RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids
RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids

Class 10: Maths Chapter 17 solutions. Complete Class 10 Maths Chapter 17 Notes.

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids

RS Aggarwal 10th Maths Chapter 17, Class 10 Maths Chapter 17 solutions

Question 1:

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 1

Radius of the cylinder = 14 m
And its height = 3 m
Radius of cone = 14 m
And its height = 10.5 m
Let l be the slant height

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 1

Curved surface area of tent
= (curved area of cylinder + curved surface area of cone)

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 1

Hence, the curved surface area of the tent = 1034
Cost of canvas = Rs.(1034 × 80) = Rs. 82720

Question 2:

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 2

For the cylindrical portion, we have radius = 52.5 m and height = 3 m
For the conical portion, we have radius = 52.5 m
And slant height = 53 m
Area of canvas = 2rh + rl = r(2h + l)

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 2

Question 3:

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 3


Height of cylinder = 20 cm
And diameter = 7 cm and then radius = 3.5 cm
Total surface area of article
= (lateral surface of cylinder with r = 3.5 cm and h = 20 cm)

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 3

Question 4:

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 4


Radius of wooden cylinder = 4.2 cm
Height of wooden cylinder = 12 cm
Lateral surface area

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 4


Radius of hemisphere = 4.2 cm
Surface area of two hemispheres

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 4


Total surface area = (100.8 + 70.56) π cm2
= 538.56 cm2
= 171.36 π
= 171.36 × 227  cm2
= 538.56 cm2
Further, volume of cylinder = πr2h = 4.2 × 4.2 × 12 π cm2
= 211.68 π cm2
Volume of two hemispheres = 2 × 23 πr3 cu.units
= 43 π  × 4.2 × 4.2 × 4.2
= 98.784 cm3
Volume of wood left = (211.68 – 98.784) π
= 112.896 π cm3
= 112.896 × 227  cm3
= 354.816 cm3

Question 5:
Radius o f cylinder = 2.5 m
Height of cylinder = 21 m
Slant height of cone = 8 m
Radius of cone = 2.5 m
Total surface area of the rocket = (curved surface area of cone + curved surface area of cylinder + area of base)

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 5

Question 6:

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 6


Height of cone = h = 24 cm
Its radius = 7 cm

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 6


Total surface area of toy

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 6
https://youtube.com/watch?v=YA0xN128QGg%3Ffeature%3Doembed

Question 7:

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 7


Height of cylindrical container h1 = 15 cm
Diameter of cylindrical container = 12 cm
Volume of container = 

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 7


Height of cone r2 = 12 cm
Diameter = 6 cm
Radius of r2 = 3 cm

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 7


Radius of hemisphere = 3 cm
Volume of hemisphere = 

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 7


Volume of cone + volume of hemisphere
= 36π + 18π = 54π
Number of cones

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 7


Number of cones that can be filled = 10

Question 8:


Diameter of cylindrical gulabjamun = 2.8 cm
Its radius = 1.4 cm
Total height of gulabjamun = AC + CD + DB = 5 cm
1.4 + CD + 1.4 = 5
2.8 + CD = 5
CD = 2.2 cm
Height of cylindrical part h = 2.2 cm
Volume of 1 gulabjamun = Volume of cylindrical part + Volume of two hemispherical parts

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 8


Volume of 45 gulabjamuns = 45 × 25.07 cm3
Quantity of syrup = 30% of volume of gulabjamuns
= 0.3 × 45 × 25.07  = 338.46 cm3

Question 9:
Diameter = 7cm, radius = = 3.5 cm
Height of cone = 14.5 cm – 3.5 cm = 11 cm

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 9


Total surface area of toy = 

Question 10:
Diameter of cylinder = 24 m
Radius of cylinder = 242 = 12 cm
Height of the cylinder = 11 m
Height of cone = (16 – 11) cm = 5 cm
Slant height of the cone l = 

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 10


Area of canvas required = (curved surface area of the cylindrical part) + (curved surface area of the conical part)

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 10

Question 11:
Radius of hemisphere = 10.5 cm
Height of cylinder = (14.5 – 10.5) cm = 4 cm
Radius of cylinder = 10.5 cm
Capacity = Volume of cylinder + Volume of hemisphere

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 11

Question 12:

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 12


Height of cylinder = 6.5 cm
Height of cone = h2 = (12.8-6.5) cm = 6.3 cm
Radius of cylinder = radius of cone
= radius of hemisphere
= 72 cm
Volume of solid = Volume of cylinder + Volume of cone + Volume of hemisphere

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 12

Question 13:

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 13


Radius of each hemispherical end = 282 = 14 cm
Height of each hemispherical part = Its Radius
Height of cylindrical part = (98 – 2 × 14) = 70 cm
Area of surface to be polished = 2(curved surface area of hemisphere) + (curved surface area of cylinder)

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 13


Cost of polishing the surface of the solid
= Rs. (0.15 × 8624)
= Rs. 1293. 60

Question 14:

