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RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities
RD Sharma Solutions Class 9 Chapter 4 Algebraic Identities Ex 4.1
Question 1.
Evaluate each of the following using identities:
(i) (2x –(frac { 1 }{ x }))2
(ii) (2x + y) (2x – y)
(iii) (a2b – b2a)2
(iv) (a – 0.1) (a + 0.1)
(v) (1.5.x2 – 0.3y2) (1.5x2 + 0.3y2)
Solution:
Question 2.
Evaluate each of the following using identities:
(i) (399)2
(ii) (0.98)2
(iii) 991 x 1009
(iv) 117 x 83
Solution:
Question 3.
Simplify each of the following:
Solution:
Question 4.
Solution:
Question 5.
Solution:
Question 6.
Solution:
Question 7.
If 9x2 + 25y2 = 181 and xy = -6, find the value of 3x + 5y.
Solution:
9x2 + 25y2 = 181, and xy = -6
(3x + 5y)2 = (3x)2 + (5y)2 + 2 x 3x + 5y
⇒ 9X2 + 25y2 + 30xy
= 181 + 30 x (-6)
= 181 – 180 = 1
= (±1 )2
∴ 3x + 5y = ±1
Question 8.
If 2x + 3y = 8 and xy = 2, find the value of 4X2 + 9y2.
Solution:
2x + 3y = 8 and xy = 2
Now, (2x + 3y)2 = (2x)2 + (3y)2 + 2 x 2x x 3y
⇒ (8)2 = 4x2 + 9y2 + 12xy
⇒ 64 = 4X2 + 9y2 + 12 x 2
⇒ 64 = 4x2 + 9y2 + 24
⇒ 4x2 + 9y2 = 64 – 24 = 40
∴ 4x2 + 9y2 = 40
Question 9.
If 3x -7y = 10 and xy = -1, find the value of 9x2 + 49y2
Solution:
3x – 7y = 10, xy = -1
3x -7y= 10
Squaring both sides,
(3x – 7y)2 = (10)2
⇒ (3x)2 + (7y)2 – 2 x 3x x 7y = 100
⇒ 9X2 + 49y2 – 42xy = 100
⇒ 9x2 + 49y2 – 42(-l) = 100
⇒ 9x2 + 49y2 + 42 = 100
∴ 9x2 + 49y2 = 100 – 42 = 58
Question 10.
Simplify each of the following products:
Solution:
Question 11.
Solution:
Question 12.
Solution:
Question 13.
Simplify each of the following products:
Solution:
Question 14.
Prove that a2 + b2 + c2 – ab – bc – ca is always non-negative for all values of a, b and c.
Solution:
∵ The given expression is sum of these squares
∴ Its value is always positive Hence the given expression is always non-negative for all values of a, b and c
Class 9 RD Sharma Solutions Chapter 4 Algebraic Identities Ex 4.2
Question 1.
Write the following in the expanded form:
Solution:
Question 2.
If a + b + c = 0 and a2 + b2 + c2 = 16, find the value of ab + be + ca.
Solution:
a + b+ c = 0
Squaring both sides,
(a + b + c)2 = 0
⇒ a2 + b2 + c2 + 2(ab + bc + ca) = 0
16 + 2(ab + bc + c) = 0
⇒ 2(ab + bc + ca) = -16
⇒ ab + bc + ca =-(frac { 16 }{ 2 }) = -8
∴ ab + bc + ca = -8
Question 3.
If a2 + b2 + c2 = 16 and ab + bc + ca = 10, find the value of a + b + c.
Solution:
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
= 16 + 2 x 10
= 16 + 20 = 36
= (±6)2
∴ a + b + c = ±6
Question 4.
If a + b + c = 9 and ab + bc + ca = 23, find the value of a2 + b2 + c2.
Solution:
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
⇒ (9)2 = a2 + b2 + c2 + 2 x 23
⇒ 81= a2 + b2 + c2 + 46
⇒ a2 + b2 + c2 = 81 – 46 = 35
∴ a2 + b2 + c2 = 35
Question 5.
Find the value of 4x2 + y2 + 25z2 + 4xy – 10yz – 20zx when x = 4, y = 3 and z = 2.
