RD Sharma solution class 8 chapter 6 Algebraic Expressions and Identity Exercise 6.3

Exercise 6.3

Page-6.13

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:

$$\begin{gathered} 5x^2\times 4x^3 \\ =\left(5\times 4\right)\times \left(x^2\times x^3\right) \end{gathered}$$
$=20x^5$                           ($\because$ $a^m\times a^n=a^{m+n}$)

Thus, the answer is $20x^5$.

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:

$$\begin{gathered} -3a^2\times 4b^4 \\ =\left(-3\times 4\right)\times \left(a^2\times b^4\right) \\ =-12a^2{b}^4 \end{gathered}$$

Thus, the answer is $-12a^2{b}^4$.

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:

$$\begin{gathered} \left(-5xy\right)\times \left(-3x^{2}yz\right) \\ =\left\{\left(-5\right)\times \left(-3\right)\right\}\times \left(x\times x^2\right)\times \left(y\times y\right)\times z \\ =15\times \left(x^{1+2}\right)\times \left(y^{1+1}\right)\times z \\ =15x^3{y}^{2}z \end{gathered}$$

Thus, the answer is $15x^3{y}^{2}z$.

Question 4:

Find each of the following product:
$\frac{1}{4}xy\times \frac{2}{3}{x}^{2}y{z}^2$

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:

$$\begin{gathered} \frac{1}{4}xy\times \frac{2}{3}{x}^{2}y{z}^2 \\ =\left(\frac{1}{4}\times \frac{2}{3}\right)\times \left(x\times x^2\right)\times \left(y\times y\right)\times z^2 \\ =\left(\frac{1}{4}\times \frac{2}{3}\right)\times \left(x^{1+2}\right)\times \left(y^{1+1}\right)\times z^2 \\ =\frac{1}{6}{x}^3{y}^2{z}^2 \end{gathered}$$

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

Page-6.14

Question 5:

Find each of the following product:
$\left(-\frac{7}{5}x{y}^{2}z\right)\times \left(\frac{13}{3}{x}^{2}y{z}^2\right)$

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:

$$\begin{gathered} \left(-\frac{7}{5}x{y}^{2}z\right)\times \left(\frac{13}{3}{x}^{2}y{z}^2\right) \\ =\left(-\frac{7}{5}\times \frac{13}{3}\right)\times \left(x\times x^2\right)\times \left(y^2\times y\right)\times \left(z\times z^2\right) \\ =\left(-\frac{7}{5}\times \frac{13}{3}\right)\times \left(x^{1+2}\right)\times \left(y^{2+1}\right)\times \left(z^{1+2}\right) \\ =-\frac{91}{15}{x}^3{y}^3{x}^3 \end{gathered}$$

Thus, the answer is $-\frac{91}{15}{x}^3{y}^3{x}^3$.
 

Question 6:

Find each of the following product:
$\left(\frac{-24}{25}{x}^{3}z\right)\times \left(-\frac{15}{16}x{z}^{2}y\right)$

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:

$$\begin{gathered} \left(-\frac{24}{25}{x}^{3}z\right)\times \left(-\frac{15}{16}x{z}^{2}y\right) \\ =\left\{\left(-\frac{24}{25}\right)\times \left(-\frac{15}{16}\right)\right\}\times \left(x^3\times x\right)\times \left(z\times z^2\right)\times y \\ =\left\{\left(-\frac{24}{25}\right)\times \left(-\frac{15}{16}\right)\right\}\times \left(x^{3+1}\right)\times \left(z^{1+2}\right)\times y \end{gathered}$$
$=\frac{9}{10}{x}^{4}y{z}^3$                                                           

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

Question 7:

Find each of the following product:
$\left(-\frac{1}{27}{a}^2{b}^2\right)\times \left(\frac{9}{2}{a}^3{b}^2{c}^2\right)$

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:

$$\begin{gathered} \left(-\frac{1}{27}{a}^2{b}^2\right)\times \left(\frac{9}{2}{a}^3{b}^2{c}^2\right) \\ =\left(-\frac{1}{27}\times \frac{9}{2}\right)\times \left(a^2\times a^3\right)\times \left(b^2\times b^2\right)\times c^2 \\ =\left(-\frac{1}{27}\times \frac{9}{2}\right)\times \left(a^{2+3}\right)\times \left(b^{2+2}\right)\times c^2 \\ =-\frac{1}{6}{a}^5{b}^4{c}^2 \end{gathered}$$

Thus, the answer is $-\frac{1}{6}{a}^5{b}^4{c}^2$.

