Hello Aspirants. Welcome to Online Quantitative Aptitude section in AffairsCloud.com. Here we are creating question sample from **Mensuration** that are important for all the competitive exams. We have included some questions that are repeatedly asked in exams !!

Questions penned by Yogit

**Perimeter of a square and an equilateral triangle is equal. If the diagonal of the square is 10√2 cm, then find the area of equilateral triangle?**

a) (400√3)/9

b) (400√3/7)

c) (200√3/7)

d) (200√3)/9

e) None of theseAnswer –**a) (400√3) /9**

**Explanation :**

Diagonal of a square = a√2 = 10√2

so a = 10, perimeter of square = 4*10 = 40 = 3x (x is the length of each side of triangle)

x = 40/3, so are of equilateral triangle = √3/4*40/3*40/3 = (400√3)/9 cm^2**Length of a rectangular field is increased by 10 meters and breadth is decreased by 4 meters, area of the field remains unchanged. If the length decreased by 5 meters and breadth is increased by 7 meters, again the area remains unchanged. Find the length and breadth of the rectangular field.**

a) 12, 11

b) 13, 12

c) 13,11

d) 14,12

e) 13,15

e) None of theseAnswer –**c) 13,11**

**Explanation :**

Length = l and breadth = b,

(l +10)*(b-4) = lb and (l-5)*(b+7) = lb

Solve both equation to get l and b**If the length of the rectangle is increased by 20%, by what percent should the width be reduced to maintain the same area?**

a) 13.37

b) 16.67

c) 21.33

d) 33.33

e) None of theseAnswer –**b) 16.67**

**Explanation :**

let length = 100 and breadth = 100

now new length = 120 and let breadth = b

so, 100*100 = 120*b

b = 250/3, so % decrease = 100 – 250/3 = 50/3 = 16.67%**A cone whose height is half of its radius is melted to from a hemi-sphere. Find the ratio of the radius of the hemi-sphere to that of cone.**

a) 1:3

b) 4:1

c) 1:4

d) 1:2Answer –**c)1:4**

**Explanation :**

volume will remains constant. So,

V = 1/3*22/7*r^2*r/2 (volume of cone) and V = 2/3*22/7*R^3 (volume of hemi-sphere)

So, R/r = 1:4**Find the number of spherical balls of radius 1 cm that can be made from a cylinder of height 8 cm and diameter 14 cm?**

a) 284

b) 294

c) 304

d) 314Answer –**b) 294**

**Explanation :**

(22/7)*7*7*8 = x*(4/3)*(22/7)*1^3 (x = number of spherical balls)**A rectangle whose sides are in the ratio 6:5 is formed by bending a circular wire of radius 21cm. Find the difference between the length and breadth of the rectangle?**

a) 6cm

b) 8cm

c) 10cm

d) 12cmAnswer –**a) 6cm**

**Explanation :**

circumference of the wire = 2*(22/7)*21 = 22*6

perimeter of rectangle = 2*11x = 22*6, so x= 6

difference = 36 -30 = 6cm**A right circular cone is exactly fitted inside a cube in such a way that the edges of the base of the cone are touching the edges of one face of the cube and the vertex is on the opposite face of the cube. If the volume of the cube is 512 cubic cm. find the approximate volume of the cone?**

a) 104 cubic cm

b) 124 cubic cm

c) 134 cubic cm

d) 144 cubic cmAnswer –**c) 134 cubic cm**

**Explanation :**

when cone is completely fitted inside the cube, then diameter of cone = side of cube and height of cone = height of cube

so, volume = (1/3)*(22/7)*4*4*8 = 134 (approx)**if the radius of a cylinder is doubled and height is halved, what is the ratio between the new volume and the previous volume?**

a) 1:2

b) 2:1

c) 2:3

d) 1:4Answer –**b) 2:1**

**Explanation :**

New volume = (22/7)*4r^2*h/2 and old volume = (22/7)*r^2*h

so ratio = 2:1**A cone of radius 12 cm and height 5 cm is mounted on a cylinder of radius 12 cm and height 19 cm. Find the total surface area of the figure thus formed?**

a) 2276

b) 2376

c) 2476

d) 2576Answer –

**Explanation :**

total surface area = curved surface area of cone + curved surface area of cylinder + base area

**A rectangular garden is 30 meter long and 20 meter broad. It has 6 meter wide pavements all around it both on its inside and outside. Find the total area of pavements?**

a) 800

b) 1000

c) 1200

d) 1600Answer –**c) 1200**

**Explanation :**

Required area = 42*32 – 18*8 = 1200

**Note:** Dear Readers if you have any doubt in any chapter in Quants you can ask here. We will clear your doubts

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