RS Aggarwal Solutions for Class 7 Maths Exercise 6C Chapter 6 Algebraic Expressions

RS Aggarwal Solutions for Class 7 Maths Exercise 6C Chapter 6 Algebraic Expressions in simple PDF are given here. Multiplication of a monomial and a binomial is the topic covered in this exercise of RS Aggarwal Solutions for Class 7 Chapter 6. The main aim is to help students understand and crack these problems. We at BYJUâ€™S have prepared the RS Aggarwal Solutions for Class 7 Maths Chapter 6 Algebraic Expressions. By practising these solutions, students can get good marks in Maths.

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Find each of the following products:

1. 4a(3a + 7b)

Solution:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over addition, we have:

P Ã— (q + r) = (p Ã— q) + (p Ã— r)

Now,

= (4a Ã— 3a) + (4a Ã— 7b)

= (12a2 + 28ab) â€¦ [âˆµ am Ã— an = am+n]

2. 5a(6a – 3b)

Solution:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over subtraction, we have:

P Ã— (q – r) = (p Ã— q) – (p Ã— r)

Now,

= (5a Ã— 6a) – (5a Ã— 3b)

= (30a2 – 15ab) â€¦ [âˆµ am Ã— an = am+n]

3. 8a2(2a + 5b)

Solution:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over addition, we have:

P Ã— (q + r) = (p Ã— q) + (p Ã— r)

Now,

= (8a2 Ã— 2a) + (8a2 Ã— 5b)

= (16a3 + 40a2b) â€¦ [âˆµ am Ã— an = am+n]

4. 9x2(5x + 7)

Solution:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over addition, we have:

P Ã— (q + r) = (p Ã— q) + (p Ã— r)

Now,

= (9x2 Ã— 5x) + (9x2 Ã— 7)

= (45x3 + 63x2) â€¦ [âˆµ am Ã— an = am+n]

5. ab(a2 â€“ b2)

Solution:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over subtraction, we have:

P Ã— (q – r) = (p Ã— q) – (p Ã— r)

Now,

= (ab Ã— a2) – (ab Ã— b2)

= (a3b + ab3) â€¦ [âˆµ am Ã— an = am+n]

6. 2x2(3x â€“ 4x2)

Solution:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over subtraction, we have:

P Ã— (q – r) = (p Ã— q) – (p Ã— r)

Now,

= (2x2 Ã— 3x) – (2x2 Ã— 4x2)

= (6x3 â€“ 8x4) â€¦ [âˆµ am Ã— an = am+n]

7.(3/5)m2n(m + 5n)

Solution:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over addition, we have:

P Ã— (q + r) = (p Ã— q) + (p Ã— r)

Now,

= ((3/5)m2n Ã— m) + ((3/5)m2n Ã— 5n)

= ((3/5)m3n + 3m2n2) â€¦ [âˆµ am Ã— an = am+n]

8. -17x2(3x â€“ 4)

Solution:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over subtraction, we have:

P Ã— (q – r) = (p Ã— q) – (p Ã— r)

Now,

= (-17x2 Ã— 3x) – (-17x2 Ã— 4)

= (-51x3 + 68x2) â€¦ [âˆµ am Ã— an = am+n]

9. (7/2)x2((4/7)x + 2)

Solution:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over addition, we have:

P Ã— (q + r) = (p Ã— q) + (p Ã— r)

Now,

= ((7/2)x2 Ã— (4/7)x) + ((7/2)x2 Ã— 2)

= ((7Ã—4)/ (2Ã—7))x3 + ((7Ã—2)/(2Ã—1))x2 â€¦ [âˆµ am Ã— an = am+n]

= ((1Ã—2)/ (1Ã—1))x3 + ((7Ã—1)/(1Ã—1))x2

= (2)x3 + (7)x2

10. -4x2y(3x2 â€“ 5y)

Solution:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over subtraction, we have:

P Ã— (q – r) = (p Ã— q) – (p Ã— r)

Now,

= (-4x2y Ã— 3x2) – (-4x2y Ã— -5y)

= (-12x4y + 20x2y2) â€¦ [âˆµ am Ã— an = am+n]

11. (-4/27)xyz((9/2)x2yz â€“ (3/4)xyz2)

Solution:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over subtraction, we have:

P Ã— (q – r) = (p Ã— q) – (p Ã— r)

Now,

= ((-4/27)xyz Ã— (9/2)x2yz) – ((-4/27)xyz Ã— (-3/4)xyz2)

= ((-4Ã—9)/ (27Ã—2))x3y2z2 + ((-4Ã—-3)/(27Ã—4))x2y2z3 â€¦ [âˆµ am Ã— an = am+n]

= ((-2Ã—1)/ (3Ã—1))x3y2z2 + ((-1Ã—-1)/(9Ã—1))x2y2z3

= (-2/3)x3y2z2 + (1/9)x2y2z3

12. 9t2(t + 7t3)

Solution:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over addition, we have:

P Ã— (q + r) = (p Ã— q) + (p Ã— r)

Now,

= (9t2 Ã— t) + (9t2 Ã— 7t3)

= (9t3 + 63t5) â€¦ [âˆµ am Ã— an = am+n]

13. 10a2(0.1a â€“ 0.5b)

Solution:-

Let p, q and r be three monomials.

Then, by distributive law of multiplication over subtraction, we have:

P Ã— (q – r) = (p Ã— q) – (p Ã— r)

Now,

= (10a2 Ã— 0.1a) – (10a2 Ã— 0.5b)

= (1a3 â€“ 5a2b) â€¦ [âˆµ am Ã— an = am+n]

Access other exercises of RS Aggarwal Solutions For Class 7 Chapter 6 – Algebraic Expressions

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