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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-2)

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) | RD Sharma Solutions for Class 8 Mathematics PDF Download

Question 1: Find each of the following product:
5x²× 4x³

Answer 1: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices. However, use of these laws are subject to their applicability in the given expressions.
In the present problem, to perform the multiplication, we can proceed as follows:
$5 x^{2} \times 4 x^{3} = (5 \times 4) \times (x^{2} \times x^{3})$
$= 20 x^{5}$ ( $∵$ $a^{m} \times a^{n} = a^{m + n}$ )
Thus, the answer is $20 x^{5}$ .

Question 2: Find each of the following product:
−3a² × 4b⁴

Answer 2: To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, $a^{m} \times a^{n} = a^{m + n}$ , wherever applicable.
We have:
$- 3 a^{2} \times 4 b^{4} = (- 3 \times 4) \times (a^{2} \times b^{4}) = - 12 a^{2} b^{4}$ Thus, the answer is $- 12 a^{2} b^{4}$ .

Question 3: Find each of the following product:
(−5xy) × (−3x²yz)

Answer 3: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, $a^{m} \times a^{n} = a^{m + n}$ , wherever applicable.
We have:
$(- 5 x y) \times (- 3 x^{2} y z) = \{(- 5) \times (- 3)\} \times (x \times x^{2}) \times (y \times y) \times z = 15 \times (x^{1 + 2}) \times (y^{1 + 1}) \times z = 15 x^{3} y^{2} z$ Thus, the answer is $15 x^{3} y^{2} z$ .

Question 4: Find each of the following product:

Answer 4: To multiply algebraic expressions, we use commutative and associative laws along with the the law of indices, that is, $a^{m} \times a^{n} = a^{m + n}$ .

We have:

Thus, the answer is 1/6 $\frac{1}{6} x^{3} y^{2} z^{2}$ $\frac{1}{6} x^{3} y^{2} z^{2}$ .

Question 5: Find each of the following product:

Answer 5: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

(x×x2)×(y2×y)×(z×z2)

×(x1+2)×(y2+1)×(z1+2)

= - 91/15 x3y3x3

Thus, the answer is = - 91/15 x3y3x3

Question 6: Find each of the following product:

Answer 6: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

×(x3×x)×(z×z2)×y

×(x3+1)×(z1+2)×y

$= \frac{9}{10} x^{4} y z^{3}$

Thus, the answer is $= \frac{9}{10} x^{4} y z^{3}$

Question 7: Find each of the following product:

Answer 7:To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:

(a2×a3)×(b2×b2)×c2

(a2+3)×(b2+2)×c2

= - 1/6 a5b4c2

Thus, the answer is = - 1/6 a5b4c2

Question 8: Find each of the following product:

Answer 8: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) | RD Sharma Solutions for Class 8 Mathematics

=(−7×1/4)×(x×x2)×(y×y)×z

=(−7×1/4)×(x1+2)×(y1+1)×z

=− 7/4 x3y2z

Thus, the answer is − 7/4 x3y2z

Question 9: Find each of the following product:
(7ab) × (−5ab²c) × (6abc²)

Answer 9: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have: (7ab)×(−5ab2c)×(6abc2)={7×(−5)×6}×(a×a×a)×(b×b2×b)×(c×c2)={7×(−5)×6}×(a1+1+1)×(b1+2+1)×(c1+2)=−210a3b4c3 Thus, the answer is $- 210 a^{3} b^{4} c^{3}$ .

Question 10: Find each of the following product:
(−5a) × (−10a²) × (−2a³)

Answer 10: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:
$(- 5 a) \times (- 10 a^{2}) \times (- 2 a^{3}) = \{(- 5) \times (- 10) \times (- 2)\} \times (a \times a^{2} \times a^{3}) = \{(- 5) \times (- 10) \times (- 2)\} \times (a^{1 + 2 + 3}) = - 100 a^{6}$ Thus, the answer is $- 100 a^{6}$ .

Question 11: Find each of the following product:
(−4x²) × (−6xy²) × (−3yz²)

Answer 11: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:
$(- 4 x^{2}) \times (- 6 x y^{2}) \times (- 3 y z^{2}) = \{(- 4) \times (- 6) \times (- 3)\} \times (x^{2} \times x) \times (y^{2} \times y) \times z^{2} = \{(- 4) \times (- 6) \times (- 3)\} \times (x^{2 + 1}) \times (y^{2 + 1}) \times z^{2} = - 72 x^{3} y^{3} z^{2}$ Thus, the answer is $- 72 x^{3} y^{3} z^{2}$ .

