Class 8 Exam  >  Class 8 Notes  >  RD Sharma Solutions for Class 8 Mathematics  >  RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 )

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics PDF Download

Question 1: Multiply:
(5x + 3) by (7x + 2)

Answer 1: To multiply, we will use distributive law as follows:
(5x+3)(7x+2)=5x(7x+2)+3(7x+2)=(5x×7x+5x×2)+(3×7x+3×2)=(35x2+10x)+(21x+6)=35x2+10x+21x+6=35x2+31x+6
Thus, the answer is 35x2+31x+635x2+31x+6. 

Question 2: Multiply:
(2x + 8) by (x − 3)

Answer 2: To multiply the expressions, we will use the distributive law in the following way:
(2x+8)(x3)=2x(x3)+8(x3)=(2x×x2x×3)+(8x8×3)=(2x26x)+(8x24)=2x26x+8x24=2x2+2x242x+8x-3=2xx-3+8x-3=2x×x-2x×3+8x-8×3=2x2-6x+8x-24=2x2-6x+8x-24=2x2+2x-2Thus, the answer is 2x2+2x242x2+2x-24. 

Question 3: Multiply:
(7x + y) by (x + 5y)

Answer 3: To multiply, we will use distributive law as follows:
(7x+y)(x+5y)=7x(x+5y)+y(x+5y)=7x2+35xy+xy+5y2=7x2+36xy+5y27x+yx+5y=7xx+5y+yx+5y=7x2+35xy+xy+5y2=7x2+36xy+5y2
Thus, the answer is 7x2+36xy+5y27x2+36xy+5y2. 

Question 4: Multiply:
(a − 1) by (0.1a2 + 3)

Answer 4: To multiply, we will use distributive law as follows:
(a1)(0.1a2+3)=0.1a2(a1)+3(a1)=0.1a30.1a2+3a3a-10.1a2+3=0.1a2a-1+3a-1=0.1a3-0.1a2+3a-3
Thus, the answer is 0.1a30.1a2+3a30.1a3-0.1a2+3a-3. 

Question 5: Multiply:
(3x2 + y2) by (2x2 + 3y2)

Answer 5: To multiply, we will use distributive law as follows:
(3x2+y2)(2x2+3y2)=3x2(2x2+3y2)+y2(2x2+3y2)=6x4+9x2y2+2x2y2+3y4=6x4+11x2y2+3y43x2+y22x2+3y2=3x22x2+3y2+y22x2+3y2=6x4+9x2y2+2x2y2+3y4=6x4+11x2y2+3yThus, the answer is 6x4+11x2y2+3y46x4+11x2y2+3y4. 

Question 6: Multiply:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

Answer 6: To multiply, we will use distributive law as follows: 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

Thus, the answer is RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

Question 7: Multiply:
(x6y6) by (x2 + y2)

Answer 7: To multiply, we will use distributive law as follows: 

(x6y6)(x2+y2)=x6(x2+y2)y6(x2+y2)=(x8+x6y2)(y6x2+y8)=x8+x6y2y6x2y8 Thus, the answer is x8+x6y2y6x2y8x8+x6y2-y6x2-y8. 

Question 8: Multiply:
(x2 + y2) by (3a + 2b)

Answer 8: To multiply, we will use distributive law as follows: 

(x2+y2)(3a+2b)=x2(3a+2b)+y2(3a+2b)=3ax2+2bx2+3ay2+2by2 Thus, the answer is 3ax2+2bx2+3ay2+2by23ax2+2bx2+3ay2+2by2. 

Question 9: Multiply:
[−3d + (−7f)] by (5d + f)

Answer 9: To multiply, we will use distributive law as follows:
[3d+(7f)](5d+f)=(3d)(5d+f)+(7f)(5d+f)=(15d23df)+(35df7f2)=15d23df35df7f2=15d238df7f2
Thus, the answer is 15d238df7f2-15d2-38df-7f2. 

