RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev

RD Sharma Solutions for Class 8 Mathematics

Created by: Abhishek Kapoor

Class 8 : RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev

The document RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev 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

Question 1: Find the following product:
2a3(3a + 5b)

Answer 1: To find the product, we will use distributive law as follows:
2a3(3a+5b)=2a3×3a+2a3×5b=(2×3)(a3×a)+(2×5)a3b=(2×3)a3+1+(2×5)a3b=6a4+10a3b2a33a+5b=2a3×3a+2a3×5b=2×3a3×a+2×5a3b=2×3a3+1+2×5a3b=6a4+10a3b
Thus, the answer is 6a4+10a3b6a4+10a3b. 

Question 2: Find the following product:
−11a(3a + 2b)

Answer 2: To find the product, we will use distributive law as follows:
11a(3a+2b)=(11a)×3a+(11a)×2b=(11×3)×(a×a)+(11×2)×(a×b)=(33)×(a1+1)+(22)×(a×b)=33a222ab-11a3a+2b=-11a×3a+-11a×2b=-11×3×a×a+-11×2×a×b=-33×a1+1+-22×a×b=-33a2-22ab
Thus, the answer is 33a222ab-33a2-22ab. 

Question 3: Find the following product:
−5a(7a − 2b)

Answer 3: To find the product, we will use distributive law as follows:
5a(7a2b)=(5a)×7a+(5a)×(2b)=(5×7)×(a×a)+(5×(2))×(a×b)=(35)×(a1+1)+(10)×(a×b)=35a2+10ab-5a7a-2b=-5a×7a+-5a×-2b=-5×7×a×a+-5×-2×a×b=-35×a1+1+10×a×b=-35a2+10ab
Thus, the answer is 35a2+10ab-35a2+10ab. 

Question 4: Find the following product:
−11y2(3y + 7)

Answer 4: To find the product, we will use distributive law as follows:

11y2(3y+7)=(11y2)×3y+(11y2)×7=(11×3)(y2×y)+(11×7)×(y2)=(33)(y2+1)+(77)×(y2)
=−33y3−77y2

Thus, the answer is −33y3−77y2

Question 5: Find the following product:
6x/5 (x3+y3)6x5(x3+y3)

Answer 5: To find the product, we will use distributive law as follows: 

6x/5 (x3+y3) 

=6x/5×x3+6x/5×y3 

=6/5×(x×x3)+6/5×(x×y3) 

=6/5×(x1+3)+6/5×(x×y3) 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev

Thus, the answer isRD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev .

Question 6: xy(x3y3)

Answer 6: To find the product, we will use the distributive law in the following way:
xy(x3y3)=xy×x3xy×y3=(x×x3)×yx×(y×y3)=x1+3yxy1+3=x4yxy4xyx3-y3=xy×x3-xy×y3=x×x3×y-x×y×y3=x1+3y-xy1+3=x4y-xy4
Thus, the answer is x4yxy4x4y-xy4. 

Question 7: Find the following product:
0.1y(0.1x5 + 0.1y)

Answer 7: To find the product, we will use distributive law as follows:
0.1y(0.1x5+0.1y)=(0.1y)(0.1x5)+(0.1y)(0.1y)=(0.1×0.1)(y×x5)+(0.1×0.1)(y×y)=(0.1×0.1)(x5×y)+(0.1×0.1)(y1+1)=0.01x5y+0.01y20.1y0.1x5+0.1y=0.1y0.1x5+0.1y0.1y=0.1×0.1y×x5+0.1×0.1y×y=0.1×0.1x5×y+0.1×0.1y1+1=0.01x5y+0.01y2
Thus, the answer is 0.01x5y+0.01y20.01x5y+0.01y2. 

Question 8: Find the following product:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev

Answer 8: To find the product, we will use distributive law as follows: 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev

={{−7/4×(−50)}(a×a2)×(b2×b2)×(c×c2)}−{(625)(−50)(a2×a2)×(b2)×(c2×c2)} 

={7/4×(50)}(a1+2b2+2c1+2){(6/25)(50)(a2+2b2c2+2)} 

=175/2 a3b4c3(12a4b2c4) 

=175/2 a3b4c3+12a4b2c4 

Thus, the answer is 175/2 a3b4c3+12a4b2c41752a3b4c3+12a4b2c4. 

Question 9: Find the following product:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev

Answer 9: To find the product, we will use the distributive law in the following way:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev (x×x)×(y×y)×(z×z2)}
RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev (x×x)×(y×y2)×(z×z3)} 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev (x1+1y1+1z1+2)}− RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev (x1+1y1+2z1+3)} 

23y+7=-11y2×3y+-11y2×7=-11×3y2×y+-11×7×y2=-33y2+1+-77×y2=-33y3-77y2

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev -RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev

=4/9 x2y2z3+2/3 x2y3z4

Thus, the answer is 4/9 x2y2z3+2/3 x2y3z4 

Question 10: Find the following product:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev

Answer 10: To find the product, we will use distributive law as follows: 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev

=2/3 x3y2z2+1/9 x2y2z3

Thus, the answer is 2/3 x3y2z2+1/9 x2y2z3 

Question 11: Find the following product:
1.5x(10x2y − 100xy2)

Answer 11: To find the product, we will use distributive law as follows:
1.5x(10x2y100xy2)=(1.5x×10x2y)(1.5x×100xy2)=(15x1+2y)(150x1+1y2)=15x3y150x2y2
Thus, the answer is 15x3y150x2y215x3y-150x2y2. 