Question 12: Find each of the following product:

Answer 12: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:

×(a4×a2)×(b×b2)

×a4+2×b1+2

×a6×b3

=− 3/5 a6b3

Thus, the answer is − 3/5 a6b3

Question 13: Find each of the following product:

Answer 13: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:

×(a×a×a2)×(b2×b)×(c2×c)

×(a1+1+2)×(b2+1)×(c2+1)

=−a4b3c3

Thus, the answer is $- a^{4} b^{3} c^{3}$ .

Question 14: Find each of the following product:

Answer 14: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:

(4/3 u2vw)×(−5uvw2)×(1/3 v2wu)

×(u2×u×u)×(v×v×v2)×(w×w2×w)

×(u2+1+1)×(v1+1+2)×(w1+2+1)

=− 20/9 u4v4w4

Thus, the answer is − 20/9 u4v4w4

Question 15: Find each of the following product:
$(0.5 x) \times (\frac{1}{3} x y^{2} z^{4}) \times (24 x^{2} y z)$

Answer 15: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:

(0.5x)×(1/3 xy2z4)×(24x2yz)

=(0.5×1/3×24)×(x×x×x2)×(y2×y)×(z4×z)

=(0.5×1/3×24)×(x1+1+2)×(y2+1)×(z4+1)

=4x4y3z5

Thus, the answer is $4 x^{4} y^{3} z^{5}$ .

Question 16: Find each of the following product:

(4/3 pq2)×(−1/4 p2r)×(16p2q2r2)43pq2×-14p2r×16p2q2r2

Answer 16: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:

(4/3 pq2)×(−1/4 p2r)×(16p2q2r2)

={4/3×(−1/4)×16}×(p×p2×p2)×(q2×q2)×(r×r2)

={4/3×(− 1/4)×16}×(p1+2+2)×(q2+2)×(r1+2)

=− 16/3p5q4r3

Thus, the answer is $- \frac{1}{3} p^{5} q^{4} r^{3}$ .

Question 17: Find each of the following product:
(2.3xy) × (0.1x) × (0.16)

Answer 17: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:
$(2.3 x y) \times (0.1 x) \times (0.16) = (2.3 \times 0.1 \times 0.16) \times (x \times x) \times y = (2.3 \times 0.1 \times 0.16) \times (x^{1 + 1}) \times y = 0.0368 x^{2} y$ Thus, the answer is $0.0368 x^{2} y$ .

Question 18: Express each of the following product as a monomials and verify the result in each case for x = 1:
(3x) × (4x) × (−5x)

Answer 18: We have to find the product of the expression in order to express it as a monomial.
To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:
$(3 x) \times (4 x) \times (- 5 x) = \{3 \times 4 \times (- 5)\} \times (x \times x \times x) = \{3 \times 4 \times (- 5)\} \times (x^{1 + 1 + 1}) = - 60 x^{3}$
Substituting x = 1 in LHS, we get:
$LHS = (3 x) \times (4 x) \times (- 5 x) = (3 \times 1) \times (4 \times 1) \times (- 5 \times 1) = - 60$ =(3×1)×(4×1)×(−5×1)

=−60
Putting x = 1 in RHS, we get:
$RHS = - 60 x^{3} = - 60 {(1)}^{3} = - 60 \times 1 = - 60$ $∵$ LHS = RHS for x = 1; therefore, the result is correct
Thus, the answer is $- 60 x^{3}$ .

Question 19: Express each of the following product as a monomials and verify the result in each case for x = 1:
(4x²) × (−3x) × $(\frac{4}{5} x^{3})$

Answer 19: We have to find the product of the expression in order to express it as a monomial.
To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

(4x2)×(−3x)×(4/5 x3)

={4×(−3)×4/5}×(x2×x×x3)

={4×(−3)×4/5}×(x2+1+3)

=− 48/5x6

$∴$ $(4 x^{2}) \times (- 3 x) \times (\frac{4}{5} x^{3}) = - \frac{48}{5} x^{6}$

$(4 x^{2}) \times (- 3 x) \times (\frac{4}{5} x^{3}) = - \frac{48}{5} x^{6}$

$(4 x^{2}) \times (- 3 x) \times (\frac{4}{5} x^{3}) = - \frac{48}{5} x^{6}$ $(4 x^{2}) \times (- 3 x) \times (\frac{4}{5} x^{3}) = - \frac{48}{5} x^{6}$

$(4 x^{2}) \times (- 3 x) \times (\frac{4}{5} x^{3}) = - \frac{48}{5} x^{6}$

Question 20: Express each of the following product as a monomials and verify the result in each case for x = 1:
(5x⁴) × (x²)³ × (2x)²

Answer 20: We have to find the product of the expression in order to express it as a monomial.