Question 10: Multiply:
(0.8a − 0.5b) by (1.5a − 3b)

Answer 10: To multiply, we will use distributive law as follows:
(0.8a0.5b)(1.5a3b)=0.8a(1.5a3b)0.5b(1.5a3b)=1.2a22.4ab0.75ab+1.5b2=1.2a23.15ab+1.5b2
Thus, the answer is 1.2a23.15ab+1.5b21.2a2-3.15ab+1.5b2. 

Question 11: Multiply:
(2x2y2 − 5xy2) by (x2y2)

Answer 11: To multiply, we will use distributive law as follows:

(2x2y25xy2)(x2y2)=2x2y2(x2y2)5xy2(x2y2)=2x4y22x2y45x3y2+5xy4 Thus, the answer is 2x4y22x2y45x3y2+5xy42x4y2-2x2y4-5x3y2+5xy4. 

Question 12: Multiply:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

Answer 12: To multiply the expressions, we will use the distributive law in the following way: 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

Thus, the answer is RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

Question 13: Multiply:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

Answer 13: To multiply, we will use distributive law as follows: 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

Thus, the answer is RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

Question 14: Multiply:

(3x2y − 5xy2) by RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

Answer 14: To multiply, we will use distributive law as follows: 

(3x2y5xy2) (1/5 x2+1/3 y2) 

1/5 x2(3x2y5xy2)+1/3 y2(3x2y5xy2) 

= 3/5 x4yx3y2+x2y35/3 xy4 

Thus, the answer is 3/5 x4yx3y2+x2y35/3 xy4 

Question 15: Multiply:
(2x2 − 1) by (4x3 + 5x2)

Answer 15: To multiply, we will use distributive law as follows:
(2x21)(4x3+5x2)=2x2(4x3+5x2)1(4x3+5x2)=8x5+10x44x35x2
Thus, the answer is 8x5+10x44x35x28x5+10x4-4x3-5x2. 

Question 16: (2xy + 3y2) (3y2 − 2)

Answer 16: To multiply, we will use distributive law as follows:
(2xy+3y2)(3y22)=2xy(3y22)+3y2(3y22)=6xy34xy+9y46y2=9y4+6xy36y24xy
Thus, the answer is 9y4+6xy36y24xy9y4+6xy3-6y2-4xy. 

Question 17: Find the following product and verify the result for x = − 1, y = − 2:
(3x − 5y) (x + y)

Answer 17: To multiply, we will use distributive law as follows:
(3x5y)(x+y)=3x(x+y)5y(x+y)=3x2+3xy5xy5y2=3x22xy5y2
 (3x5y)(x+y)=3x22xy5y23x-5yx+y=3x2-2xy-5y2. 

Now, we put x = -1 and y = -2 on both sides to verify the result.

LHS=(3x5y)(x+y)={3(1)5(2)}{1+(2)}=(3+10)(3)=(7)(3)=21 RHS=3x22xy5y2=3(1)22(1)(2)5(2)2=3×145×4=3420=21 Because LHS is equal to RHS, the result is verified.
Thus, the answer is 3x22xy5y23x2-2xy-5y2. 
 

Question 18: Find the following product and verify the result for x = − 1, y = − 2:
(x2y 1) (3 2x2y)

Answer 18: To multiply, we will  use distributive law as follows: 

(x2y1)(32x2y)=x2y(32x2y)1×(32x2y)=3x2y2x4y23+2x2y=5x2y2x4y23  (x2y1)(32x2y)=5x2y2x4y23x2y-13-2x2y=5x2y-2x4y2-3

Now, we put x = -1 and y = -2 on both sides to verify the result. 

LHS = (x2y1)(32x2y)=[(1)2(2)1][32(1)2(2)]=[1×(2)1][32×1×(2)]=(21)(3+4)=3×7=21RHS=5x2y2x4y23=5(1)2(2)2(1)4(2)23=[5×1×(2)][2×1×4]3=1083=21  Because LHS is equal to RHS, the result is verified.
Thus, the answer is 5x2y2x4y235x2y-2x4y2-3. 

