Students can refer to RD Sharma Solutions for Class 8 Maths Exercise 6.3 of Chapter 6 Algebraic Expressions and Identities which are available here in simple PDF. RD Sharma Solutions PDF can be downloaded easily from the links provided below. BYJU’S expert tutors have designed solutions uniquely which will help students come up with flying colours in their examinations. In Exercise 6.3 of Chapter 6, Algebraic Expressions and Identities, we will study problems based on the multiplication of algebraic expressions and the multiplication of two monomials.
RD Sharma Solutions for Class 8 Maths Exercise 6.3 Chapter 6 Algebraic Expressions and Identities
Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 6.3 Chapter 6 Algebraic Expressions and Identities
EXERCISE 6.3 PAGE NO: 6.13
Find each of the following products:
1. 5x2 × 4x3
Solution:
Let us simplify the given expression
5 × x × x × 4 × x × x × x
5 × 4 × x1+1+1+1+1
20 × x5
20x5
2. -3a2 × 4b4
Solution:
Let us simplify the given expression
– 3 × a2 × 4 × b4
-12 × a2 × b4
-12a2b4
3. (-5xy) × (-3x2yz)
Solution:
Let us simplify the given expression
(-5) × (-3) × x × x2 × y × y × z
15 × x1+2 × y1+1 × z
15x3y2z
4. 1/2xy × 2/3x2yz2
Solution:
Let us simplify the given expression
1/2 × 2/3 × x × x2 × y × y × z2
1/3 × x1+2 × y1+1 × z2
1/3x3y2z2
5. (-7/5xy2z) × (13/3x2yz2)
Solution:
Let us simplify the given expression
-7/5 × 13/3 × x × x2 × y2 × y × z × z2
-91/15 × x1+2 × y2+1 × z1+2
-91/15x3y3z3
6. (-24/25x3z) × (-15/16xz2y)
Solution:
Let us simplify the given expression
-24/25 × -15/16 × x3 × x × z × z2 × y
18/20 × x3+1 × z1+2 × y
9/10x4z3y
7. (-1/27a2b2) × (9/2a3b2c2)
Solution:
Let us simplify the given expression
-1/27 × 9/2 × a2 × a3 × b2 × b2 × c2
-1/6 x a2+3 × b2+2 × c2
-1/6a5b4c2
8. (-7xy) × (1/4x2yz)
Solution:
Let us simplify the given expression
-7 × 1/4 × x × y × x2 × y × z
-7/4 × x1+2 × y1+1 × z
-7/4x3y2z
9. (7ab) × (-5ab2c) × (6abc2)
Solution:
Let us simplify the given expression
7 × -5 × 6 × a × a × a × b × b2 × b × c × c2
210 × a1+1+1 × b1+2+1 × c1+2
210a3b4c3
10. (-5a) × (-10a2) × (-2a3)
Solution:
Let us simplify the given expression
(-5) × (-10) × (-2) × a × a2 × a3
-100 × a1+2+3
-100a6
11. (-4x2) × (-6xy2) × (-3yz2)
Solution:
Let us simplify the given expression
(-4) × (-6) – (-3) × x2 × x × y2 × y × z2
– 72 × x2+1 × y2+1 × z2
-72x3y3z2
12. (-2/7a4) × (-3/4a2b) × (-14/5b2)
Solution:
Let us simplify the given expression
-2/7 × -3/4 × -14/5 × a4 × a2 × b × b2
-6/10 × a4+2 × b1+2
-3/5a6b3
13. (7/9ab2) × (15/7ac2b) × (-3/5a2c)
Solution:
Let us simplify the given expression
7/9 × 15/7 × -3/5 × a × a × a2 × b2 × b × c2 × c
– a1+1+2 × b2+1 × c2+1
-a4b3c3
14. (4/3u2vw) × (-5uvw2) × (1/3v2wu)
Solution:
Let us simplify the given expression
4/3 × -5 × 1/3 × u2 × u × u × v × v × v2 × w × w2 × w
-20/9 × u2+1+1 × v1+1+2 × w1+2+1
-20/9u4v4w4
15. (0.5x) × (1/3xy2z4) × (24x2yz)
Solution:
Let us simplify the given expression
0.5 × 1/3 × 24 × x × x × y2 × y × x2 × z4 × z
12/3 × x1+1+2 × y2+1 × z4+1
4x4 × y3 × z5
4x4y3z5
16. (4/3pq2) × (-1/4p2r) × (16p2q2r2)
Solution:
Let us simplify the given expression
4/3 × 1/4 × 16 × p × p2 × p2 × q2 × q2 × r × r2
-16/3 × p1+2+2 × q2+2 × r1+2
-16/3p5q4r3
17. (2.3xy) × (0.1x) × (0.16)
Solution:
Let us simplify the given expression
2.3 × 0.1 × 0.16 × x × x × y
0.0368 × x1+1 × y
0.0368x2y
Express each of the following products as monomials and verify the result in each case for x=1:
18. (3x) × (4x) × (-5x)
Solution:
Let us simplify the given expression
3 × 4 × -5 × x × x × x
-60 × x1+1+1
-60x3
Verification
LHS = (3 × 1) × (4 × 1) × (-5 × 1)
= 3 × 4 × – 5
= – 60
RHS = -60 (1)3 = – 60
Therefore, LHS = RHS.
19. (4x2) × (-3x) × (4/5x3)
Solution:
Let us simplify the given expression
4 × -3 × 4/5 × x2 × x × x3
-48/5 × x2+1+3
-48/5x6
Verification
LHS = 4 × 12 × – 3 × 1 × 4/5 × 13
= – 48/5
RHS = – 48/5 × 16 = – 48/5
Therefore, LHS = RHS.