Solution:
x = 4, y – 3, z = 2
⇒ 4x2 + y2 + 25z2 + 4xy – 10yz – 20zx
= (2x)2 + (y)2 + (5z)2 + 2 x2 x x y-2 x y x 5z – 2 x 5z x 2x
= (2x + y- 5z)2
= (2 x 4 + 3- 5 x 2)2
= (8 + 3- 10)2
= (11 – 10)2
= (1)2 = 1
Question 6.
Simplify:
(i) (a + b + c)2 + (a – b + c)2
(ii) (a + b + c)2 – (a – b + c)2
(iii) (a + b + c)2 + (a – b + c)2 + (a + b – c)2
(iv) (2x + p – c)2 – (2x – p + c)2
(v) (x2 + y2 – z2)2 – (x2 – y2 + z2)2
Solution:
Question 7.
Simplify each of the following expressions:
Solution:
Class 9 RD Sharma Solutions Chapter 4 Algebraic Identities Ex 4.3
Question 1.
Find the cube of each of the following binomial expressions:
Solution:
Question 2.
If a + b = 10 and ab = 21, find the value of a3 + b3.
Solution:
a + b = 10, ab = 21
Cubing both sides,
(a + b)3 = (10)3
⇒ a3 + 63 + 3ab (a + b) = 1000
⇒ a3 + b3 + 3 x 21 x 10 = 1000
⇒ a3 + b3 + 630 = 1000
⇒ a3 + b3 = 1000 – 630 = 370
∴ a3 + b3 = 370
Question 3.
If a – b = 4 and ab = 21, find the value of a3-b3.
Solution:
a – b = 4, ab= 21
Cubing both sides,
⇒ (a – A)3 = (4)3
⇒ a3 – b3 – 3ab (a – b) = 64
⇒ a3-i3-3×21 x4 = 64
⇒ a3 – 63 – 252 = 64
⇒ a3 – 63 = 64 + 252 =316
∴ a3 – b3 = 316
Question 4.
Solution:
Question 5.
Solution:
Question 6.
Solution:
Question 7.
Solution:
Question 8.
Solution:
Question 9.
If 2x + 3y = 13 and xy = 6, find the value of 8x3 + 21y3.Solution:
2x + 3y = 13, xy = 6
Cubing both sides,
(2x + 3y)3 = (13)3
⇒ (2x)3 + (3y)3 + 3 x 2x x 3X2x + 3y) = 2197
⇒ 8x3 + 27y3 + 18xy(2x + 3y) = 2197
⇒ 8x3 + 27y3 + 18 x 6 x 13 = 2197
⇒ 8X3 + 27y3 + 1404 = 2197
⇒ 8x3 + 27y3 = 2197 – 1404 = 793
∴ 8x3 + 27y3 = 793
Question 10.
If 3x – 2y= 11 and xy = 12, find the value of 27x3 – 8y3.
Solution:
3x – 2y = 11 and xy = 12 Cubing both sides,
(3x – 2y)3 = (11)3
⇒ (3x)3 – (2y)3 – 3 x 3x x 2y(3x – 2y) =1331
⇒ 27x3 – 8y3 – 18xy(3x -2y) =1331
⇒ 27x3 – 8y3 – 18 x 12 x 11 = 1331
⇒ 27x3 – 8y3 – 2376 = 1331
⇒ 27X3 – 8y3 = 1331 + 2376 = 3707
∴ 2x3 – 8y3 = 3707
Question 11.