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 14


Radius of cylinder r1 = 5 cm
And height of cylinder h1 = 9.8 cm
Radius of cone r = 2.1 cm
And height of cone h2 = 4 cm
Volume of water left in tub = (volume of cylindrical tub – volume of solid)

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 14

Question 15:
(i) Radius of cylinder = 6 cm
Height of cylinder = 8 cm


Volume of cylinder

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 15


Volume of cone removed

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 15


(ii) Surface area of cylinder = 2π = 2π × 6 × 8 cm2 = 96 π cm2

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 15

Question 16:

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 16


Diameter of spherical part of vessel = 21 cm

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 16

Question 17:


Height of cylindrical tank = 2.5 m
Its diameter = 12 m, Radius = 6 m
Volume of tank = 

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 17


Water is flowing at the rate of 3.6 km/ hr = 3600 m/hr
Diameter of pipe = 25 cm, radius = 0.125 m
Volume of water flowing per hour

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 17

Question 18:

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 18


Diameter of cylinder = 5 cm
Radius = 2.5 cm
Height of cylinder = 10 cm
Volume of cylinder = πr2h  cu.units = 3.14 × 2.5 × 2.5 × 10  cm= 196.25 cm3
Apparent capacity of glass = 196.25
Radius of hemisphere = 2.5 cm
Volume of hemisphere

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Question 18


Actual capacity of glass = ( 196.25 – 32.608 ) cm3 = 163.54 cm3

Exercise 17B Solutions

https://youtube.com/watch?v=6KpStN_0mjE%3Ffeature%3Doembed

Question 1:

Radius of the cone = 12 cm and its height = 24 cm
Volume of cone = 13 πr3 h = (\frac { 1 }{ 3 } \times 12\times 12\times 24) π cm3  = (48 × 24 )π cm3

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 1

Question 2:
Internal radius = 3 cm and external radius = 5 cm

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 2


Hence, height of the cone = 4 cm

Question 3:
Inner radius of the bowl = 15 cm
Volume of liquid in it =

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 3


Radius of each cylindrical bottle = 2.5 cm and its height = 6 cm
Volume of each cylindrical bottle

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 3


Required number of bottles = 

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 3
RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 3



Hence, bottles required = 60

Question 4:
Radius of the sphere = 212 cm

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 4


Let the number of cones formed be n, then

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 4


Hence, number of cones formed = 504

Question 5:
Radius of the cannon ball = 14 cm
Volume of cannon ball = 

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 5


Radius of the cone = 352 cm
Let the height of cone be h cm
Volume of cone = 

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 5
RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 5

Hence, height of the cone = 35.84 cm

Question 6:
Let the radius of the third ball be r cm, then,
Volume of third ball = Volume of spherical ball – volume of 2 small balls

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 6

Question 7:
External radius of shell = 12 cm and internal radius = 9 cm
Volume of lead in the shell = 

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 7

Let the radius of the cylinder be r cm
Its height = 37 cm
Volume of cylinder = πr2h = ( πr2 × 37 )

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 7


Hence diameter of the base of the cylinder = 12 cm

Question 8:
Volume of hemisphere of radius 9 cm

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 8


Volume of circular cone (height = 72 cm)

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 8


Volume of cone = Volume of hemisphere

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 8

Hence radius of the base of the cone = 4.5 cm

Question 9:
Diameter of sphere = 21 cm
Hence, radius of sphere = 192 cm
Volume of sphere = 43 πr3 = (43×227×212×212×212)
Volume of cube = a3 = (1 × 1 × 1)
Let number of cubes formed be n
∴ Volume of sphere = n × Volume of cube

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 9


Hence, number of cubes is 4851.

Question 10:
Volume of sphere (when r = 1 cm) = 43 πr3 = (\frac { 4 }{ 3 } \times 1\times 1\times 1) π cm3
Volume of sphere (when r = 8 cm) = 43 πr3 = (\frac { 4 }{ 3 } \times 8\times 8\times 8) π cm3
Let the number of balls = n

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 10

Question 11:
Radius of marbles = Diameter2=1.42cm

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 11


Let the number of marbles be n
∴ n × volume of marble = volume of rising water in beaker

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 11

Question 12:
Radius of sphere = 3 cm
Volume of sphere = 43 πr3 = (\frac { 4 }{ 3 } \times 3\times 3\times 3) π cm3  = 36π cm3
Radius of small sphere = 0.62 cm = 0.3 cm
Volume of small sphere = (\frac { 4 }{ 3 } \times 0.3\times 0.3\times 0.3) π cm3

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 12

Let number of small balls be n

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 12


Hence, the number of small balls = 1000.

Question 13:
Diameter of sphere = 42 cm
Radius of sphere = 422 cm = 21 cm
Volume of sphere = 43 πr3 = (\frac { 4 }{ 3 } \times 21\times 21\times 21) π cm3
Diameter of cylindrical wire = 2.8 cm
Radius of cylindrical wire = 2.82 cm = 1.4 cm
Volume of cylindrical wire =  πr2h = ( π × 1.4 × 1.4 × h ) cm3 = ( 1.96πh ) cm3
Volume of cylindrical wire = volume of sphere

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 13


Hence length of the wire 63 m.