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n} and {(a^{m})}^{n} = a^{m n}$ .
We have:
$(5 x^{4}) \times {(x^{2})}^{3} \times {(2 x)}^{2} = (5 x^{4}) \times (x^{6}) \times (2^{2} \times x^{2}) = (5 \times 2^{2}) \times (x^{4} \times x^{6} \times x^{2}) = (5 \times 2^{2}) \times (x^{4 + 6 + 2}) = 20 x^{12}$ ∴ (5x4)×(x²)³×(2x)² =20x¹²

Substituting x = 1 in LHS, we get:
$LHS = (5 x^{4}) \times {(x^{2})}^{3} \times {(2 x)}^{2} = (5 \times 1^{4}) \times {(1^{2})}^{3} \times {(2 \times 1)}^{2} = (5 \times 1) \times (1^{6}) \times {(2)}^{2} = 5 \times 1 \times 4 = 20$ =20

Put x =1 in RHS, we get $LHS = (5 x^{4}) \times {(x^{2})}^{3} \times {(2 x)}^{2} = (5 \times 1^{4}) \times {(1^{2})}^{3} \times {(2 \times 1)}^{2} = (5 \times 1) \times (1^{6}) \times {(2)}^{2} = 5 \times 1 \times 4 = 20$ $∵$ LHS = RHS for x = 1; therefore, the result is correct.

Thus, the answer is $20 x^{12}$ .

Question 21: Express each of the following product as a monomials and verify the result in each case for x = 1:
(x²)³ × (2x) × (−4x) × (5)

Answer 21: We have to find the product of the expression in order to express it as a monomial.
To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n} and {(a^{m})}^{n} = a^{m n}$ .
We have:

(x2)3×(2x)×(−4x)×5=(x6)×(2x)×(−4x)×5={2×(−4)×5}×(x6×x×x)={2×(−4)×5}×(x6+1+1)=−40x8

$∴$ ${(x^{2})}^{3} \times (2 x) \times (- 4 x) \times 5 = - 40 x^{8}$

Substituting x = 1 in LHS, we get:

LHS =(x2)3×(2x)×(−4x)×5=(12)3×(2×1)×(−4×1)×5=16×2×(−4)×5=1×2×(−4)×5=−40 Putting x = 1 in RHS, we get:

RHS=−40x8=−40(1)8=−40×1=−40 $∵$ LHS = RHS for x = 1; therefore, the result is correct

Thus, the answer is −40x⁸ .

Question 22: Write down the product of −8x²y⁶ and −20xy. Verify the product for x = 2.5, y = 1.

Answer 22: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

(−8x2y6)×(−20xy)={(−8)×(−20)}×(x2×x)×(y6×y)={(−8)×(−20)}×(x2+1)×(y6+1)=−160x3y7 $∴$ $(- 8 x^{2} y^{6}) \times (- 20 x y) = - 160 x^{3} y^{7}$

Substituting x = 2.5 and y = 1 in LHS, we get:

LHS=(−8x2y6)×(−20xy)={−8(2.5)2(1)6}×{−20(2.5)(1)}={−8(6.25)(1)}×{−20(2.5)(1)}=(−50)×(−50)=2500 Substituting x = 2.5 and y = 1 in RHS, we get:

RHS=−160x3y7=−160(2.5)3(1)7=−160(15.625)×1=−2500 Because LHS is equal to RHS, the result is correct.

Thus, the answer is $- 160 x^{3} y^{7}$ .