Evaluate each of the following:
(i) (103)3
(ii) (98)3
(iii) (9.9)3
(iv) (10.4)3
(v) (598)3
(vi) (99)3
Solution:
We know that (a + bf = a3 + b3 + 3ab(a + b) and (a – b)3= a3 – b3 – 3 ab(a – b)
Therefore,
(i) (103)3 = (100 + 3)3
= (100)3 + (3)3 + 3 x 100 x 3(100 + 3) {∵ (a + b)3 = a3 + b3 + 3ab(a + b)}
= 1000000 + 27 + 900 x 103
= 1000000 + 27 + 92700
= 1092727
(ii) (98)3 = (100 – 2)3
= (100)3 – (2)3 – 3 x 100 x 2(100 – 2)
= 1000000 – 8 – 600 x 98
= 1000000 – 8 – 58800
= 1000000-58808
= 941192
(iii) (9.9)3 = (10 – 0.1)3
= (10)3 – (0.1)3 – 3 X 10 X 0.1(10 – 0.1)
= 1000 – 0.001 – 3 x 9.9
= 1000 – 0.001 – 29.7
= 1000 – 29.701
= 970.299
(iv) (10.4)3 = (10 + 0.4)3
= (10)3 + (0.4)3 + 3 x 10 x 0.4(10 + 0.4)
= 1000 + 0.064 + 12(10.4)
= 1000 + 0.064 + 124.8 = 1124.864
(v) (598)3 = (600 – 2)3
= (600)3 – (2)3 – 3 x 600 x 2 x (600 – 2)
= 216000000 – 8 – 3600 x 598
= 216000000 – 8 – 2152800
= 216000000 – 2152808
= 213847192
(vi) (99)3 = (100 – 1)3
= (100)3 – (1)3 – 3 x 100 x 1 x (100 – 1)
= 1000000 – 1 – 300 x 99
= 1000000 – 1 – 29700
= 1000000 – 29701
= 970299
Question 12.
Evaluate each of the following:
(i) 1113 – 893
(ii) 463 + 343
(iii) 1043 + 963
(iv) 933 – 1073
Solution:
We know that a3 + b3 = (a + bf – 3ab(a + b) and a3 – b3 = (a – bf + 3 ab(a – b)
(i) 1113 – 893
= (111 – 89)3 + 3 x ill x 89(111 – 89)
= (22)3 + 3 x 111 x 89 x 22
= 10648 + 652014 = 662662
(OR)
(a + b)3 – (a – b)3 = 2(b3 + 3a2b)
= 1113 – 893 = (100 + 11)3 – (100 – 11)3
= 2(113 + 3 x 1002 x 11]
= 2(1331 + 330000]
= 331331 x 2 = 662662
(a + b)3 + (a- b)3 = 2(b3 + 3ab2)
(ii) 463 + 343 = (40 + 6)3 + (40 – 6)3
= 2[(40)3 + 3 x 40 x 62]
= 2[64000 + 3 x 40 x 36]
= 2[64000 + 4320]
= 2 x 68320 = 136640
(iii) 1043 + 963 = (100 + 4)3 + (100 – 96)3
= 2 [a3 + 3 ab2]
= 2[(100)3 + 3 x 100 x (4)2]
= 2[ 1000000 + 300 x 16]
= 2[ 1000000 + 4800]
= 1004800 x 2 = 2009600
(iv) 933 – 1073 = -[(107)3 – (93)3]
= -[(100 + If – (100 – 7)3]
= -2[b3 + 3a2b)]
= -2[(7)3 + 3(100)2 x 7]
= -2(343 + 3 x 10000 x 7]
= -2[343 + 210000]
= -2[210343] = -420686
Question 13.
Solution:
Question 14.
Find the value of 27X3 + 8y3 if
(i) 3x + 2y = 14 and xy = 8
(ii) 3x + 2y = 20 and xy = (frac { 14 }{ 9 })
Solution:
Question 15.
Find the value of 64x3 – 125z3, if 4x – 5z = 16 and xz = 12.
Solution:
4x – 5z = 16, xz = 12
Cubing both sides,
(4x – 5z)3 = (16)3
⇒ (4x)3 – (5y)3 – 3 x 4x x 5z(4x – 5z) = 4096
⇒ 64x3 – 125z3 – 3 x 4 x 5 x xz(4x – 5z) = 4096
⇒ 64x3 – 125z3 – 60 x 12 x 16 = 4096
⇒ 64x3 – 125z3 – 11520 = 4096
⇒ 64x3 – 125z3 = 4096 + 11520 = 15616
Question 16.
Solution:
Question 17.
Simplify each of the following:
Solution:
Question 18.
Solution:
Question 19.
Solution:
RD Sharma Mathematics Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.4
Question 1.
Find the following products:
(i) (3x + 2y) (9X2 – 6xy + Ay2)
(ii) (4x – 5y) (16x2 + 20xy + 25y2)
(iii) (7p4 + q) (49p8 – 7p4q + q2)
Solution:
Question 2.
