# Mensuration Questions and Answers updated daily – Aptitude

## Mensuration Questions

**Volume of the cylinder = πr**= π * 6 * 6 * 21

^{2}hGiven that, diameter of bullet = 1.5 cm

--> radius of bullet = Diameter/2 = 1.5/2 = 0.75 cm = 3/4 cm

**Volume of 1 bullet = Volume of Sphere = (4/3)πr**

^{3}= (4/3) * π * (3/4) * (3/4) * (3/4)

Here,

**Number of Bullets = Volume of the cylinder/Volume of 1 bullet**

= [π * 6 * 6 * 21]/ [(4/3) * π * (3/4) * (3/4) * (3/4)]

= [6 * 6 * 21 * 3 * 4 * 4 * 4]/ [4 * 3 * 3 * 3]

=

**1344**

In âŠ¿ABC

**AC = √((AB)² + (BC)²)**

AC = √(16² + 12² )

AC = √400

AC = 20 cm

Therefore, diameter of circumcircle, AC = 20 cm

Radius of circumcircle, r = 10 cm

WKT,

**Area of Circumcircle = πr²**

Area of circumcircle = π10²

=

**100π cm²**

^{3}

Cylindrical pencil of diameter = 1.2 cm

Height of the pencil, h = 1.4 cm

Radius of Cylindrical pencil, r = 1.2/2 = 0.6 cm

WKT,

**Volume of cylinder = πr² h**

**Volume of cone = (1/3)πr² h**

**Volume of material removed = πr² h - (1/3)πr² h**

= (2/3)πr² h

= (2/3)(22/7)(0.6 x 0.6 x 1.4)

= 1.056 cm

^{3}

Perimeter of an equilateral triangle = √3 Area of that triangle.

WKT,

**Perimeter of an equilateral triangle = 3a**

Area of an equilateral triangle = (√3/4)a²

Therefore,

3a = √3 (√3/4)a²

**a = 4 unit.**

On keeping cans each having diameter of 14 cm, erect in the rectangular box, the height occupied by the cans is 30 cm. So, some cans can be placed horizontally on the base.

**Number of cans that can be placed erect in a row**= Length of the box/Diameter of can

= 76/14

= 5.43

= 5 cans (Remaining space will be vacant)

**Number of such rows that can be placed**= Breadth of the box/Diameter of can

= 46/14

= 3.3

= 3 rows

Hence,

**Number of cans placed in an erect position**= 5 cans * 3 rows

=

**15 cans**

So far, only 30 cm of the height of the box has been filled with cans. Given, Total height of box = 45cm

Remaining Height of the box = 45 - 30 = 15 cm

which is greater than Diameter of the can (14 cm)

So, in remaining 15 cm of height of the box, we can fill some cans horizontally on the base.

**Number of cans that can be placed horizontally in a row**= Length of the box/Height of can

= 76/30

= 2.5

= 2 cans (Remaining space will be vacant)

**Number of such rows that can be placed**= Breadth of the box/Diameter of can

= 46/14

= 3.3

= 3 rows

Hence,

**Number of cans placed horizontally**= 2 cans * 3 rows

=

**6 cans**

Therefore,

**Maximum number of cans that can fit in the box**= 15 + 6 =

**21 cans**

radius (r) = diameter/2 = 49/2 dm

--> converting into cm

radius = (49/2) x 10 cm

--->

**r = 245 cm**

Height h = 36 dm

--> converting into cm

h = 36 x 10 cm

-->

**h = 360 cm**

**Volume of Cylinder**= π r

^{2}h

= (22/7) * 245 * 245 * 360

=

**22 * 35 * 245 * 360 cm**

^{3}**Volume of tin = (22 * 18 * 49) cm**

^{3}**Number of tins**to be filled with kerosene from the drum= (Volume of Cylinder/Volume of tin)

= (22 * 35 * 245 * 360) / (22 * 18 * 49)

=

**3500**

**Perimeter of a square = 2 * perimeter of a rectangle**

W.K.T:-

**Perimeter of a square = 4 * side**

**Perimeter of a rectangle = 2(length + breadth)**

Given Rectangle of length = 8 cm and breadth = 7 cm

Perimeter of a rectangle = 2(8 + 7)

= 2 * 15

= 30 cm

So, Perimeter of a square = 4 * side = 2 * perimeter of a rectangle

--> 4 * side = 2 * 30

-->

**side of the square = 15 cm**

Given, Diameter of semicircle (D) = side of the square = 15 cm

So,

**Circumference/perimeter of the semicircle = πR + 2R = (πD/2) + D**

= (3.14 * 15/2) + 15

= 3.14 * 7.5 + 15

= 23.55 + 15

=

**38.55 cm**

Width of the hall = 8 meters

If 0.5 meters of space is left around the room, then

Length of the hall = 10 - 0.5 - 0.5 =

**9 meters**

Width of the hall = 8 - 0.5 - 0.5 =

**7 meters**

Required

**Area**of the hall for carpeting = Length * Width

= 9 * 7

=

**63 sq.meter**

Required Length of carpet = 63/0.25 = 252 meter

**Required cost of carpeting the hall**= Length of carpet * cost of carpet per meter

= 252 * 20

=

**Rs. 5,040**

Let, the radius of cylinder = r cm

height = 3r cm

**Volume of cylinder = π * r**

^{2}* h--> 25872 = (22/7) * r

^{2}* 3r

---> 25872 = (66/7) * r

^{3}

---> 25872 * (7/66) = r

^{3}

---> 392 * 7 = r

^{3}

---> 2744 = r

^{3}

---> âˆ›2744 = r

--->

**r = 14**

**Area of the base of cylinder = Area of circle = π * r**

^{2}= (22/7) * 14 * 14

= 44 * 14

=

**616 cm square.**

^{2}

**Circumference of a circle (C) = π D**

= π 42

= (22/7) x 42

= 22 x 6

= 132 cm

Let the length of the rectangle (l) = 6x

and breadth (b) = 5x cm

**Perimeter of rectangle = 2 * (l + b)**

= 2 * (6x + 5x)

= 2(11x)

= 22x

**Perimeter of rectangle = Circumference of a circle**

---> 22x = 132

---> x = 6

The length of the rectangle (l) = 6x = 6 * 6 = 36 cm

and breadth (b) = 5x = 5 * 6 = 30 cm

**Area of rectangle = l x b**

= 36 * 30

=

**1080 cm**

^{2}^{2}

7 meter wide road is surrounding a circular park, having circumference of 176 meters.