Question 23:Evaluate (3.2x⁶y³) × (2.1x²y²) when x = 1 and y = 0.5
Ans
First multiply the expressions and then substitute the values for the variables.
To multiply algebric experssions use the commutative and the associative laws along with the law of indices, $a^{m} \times a^{n} = a^{m + n}$ .
We have,
$(3.2 x^{6} y^{3}) \times (2.1 x^{2} y^{2}) = (3.2 \times 2.1) \times (x^{6} \times x^{2}) \times (y^{3} \times y^{2}) = 6.72 x^{8} y^{5}$ Hence, $(3.2 x^{6} y^{3}) \times (2.1 x^{2} y^{2}) = 6.72 x^{8} y^{5}$
Now, substitute 1 for x and 0.5 for y in the result.
6.72x⁸y⁵

=6.72(1)⁸(0.5)⁵

=6.72×1×0.03125

=0.21 $6.72 x^{8} y^{5} = 6.72 {(1)}^{8} {(0.5)}^{5} = 6.72 \times 1 \times 0.03125 = 0.21$ Hence, the answer is $0.21$ .

Answer 23: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:
$(3.2 x^{6} y^{3}) \times (2.1 x^{2} y^{2}) = (3.2 \times 2.1) \times (x^{6} \times x^{2}) \times (y^{3} \times y^{2}) = (3.2 \times 2.1) \times (x^{6 + 2}) \times (y^{3 + 2}) = 6.72 x^{8} y^{5}$
$∴$ $(3.2 x^{6} y^{3}) \times (2.1 x^{2} y^{2}) = 6.72 x^{8} y^{5}$
Substituting x = 1 and y = 0.5 in the result, we get:
6.72x⁸y⁵

=6.72(1)⁸(0.5)⁵

=6.72×1×0.03125

=0.21 $6.72 x^{8} y^{5} = 6.72 {(1)}^{8} {(0.5)}^{5} = 6.72 \times 1 \times 0.03125 = 0.21$
Thus, the answer is $0.21$ .

Question 24: Find the value of (5x⁶) × (−1.5x²y³) × (−12xy²) when x = 1, y = 0.5.

Answer 24: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:
(5x⁶)×(−1.5x²y³)×(−12xy²)

={5×(−1.5)×(−12)}×(x⁶×x²×x)×(y³×y²)

=5×(−1.5)×(−12)×(x⁶⁺²⁺¹)×(y³⁺²)

=90x⁹y⁵
Substituting x = 1 and y = 0.5 in the result, we get:
$90 x^{9} y^{5} = 90 {(1)}^{9} {(0.5)}^{5} = 90 \times 1 \times 0.03125 = 2.8125$
Thus, the answer is 2.8125.

Question 25: Evaluate (2.3a⁵b²) × (1.2a²b²) when a = 1 and b = 0.5.

Answer 25: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:
(2.3a⁵b²)×(1.2a²b²)

=(2.3×1.2)×(a⁵×a²)×(b²×b²)

=(2.3×1.2)×(a⁵⁺²)×(b²⁺²)

=2.76a⁷b⁴
∴∴ (2.3a⁵b²)×(1.2a²b²)=2.76a⁷b⁴
Substituting a =1 and b = 0.5 in the result, we get:
$2.76 a^{7} b^{4} = 2.76 {(1)}^{7} {(0.5)}^{4} = 2.76 \times 1 \times 0.0625 = 0.1725$
Thus, the answer is $0.1725$ .

Question 26: Evaluate (−8x²y⁶) × (−20xy) for x = 2.5 and y = 1.

Answer 26: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:
$(- 8 x^{2} y^{6}) \times (- 20 x y) = \{(- 8) \times (- 20)\} \times (x^{2} \times x) \times (y^{6} \times y) = \{(- 8) \times (- 20)\} \times (x^{2 + 1}) \times (y^{6 + 1}) = 160 x^{3} y^{7}$
$∴$ $(- 8 x^{2} y^{6}) \times (- 20 x y) = 160 x^{3} y^{7}$
Substituting x = 2.5 and y = 1 in the result, we get:
$160 x^{3} y^{7} = 160 {(2.5)}^{3} {(1)}^{7} = 160 \times 15.625 = 2500$
Thus, the answer is $2500$ .