If x = 3 and y = -1, find the values of each of the following using in identity:
Solution:
Question 3.
If a + b = 10 and ab = 16, find the value of a2 – ab + b2 and a2 + ab + b2.
Solution:
a + b = 10, ab = 16 Squaring,
(a + b)2 = (10)2
⇒ a2 + b2 + lab = 100
⇒ a2 + b2 + 2 x 16 = 100
⇒ a2 + b2 + 32 = 100
∴ a2 + b2 = 100 – 32 = 68
Now, a2 – ab + b2 = a2 + b2 – ab = 68 – 16 = 52
and a2 + ab + b2 = a2 + b2 + ab = 68 + 16 = 84
Question 4.
If a + b = 8 and ab = 6, find the value of a3 + b3.
Solution:
a + b = 8, ab = 6
Cubing both sides,
(a + b)3 = (8)3
⇒ a3 + b3 + 3 ab{a + b) = 512
⇒ a3 + b3 + 3 x 6 x 8 = 512⇒ a3 + b3 + 144 = 512
⇒ a3 + b3 = 512 – 144 = 368
∴ a3 + b3 = 368
Question 5.
If a – b = 6 and ab = 20, find the value of a3-b3.
Solution:
a – b = 6, ab = 20
Cubing both sides,
(a – b)3 = (6)3
⇒ a3 – b3 – 3ab(a – b) = 216
⇒ a3 – b3 – 3 x 20 x 6 = 216
⇒ a3 – b3 – 360 = 216
⇒ a3 -b3 = 216 + 360 = 576
∴ a3 – b3 = 576
Question 6.
If x = -2 and y = 1, by using an identity find the value of the following:
Solution:
RD Sharma Solutions Class 9 Chapter 4 Algebraic Identities Ex 4.5
Question 1.
Find the following products:
(i) (3x + 2y + 2z) (9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx)
(ii) (4x -3y + 2z) (16x2 + 9y2+ 4z2 + 12xy + 6yz – 8zx)
(iii) (2a – 3b – 2c) (4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca)
(iv) (3x -4y + 5z) (9x2 + 16y2 + 25z2 + 12xy- 15zx + 20yz)
Solution:
(i) (3x + 2y + 2z) (9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx)
= (3x + 2y + 2z) [(3x)2 + (2y)2 + (2z)2 – 3x x 2y + 2y x 2z + 2z x 3x]
= (3x)3 + (2y)3 + (2z)3 – 3 x 3x x 2y x 2z
= 27x3 + 8y3 + 8Z3 – 36xyz
(ii) (4x – 3y + 2z) (16x2 + 9y2 + 4z2 + 12xy + 6yz – 8zx)
= (4x -3y + 2z) [(4x)2 + (-3y)2 + (2z)2 – 4x x (-3y + (3y) x (2z) – (2z x 4x)]
= (4x)3 + (-3y)3 + (2z)3 – 3 x 4x x (-3y) x (2z)
= 64x3 – 27y3 + 8z3 + 72xyz
(iii) (2a -3b- 2c) (4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca)
= (2a -3b- 2c) [(2a)2 + (3b)2 + (2c)2 – 2a x (-3b) – (-3b) x (-2c) – (-2c) x 2a]
= (2a)3 + (3b)3 + (-2c)3 -3x 2a x (-3 b) (-2c)
= 8a3 – 21b3 -8c3 – 36abc
(iv) (3x – 4y + 5z) (9x2 + 16y2 + 25z2 + 12xy – 15zx + 20yz)
= [3x + (-4y) + 5z] [(3x)2 + (-4y)2 + (5z)2 – 3x x (-4y) -(-4y) (5z) – 5z x 3x]
= (3x)3 + (-4y)3 + (5z)3 – 3 x 3x x (-4y) (5z)
= 27x3 – 64y3 + 125z3 + 180xyz
Question 2.
Evaluate:
Solution:
Question 3.
If x + y + z = 8 and xy + yz+ zx = 20, find the value of x3 + y3 + z3 – 3xyz.
