Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

Class 10 Mathematics by VP Classes

Class 10 : Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

The document Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.
All you need of Class 10 at this link: Class 10

Q1. Rohan is in the business of supplying water. He has three types of tankers of inner diameter 1.0 m to supply water to the customers. The length of the tankers is 6 m. [use π = 3.14] He decided to serve his customers with type ‘A’ tankers.

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev
Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev
Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

(a) Find the volume of tanker of type A.
(b) Which tanker has the minimum capacity?
(c) Which mathematical concept is used in the above problem?
(d) By choosing a tanker of ‘type-A’, which value is depicted by Rohan?

Sol. ∵ Diameter = 1.0 m

∴ Radius = 1/2 m 

Length of the tank = 6 m ⇒ h = 6 m
(a) Volume of the tanker of type-A
= πr2h

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev
(b) Volume of the tanker of ‘type-B’
= [Volume of the tanker-A] – [Volume of hemisphere]
Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev
Volume of the tanker of ‘type-C’ =  [Volume of the tanker-A] – [Volume of the cone]
Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

=  [4.71 – 0.157] cu. m = [0.157] cu. m
=  4.553 cu. m
Thus, the tanker of ‘type-B’ has the minimum capacity of 4.45 cu. m.
(c) Mensuration [volume of solid figures]
(d) Honesty

Q2. Shankar, a ‘Kulfi-vendor’ has three types (spherical, conical and cuboidal) of containers for making Kulfi :

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRevValue Based Questions- Surface Areas and Volumes Class 10 Notes | EduRevValue Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

He decided to serve the customers with ‘typeI’ container.

(a) Find the volume of the container ‘type-I’.
 (b) Which container has the minimum capacity?
 (c) Which mathematical concept is used in this problem?
 (d) By choosing to prepare to sell Kulfi using container of ‘type-I’, which value is depicted by Shankar?

Sol. (a) Finding the volume of Type-I container:
∵ Diameter = 7 cm
⇒ Volume of the container of ‘type-I’
= 4/3 πr3

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

(b) Volume of conical (type-II) container

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev
Volume of cuboidal (type-III) container
Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

Obviously, the capacity of the cuboidal (type-III) container is minimum, which is 122.5 cu. cm.
(c) Volume of solid bodies (mensuration)
(d) Honesty

Q3. Prashant has undertaken a contract to build a wall of 9m long, 2.5m thick and 6m high. His labour is to be calculated according to the number of bricks used to complete the wall. In the market three types of bricks are available.

Type-I :Each measuring 25cm × 11.25cm × 6cm
 Type-II : Each measuring 20cm × 8cm × 10cm
 Type-III : Each measuring 25cm × 10cm × 9cm
 Prashant used bricks of type-III.

(a) Find the number of bricks of type-III required to build the wall.
 (b) In which case, maximum number of bricks will be used?
 (c) Which mathematical concept is used in the above problem?
 (d) By using the bricks of type-III, which value is depicted by Prashant?

Sol. Volume of a brick of type-I =l × b × h

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

= 25 cm × 11.25 cm × 6 cm = 1687.50 cm3
Volume of a brick of type-II
= l × b × h = 20 cm × 8 cm × 10 cm
= 1600 cm3
Volume of a brick of type-III
= l × b × h = 25 cm × 10 cm × 9 cm
= 2250 cm3
Volume of the wall
= l × b × h = 9 × 2.5 × 6 m3
= 135 × 106 cm3
(a) ∴ Number of bricks of type-III required for building the wall

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev
(b) Number of bricks required for building the wall using type-I bricks:
Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev
Number of bricks of type-II required for building the wall

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

⇒ In case of type-II, the maximum number of bricks will be required.
(c) Volume of solid bodies (mensuration)
(d) Honesty

Q4. Sampat has set up his juice shop. He has three types of cylindrical glasses as given below :

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev   A cylindrical glass with inner diameter 7 cm and height as 10 cm.

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev A cylindrical glass with inner diameter 4 cm and height as 14 cm.

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev  A cylindrical glass with inner diameter 14 cm and height as 4 cm.

He decided to serve the customers in ‘type-I’ of glasses.
 (a) Find the volume of the glass of type-I.
 (b) Which glass has the minimum capacity?
 (c) Which mathematical concept is used in above problem?
 (d) By choosing a glass of type-I, which value is depicted by juice seller Sampat?