Question 12: Find the following product:
4.1xy(1.1xy)

Answer 12: To find the product, we will use distributive law as follows:
4.1xy(1.1xy)=(4.1xy×1.1x)(4.1xy×y)={(4.1×1.1)×xy×x}(4.1xy×y)=(4.51x1+1y)(4.1xy1+1)=4.51x2y4.1xy2
Thus, the answer is 4.51x2y4.1xy24.51x2y-4.1xy2. 

Question 13: Find the following product:

250.5xy (xz+y/10)xz+y10 

Answer 13: To find the product, we will use distributive law as follows: 

250.5xy(xz+y/10)=250.5xy×xz+250.5xy×y/10=250.5x1+1yz+25.05xy1+1=250.5x2yz+25.05xy2Thus, the answer is 250.5x2yz+25.05xy2250.5x2yz+25.05xy2. 

Question 14: Find the following product: 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev

Answer 14: To find the product, we will use distributive law as follows: 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev

Thus, the answer is RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) Class 8 Notes | EduRev

Question 15: Find the following product:

4/3 a(a2 + b2  3c2)43a(a2 + b2 - 3c2) 

Answer 15: To find the product, we will use distributive law as follows: 

4/3 a(a2+b23c2)=4/3 a×a2+4/3 a×b24/3 a×3c2=4/3 a1+2+4/3 ab24ac2=4/3 a3+4/3 ab24ac2

Thus, the answer is 4/3 a3+4/3 ab24ac2

Question 16: Find the product 24x2 (1 − 2x) and evaluate its value for x = 3.

Answer 16: To find the product, we will use distributive law as follows:
24x2(12x)=24x2×124x2×2x=24x248x1+2=24x248x324x21-2x=24x2×1-24x2×2x=24x2-48x1+2=24x2-48x3
Substituting  x = 3 in the result, we get: 

24x248x3=24(3)248(3)3=24×948×27=2161296=1080 Thus, the product is (24x248x3) and its value for x = 3 is (1080)(24x2-48x3) and its value for x = 3 is (-1080). 

Question 17: Find the product −3y(xy + y2) and find its value for x = 4 and y = 5.

Answer 17: To find the product, we will use distributive law as follows:
3y(xy+y2)=3y×xy+(3y)×y2=3xy1+13y1+2=3xy23y3
Substituting x = 4 and y = 5 in the result, we get:

3xy23y3=3(4)(5)23(5)3=3(4)(25)3(125)=300375=675 Thus, the product is (3xy23y3-3xy2-3y3), and its value for x = 4 and y = 5 is (-675). 

Question 18: Multiply 3/2 x2y3 by (2x  y)-32x2y3 by (2x - y) and verify the answer for x = 1 and y = 2.

Answer 18: To find the product, we will use distributive law as follows: 

3/2 x2y3×(2xy)=(3/2 x2y3×2x)(3/2 x2y3×y)=(3x2+1y3)(3/2 x2y3+1)=3x3y3+3/2 x2y4

Substituting x = 1 and y = 2 in the result, we get: 

3x3y3+3/2 x2y4=3(1)3(2)3+3/2 (1)2(2)4=3×1×8+3/2 ×1×16=24+24=0Thus, the product is 3x3y3+3/2 x2y4-3x3y3+32x2y4, and its value for x = 1 and y = 2 is 0. 

Question 19: Multiply the monomial by the binomial and find the value of each for x = −1, y = 0.25 and z = 0.05:
(i) 15y2(2 − 3x)
(ii) −3x(y2 + z2)
(iii) z2(xy)
(iv) xz(x2 + y2)

Answer 19: (i) To find the product, we will use distributive law as follows:
15y2(23x)=15y2×215y2×3x=30y245xy2
Substituting x = -1 and y = 0.25 in the result, we get: 30y245xy2=30(0.25)245(1)(0.25)2=30×0.0625{45×(1)×0.0625}=30×0.0625{45×(1)×0.0625}=1.875(2.8125)=1.875+2.8125=4.6875 (ii) To find the product, we will use distributive law as follows:
3x(y2+z2)=3x×y2+(3x)×z2=3xy23xz2 Substituting x = -1, y = 0.25 and z = 0.05 in the result, we get: 

3xy23xz2=3(1)(0.25)23(1)(0.05)2=3(1)(0.0625)3(1)(0.0025)=01875+0.0075=0.195 (iii) To find the product, we will use distributive law as follows:
z2(xy)=z2×xz2×y=xz2yz2z2x-y=z2×x-z2×y=xz2-yzSubstituting x = -1, y = 0.25 and z = 0.05 in the result, we get:

xz2yz2=(1)(0.05)2(0.25)(0.05)2=(1)(0.0025)(0.25)(0.0025)=0.00250.000625=0.003125 (iv) To find the product, we will use distributive law as follows: 

xz(x2+y2)=xz×x2+xz×y2=x3z+xy2z Substituting x = -1, y = 0.25 and z = 0.05 in the result, we get:
x3z+xy2z=(1)3(0.05)+(1)(0.25)2(0.05)=(1)(0.05)+(1)(0.0625)(0.05)=0.050.003125=0.053125 

Question 20: Simplify:
(i) 2x2(x3x) − 3x(x4 + 2x) − 2(x4 − 3x2)
(ii) x3y(x2 − 2x) + 2xy(x3x4)
(iii) 3a2 + 2(a + 2) − 3a(2a + 1)
(iv) x(x + 4) + 3x(2x2 − 1) + 4x2 + 4
(v) a(c) − b(ca) − c(ab)
(vi) a(bc) + b(ca) + c(ab)
(vii) 4ab(ab) − 6a2(bb2) − 3b2<