Question 19: Find the following product and verify the result for x = − 1, y = − 2:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

Answer 19: To multiply, we will use distributive law as follows:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

Now, we will put x = -1 and y = -2 on both the sides to verify the result. 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

=- 119/225

RHS= RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

= - 119/225

Because LHS is equal to RHS, the result is verified. 

Thus, the answer is RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

Question 20: Simplify:
x2(x + 2y) (x − 3y)

Answer 20: To simplify, we will proceed as follows: 

x2(x+2y)(x3y)=[x2(x+2y)](x3y)=(x3+2x2y)(x3y)=x3(x3y)+2x2y(x3y)=x43x3y+2x3y6x2y2=x4x3y6x2y2 Thus, the answer is x4x3y6x2y2x4-x3y-6x2y2. 

Question 21: Simplify:
(x2 − 2y2) (x + 4y) x2y2

Answer 21: To simplify, we will proceed as follows: 

(x22y2)(x+4y)x2y2=[x2(x+4y)2y2(x+4y)]x2y2=(x3+4x2y2xy28y3)x2y2=x5y2+4x4y32x3y48x2y5 Thus, the answer is x5y2+4x4y32x3y48x2y5x5y2+4x4y3-2x3y4-8x2y5. 

Question 22: Simplify:
a2b2(a + 2b)(3a + b)

Answer 22: To simplify, we will proceed as follows: 

a2b2(a+2b)(3a+b)=[a2b2(a+2b)](3a+b)=(a3b2+2a2b3)(3a+b)=3a(a3b2+2a2b3)+b(a3b2+2a2b3)=3a4b2+6a3b3+a3b3+2a2b4=3a4b2+7a3b3+2a2b4 Thus, the answer is 3a4b2+7a3b3+2a2b43a4b2+7a3b3+2a2b4. 

Question 23: Simplify:
x2(xy) y2(x + 2y)

Answer 23: To simplify, we will proceed as follows: 

x2(xy)y2(x+2y)=[x2(xy)][y2(x+2y)]=(x3x2y)(xy2+2y3)=x3(xy2+2y3)x2y(xy2+2y3)=x4y2+2x3y3[x3y3+2x2y4]=x4y2+2x3y3x3y32x2y4=x4y2+x3y32x2y4 Thus, the answer is x4y2+x3y32x2y4x4y2+x3y3-2x2y4. 

Question 24: Simplify:
(x3 − 2x2 + 5x − 7)(2x − 3)

Answer 24: To simplify, we will proceed as follows: 

(x32x2+5x7)(2x3)=2x(x32x2+5x7)3(x32x2+5x7)=2x44x3+10x214x3x3+6x215x+21 =2x44x33x3+10x2+6x214x15x+21=2x4-4x3-3x3+10x2+6x2-14x-15x+21     (Rearranging)
=2x47x3+16x229x+21=2x4-7x3+16x2-29x+21                              (Combining like terms)

Thus, the answer is 2x47x3+16x229x+212x4-7x3+16x2-29x+21. 

Question 25: Simplify:
(5x + 3)(x − 1)(3x − 2)

Answer 25: To simplify, we will proceed as follows:
(5x+3)(x1)(3x2)=[(5x+3)(x1)](3x2)
=[5x(x1)+3(x1)](3x2)=5xx-1+3x-13x-2              (Distributive law)
=[5x25x+3x3](3x2)=[5x22x3](3x2)=3x(5x22x3)2(5x22x3)=15x36x29x[10x24x6]=15x36x29x10x2+4x+6=5x2-5x+3x-33x-2=5x2-2x-33x-2=3x5x2-2x-3-25x2-2x-3=15x3-6x2-9x-10x2-4x-6=15x3-6x2-9x-10x2+4x+=15x36x210x29x+4x+6=15x3-6x2-10x2-9x+4x+6              (Rearranging)
=15x316x25x+6=15x3-16x2-5x+6                              (Combining like terms) 

Thus, the answer is 15x316x25x+615x3-16x2-5x+6. 