Question 8:

Find each of the following product:
$(-7xy)\times \left(\frac{1}{4}{x}^{2}yz\right)$

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:

$$\begin{gathered} \left(-7xy\right)\times \left(\frac{1}{4}{x}^{2}yz\right) \\ =\left(-7\times \frac{1}{4}\right)\times \left(x\times x^2\right)\times \left(y\times y\right)\times z \\ =\left(-7\times \frac{1}{4}\right)\times \left(x^{1+2}\right)\times \left(y^{1+1}\right)\times z \\ =-\frac{7}{4}{x}^3{y}^{2}z \end{gathered}$$

Thus, the answer is $-\frac{7}{4}{x}^3{y}^{2}z$.

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:

$$\begin{gathered} \left(7ab\right)\times \left(-5a{b}^{2}c\right)\times \left(6ab{c}^2\right) \\ =\left\{7\times \left(-5\right)\times 6\right\}\times \left(a\times a\times a\right)\times \left(b\times b^2\times b\right)\times \left(c\times c^2\right) \\ =\left\{7\times \left(-5\right)\times 6\right\}\times \left(a^{1+1+1}\right)\times \left(b^{1+2+1}\right)\times \left(c^{1+2}\right) \\ =-210a^3{b}^4{c}^3 \end{gathered}$$

Thus, the answer is $-210a^3{b}^4{c}^3$.

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:

$$\begin{gathered} \left(-5a\right)\times \left(-10a^2\right)\times \left(-2a^3\right) \\ =\left\{\left(-5\right)\times \left(-10\right)\times \left(-2\right)\right\}\times \left(a\times a^2\times a^3\right) \\ =\left\{\left(-5\right)\times \left(-10\right)\times \left(-2\right)\right\}\times \left(a^{1+2+3}\right) \\ =-100a^6 \end{gathered}$$

Thus, the answer is $-100a^6$.

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:

$$\begin{gathered} \left(-4x^2\right)\times \left(-6x{y}^2\right)\times \left(-3y{z}^2\right) \\ =\left\{\left(-4\right)\times \left(-6\right)\times \left(-3\right)\right\}\times \left(x^2\times x\right)\times \left(y^2\times y\right)\times z^2 \\ =\left\{\left(-4\right)\times \left(-6\right)\times \left(-3\right)\right\}\times \left(x^{2+1}\right)\times \left(y^{2+1}\right)\times z^2 \\ =-72x^3{y}^3{z}^2 \end{gathered}$$

Thus, the answer is $-72x^3{y}^3{z}^2$.

Question 12:

Find each of the following product:
$\left(-\frac{2}{7}{a}^4\right)\times \left(-\frac{3}{4}{a}^{2}b\right)\times \left(-\frac{14}{5}{b}^2\right)$

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:

$$\begin{gathered} \left(-\frac{2}{7}{a}^4\right)\times \left(-\frac{3}{4}{a}^{2}b\right)\times \left(-\frac{14}{5}{b}^2\right) \\ =\left\{\left(-\frac{2}{7}\right)\times \left(-\frac{3}{4}\right)\times \left(-\frac{14}{5}\right)\right\}\times \left(a^4\times a^2\right)\times \left(b\times b^2\right) \\ =\left\{-\left(\frac{2}{7}\times \frac{3}{4}\times \frac{14}{5}\right)\right\}\times a^{4+2}\times b^{1+2} \\ =\left\{-\left(\frac{\cancel{2}}{\cancel{7}}\times \frac{3}{{\cancel{4}}_{\cancel{2}}}\times \frac{{\cancel{14}}^{{\cancel{2}}^1}}{5}\right)\right\}\times a^6\times b^3 \\ =-\frac{3}{5}{a}^6{b}^3 \end{gathered}$$

Thus, the answer is $-\frac{3}{5}{a}^6{b}^3$.