Solution:
We know that
x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 -xy -yz – zx)
Now, x + y + z = 8
Squaring, we get
(x + y + z)2 = (8)2
x2 + y2 + z2 + 2(xy + yz + zx) = 64
⇒ x2 + y2 + z2 + 2 x 20 = 64
⇒ x2 + y2 + z2 + 40 = 64
⇒ x2 + y2 + z2 = 64 – 40 = 24
Now,
x3 + y3 + z3 – 3xyz = (x + y + z) [x2 + y2 + z2 – (xy + yz + zx)]
= 8(24 – 20) = 8 x 4 = 32
Question 4.
If a +b + c = 9 and ab + bc + ca = 26, find the value of a3 + b3 + c3 – 3abc.
Solution:
a + b + c = 9, ab + be + ca = 26
Squaring, we get
(a + b + c)2 = (9)2
a2 + b2 + c2 + 2 (ab + be + ca) = 81
⇒ a2 + b2 + c2 + 2 x 26 = 81
⇒ a2 + b2 + c2 + 52 = 81
∴ a2 + b2 + c2 = 81 – 52 = 29
Now, a3 + b3 + c3 – 3abc = (a + b + c) [(a2 + b2 + c2 – (ab + bc + ca)]
= 9[29 – 26]
= 9 x 3 = 27
Question 5.
If a + b + c = 9, and a2 + b2 + c2 = 35, find the value of a3 + b3 + c3 – 3abc.
Solution:
a + b + c = 9
Squaring, we get
(a + b + c)2 = (9)2
⇒ a2 + b2 + c2 + 2 (ab + be + ca) = 81
⇒ 35 + 2(ab + bc + ca) = 81
2(ab + bc + ca) = 81 – 35 = 46
∴ ab + bc + ca = (frac { 46 }{ 2 }) = 23
Now, a3 + b3 + c3 – 3abc
= (a + b + c) [a2 + b2 + c2 – (ab + bc + ca)]
= 9[35 – 23] = 9 x 12 = 108
Class 9 Maths Chapter 4 Algebraic Identities RD Sharma Solutions VSAQS
Question 1.
Solution:
Question 2.
Solution:
Question 3.
If a + b = 7 and ab = 12, find the value of a2 + b2.
Solution:
a + b = 7, ab = 12
Squaring both sides,
(a + b)2 = (7)2
⇒ a2 + b2 + 2ab = 49
⇒ a2 + b2 + 2 x 12 = 49
⇒ a2 + b2 + 24 = 49
⇒ a2 + b2 = 49 – 24 = 25
∴ a2 + b2 = 25
Question 4.
If a – b = 5 and ab = 12, find the value of a2 + b2 .
Solution:
a – b = 5, ab = 12
Squaring both sides,
⇒ (a – b)2 = (5)2
⇒ a2 + b2 – 2ab = 25
⇒ a2 + b2 – 2 x 12 = 25
⇒ a2 + b2 – 24 = 25
⇒ a2 + b2 = 25 + 24 = 49
∴ a2 + b2 = 49
Question 5.
Solution:
Question 6.
Solution:
Question 7.
Solution:
Algebraic Identities Class 9 RD Sharma Solutions MCQS
Question 1.
Solution:
Question 2.
Solution:
Question 3.
Solution:
Question 4.
Solution:
Question 5.
Solution:
Question 6.
Solution:
Question 7.
Solution:
Question 8.
If a + b + c = 9 and ab + bc + ca = 23, then a2 + b2 + c2 =
(a) 35
(b) 58
(c) 127
(d) none of these
Solution:
a + b + c = 9, ab + bc + ca = 23
Squaring,
(a + b+ c) = (9)2
a2 + b2 + c2 + 2 (ab + bc + ca) = 81
⇒ a2 + b2 + c2 + 2 x 23 = 81
⇒ a2 + b2+ c2 + 46 = 81
⇒ a2 + b2+ c2 = 81 – 46 = 35 (a)
Question 9.
(a – b)3 + (b – c)3 + (c – a)3 =
(a) (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
(d) (a -b)(b- c) (c – a)
(c) 3(a – b) (b – c) (c – a)
(d) none of these
Solution:
(a – b)3 + (b- c)3 + (c- a)3
∵ a – b + b – c + c – a = 0
∴ (a – b)3 + (b – c)3 + (c – a)3
= 3 (a -b)(b- c) (c – a) (c)
Question 10.