Question 27: Express each of the following product as a monomials and verify the result for x = 1, y = 2:
(−xy³) × (yx³) × (xy)

Answer 27: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:
$(- x y^{3}) \times (y x^{3}) \times (x y) = (- 1) \times (x \times x^{3} \times x) \times (y^{3} \times y \times y) = (- 1) \times (x^{1 + 3 + 1}) \times (y^{3 + 1 + 1}) = - x^{5} y^{5}$
To verify the result, we substitute x = 1 and y = 2 in LHS; we get:
$LHS = (- x y^{3}) \times (y x^{3}) \times (x y) = \{(- 1) \times 1 \times 2^{3}\} \times (2 \times 1^{3}) \times (1 \times 2) = \{(- 1) \times 1 \times 8\} \times (2 \times 1) \times 2 = (- 8) \times 2 \times 2 = - 32$
Substituting x = 1 and y = 2 in RHS, we get:
$RHS = - x^{5} y^{5} = (- 1) {(1)}^{5} {(2)}^{5} = (- 1) \times 1 \times 32 = - 32$
Because LHS is equal to RHS, the result is correct.
Thus, the answer is $- x^{5} y^{5}$ .

Question 28: Express each of the following product as a monomials and verify the result for x = 1, y = 2:

Answer 28: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

×(x2×x4×x)×(y4×y2×y)

×(x2+4+1)×(y4+2+1)

= 5/32 x7y7

To verify the result, we substitute x = 1 and y = 2 in LHS; we get:

LHS= ×(xy)×5

={1/8×(1)2×(2)4}×{1/4×(1)4×(2)2}×(1×2)×5

=(1/8×1×16)×(1/4 ×1×4)×(1×2)×5

=2×1×2×5=20 Substituting x = 1 and y = 2 in RHS, we get:

RHS=5/32 x7y7

=5/32 (1)7(2)7

=20

Because LHS is equal to RHS, the result is correct.
Thus, the answer is 5/32 $\frac{5}{32} x^{7} y^{7}$ .

Question 29: Express each of the following product as a monomials and verify the result for x = 1, y = 2:

(2/5 a2b)×(−15b2ac)×(−1/2 c2)25a2b×-15b2ac×-12c2

Answer 29: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

(2/5 a2b)×(−15b2ac)×(−1/2 c2)

={2/5 ×(−15)×(−1/2)}×(a2×a)×(b×b2)×(c×c2)

={2/5 ×(−15)×(−1/2)}×(a2+1)×(b1+2)×(c1+2)

=3a3b3c3

$∵$ The expression doesn't consist of the variables x and y.

$∴$ The result cannot be verified for x = 1 and y = 2

Thus, the answer is $3 a^{3} b^{3} c^{3}$ .

Question 30: Express each of the following product as a monomials and verify the result for x = 1, y = 2:

(−4/7 a2b)×(−2/3 b2c)×(−7/6 c2a)-47a2b×-23b2c×-76c2a

Answer 30: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

(−4/7 a2b)×(−2/3 b2c)×(−7/6 c2a)

={(−4/7)×(−2/3)×(−7/6)}×(a2×a)×(b×b2)×(c×c2)

={(−4/7)×(−2/3)×(−7/6)}×(a2+1)×(b1+2)×(c1+2)

=−4/9a3b3c3

$∵$ The expression doesn't consist of the variables x and y.
$∴$ The result cannot be verified for x = 1 and y = 2.

Thus, the answer is $- \frac{4}{9} a^{3} b^{3} c^{3}$ .

Question 31: Express each of the following product as a monomials and verify the result for x = 1, y = 2:

(4/9 abc3)×(−27/5a3b2)×(−8b3c)49abc3×-275a3b2×-8b3c

Answer 31: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

(4/9 abc3)×(−27/5 a3b2)×(−8b3c)

={(4/9)×(−27/5)×(−8)}×(a×a3)×(b×b2×b3)×(c3×c)

={(4/9)×(−27/5)×(−8)}×(a1+3)×(b1+2+3)×(c3+1)

=96/5 a4b6c4

Thus, the answer is 96/5 $\frac{96}{5} a^{4} b^{6} c^{4}$ .

$∵$ The expression doesn't consist of the variables x and y.

$∴$ The result cannot be verified for x = 1 and y = 2

Question 32: Evaluate each of the following when x = 2, y = −1.

(2xy)××(x2)×(y2)(2xy)×x2y4×x2×y2

Answer 32: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

(2xy)××(x2)×(y2)

=(2×1/4)×(x×x2×x2)×(y×y×y2)

(2×1/4)×(x1+2+2)×(y1+1+2)

=1/2 x5y4

$∴$ $(2 x y) \times (\frac{x^{2} y}{4}) \times (x^{2}) \times (y^{2}) = \frac{1}{2} x^{5} y^{4}$

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

1/2 x5y4

=1/2 (2)5(−1)4

=1/2×32×1

= 16

Thus, the answer is $16$ .

