NCERT Maths Specific Questions Of Mensuration for 9 and 10 grade - Future Study Point

NCERT Maths Specific Questions Of Mensuration for 9 and 10 grade

Specific mensuration questions

NCERT Maths Specific Questions Of Mensuration for 9 and 10 grade

Mensuration is the part of mathematics that was born in the lap of geometry in which we study physical quantity like area, volume, perimeter and another related quantity of all kinds of geometrical figures. The questions of mensuration obligatorily are a piece of the maths study material of each class and the questions of mensuration are likewise asked in each competitive exam additionally so this post of Future Study Point is helpful for everyone. Mensuration is ordered into two sections (i) Two-dimensional geometrical figure (ii) Three-dimensional geometrical figure, here we will study specific questions of mensuration for which students commonly required assistance from his mentor or educator. These are the specific questions of mensuration that are posed in the CBSE board exam and in the competitive exams. We trust you will get support from the way of solutions for illuminating the questions. Questions are clarified through a step-by-step way by a specialist in maths so that everyone could comprehend the way of solution.

Specific mensuration questions


This post consists of a few examples with the solutions and some of the unsolved questions which are taken from the NCERT and different maths CBSE guides.

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NCERT Maths Specific Questions Of Mensuration for 9 and 10 grade

Example(1)-Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs. 4 for 1000 cm2, find the cost of cardboard required for supplying 250 boxes of each kind.


We have to determine total cost of cardboard required to supply 250 boxes of each type

The cardboard required to prepare bigger box  = TSA of box + 5% of TSA of the box

TSA of box  = 2(lb + bh + hl) = 2(25×20 + 20 × 5 + 5×25)

⇒2×(500 + 100 + 125) = 2 ×725 =14 50

Additional cardboard is required  to buid the box  = 5 % of 1450 =5/100 ×(1450)=72.5

Cardboard required to build bigger box = 1450 + 72.5 =1522 .5

Cardboard required for 250 bigger boxes = 250× 1522.5 = 380625

Therefore 380625 cardboard is required to prepare 250 bigger boxes

TSA of small box of the dimension = 2(15×12 + 12×5 + 15× 5) = 2 ×(180 + 60 +75) = 2×315= 630

Additional cardboard is required  to build smaller box  = 5% of 630 = 5/100 ×(630) = 31.5

Cardboard required to build this box = 630 + 31.5 = 661.5

Cardboard required for 250 smaller boxes = 250× 661.5 = 165375

Therefore 165375 cardboard is required to prepare 250 boxes

Cardboard required to prepare 250 boxes of both kinds = 380625 + 165375 = 546000

The cost of 1000 cardboard is = Rs 4

Cost of 546000 cardboard will be

= (4/1000)×546000 = 2184

Therefore the cost of cardboard required to prepare 250 boxes of each type is Rs 2184

Example 2-   A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm andlength is 1 m.


The length of road roller(h) = 1 m

The diameter of road roller = 84 cm

Its radius = 42 cm

The area occupied by it on the road in one revolution

= The curved surface area of the road

= 2πrh

= 2×(22/7) ×0.42×1 = 2.64

Therefore area occupied by it on the road in one revolution =2.64 sq.m

In 750 revolution the whole of the road is covered

In 750 revolution the roller will cover  the area

=2.64 × 750=1980

Therefore the area of road = 1980 sq.m

Example 3. A cubical block of the side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of solid.

Ans. The greatest diameter of the hemisphere surmounted on one of the face of a cube is the side of the cube

The side of the cube = 7 cm

The surface area of solid =TSA of cube + CSA of hemisphere –area of the circular base of the hemisphere

⇒ 6a² + 2πr² – πr²= 6a² + πr²

a = 7 cm, r = 7/2 = 3.5 cm

⇒ 6 × a² + π ×r²

= 6×7²+(22/7)×3.5²

=6×49 +22×0.5×3.5

= 294 + 38.5

= 332.5

The surface area of solid = 332.5

Q4.A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends(see fig.). The length of the entire capsule is  14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Ans. The length of the entire capsule = 14 mm

The diameter of the capsule = 5 mm

Radius of the capsule = 2.5 mm

The length of the cylindrical part of the capsule = 14 –(2.5 + 2.5) = 9 mm

The surface area of capsule =  The curved surface area of its cylindrical part + curved surface area of its two hemispherical ends

⇒ 2πrh + 2×2πr²

=2πr(h +2r)

=2(22/7)× 2.5 (9+2×2.5)


Therefore 220 is the entire surface area of the capsule

Q5.A gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with a length 5 cm and a diameter of 2.8 cm.




The length of gulab jamun = 5 cm

The length of its cylindrical part (h) = 5 –(1.4 + 1.4) =5 –2.8=2.2

Therefore the length(h) of  cylindrical part is 2.2 cm

Diameter of gulam jamun = 2.8 cm

Radius of gulab jamun = 2.8/2 = 1.4 cm

Volume of gulab jamun(V) = Volume of cylinderical part + Volume of two hemispherical part

V = πr²h +2×(2/3)πr³


=13.552+(34.496/3) + (4/3)×(22/7)×1.4×1.4×1.4


=(40.656+34.496)/3 =75.152/3

The volume of 45 gulab jamun = 45×(75.152/3) =1127.28

45 gulab jamun contain sugar syrup =30 % of 1127.28 

= .30 ×1127.28 = 338.184 ≈ 338 

Therefore 338 sugar syrup contained in 45 gulab jamun.