Question 13:

Find each of the following product:
$\left(\frac{7}{9}a{b}^2\right)\times \left(\frac{15}{7}a{c}^{2}b\right)\times \left(-\frac{3}{5}{a}^{2}c\right)$

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:

$$\begin{gathered} \left(\frac{7}{9}a{b}^2\right)\times \left(\frac{15}{7}a{c}^{2}b\right)\times \left(-\frac{3}{5}{a}^{2}c\right) \\ =\left\{\frac{7}{9}\times \frac{15}{7}\times \left(-\frac{3}{5}\right)\right\}\times \left(a\times a\times a^2\right)\times \left(b^2\times b\right)\times \left(c^2\times c\right) \\ =\left\{\frac{{\cancel{7}}^1}{{\cancel{9}}_3}\times \frac{{\cancel{15}}^3}{\cancel{7}}\times \left(-\frac{{\cancel{3}}^1}{\cancel{5}}\right)\right\}\times \left(a\times a\times a^2\right)\times \left(b^2\times b\right)\times \left(c^2\times c\right) \\ =\left\{\frac{{\cancel{7}}^1}{{\cancel{9}}_{\cancel{3}}}\times \frac{{\cancel{15}}^{{\cancel{3}}^1}}{\cancel{7}}\times \left(-\frac{{\cancel{3}}^1}{\cancel{5}}\right)\right\}\times \left(a^{1+1+2}\right)\times \left(b^{2+1}\right)\times \left(c^{2+1}\right) \\ =-a^4{b}^3{c}^3 \end{gathered}$$

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

Question 14:

Find each of the following product:
$\left(\frac{4}{3}{u}^{2}vw\right)\times \left(-5uv{w}^2\right)\times \left(\frac{1}{3}{v}^{2}wu\right)$

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:

$$\begin{gathered} \left(\frac{4}{3}{u}^{2}vw\right)\times \left(-5uv{w}^2\right)\times \left(\frac{1}{3}{v}^{2}wu\right) \\ =\left\{\frac{4}{3}\times \left(-5\right)\times \frac{1}{3}\right\}\times \left(u^2\times u\times u\right)\times \left(v\times v\times v^2\right)\times \left(w\times w^2\times w\right) \\ =\left\{\frac{4}{3}\times \left(-5\right)\times \frac{1}{3}\right\}\times \left(u^{2+1+1}\right)\times \left(v^{1+1+2}\right)\times \left(w^{1+2+1}\right) \\ =-\frac{20}{9}{u}^4{v}^4{w}^4 \end{gathered}$$

Thus, the answer is $-\frac{20}{9}{u}^4{v}^4{w}^4$.

Question 15:

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

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:
$$\begin{gathered} \left(0.5x\right)\times \left(\frac{1}{3}x{y}^2{z}^4\right)\times \left(24x^{2}yz\right) \\ =\left(0.5\times \frac{1}{3}\times 24\right)\times \left(x\times x\times x^2\right)\times \left(y^2\times y\right)\times \left(z^4\times z\right) \\ =\left(0.5\times \frac{1}{3}\times 24\right)\times \left(x^{1+1+2}\right)\times \left(y^{2+1}\right)\times \left(z^{4+1}\right) \\ =4x^4{y}^3{z}^5 \end{gathered}$$

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

Question 16:

Find each of the following product:
$\left(\frac{4}{3}p{q}^2\right)\times \left(-\frac{1}{4}{p}^{2}r\right)\times \left(16p^2{q}^2{r}^2\right)$

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:

$$\begin{gathered} \left(\frac{4}{3}p{q}^2\right)\times \left(-\frac{1}{4}{p}^{2}r\right)\times \left(16p^2{q}^2{r}^2\right) \\ =\left\{\frac{4}{3}\times \left(-\frac{1}{4}\right)\times 16\right\}\times \left(p\times p^2\times p^2\right)\times \left(q^2\times q^2\right)\times \left(r\times r^2\right) \\ =\left\{\frac{4}{3}\times \left(-\frac{1}{4}\right)\times 16\right\}\times \left(p^{1+2+2}\right)\times \left(q^{2+2}\right)\times \left(r^{1+2}\right) \\ =-\frac{16}{3}{p}^5{q}^4{r}^3 \end{gathered}$$

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:
$$\begin{gathered} \left(2.3xy\right)\times \left(0.1x\right)\times \left(0.16\right) \\ =\left(2.3\times 0.1\times 0.16\right)\times \left(x\times x\right)\times y \\ =\left(2.3\times 0.1\times 0.16\right)\times \left(x^{1+1}\right)\times y \\ =0.0368x^{2}y \end{gathered}$$

Thus, the answer is $0.0368x^{2}y$.