Question 33: Evaluate each of the following when x = 2, y = −1.

Answer 33: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

We have:

×(x2×x×x2)×(y×y2×y2)

×(x2+1+2)×(y1+2+2)

=− 7/4 x5y5

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

−7/4 x5y5

=−7/4 (2)5(−1)5

=(−7/4)×32×(−1)

= 56

Thus, the answer is 56. $(- 5 a) \times (- 10 a^{2}) \times (- 2 a^{3}) = \{(- 5) \times (- 10) \times (- 2)\} \times (a \times a^{2} \times a^{3}) = \{(- 5) \times (- 10) \times (- 2)\} \times (a^{1 + 2 + 3}) = - 100 a^{6}$

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RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) | RD Sharma Solutions for Class 8 Mathematics PDF Download

Question 1: Find each of the following product:5x2 × 4x3

Question 2: Find each of the following product:−3a2 × 4b4

Answer 2: To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, am×an=am+nam×an=am+n, wherever applicable.We have:−3a2×4b4=(−3×4)×(a2×b4)=−12a2b4-3a2×4b4=-3×4×a2×b4=-12a2Thus, the answer is −12a2b4-12a2b4.

Question 3: Find each of the following product:(−5xy) × (−3x2yz)

Question 4: Find each of the following product:

Answer 4: To multiply algebraic expressions, we use commutative and associative laws along with the the law of indices, that is, am×an=am+nam×an=am+n.

Question 5: Find each of the following product:

Answer 5: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+nam×an=am+n.

Question 6: Find each of the following product:

Answer 6: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+nam×an=am+n.

Question 7: Find each of the following product:

Answer 7:To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+nam×an=am+n.We have:

Question 8: Find each of the following product:

Answer 8: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+nam×an=am+n.We have:

Question 9: Find each of the following product:(7ab) × (−5ab2c) × (6abc2)

Question 10: Find each of the following product:(−5a) × (−10a2) × (−2a3)

Question 11: Find each of the following product:(−4x2) × (−6xy2) × (−3yz2)

Question 12: Find each of the following product:

Answer 12: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+nam×an=am+n.We have:

Question 13: Find each of the following product:

Answer 13: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+nam×an=am+n.We have:

Question 14: Find each of the following product:

Answer 14: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e.,am×an=am+nam×an=am+n.We have:

(4/3 u2vw)×(−5uvw2)×(1/3 v2wu)

Question 15: Find each of the following product:(0.5x)×(1/3 xy2z4)×(24x2yz)0.5x×13xy2z4×24x2yz

Answer 15: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+nam×an=am+n.We have:

Question 16: Find each of the following product:

Answer 16: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+nam×an=am+n.We have:

Question 17: Find each of the following product:(2.3xy) × (0.1x) × (0.16)

Answer 17: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+nam×an=am+n.We have:(2.3xy)×(0.1x)×(0.16)=(2.3×0.1×0.16)×(x×x)×y=(2.3×0.1×0.16)×(x1+1)×y=0.0368x2yThus, the answer is 0.0368x2y0.0368x2y.

Question 18: Express each of the following product as a monomials and verify the result in each case for x = 1:(3x) × (4x) × (−5x)

=−60Putting x = 1 in RHS, we get:RHS =−60x3=−60(1)3=−60×1=−60∵∵ LHS = RHS for x = 1; therefore, the result is correctThus, the answer is −60x3-60x3.

Question 19: Express each of the following product as a monomials and verify the result in each case for x = 1:(4x2) × (−3x) × (4/5 x3)

Answer 19: We have to find the product of the expression in order to express it as a monomial.To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+nam×an=am+n.

Question 20: Express each of the following product as a monomials and verify the result in each case for x = 1:(5x4) × (x2)3 × (2x)2

Question 21: Express each of the following product as a monomials and verify the result in each case for x = 1:(x2)3 × (2x) × (−4x) × (5)

Answer 21: We have to find the product of the expression in order to express it as a monomial.To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,am×an=am+n and (am)n=amn am×an=am+n and amn=amn .We have:

(x2)3×(2x)×(−4x)×5=(x6)×(2x)×(−4x)×5={2×(−4)×5}×(x6×x×x)={2×(−4)×5}×(x6+1+1)=−40x8

∴∴ (x2)3×(2x)×(−4x)×5=−40x8x23×2x×-4x×5=-40x8

Substituting x = 1 in LHS, we get:

LHS =(x2)3×(2x)×(−4x)×5=(12)3×(2×1)×(−4×1)×5=16×2×(−4)×5=1×2×(−4)×5=−40 Putting x = 1 in RHS, we get:

RHS=−40x8=−40(1)8=−40×1=−40 ∵∵ LHS = RHS for x = 1; therefore, the result is correct

Question 22: Write down the product of −8x2y6 and −20xy. Verify the product for x = 2.5, y = 1.