Q6.A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see figure)



Ans. The volume of wood in the pen stand = Volume of the wooden cuboid – Volume of  4 depressions

The volume of the wooden cuboid = l × b × h

l = 15 cm, b = 10 cm, h = 3.5 cm

⇒The volume of the wooden cuboid = 15 × 10 × 3.5 = 525

The volume of 4 conical depression of each of radius(r) 0.5 cm and height(h) 1.4 cm

Since volume of the cone = (1/3)πr²h

The volume of 4 conical depression = 4×(1/3)×(22/7)×0.5×0.5×1.4

= 1.47

∴ The volume of wood in the pen stand = 525 – 1.47 =523.53

Therefore 523.53 is the volume of wood in the pen stand

Solve the following important questions generally asked in the CBSE board exams and competitive entrance exams.

Q1. The water is flowing within a circular pipe of diameter 1.4 cm at the speed of 1.8 km/h, how much time it will take to fill a cuboidal tank of 30000 liters.

Q2. A solid metallic sphere of the diameter of 16 cm is melted and recast into spherical bullets, how many bullets can be formed of the diameter 4 mm.

Q3. A solid metallic cylinder of the diameter 40 cm and height 50 cm is melted and recast into a wire of  4 mm diameter, find the length of wire so formed.

Q4. What will be the ratio of radius and height of a circular cylinder if its volume and radius is equal to the volume and radius of a sphere?

Q5. If the ratio of the volume of two-sphere is 27: 64, find the ratio between their surface areas.

Q6. The level of water in a cylindrical vessel of the 9 cm is raised by 12 cm when a certain ball is dropped inside it and submerged completely, finds the diameter of the ball.

We will be posting similar questions continually for your help, please subscribe to our website for the update of the information

NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for class 9 maths

Chapter 1- Number SystemChapter 9-Areas of parallelogram and triangles
Chapter 2-PolynomialChapter 10-Circles
Chapter 3- Coordinate GeometryChapter 11-Construction
Chapter 4- Linear equations in two variablesChapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s GeometryChapter 13-Surface Areas and Volumes
Chapter 6-Lines and AnglesChapter 14-Statistics
Chapter 7-TrianglesChapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions for class 9 science 

Chapter 1-Matter in our surroundingsChapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?Chapter 10- Gravitation
Chapter3- Atoms and MoleculesChapter 11- Work and Energy
Chapter 4-Structure of the AtomChapter 12- Sound
Chapter 5-Fundamental unit of lifeChapter 13-Why do we fall ill ?
Chapter 6- TissuesChapter 14- Natural Resources
Chapter 7- Diversity in living organismChapter 15-Improvement in food resources
Chapter 8- MotionLast years question papers & sample papers

NCERT Solutions for class 10 maths

Chapter 1-Real numberChapter 9-Some application of Trigonometry
Chapter 2-PolynomialChapter 10-Circles
Chapter 3-Linear equationsChapter 11- Construction
Chapter 4- Quadratic equationsChapter 12-Area related to circle
Chapter 5-Arithmetic ProgressionChapter 13-Surface areas and Volume
Chapter 6-TriangleChapter 14-Statistics
Chapter 7- Co-ordinate geometryChapter 15-Probability
Chapter 8-Trigonometry

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Chapter 1- Chemical reactions and equationsChapter 9- Heredity and Evolution
Chapter 2- Acid, Base and SaltChapter 10- Light reflection and refraction
Chapter 3- Metals and Non-MetalsChapter 11- Human eye and colorful world
Chapter 4- Carbon and its CompoundsChapter 12- Electricity
Chapter 5-Periodic classification of elementsChapter 13-Magnetic effect of electric current
Chapter 6- Life ProcessChapter 14-Sources of Energy
Chapter 7-Control and CoordinationChapter 15-Environment
Chapter 8- How do organisms reproduce?Chapter 16-Management of Natural Resources

NCERT Solutions for class 11 maths

Chapter 1-SetsChapter 9-Sequences and Series
Chapter 2- Relations and functionsChapter 10- Straight Lines
Chapter 3- TrigonometryChapter 11-Conic Sections
Chapter 4-Principle of mathematical inductionChapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbersChapter 13- Limits and Derivatives
Chapter 6- Linear InequalitiesChapter 14-Mathematical Reasoning
Chapter 7- Permutations and CombinationsChapter 15- Statistics
Chapter 8- Binomial Theorem Chapter 16- Probability

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NCERT Solutions for Class 11 Physics

Chapter 1- Physical World

chapter 3-Motion in a Straight Line

NCERT Solutions for Class 11 Chemistry

Chapter 1-Some basic concepts of chemistry

Chapter 2- Structure of Atom

NCERT Solutions for Class 11 Biology

Chapter 1 -Living World

NCERT solutions for class 12 maths

Chapter 1-Relations and FunctionsChapter 9-Differential Equations
Chapter 2-Inverse Trigonometric FunctionsChapter 10-Vector Algebra
Chapter 3-MatricesChapter 11 – Three Dimensional Geometry
Chapter 4-DeterminantsChapter 12-Linear Programming
Chapter 5- Continuity and DifferentiabilityChapter 13-Probability
Chapter 6- Application of DerivationCBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals

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