Solution:
Question 11.
If a – b = -8 and ab = -12 then a3 – b3 =
(a) -244
(b) -240
(c) -224
(d) -260
Solution:
a – b = -8, ab = -12
(a – b)3 = a3 – b3 – 3ab (a – b)
(-8)3 = a3 – b3 – 3 x (-12) (-8)
-512 = a3-b3– 288
a3 – b3 = -512 + 288 = -224 (c)
Question 12.
If the volume of a cuboid is 3x2 – 27, then its possible dimensions are
(a) 3, x2, -27x
(b) 3, x – 3, x + 3
(c) 3, x2, 27x
(d) 3, 3, 3
Solution:
Volume = 3x2 -27 = 3(x2 – 9)
= 3(x + 3) (x – 3)
∴ Dimensions are = 3, x – 3, x + 3 (b)
Question 13.
75 x 75 + 2 x 75 x 25 + 25 x 25 is equal to
(a) 10000
(b) 6250
(c) 7500
(d) 3750
Solution:
Question 14.
(x – y) (x + y)(x2 + y2) (x4 + y4) is equal to
(a) x16 – y16
(b) x8 – y8
(c) x8 + y8
(d) x16 + y16
Solution:
Question 15.
Solution:
Question 16.
Solution:
Question 17.
Solution:
Question 18.
Solution:
Question 19.
If a2 + b2 + c2 – ab – bc – ca = 0, then
(a) a + b = c
(b) b + c = a
(c) c + a = b
(d) a = b = c
Solution:
a2 + b2 + c2 – ab – bc – ca = 0
2 {a2 + b2 + c2 – ab – be – ca) = 0 (Multiplying by 2)
⇒ 2a2 + 2b2 + 2c2– 2ab – 2bc – 2ca = 0
⇒ a2 + b2 – 2ab + b2 + c2 – 2bc + c2 + a2 – 2ca = 0
⇒ (a – b)2 + (b – c)2 + (c – a)2 = 0
(a – b)2 = 0, then a – b = 0
⇒ a = b
Similarly, (b – c)2 = 0, then
b-c = 0
⇒ b = c
and (c – a)2 = 0, then c-a = 0
⇒ c = a
∴ a = b – c (d)
Question 20.
Solution:
Question 21.
Solution:
Question 22.
If a + b + c = 9 and ab + bc + ca = 23, then a3 + b3 + c3 – 3 abc =
(a) 108
(b) 207
(c) 669
(d) 729
Solution:
a3 + b3 + c3 – 3abc
= (a + b + c) [a2 + b2 + c2 – (ab + bc + ca)
Now, a + b + c = 9
Squaring,
a2 + b2 + c2 + 2 (ab + be + ca) = 81
⇒ a2 + b2 + c2 + 2 x 23 = 81
⇒ a2 + b2 + c2 + 46 = 81
⇒ a2 + b2 + c2 = 81 – 46 = 35
Now, a3 + b3 + c3 – 3 abc = (a + b + c) [(a2 + b2 + c2) – (ab + bc + ca)
= 9[35 -23] = 9 x 12= 108 (a)
Question 23.
Solution:
Question 24.
The product (a + b) (a – b) (a2 – ab + b2) (a2 + ab + b2) is equal to
(a) a6 + b6
(b) a6 – b6
(c) a3 – b3
(d) a3 + b3
Solution:
(a + b) (a – b) (a2 – ab + b2) (a2 + ab +b2)
= (a + b) (a2-ab + b2) (a-b) (a2 + ab + b2)
= (a3 + b3) (a3 – b3)
= (a3)2 – (b3)2 = a6 – b6 (b)
Question 25.
The product (x2 – 1) (x4 + x2 + 1) is equal to
(a) x8 – 1
(b) x8 + 1
(c) x6 – 1
(d) x6 + 1
Solution:
(x2 – 1) (x4 + x2 + 1)
= (x2)3 – (1)3 = x6 – 1 (c)
Question 26.
Solution:
Question 27.
Solution:
All Chapter RD Sharma Solutions For Class 9 Maths
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