Question 26: Simplify:
(5 − x)(6 − 5x)( 2 − x)

Answer 26: To simplify, we will proceed as follows: 

(5x)(65x)(2x)=[(5x)(65x)](2x) =[5(65x)x(65x)](2x)=56-5x-x6-5x2-x                (Distributive law)
=(3025x6x+5x2)(2x)=(3031x+5x2)(2x)=2(3031x+5x2)x(3031x+5x2)=6062x+10x230x+31x25x3=30-25x-6x+5x22-x=30-31x+5x22-x=230-31x+5x2-x30-31x+5x2=60-62x+10x2-30x+31x2-5x=6062x30x+10x2+31x25x3=60-62x-30x+10x2+31x2-5x3              (Rearranging)
=6092x+41x25x3=60-92x+41x2-5x3                                (Combining like terms) 

Thus, the answer is 6092x+41x25x360-92x+41x2-5x3. 

Question 27: Simplify:
(2x2 + 3x − 5)(3x2 − 5x + 4)

Answer 27: To simplify, we will proceed as follows: 

(2x2+3x5)(3x25x+4)2x2+3x-53x2-5x+4
=2x2(3x25x+4)+3x(3x25x+4)5(3x25x+4)=2x23x2-5x+4+3x3x2-5x+4-53x2-5x+4           (Distributive law)
=6x410x3+8x2+9x315x2+12x15x2+25x20=6x4-10x3+8x2+9x3-15x2+12x-15x2+25x-20
=6x410x3+9x3+8x215x215x2+12x+25x20=6x4-10x3+9x3+8x2-15x2-15x2+12x+25x-20              (Rearranging)
=6x4x322x2+36x20=6x4-x3-22x2+36x-20                                                     (Combining like terms) 

Thus, the answer is 6x4x322x2+36x206x4-x3-22x2+36x-20. 

Question 28: Simplify:
(3x − 2)(2x − 3) + (5x − 3)(x + 1)

Answer 28: To simplify, we will proceed as follows: 

(3x2)(2x3)+(5x3)(x+1)=[(3x2)(2x3)]+[(5x3)(x+1)] =[3x(2x3)2(2x3)]+[5x(x+1)3(x+1)]=3x2x-3-22x-3+5xx+1-3x+1           (Distributive law)
=6x29x4x+6+5x2+5x3x3=6x2-9x-4x+6+5x2+5x-3x-3
=6x2+5x29x4x+5x3x3+6=6x2+5x2-9x-4x+5x-3x-3+6                                  (Rearranging)
=11x211x+3=11x2-11x+3                                                                 (Combining like terms) 

Thus, the answer is 11x211x+311x2-11x+3. 

Question 29: Simplify:
(5x − 3)(x + 2) − (2x + 5)(4x − 3)

Answer 29: To simplify, we will proceed as follows: 

(5x3)(x+2)(2x+5)(4x3)=[(5x3)(x+2)][(2x+5)(4x3)] =[5x(x+2)3(x+2)][2x(4x3)+5(4x3)]=5xx+2-3x+2-2x4x-3+54x-3            (Distributive law)
=5x2+10x3x68x2+6x20x+15=5x2+10x-3x-6-8x2+6x-20x+15
=5x28x2+10x3x+6x20x6+15=5x2-8x2+10x-3x+6x-20x-6+15                              (Rearranging)
=5x28x2+10x3x+6x20x6+15=3x27x+9=5x2-8x2+10x-3x+6x-20x-6+15=-3x2-7x+9(Combining like terms) 

Hence, the answer is 3x27x+9-3x2-7x+9. 

