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Class 8 RD Sharma Solutions – Chapter 6 Algebraic Expressions And Identities – Exercise 6.4 | Set 1

  • Last Updated : 07 Apr, 2021

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

Solution:

Using Distributive law,

2a3 (3a + 5b) = 2a3 × 3a + 2a3 × 5b

= 6a3+1 + 10a3b = 6a4 + 10a3b

Hence, the product is 6a4 + 10a3b



Question 2. Find the product -11a(3a + 2b)

Solution:

Using Distributive law,

-11a (3a + 2b) = -11a × 3a + (-11a) × 2b

= -33a1+1 – 22ab = -33a2 – 22ab

Hence, the product is -33a2 – 22ab

Question 3. Find the product -5a(7a – 2b)

Solution:

Using Distributive law,

-5a (7a – 2b) = -5a × 7a – (-5a) × 2b



= -35a1+1 + 10ab = -35a2 + 10ab

Hence, the product is -35a2 + 10ab

Question 4. Find the product -11y2(3y + 7)

Solution:

Using Distributive law,

-11y2 (3y + 7) = -11y2 × 3y + (-11y2) × 7

= -33y2+1 – 77y2 = -33y3 – 77y2

Hence, the product is -33y3 – 77y2

Question 5. Find the product of 6x/5(x3+y3)

Solution:

Using Distributive law,

6x/5 (x3+y3) = 6x/5 × x3 + 6x/5 × y3



= 6x3+1/5 + 6xy3/5 = 6x4/5 + 6xy3/5

Hence, the product is 6x4/5 + 6xy3/5

Question 6. Find the product of xy(x3 – y3)

Solution:

Using Distributive law,

xy (x3-y3) = xy × x3 – xy × y3

= x3+1y – xy3+1 = x4y – xy4

Hence, the product is x4y – xy4

Question 7. Find the product of 0.1y (0.1x5 + 0.1y)

Solution:

Using Distributive law,

0.1y (0.1x5 + 0.1y) = 0.1y × 0.1x5 + 0.1y × 0.1y 

= 0.01x5y + 0.01y2

Hence, the product is 0.01x5y + 0.01y2

Question 8. Find the product of (-7ab2c/4 – 6a2c2/25) (-50a2b2c2)

Solution:

Using Distributive law,

(-7ab2c/4 – 6a2c2/25) (-50a2b2c2) = (-50a2b2c2) × (-7ab2c/4) – (-50a2b2c2) × (6a2c2/25)

= 175a2+1b2+2c2+1/2 + 12a2+2b2c2+2 

= 175a3b4c3/2 + 12a4b2c4

Hence, the product is 175a3b4c3/2 + 12a4b2c4

Question 9. Find the product of -8xyz/27 (3xyz2/2 – 9xy2z3/4)

Solution:

Using Distributive law,



-8xyz/27 (3xyz2/2 – 9xy2z3/4) = (-8xyz/27) × (3xyz2/2) – (-8xyz/27) × (9xy2z3/4)

= -4x1+1y1+1z1+2/9 + 2x1+1y1+2z1+3/3
= -4x2y2z3/9 + 2x2y3z4/3

Hence, the product is -4x2y2z3/9 + 2x2y3z4/3

Question 10. Find the product of (-4xyz/27) [9x2yz/2 – 3xyz2/4]

Solution:

Using Distributive law,

(-4xyz/27) [9x2yz/2 – 3xyz2/4] = (-4xyz/27) × (9x2yz/2) + (-4xyz/27) × (3xyz2/4)

= -2x1+2y1+1z1+1/3 + 1x1+1y1+1z1+2/9 

= -2x3y2z2/3 + 1x2y2z3/9 

Hence, the product is 2x3y2z2/3 + 1x2y2z3/9 

 Chapter 6 Algebraic Expression and Identities – Exercise 6.4 | Set 2

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.

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