Answer 22: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., am×an=am+nam×an=am+n.

=6.72(1)8(0.5)5

=6.72×1×0.03125

=0.216.72x8y5=6.72180.55=6.72×1×0.03125=0.2Hence, the answer is 0.210.21.

=6.72(1)8(0.5)5

=6.72×1×0.03125

=0.2172x8y5=6.72180.55=6.72×1×0.03125=0.21Thus, the answer is 0.210.21.

Question 24: Find the value of (5x6) × (−1.5x2y3) × (−12xy2) when x = 1, y = 0.5.

Answer 24: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., am×an=am+nam×an=am+n.We have:(5x6)×(−1.5x2y3)×(−12xy2)

={5×(−1.5)×(−12)}×(x6×x2×x)×(y3×y2)

=5×(−1.5)×(−12)×(x6+2+1)×(y3+2)

=90x9y5Substituting x = 1 and y = 0.5 in the result, we get:90x9y5=90(1)9(0.5)5=90×1×0.03125=2.812590x9y5=90190.55=90×1×0.03125=2.8125Thus, the answer is 2.8125.

Question 25: Evaluate (2.3a5b2) × (1.2a2b2) when a = 1 and b = 0.5.

Answer 25: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., am×an=am+nam×an=am+n.We have:(2.3a5b2)×(1.2a2b2)

=(2.3×1.2)×(a5×a2)×(b2×b2)

=(2.3×1.2)×(a5+2)×(b2+2)

=2.76a7b4∴∴ (2.3a5b2)×(1.2a2b2)=2.76a7b4Substituting a =1 and b = 0.5 in the result, we get:2.76a7b4=2.76(1)7(0.5)4=2.76×1×0.0625=0.17252.76a7b4=2.76170.54=2.76×1×0.0625=0.1725Thus, the answer is 0.17250.1725.

Question 26: Evaluate (−8x2y6) × (−20xy) for x = 2.5 and y = 1.

Question 27: Express each of the following product as a monomials and verify the result for x = 1, y = 2:(−xy3) × (yx3) × (xy)

Question 28: Express each of the following product as a monomials and verify the result for x = 1, y = 2:

Answer 28: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., am×an=am+nam×an=am+n.

Question 29: Express each of the following product as a monomials and verify the result for x = 1, y = 2:

Answer 29: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., am×an=am+nam×an=am+n.

Question 30: Express each of the following product as a monomials and verify the result for x = 1, y = 2:

Answer 30: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., am×an=am+nam×an=am+n.

Thus, the answer is −4/9a3b3c3-49a3b3c3.

Question 31: Express each of the following product as a monomials and verify the result for x = 1, y = 2:

Answer 31: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., am×an=am+nam×an=am+n.

Question 32: Evaluate each of the following when x = 2, y = −1.

Answer 32: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., am×an=am+nam×an=am+n.

Question 33: Evaluate each of the following when x = 2, y = −1.

Answer 33: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., am×an=am+nam×an=am+n.

RD Sharma Solutions for Class 8 Mathematics

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RD Sharma Solutions for Class 8 Mathematics

Top Courses for Class 8

Question 1: Find each of the following product:
5x²× 4x³

Question 2: Find each of the following product:
−3a² × 4b⁴

Question 3: Find each of the following product:
(−5xy) × (−3x²yz)

Answer 4: To multiply algebraic expressions, we use commutative and associative laws along with the the law of indices, that is, $a^{m} \times a^{n} = a^{m + n}$ .