Question 30: Simplify:
(3x + 2y)(4x + 3y) − (2xy)(7x − 3y)

Answer 30: To simplify, we will proceed as follows:

(3x+2y)(4x+3y)(2xy)(7x3y)=[(3x+2y)(4x+3y)][(2xy)(7x3y)] =[3x(4x+3y)+2y(4x+3y)][2x(7x3y)y(7x3y)]=3x4x+3y+2y4x+3y-2x7x-3y-y7x-3y            (Distributive law)
=12x2+9xy+8xy+6y2[14x26xy7xy+3y2]=12x2+9xy+8xy+6y214x2+6xy+7xy3y2

=12x214x2+9xy+8xy+6xy+7xy+6y23y2=12x2-14x2+9xy+8xy+6xy+7xy+6y2-3y2                              (Rearranging)
=2x2+30xy+3y2=-2x2+30xy+3y2                                                                       (Combining like terms) 

Thus, the answer is 2x2+30xy+3y2-2x2+30xy+3y2. 

Question 31: Simplify:
(x2 − 3x + 2)(5x − 2) − (3x2 + 4x − 5)(2x − 1)

Answer 31: To simplify, we will  proceed as follows: 

(x23x+2)(5x2)(3x2+4x5)(2x1)=[(x23x+2)(5x2)][(3x2+4x5)(2x1)]x2-3x+25x-2-3x2+4x-52x-1=x2-3x+25x-2-3x2+4x-52x-1
=[5x(x23x+2)2(x23x+2)][2x(3x2+4x5)1×(3x2+4x5)]=5xx2-3x+2-2x2-3x+2-2x3x2+4x-5-1×3x2+4x-5            (Distributive law)
=[5x315x2+10x(2x26x+4)][6x3+8x210x3x24x+5]=[5x315x2+10x2x2+6x4][6x3+8x210x3x24x+5]=5x315x2+10x2x2+6x46x38x2+10x+3x2+4x5=5x3-15x2+10x-2x2-6x+4-6x3+8x2-10x-3x2-4x+5=5x3-15x2+10x-2x2+6x-4-6x3+8x2-10x-3x2-4x+5=5x3-15x2+10x-2x2+6x-4-6x3-8x2+10x+3x2+4x-5
=5x36x315x22x28x2+3x2+10x+6x+10x+4x54=5x3-6x3-15x2-2x2-8x2+3x2+10x+6x+10x+4x-5-4                                (Rearranging)
=x322x2+30x22x2+30x-9                          (Combining like terms) 

Thus, the answer is x322x2+30x9-x3-22x2+30x-9. 

Question 32: Simplify:
(x3 − 2x2 + 3x − 4) (x −1) − (2x − 3)(x2x + 1)

Answer 32: To simplify,we will proceed as follows: 

(x32x2+3x4)(x1)(2x3)(x2x+1)=[(x32x2+3x4)(x1)][(2x3)(x2x+1)] =[x(x32x2+3x4)1(x32x2+3x4)][2x(x2x+1)3(x2x+1)]=xx3-2x2+3x-4-1x3-2x2+3x-4-2xx2-x+1-3x2-x+1            (Distributive law) 

=[x(x32x2+3x4)1(x32x2+3x4)][2x(x2x+1)3(x2x+1)]=x42x3+3x24xx3+2x23x+4[2x32x2+2x3x2+3x3]=x42x3+3x24xx3+2x23x+42x3+2x22x+3x23x+3 =x42x32x3x3+3x2+2x2+2x2+3x24x3x2x3x+4+3=x4-2x3-2x3-x3+3x2+2x2+2x2+3x2-4x-3x-2x-3x+4+3                             (Rearranging)
=x45x3+10x212x+7=x4-5x3+10x2-12x+7                                               (Combining like terms) 

Thus, the answer is x45x3+10x212x+7x4-5x3+10x2-12x+7. 

The document RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
All you need of Class 8 at this link: Class 8
88 docs

Top Courses for Class 8

88 docs
Download as PDF
Explore Courses for Class 8 exam

Top Courses for Class 8

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

video lectures

,

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

,

MCQs

,

Free

,

past year papers

,

Summary

,

study material

,

practice quizzes

,

mock tests for examination

,

Objective type Questions

,

Exam

,

Semester Notes

,

ppt

,

Viva Questions

,

shortcuts and tricks

,

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

,

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

,

Sample Paper

,

pdf

,

Important questions

,

Previous Year Questions with Solutions

,

Extra Questions

;