All popular competitive exams ask questions on Volume and Surface Area making it a must for you to master this topic. Read on to find out how to solve questions on Volume and Surface Area.
Mensuration
as a whole is a formula oriented topic where most mensuration problems can be
solved by using the appropriate formulas. After discussing introduction to mensuration with a list of mensuration formulas and how to solve mensuration problems on area and perimeter, it’s time to move to the third and last part of this series on
mensuration where we discuss mensuration problems on volume and surface area.
In the third post in series on mensuration problems we will be discussing volume
and surface area based mensuration problems on 3D shapes. So go through the
list of formulas and gear up for questions on volume and surface area that are asked in SBI PO, IBPS and SSC CGL Exam.
The
Mensuration problems on volume and surface area of 3D shapes is divided in 3
parts Basic Problems, Problems on Lateral and Curved Surface Area and finally,
Problems on Proportionality of Volume.
Set I: Basic Mensuration Problems on Volume and Surface Area
Questions
discussed in this part involve basic usage of the volume and surface area formula
with a little bit of modification.
Problem 1: The volume
of a circular cone is 100p cm^{3} and its height is 12cm. What is the slant
height?
Solution 1:
Surely you
know the formula to find the slant height and the volume of a circular cone
From the
diagram we know that ‘l’ is the slant height. To find ‘l’ we need to know the
radius and the height of the cone. Once we have these two, we can easily use
the hypotenuse formula get the value of ‘l’ since the three for a right angled
triangle. We already have the height of the cone so we just need to find the
radius of the cone.
Step 1
We know the
formula for the volume of a cone
Volume of a
Cone = (1/3)pr^{2} h cubic Units (i)
Volume of a
Cone = 100p cm^{3} (ii)
Step 2
Both
Equation (i) and (ii), since both of them
(1/3)pr^{2} h = 100p
Putting the
value of height and cancelling p on both the ends we get –
(1/3)r^{2}
x 12 = 100
r^{2 }=
(100 x 3) / 12
r^{2}
= 25
r = 5
So the
radius of the circular cone is 5cm
Step 3
Now that we
know the value of both, height and radius, we can simply put the values in the
formula and get the value of ‘l’.
Slant Height
of cone, l = Ö{h^{2}+r^{2}}
Slant Height
of cone, l = Ö{12^{2}+5^{2}}
Slant Height
of cone, l = Ö{144+25}
Slant Height
of cone, l = Ö169
Slant Height
of cone, l = 13
So the slant
height of the circular cone is 13cm.
So every
time you see a question like this, go through the list of formulas of volume
and surface area of the given shape in your head and solve such mensuration
problems.
Set II: Mensuration Problems on Lateral and Curved Surface Area
Volume and
surface area problems discussed in this set are about the interiors of closed
3D shapes. The trick to solve such questions on volume and surface area, lies
in being able to identify the right formula and then use in to get values of
the needed parameters based on the values that you and have and the finally get
the correct answer.
Problem 1: All the four
lateral walls and the ceiling of a room of length 12 feet, breadth 10 feet and
height 8 feet are to be painted. Find the total cost, if the cost of painting
is Rs 15/sq. feet and the doors and windows in the room occupy 40 sq. feet of
the area.
Solution 1:
From the
information given in the question we infer that the 3D shape here is a cuboid.
The four walls and the ceiling need to be painted sans the area occupied by the
doors and windows. We can have two different approaches to find this area.

The
Lateral Surface Area will include the area covered by the walls on all the four
sides. We can find the Lateral Surface Area, add the area of the roof and from
this total subtract the area occupied by doors and windows.

The
Total Surface Area will include the area of all the four walls, the floor and
the roof. We can take the total surface area of the room and from that subtract
the area of the floor and the area occupied by doors and windows.
In both
these cases we will get the total area in the room that is to be painted.