20. (5x4) × (x2)3 × (2x) 2
Solution:
Let us simplify the given expression
5 × x4 × x6 × 4 × x2
5 × 4 × x4 × x6 × x2
20 × x4+6+2
20x12
Verification
LHS = (5 × 14) × (12)3 × (2 × 1)2
= 5 × 4
= 20
RHS = 20 × 112 = 20
Therefore, LHS = RHS.
21. (x2)3 × (2x) × (-4x) × (5)
Solution:
Let us simplify the given expression
x6 × 2 × x × -4 × x × 5
2 × -4 × 5 × x6 × x × x
-40 × x6+1+1
-40x8
Verification
LHS = (12)3 × (2 × 1) × (-4 × 1) × 5
= – 40
RHS = – 40 × 18 = – 40
Therefore, LHS = RHS.
22. Write down the product of -8x2y6 and -20xy verify the product for x = 2.5, y = 1
Solution:
Let us simplify the given expression
-8 × -20 × x2 × x × y6 × y
160 × x2+1 × y6+1
160x3y7
Now let us verify when, x = 2.5 and y = 1
For 160x3y7
160 (2.5)3 × (1)7
160 × 15.625
2500
For -8x2y6 and -20xy
-8 × 2.52 × 16 × -20 × 1 × 2.5
2500
Hence, the given expression is verified.
23. Evaluate (3.2x6y3) × (2.1x2y2) when x = 1 and y = 0.5
Solution:
Let us simplify the given expression
3.2 × 2.1 × x6 × x2 × y3 × y2
6.72 × x6+2 × y3+2
6.72x8y5
Now let us substitute when, x = 1 and y = 0.5
For 6.72x8y5
6.72 × 18 × 0.55
0.21
24. Find the value of (5x6) × (-1.5x2y3) × (-12xy2) when x = 1, y = 0.5
Solution:
Let us simplify the given expression
5 × -1.5 × -12 × x6 × x2 × x × y3 × y2
90 × x6+2+1 × y3+2
90x9y5
Now let us substitute when, x = 1 and y = 0.5
For 90x9y5
90 × (1)9× (0.5)5
2.8125
45/16
25. Evaluate (2.3a5b2) × (1.2a2b2) when a = 1 and b = 0.5
Solution:
Let us simplify the given expression
2.3a5b2 × 1.2a2b2
2.3 × 1.2 × a5 × a2 × b2 × b2
2.76 × a5+2 × b2+2
2.76a7b4
Now let us substitute when, a = 1 and b = 0.5
For 2.76 a7 b4
2.76 (1)7 (0.5)4
2.76 × 1 × 0.0025
0.1725
6.9/40
26. Evaluate (-8x2y6) × (-20xy) for x = 2.5 and y = 1
Solution:
Let us simplify the given expression
-8 × – 20 × x2 × x × y6 × y
160x2+1y6+1
160x3y7
Now let us substitute when, x = 2.5 and y = 1
160x3y7
160 × (2.5)3 × (1)7
2500
Express each of the following products as monomials and verify the result for x = 1, y = 2:
27. (-xy3) × (yx3) × (xy)
Solution:
Let us simplify the given expression
-x × y3 × y × x3 × x × y
-x1+3+1 × y3+1+1
-x5y5
Now let us substitute when, x = 1 and y = 2
-x5y5
-15 × 25
-32
28. (1/8x2y4) × (1/4x4y2) × (xy) × 5
Solution:
Let us simplify the given expression
1/8 × 1/4 × 5 × x2 × x4 × x × y4 × y2 × y
5/32 × x2+4+1 × y4+2+1
5/32x7y7
Now let us substitute when, x = 1 and y = 2
5/32 × 17 × 27
5/32 × 128
5 × 4
20
29. (2/5a2b) × (-15b2ac) × (-1/2c2)
Solution:
Let us simplify the given expression
2/5 × -15 × -1/2 × a2 × a × b × b2 × c × c2
3 × a2+1 × b1+2 × c1+2
3a3b3c3
30. (-4/7a2b) × (-2/3b2c) × (-7/6c2a)
Solution:
Let us simplify the given expression
-4/7 × -2/3 × -7/6 × a2 × a × b × b2 × c × c2
-4/9 × a2+1 × b2+1 × c1+2
-4/9a3b3c3
31. (4/9abc3) × (-27/5a3b2) × (-8b3c)
Solution:
Let us simplify the given expression
4/9 × -27/5 × -8 × a × a3 × b × b2 × b3 × c3 × c
96/5 × a1+3 × b1+2+3 × c3+1
96/5a4b6c4
Evaluate each of the following when x = 2, y = -1.
32. (2xy) × (x2y/4) × (x2) × (y2)
Solution:
Let us simplify the given expression
2 × 1/4 × x × x2 × x2 × y × y2 × y
1/2x1+2+2y1+2+1
1/2x5y4
Now let us substitute when, x = 2 and y = -1
For 1/2x5y4
1/2 × (2)5 × (-1)4
1/2 × 32 × 1
16
33. (3/5x2y) × (-15/4xy2) × (7/9x2y2)
Solution:
Let us simplify the given expression
3/5 × -15/4 × 7/9 × x2 × x × x2 × y × y2 × y2
-7/4 × x2+1+2 × y1+2+2
7/4x5y5
Now let us substitute when, x = 2 and y = -1
For -7/4x5y5
-7/4 × (2)5 (-1)5
-7/4 × 32 × -1
56
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