Answer 5: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

Answer 6: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

Answer 7:To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:

Answer 8: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:

Question 9: Find each of the following product:
(7ab) × (−5ab²c) × (6abc²)

Question 10: Find each of the following product:
(−5a) × (−10a²) × (−2a³)

Question 11: Find each of the following product:
(−4x²) × (−6xy²) × (−3yz²)

Answer 12: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:

Answer 13: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:

Answer 14: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:

Question 15: Find each of the following product:
$(0.5 x) \times (\frac{1}{3} x y^{2} z^{4}) \times (24 x^{2} y z)$

Answer 15: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:

Answer 16: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:

Question 17: Find each of the following product:
(2.3xy) × (0.1x) × (0.16)

Question 18: Express each of the following product as a monomials and verify the result in each case for x = 1:
(3x) × (4x) × (−5x)

=−60
Putting x = 1 in RHS, we get:
$RHS = - 60 x^{3} = - 60 {(1)}^{3} = - 60 \times 1 = - 60$ $∵$ LHS = RHS for x = 1; therefore, the result is correct
Thus, the answer is $- 60 x^{3}$ .

Question 19: Express each of the following product as a monomials and verify the result in each case for x = 1:
(4x²) × (−3x) × $(\frac{4}{5} x^{3})$

Answer 19: We have to find the product of the expression in order to express it as a monomial.
To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

Question 20: Express each of the following product as a monomials and verify the result in each case for x = 1:
(5x⁴) × (x²)³ × (2x)²

Question 21: Express each of the following product as a monomials and verify the result in each case for x = 1:
(x²)³ × (2x) × (−4x) × (5)

Answer 21: We have to find the product of the expression in order to express it as a monomial.
To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n} and {(a^{m})}^{n} = a^{m n}$ .
We have:

$∴$ ${(x^{2})}^{3} \times (2 x) \times (- 4 x) \times 5 = - 40 x^{8}$

RHS=−40x8=−40(1)8=−40×1=−40 $∵$ LHS = RHS for x = 1; therefore, the result is correct

Question 22: Write down the product of −8x²y⁶ and −20xy. Verify the product for x = 2.5, y = 1.

Answer 22: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

=6.72(1)⁸(0.5)⁵

=0.21 $6.72 x^{8} y^{5} = 6.72 {(1)}^{8} {(0.5)}^{5} = 6.72 \times 1 \times 0.03125 = 0.21$ Hence, the answer is $0.21$ .

=6.72(1)⁸(0.5)⁵

=0.21 $6.72 x^{8} y^{5} = 6.72 {(1)}^{8} {(0.5)}^{5} = 6.72 \times 1 \times 0.03125 = 0.21$
Thus, the answer is $0.21$ .

Question 24: Find the value of (5x⁶) × (−1.5x²y³) × (−12xy²) when x = 1, y = 0.5.

Answer 24: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:
(5x⁶)×(−1.5x²y³)×(−12xy²)

={5×(−1.5)×(−12)}×(x⁶×x²×x)×(y³×y²)

=5×(−1.5)×(−12)×(x⁶⁺²⁺¹)×(y³⁺²)

=90x⁹y⁵
Substituting x = 1 and y = 0.5 in the result, we get:
$90 x^{9} y^{5} = 90 {(1)}^{9} {(0.5)}^{5} = 90 \times 1 \times 0.03125 = 2.8125$
Thus, the answer is 2.8125.

Question 25: Evaluate (2.3a⁵b²) × (1.2a²b²) when a = 1 and b = 0.5.

Answer 25: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .
We have:
(2.3a⁵b²)×(1.2a²b²)

=(2.3×1.2)×(a⁵×a²)×(b²×b²)

=(2.3×1.2)×(a⁵⁺²)×(b²⁺²)

=2.76a⁷b⁴
∴∴ (2.3a⁵b²)×(1.2a²b²)=2.76a⁷b⁴
Substituting a =1 and b = 0.5 in the result, we get:
$2.76 a^{7} b^{4} = 2.76 {(1)}^{7} {(0.5)}^{4} = 2.76 \times 1 \times 0.0625 = 0.1725$
Thus, the answer is $0.1725$ .

Question 26: Evaluate (−8x²y⁶) × (−20xy) for x = 2.5 and y = 1.

Question 27: Express each of the following product as a monomials and verify the result for x = 1, y = 2:
(−xy³) × (yx³) × (xy)

Answer 28: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

Answer 29: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

Answer 30: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

Thus, the answer is $- \frac{4}{9} a^{3} b^{3} c^{3}$ .

Answer 31: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

Answer 32: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .

Answer 33: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., $a^{m} \times a^{n} = a^{m + n}$ .