Step 1
Area to be
painted is = (LSA + Area of the Ceiling) – Area of Doors and Windows
Area to be
painted is = [2h(l+b) + (lxb)] – 40
Area to be
painted is = [2x8(12+10) + (12x10)] – 40
Area to be
painted is = [16x 22 + 120] – 40
Area to be
painted is = 352 + 120 – 40
Area to be
painted is = 432 sq. feet
Step 2
Now to find
the total cost of painting the room we will multiply the total area to be
painted with the cost of painting per sq. unit.
Cost of
Painting the Room = Total Area to be Painted x Cost of Painting/ sq. unit
Cost of Painting the Room = 432 x 15
Cost of
Painting the Room = 6480
Therefore
the total cost of painting the room is Rs 6480
Alternatively,
as mentioned the area to be painted can also be calculated by the following
method –
Area to be
Painted = TSA – Area of the roof – Area of Doors and Windows
Cost of
Painting the Room = 2(lb+bh+hl) – (lxb) – 40
Remember
every time you solve volume and surface area based problems where you have to
find the area to be painted, tiled or anything or that sort to eliminate parts
of the cube/ cuboid that will not be painted or tiled.
Problem 2: What is the
approximate area of the canvas required to make a conical tent of radius 30
feet and height 18 feet?
Solution 2:
In the above
question, the area of the canvas is actually the cloth that is used to make
this tent that is in the shape of a conical tent. The tent will look like the
diagram below.
When we make
this tent, the cloth will not be needed for the base, it will only be needed for
the Curved Surface Area. Therefore the area of the cloth will be equal to the
CSA of the cone.
Step 1
To Find CSA
we need the value of the radius and the slant length. From the information
given, we have only the radius and the height but have to find the slant
length.
We know the
formula to calculate the slant length.
Slant Height
of cone, l = Ö{h^{2}+r^{2}}
Replacing
values in this formula
L = Ö{30^{2}+18^{2}}
L = Ö{900+324}
L = Ö1224
L = 35
approx. (35^{2} = 1235,
which is very close to 1224, therefore we can approximate this value)
Step 2
Substituting
the values in the CSA formula
CSA = prl
CSA = 22/7 x
30 x 35
CSA = 22 x
30 x 5
CSA = 3300
sq. feet
Therefore,
the canvas required to make this conical tent is 3300 sq. feet.
The trick in
these volume and surface area problems lies in being able to identify the
correct formula and then decide what to add or eliminate based on the
conditions to reach the correct answer for the given mensuration problem.
Set III: Mensuration Problems on Proportionality of Volume
Problems
form volume and surface area discussed in this section are really important for
SBI PO, IBPS and SSC CGL Exam. In such questions there is one solid which is
usually recast into one or many solids. But the trick is to keep in mind that
from volume and surface area, one value remains the same while the other
changes. What remains constant is volume and surface area is what changes.
Whether it is being cast into a totally different 3d shape or the same shape of
smaller sizes, the only parameter that remains constant is the volume and all
other parameters change.
Problem 1: A metal
sphere of diameter 16cm is melted and small spheres of radius 2cm each are cast
from the molten metal. How many such spheres will be formed?
Solution 1:
In volume
and surface area questions like this, where one 3D shape is being recast into
another 3D shape, the volume always remains the same. So in this question, the
volume of the bigger sphere will be equal to the volume of the smaller spheres.
We need to find how many spheres of radius 2m will be formed when the bigger
sphere of 16cm diameter is melted.
Step 1
Let’s start
with the initial volume; the initial volume is actually the volume of the
sphere with the diameter 16cm.
Radius = D/2
Radius =
16/2 = 8cm
Volume of
the sphere = (4/3) pr^{3}
Initial
Volume = 4/3 x p x 8^{3}
(i)
Step 2
The final
volume will be the volume of each of the small sphere multiplied by the number
of spheres. Let us assume the number of spheres to be ‘n’ and we already know
the radius of the smaller spheres. From this we can easily form an equation for
the final volume.
Volume of
the sphere = (4/3) pr^{3}
Final Volume
= n x (4/3) p2^{3}
(ii)
Step 3
Now we can
equate the initial and final volume to get the answer.
