All popular competitive exams like IBPS PO 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 IBPS PO 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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