Formula: **Circumference of circle = 2πr **

----> 2πr = 176

----> 2 x (22/7) x r = 176

----> r = (176 x 7) / (2 x 22)

----> r = 28 meter

Therefore, radius of circular park = 28 meter

Area of circular park = πr^{2}

= π (28)^{2}

**Area of the road = π(radius of circular park radius of road) ^{2} - Area of circular park**

= π (28 + 7)

^{2}- π (28)

^{2}

= π {35

^{2}- (28)

^{2}}

Since, a

^{2}- b

^{2}= (a + b) (a - b)

= π {35 + 28} {35 - 28}

= π x 63 x 7

= (22/7) x 63 x 7

= 22 x 63

= 1386 m

^{2}

Thus,

**Area of the road = 1386 m**

^{2}volume = a³

Let aâ‚ and aâ‚‚ are the sides of two cubes.

aâ‚³/aâ‚‚³ = 27/64 = (3/4)³

aâ‚ : aâ‚‚ = 3 : 4

The total surface area = 6a²

Ratio of their total surface area = 6aâ‚²/6aâ‚‚²

(aâ‚/aâ‚‚)² = (3/4)² = 9 : 16

Therefore, the ratio of their total surface area = 9 : 16

Area of four walls = 2h*(l + b)

Area of base = lb

Area of wet surface = Area of four walls + Area of base

= 2h*(l + b) + lb

=2*1.25(6+4) + (6*4)

=25 + 24 = 49 m²

Therefore, The total area of the wet surface = 49 m²

^{2}

Given, Perimeter of the semi-circle = 72 cm

WKT, Perimeter of the semi-circle = πr + 2r

r[(22/7) + 2] = 72

r * [(22 + 14)/7] = 72

r * (36/7) = 72

r = (72*7)/36

r = 14 cm

Therefore, area of semi circle = (πr

^{2})/2

= (22/7)*(14*14)/2

Area of sem-circle = 308 cm

^{2}.

^{2}. If the same wire is bent into the form of a circle, then the area of the circle is (use π = 22/7)

^{2}

^{2}

Side of the square, a

^{2}= 121

a = √121 = 11 cm

Length of the wire = 4*a = 4*11 = 44 cm

Circumference of the circle = Length of the wire

2πr = 44

2*(22/7)*r = 44

r = (44*7)/(2*22)

r = 7 cm

Therefore, area of circle = πr

^{2}

= (22/7)*(7)

^{2}

Area of circle = 154 cm

^{2}.

Given, a wooden box 'A' with dimensions = 8m * 7m * 6m

= 800cm *700cm * 600cm

Also given, a rectangular box 'B' with dimensions = 8cm * 7cm * 6cm

The maximum number of type B rectangular boxes that can be carried in 1 wooden box of type A is

= Volume of box A/Volume of box B

= (800*700*600)/(8*7*6)

= 10,00,000.

Number of type B rectangular boxes = 10,00,000.

WKT, Volume of cube = a

^{3}

Volume of cube 1, V

_{1}= a

_{1}

^{3}

Volume of cube 2, V

_{2}= a

_{2}

^{3}

V

_{1}/V

_{2}= a

_{1}

^{3}/a

_{2}

^{3}= 8/27

a

_{1}/ a

_{2}= 2/3

a

_{1}: a

_{2}= 2 : 3

Surface area of cube 1, S

_{1}= 6a

_{1}

^{2}

Surface area of cube 2, S

_{2}= 6a

_{2}

^{2}

S

_{1}/S

_{2}= 6a

_{1}

^{2}/6a

_{2}

^{2}

= 2

^{2}/3

^{2}

= 4 /9

Therefore, required ratio = 4:9.

Here, r - inradius of incircle.

Given, perimeter of triangle = 15 cm

Therefore, semi-perimeter (S) of triangle = 15/2 cm

Radius of incircle = 3 cm

Inradius of triangle, r = Area of triangle/Semi-perimeter of triangle

3 = Area of triangle/(15/2)

Area of triangle = 3*(15/2) = 45/2 cm²

Given that, volume of the prism = 270 cm³

Volume of the prism = Base*Height

270 = (45/2)*h

h = (270*2)/45

h = 12 cm

Therefore, the area of ground = 1000/0.25 = 4000 m²

Breadth = 50 m

Length of rectangle = Area/Breadth

Length = 4000/50 = 80 m

When the length of the ground is increased by 20m,

New length = 80 + 20 = 100m

New area of ground = 100 * 50 = 5000 m²

So, Expenditure of making the ground at same rate = 5000*0.25

= Rs. 1250

Perimeter of rectangle = 2(16x + 9x) = 100

25x = 100

x = 2

Area of rectangle = 32*18 = 576 m

^{2}

Given that, Area of rectangle = Area of square

Area of square, a

^{2}= 576 m

^{2}

Side of the square, a = 576 = 25m

Perimeter of the square = 4a = 4*24 = 96m.