Equating (i)
and (ii) we get
4/3 x p x 8^{3} = n x (4/3) p2^{3}
8^{3}
= n2^{3}
n = 8^{3}/
2^{3}
n = 64
Therefore we
can say 64 small spheres will be formed form the bigger sphere.
The bottom
line of solving such mensuration problems in volume and surface area, is that
volume is constant when a solid is converted from one form to the other.
Problem 2: Find the
number of bricks, each measuring 25cm x 12.5cm x 7.5cm, required to construct a
wall, 12m long, 5m high and 0.25m thick. While the sand and cement mixture
occupies 5% of the total volume.
Solution 2:
Questions
like this on volume and surface area are asked in most competitive exams. This
question is pretty similar to the previous question, just that it is a reverse.
Here smaller 3D are coming together to form a bigger 3D shape. We all know that
walls are made of bricks along with a mixture of sand and cement. Bricks are
nothing but small cuboids that come together to form a large cuboid that is the
wall. So here also the volume is constant. But there is a small catch that 5%
of the total volume of the wall is the sandcement mixture.
This is
actually the volume of bricks + sandcement mixture and the total volume of the
wall.
Step 1
We know that
the wall is a made up of bricks and cementsand mixture, where cementsand
mixture is 5% of the wall. This implies that the wall is 100% .
Volume of
Bricks + Volume of sandcement mixture = Volume of the wall
Volume of
Bricks + 5% of the Volume of the Wall = 100% of the Wall
Volume of
Bricks = 95% of the Wall
Step 2
Let’s
calculate the volume of the bricks and also assume the number of needed to form
the wall be ‘n’
Volume of
Bricks = n x l x b x h
Volume of
Bricks = n x 25 x 12.5 x 7.5 (i)
Now note
that all the given measurements of the brick are in ‘cm’, but the measurements
of the wall are in ‘m’, so we will convert all the units in ‘m’
Volume of
Bricks = n x 25/100 x 12.5/100 x 7.5/100
(1m = 100cm)
Step 3
Volume of
the Wall = l x b x h
Volume of
the Wall = 12 x 5 x 0.25 (ii)
Step 4
Equating (i)
and (ii) we get
n x 25/100 x
12.5/100 x 7.5/100 = 12 x 5 x 0.25
n = 6080
Therefore
6080 bricks are needed to construct this wall.
This is how
you should approach such questions in volume and surface area, when two types
or solids or number of solids are coming together to form another solid.
Remember the volume remains constant.
Mensuration Problems on Volume and Surface Area
Try these mensuration
problems on Volume and Surface Area and leave your answers in the comments
section below
Question 1: The volume of a circular
cone is 100 𝜋 cubic cm and its height is 12 cm. What is the
slant length of the cone?
1) 8 cm 2) 7 cm 3)
13 cm 4) 9 cm 5) None of these
Question 2: All the four lateral walls
and the ceiling of a room of length 12 ft, breadth 10 ft and height 8 ft are to
be painted. Find the total cost if the cost of painting is ` 15 per sq ft and
the doors and windows occupy 40 sq ft of area
1) ` 1720 2) ` 6480 3) ` 6440 4) ` 3960 5) None of these
Question 3: The height and length of a
wall of a room is 6 m and 4 m respectively. If the rate of painting is ` 75 per
sq m then, what will be the total cost if three walls (having no windows and
doors) are to be painted?
1) ` 5100 2) ` 5400 3) ` 4800 4) ` 5200 5) None of these
Question 4: A large sphere of diameter
18 cm is melted, and cast into small balls of radius 3 cm each. How many small
balls will be formed?
1) 3 2) 9 3) 27 4) 81 5) None of these
Question 5: There cubes of sides 3 cm,
4 cm and 5 cm respectively are merged together to form a single large cube.
What is the side of the cube so formed?
1) 12 cm 2) 6 cm 3) 24 cm 4) 60 cm 5) None of these
Remember most mensuration problems in
SBI PO, IBPS, SSC GL Exam are actually from volume and surface area.
To practice
more mensuration problems on volume and surface area download free ebooks.
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