × 150 m through a rectangular pipe of 1.5m × 1.25 m @ 20 kmph. In what time (in minutes) will the water rise by 2 metres?
A. 48 min B. 96 min C. 108 min D. 36 min

B
We know that, formula:
Volume of any Cuboid = Length × Breadth × Height
According to the given question:
$\Rightarrow$ Volume of the water required in the tank = (200 × 150 × 2) ${m}^{3}$ = 60000 ${m}^{3}$
∴ Length of water column flown in1 min = $\frac {(20 × 1000)}{60} m = \frac {1000}{3}$ m
∴ Volume flown per minute = $1.5 × 1.25 × (\frac {1000}{3}) {m}^{3} = 625 {m}^{3}$
∴ Required time = $(\frac {60000}{625}) min = 96 min.$
Hence, the required answer is 96 minutes.

### Q2

Area of a square is given as 12800 sq miles, find the time taken by a lady to move across a diagonal with speed of 15 miles per hour.
A. 8 hours B. 9 hours C. 8.5 hours D. 10.66 hours

D
We have, area of a square = 12800 sq miles.
Hence ${(side)}^{2} = 12800$
Side = $\sqrt {12800}$
$\Rightarrow$ Side = 80$\sqrt {2}$
We know that diagonal of a square = $\sqrt {2}$× side
= $\sqrt {2}$ × 80 $\sqrt {2}$
= 160 miles.
Hence we know that speed = $\frac {distance}{time}$
Time = $\frac {distance}{speed}$
Time = $\frac {160}{15}$
= 10.66 miles/hr.

### Q3

A sphere of radius 2 cms is dropped into a cylinder of radius 4 cms containing water upto 2.2 cms. The raise in the water level is?
A. 0.67 cms B. 0.90 cms C. 0.77 cms D. 0.80 cms

A
We know that
Volume of sphere = $(\frac {4}{3})π{r}^{3}$
Where r = radius of sphere
Volume of cylinder = $π{R}^{2}H$
Where R = radius of cylinder
H = height of cylinder
Let h = raise height of water in cylinder
∵ Volume of water displaced in cylinder = volume of sphere
∴$π{R}^{2}H = (\frac {4}{3})π{r}^{3}$
$\Rightarrow {4}^{2} × h = \frac {4}{3} × {2}^{3}$
$\Rightarrow h = \frac {2}{3} cms = 0.67$cms

### Q4

The volume and the radius of both cone and sphere are equal, then find the ratio of height of the cone to the diameter of the sphere?
A. 2 : 3 B. 3 : 2 C. 2 : 1 D. 1 : 2

C
Volume of cone = $(\frac {1}{3})π{r}^{2}h$
Where r = radius of cone
Volume of sphere = $(\frac {4}{3})π{r}^{3}$
Where R = radius of sphere
According to the question
R = r and
$(\frac {1}{3})π{r}^{2}h = (\frac {4}{3})π{r}^{3}$
$\Rightarrow$ h = 4 × R = 2 × D (∵ 2 × R = diameter of sphere = D)
$\Rightarrow \frac{h}{D} = \frac {2}{1}$

### Q5

A rectangular carpet has an area of 120 sq metres and a perimeter of 46 metres. The length of its diagonal (in metres) is:
A. 11 B. 13 C. 15 D. 17

D
We know that
Perimeter of rectangle = 2 (l + b)
Area of rectangle = l × b
Diagonal of rectangle = $\sqrt{ ({l}^{2} + {b}^{2})}$
Where l = length of rectangle
According to the question
2 (l + b) = 46 m
$\Rightarrow$ l + b = 23 m ………… (1)
And
l × b = 120 ${m}^{2}$ ………… (2)
Solving equation (1) and (2) we get
l = 15 m
b = 8 m
∴ Diagonal of rectangle = $\sqrt{ ({15}^{2} + {8}^{2})} = \sqrt{ (225 + 64)}$
= $\sqrt {289} = 17$m