# Mensuration Questions and Answers updated daily – Aptitude

Mensuration Questions: Solved 782 Mensuration Questions and answers section with explanation for various online exam preparation, various interviews, Aptitude Category online test. Category Questions section with detailed description, explanation will help you to master the topic.

## Mensuration Questions

81. How many Spherical bullets can be made out of a lead cylinder 21 cm high and 6 cm radius, each bullet being 1.5 cm in diameter?

Correct Ans:1344
Explanation:
Given, cylinder height = 21 cm Radius = 6 cm
Volume of the cylinder = πr2h = π * 6 * 6 * 21
Given 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)πr3
= (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
Workspace

82. If a circle circumscribes a rectangle with side 16 cm and 12 cm, then what is the area of the circle ?

Correct Ans:100π sq cm
Explanation:

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
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83. A cylindrical pencil of diameter 1.2 cm has one of its end sharpened into a conical shape of height 1.4 cm. The volume of the material removed is

Correct Ans:1.056 cm3
Explanation:
Given:
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 cm3
Workspace

84. If the numerical value of the perimeter of an equilateral triangle is times the area of it, then the length of each side of the triangle is

Correct Ans:4 unit
Explanation:
Given:
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.
Workspace

85. Cylindrical cans of cricket balls are to be packed in a box. Each can has a radius of 7 cm and height of 30 cm. Dimension of the box is l = 76 cm, b = 46 cm, h = 45 cm. What is the maximum number of cans that can fit in the box?

Correct Ans:21
Explanation:
Since, both the box and cans are hard solids, simply dividing the volume wonâ€™t work because the shape canâ€™t be deformed.

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
Workspace

86. The diameter of the base of a cylindrical drum is 49 dm and its height is 36 dm. It is full of kerosene. How many tins, each of size 22 cm x 18 cm x 49 cm can be filled with kerosene from the drum?

Correct Ans:3,500
Explanation:
Given, Diameter of the base of a cylindrical drum = 49 dm
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 r2h
= (22/7) * 245 * 245 * 360
= 22 * 35 * 245 * 360 cm3

Volume of tin = (22 * 18 * 49) cm3

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
Workspace

87. The perimeter of a square is equal to twice the perimeter of a rectangle of length of 8 cm and breadth 7 cm. What is the circumference of a semicircle whose diameter is equal to the side of the square? (Rounded off to the two decimal places) (take π = 3.14)

Correct Ans:38.55 cms
Explanation:
Given, 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
Workspace

88. A hall is 10 meters long and 8 meters wide. What will be the cost of carpeting the room if 0.5 meters of space is left around the room? The rate of 0.25 meter wide carpet is Rs. 20 per meter.

Correct Ans:Rs. 5040
Explanation:
Given, Length of the hall (rectangular hall) = 10 meters
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
Workspace

89. The capacity of a cylindrical vessel is 25,872 litres. If the height of the cylinder is three times the radius of its base, what is the area of the base?

Correct Ans:616 cm square
Explanation:
Given, Volume of cylindrical vessel = 25,872 litres
Let, the radius of cylinder = r cm
height = 3r cm

Volume of cylinder = π * r2 * h
--> 25872 = (22/7) * r2 * 3r
---> 25872 = (66/7) * r3
---> 25872 * (7/66) = r3
---> 392 * 7 = r3
---> 2744 = r3
---> âˆ›2744 = r
---> r = 14

Area of the base of cylinder = Area of circle = π * r2
= (22/7) * 14 * 14
= 44 * 14
= 616 cm square.
Workspace

90. A circular wire of diameter 42 cm is bent in the form a rectangle whose sides are in the ratio 6 : 5. The area of the rectangle is (Use π = 22/7)

Correct Ans:1080 cm2
Explanation:
Given, Diameter of circle (D) = 42 cm

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 cm2
Workspace

91. A 7 meter wide road runs outside around a circular park, whose circumference is 176 meters. The area of the road is:

Correct Ans:1386 m2
Explanation:

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 = πr2
= π (28)2

= π (28 + 7)2 - π (28)2
= π {352 - (28)2}
Since, a2 - b2 = (a + b) (a - b)
= π {35 + 28} {35 - 28}
= π x 63 x 7
= (22/7) x 63 x 7
= 22 x 63
= 1386 m2

Thus, Area of the road = 1386 m2

Workspace

92. If the volume of two cubes are in the ratio 27 : 64 and then the ratio of their total surface area is?

Correct Ans:9 :16
Explanation:
Vâ‚/Vâ‚‚ = 27/64
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
Workspace

93. A cistern 6m long and 4m wide, contains water up to the depth of 1m 25cm. The total area of the wet surface is ?

Correct Ans:49 m.sq
Explanation:
We know that,
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
Workspace

94. If the perimeter of a semi-circle is 72 cm. What will be the area of that semi-circle?

Correct Ans:308 cm 2
Explanation:
Let the radius of semi-circle be r cm.
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 = (πr2)/2
= (22/7)*(14*14)/2
Area of sem-circle = 308 cm2.
Workspace

95. A wire, when bent in the form of a square, encloses a region of area 121 cm2. If the same wire is bent into the form of a circle, then the area of the circle is (use π = 22/7)

Correct Ans:154 cm2
Explanation:
Given area of square = 121cm2
Side of the square, a2 = 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 = πr2
= (22/7)*(7)2
Area of circle = 154 cm2.
Workspace

96. A wooden box of dimension 8m * 7m * 6m is to carry rectangular boxes of dimensions 8cm * 7cm * 6cm. The maximum number of boxes that can be carried in 1 wooden box is

Correct Ans:10,00,000
Explanation:

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.
Workspace

97. Volumes of two cubes are in the ratio 8 : 27. Ratio of their surface area is

Correct Ans:4:9
Explanation:
Volumes of two cubes are in the ratio 8 : 27
WKT, Volume of cube = a3
Volume of cube 1, V1 = a13
Volume of cube 2, V2 = a23
V1/V2 = a13/a23 = 8/27
a1/ a2= 2/3
a1 : a2 = 2 : 3

Surface area of cube 1, S1 = 6a12
Surface area of cube 2, S2 = 6a22
S1/S2 = 6a12/6a22
= 22/32
= 4 /9
Therefore, required ratio = 4:9.
Workspace

98. The perimeter of the triangular base of a right prism is 15 cm and radius of the incircle of the triangular base is 3 cm. If the volume of the prism be 270 cm³ then the height of the prism is

Correct Ans:12 cm
Explanation:

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

Workspace

99. A playground is in the shape of a rectangle. A sum of Rs. 1000 was spent to make the ground usable at the rate of 25 paise per sq. m. The breadth of the ground is 50m. If the length of the ground is increased by 20 m, what will be the expenditure in making this ground usable at the same rate?

Correct Ans:Rs. 1250
Explanation:
Given, Rs. 1000 spent to make the ground at the rate of 25 paise sq.m.
Therefore, the area of ground = 1000/0.25 = 4000 m²
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
Workspace

100. Area of a square is equal to the area of a rectangle. The ratio of length and breadth of the rectangle is 16 : 9 then find the perimeter of the square if the perimeter of the rectangle is 100 m.

Correct Ans:96 m
Explanation:
Perimeter of rectangle = 2(l + b)
Perimeter of rectangle = 2(16x + 9x) = 100
25x = 100
x = 2
Area of rectangle = 32*18 = 576 m2
Given that, Area of rectangle = Area of square
Area of square, a2 = 576 m2
Side of the square, a = 576 = 25m
Perimeter of the square = 4a = 4*24 = 96m.
Workspace

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