Question 14:
Diameter of sphere = 6 cm
Radius of sphere = 62 cm = 3 cm
Volume of sphere = 43 πr3 = (\frac { 4 }{ 3 } \times 3\times 3\times 3) π cm3  = 36π cm3
Radius of wire = 22 mm = 1 mm = 0.1 cm
Volume of wire = πr2l = ( π × 0.1 × 0.1 × l ) cm2 = ( 0.01 πl ) cm2
36π = 0.01 π l
∴ l=360.01=3600 cm
Length of wire = 3600100 m = 36 m

Question 15:
Diameter of sphere = 18 cm
Radius of copper sphere = 3600100 m = 36 m

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 15


Length of wire = 108 m = 10800 cm
Let the radius of wire be r cm
= πr2l cm3 = ( πr2 × 10800 ) cm3
But the volume of wire = Volume of sphere

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 15


Hence the diameter = 2r = (0.3 × 2) cm = 0.6 cm

Question 16:
The radii of three metallic spheres are 3 cm, 4 cm and 5 cm respectively.
Sum of their volumes 

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 16
RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 16



Let r be the radius of sphere whose volume is equal to the total volume of three spheres.

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17B Question 16

Exercise 17C

Question 1:
Here h = 42 cm, R = 16 cm, and r = 11 cm
Capacity = 

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 1
RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 1

Question 2:
Here R = 33 cm, r = 27 cm and l = 10 cm

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 2


Capacity of the frustum

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 2
RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 2

Total surface area

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 2
RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 2

Question 3:
Height = 15 cm, R = 562 cm = 28 cm and r = 422 cm = 21 cm
Capacity of the bucket =

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 3
RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 3



Quantity of water in bucket = 28.49 litres

Question 4:
R = 20 cm, r = 8 cm and h = 16 cm

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 4


Total surface area of container = πl (R+r) + πr2

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 4

Cost of metal sheet used = Rs. (1959.36×15100) = Rs. 293.90

Question 5:
R = 15 cm, r = 5 cm and h = 24 cm

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 5


(i) Volume of bucket =

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 5
RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 5

Cost of milk = Rs. (8.164 × 20) = Rs. 163.28
(ii) Total surface area of the bucket

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 5


Cost of sheet = (1711.3×10100) = Rs. 171.13

Question 6:


R = 10cm, r = 3 m and h = 24 m
Let l be the slant height of the frustum, then

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 6


Quantity of canvas = (Lateral surface area of the frustum) + (lateral surface area of the cone)

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 6

Question 7:

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 7


ABCD is the frustum in which upper and lower radii are EB = 7 m and FD = 13 m
Height of frustum = 8 m
Slant height l1 of frustum

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 7

Radius of the cone = EB = 7 m
Slant height l2 of cone = 12 m
Surface area of canvas required

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 7

Question 8:

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 8


In the given figure, we have
∠COD = 30°, OC = 10 cm, OE = 20 cm
Let CD = r cm and EB = R cm


Also, CE = 10 cm
Thus, ABDF is the frustum of a cone in which


Volume of wire of radius r and length l

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 8


Volume of wire = Volume of frustum

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 8


Length of the wire is 7964.44 m

Question 9:

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 9


Radii of upper and lower end of frustum are r = 8 cm, R = 32 cm
Height of frustum h = 18 cm

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids Exercise 17C Question 9


Cost of milk at Rs 20 per litre = Rs. 25.344 × 20 = Rs. 506. 88

RS Aggarwal Solutions for Class 10 Maths Chapter 17: Download PDF

RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids

Download PDF: RS Aggarwal Solutions for Class 10 Maths Chapter 17–Volume and Surface Areas of Solids PDF

Chapterwise RS Aggarwal Solutions for Class 10 Maths :

About RS Aggarwal Class 10 Book

Investing in an R.S. Aggarwal book will never be of waste since you can use the book to prepare for various competitive exams as well. RS Aggarwal is one of the most prominent books with an endless number of problems. R.S. Aggarwal’s book very neatly explains every derivation, formula, and question in a very consolidated manner. It has tonnes of examples, practice questions, and solutions even for the NCERT questions.

He was born on January 2, 1946 in a village of Delhi. He graduated from Kirori Mal College, University of Delhi. After completing his M.Sc. in Mathematics in 1969, he joined N.A.S. College, Meerut, as a lecturer. In 1976, he was awarded a fellowship for 3 years and joined the University of Delhi for his Ph.D. Thereafter, he was promoted as a reader in N.A.S. College, Meerut. In 1999, he joined M.M.H. College, Ghaziabad, as a reader and took voluntary retirement in 2003. He has authored more than 75 titles ranging from Nursery to M. Sc. He has also written books for competitive examinations right from the clerical grade to the I.A.S. level.

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