Sol. (a) Diameter of glass of type-I = 7 cm

∴  Radius = 7/2 cm

Height = 10 cm
⇒ Volume = πr2h

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev
(b) Diameter of glass of type-II
= 4cm
∴ Radius = 2cm
Height = 14 cm
⇒ Volume = πr2h

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

= 176 cm3
Diameter of glass of type-III
= 14 cm

∴ Radius = 14/2 cm = 7 cm
Height = 4 cm
⇒ Volume = πr2h

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev
= 22×28 cm3
= 308 cm3

⇒ The glass of type-II has the minimum capacity.
(c) Mensuration [volume of solid bodies]
(d) Honesty.

Q5. A community-well with 10 m inside diameter is dug 14 m deep. For spreading the earth taken out of it, there are two options :  
 (i) It is spread all around the well to awidth of 6 m to form an embankment.
 (ii) It is spread evenly on a rectangular surface of 25m × 11m.

The contractor charges according to the height of the spreadover mud. He charges by choosing the option-(i).
 (a) Find the height of the spreadover mud in both options.
 (b) Which mathematical concept is involved in the above problem?
 (c) In which case cost of digging the well is less?
 (d) Which values are depicted by the contractor by charging according to option-(i)?

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

Sol. Inside diameter of the well = 10 m
⇒ Inside radius of the well = 5 m

Depth of the well = 14 m
∵ Volume of the earth dugout = πr2h

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev
= 22 × 5 × 5 × 2 m3
= 1100 m3

(a) Height of the spreadover mud: option-(i)  
Area of the shaded-region (base of thespreadover mud)

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev
= πR2 – πr2
= π [R2 – r2]
[Here: R = 11 m and r = 5 m]
= π [(R – r) (R + r)]
Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

Let height of the spreadover mud (embankment) be h1,

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev
Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

option-(ii)
Let the height of the spreadover mud = h2,

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

(b) Mensuration [volume of solid bodies]
(c) ∵ The height of spreadover mud is less  in option-(i).
∴ The cost of digging the well is less in option-(i).
(d) Community-service and honesty.

Q6. A contractor is entrusted to erect a tent for flood-victims. He is allowed a fixed amount for this task. He has two options :
 (i) to erect a tent which is cylindrical upto a height of 3 m and conical above it. The diameter of the base is 105 m and slant height of the conical part is 53 m.

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

(ii) to erect the tent as described in option(i), only replacing the conical part as a hemispherical part.

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

The contractor chooses the option-(ii) and decides to donate the extra (difference) canvas to be used in this case.
(a) How much canvas is donated by the contractor?
(b) Which mathematical concept is used in the above problem?
(c) By choosing the option-(ii) to erect the tent, which value is depicted by the contractor?

Sol. (a)

Total canvas used in option-(i) = [Curved surface area +  [Curved surface area  of the cylindrical part] of the conical part] = [2πrh + πrl]

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

Total canvas used in option-(ii) = [Curved surface area  +  [Curved surface area of cylindrical part] of hemispherical part]=[2πrh + 2πr2]

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRevValue Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

Difference in areas of canvas required in the above two options.

= 18315 – 9733 m2
= 8582 m2
⇒ The contractor donated 8582 m2 of canvas.
(b) Mensuration [volume of solid bodies]
(c) Patriotism

Q7. Sushant has a vessel, of the form of an inverted cone, open at the top, of height 11 cm and radius of top as 2.5 cm and is full of water. Metallic spherical balls each of diameter 0.5 cm are put in the vessel due to which 2/5 th of the water in the vessel flows out. Find how many balls were put in the vessel. Sushant made the arrangement so that the water that flows out irrigates the flower beds. What value has been shown by sushant?

Sol. Base radius of the cone (r) = 2.5cm
Height of the concial part (h) = 11cm

Using V = 1/3 πr2h, , the volume of the conical vessel.

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev of water is flown out due to ‘n’ spherical balls each of radius = 0.5 cm

∵ Volume of n balls =    Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

⇒ n[volume of one ball]

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev 

⇒ Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev
Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev
Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

Thus, the required number of leadshots  = 55
Value : To keep the plants green for pollution free surroundings.

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Complete Syllabus of Class 10

Dynamic Test

Content Category

Related Searches

Semester Notes

,

study material

,

practice quizzes

,

Sample Paper

,

Objective type Questions

,

pdf

,

ppt

,

Previous Year Questions with Solutions

,

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

,

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

,

video lectures

,

shortcuts and tricks

,

Viva Questions

,

Exam

,

mock tests for examination

,

Extra Questions

,

Important questions

,

past year papers

,

Summary

,

Value Based Questions- Surface Areas and Volumes Class 10 Notes | EduRev

,

Free

,

MCQs

;