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:

$$\begin{gathered} \left(3x\right)\times \left(4x\right)\times \left(-5x\right) \\ =\left\{3\times 4\times \left(-5\right)\right\}\times \left(x\times x\times x\right) \\ =\left\{3\times 4\times \left(-5\right)\right\}\times \left(x^{1+1+1}\right) \\ =-60x^3 \end{gathered}$$

Substituting x = 1 in LHS, we get:

$$\begin{gathered} \text{LHS }=\left(3x\right)\times \left(4x\right)\times \left(-5x\right) \\ =\left(3\times 1\right)\times \left(4\times 1\right)\times \left(-5\times 1\right) \\ =-60 \end{gathered}$$

Putting x = 1 in RHS, we get:
$$\begin{gathered} \text{RHS }=-60x^3 \\ =-60{\left(1\right)}^3 \\ =-60\times 1 \\ =-60 \end{gathered}$$


$\because$ LHS = RHS for x = 1; therefore, the result is correct

Thus, the answer is $-60x^3$.

Question 19:

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

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:

$$\begin{gathered} \left(4x^2\right)\times \left(-3x\right)\times \left(\frac{4}{5}{x}^3\right) \\ =\left\{4\times \left(-3\right)\times \frac{4}{5}\right\}\times \left(x^2\times x\times x^3\right) \\ =\left\{4\times \left(-3\right)\times \frac{4}{5}\right\}\times \left(x^{2+1+3}\right) \\ =-\frac{48}{5}{x}^6 \end{gathered}$$

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

Substituting x = 1 in LHS, we get:

$$\begin{gathered} \text{LHS}=\left(4x^2\right)\times \left(-3x\right)\times \left(\frac{4}{5}{x}^3\right) \\ =\left(4\times 1^2\right)\times \left(-3\times 1\right)\times \left(\frac{4}{5}\times 1^3\right) \\ =4\times \left(-3\right)\times \frac{4}{5} \\ =-\frac{48}{5} \end{gathered}$$

Putting x = 1 in RHS, we get:

$$\begin{gathered} \text{RHS}=-\frac{48}{5}{x}^6 \\ =-\frac{48}{5}\times 1^6 \\ =-\frac{48}{5} \end{gathered}$$

$\because$ LHS = RHS for x = 1; therefore, the result is correct

Thus, the answer is $-\frac{48}{5}{x}^6$.

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., $a^m\times a^n=a^{m+n} \text{and} {\left(a^m\right)}^n=a^{mn}$.

We have:

$$\begin{gathered} \left(5x^4\right)\times {\left(x^2\right)}^3\times {\left(2x\right)}^{2 } \\ =\left(5x^4\right)\times \left(x^6\right)\times \left(2^2\times x^2\right) \\ =\left(5\times 2^2\right)\times \left(x^4\times x^6\times x^2\right) \\ =\left(5\times 2^2\right)\times \left(x^{4+6+2}\right) \\ =20x^{12} \end{gathered}$$
$\therefore$ $\left(5x^4\right)\times {\left(x^2\right)}^3\times {\left(2x\right)}^{2 }=20x^{12}$

Substituting x = 1 in LHS, we get:

$$\begin{gathered} \mathrm{LHS}=\left(5x^4\right)\times {\left(x^2\right)}^3\times {\left(2x\right)}^{2 } \\ =\left(5\times 1^4\right)\times {\left(1^2\right)}^3\times {\left(2\times 1\right)}^{2 } \\ =\left(5\times 1\right)\times \left(1^6\right)\times {\left(2\right)}^{2 } \\ =5\times 1\times 4 \\ =20 \end{gathered}$$

Put x =1 in RHS, we get:

$$\begin{gathered} \text{RHS }=20x^{12} \\ =20\times {\left(1\right)}^{12} \\ =20\times 1 \\ =20 \end{gathered}$$

$\because$ LHS = RHS for x = 1; therefore, the result is correct.

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

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.,​$a^m\times a^n=a^{m+n} \text{and} {\left(a^m\right)}^n=a^{mn}$.

We have:

$$\begin{gathered} {\left(x^2\right)}^3\times \left(2x\right)\times \left(-4x\right)\times 5 \\ =\left(x^6\right)\times \left(2x\right)\times \left(-4x\right)\times 5 \\ =\left\{2\times \left(-4\right)\times 5\right\}\times \left(x^6\times x\times x\right) \\ =\left\{2\times \left(-4\right)\times 5\right\}\times \left(x^{6+1+1}\right) \\ =-40x^8 \end{gathered}$$

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

Substituting x = 1 in LHS, we get:​

$$\begin{gathered} \text{LHS }={\left(x^2\right)}^3\times \left(2x\right)\times \left(-4x\right)\times 5 \\ ={\left(1^2\right)}^3\times \left(2\times 1\right)\times \left(-4\times 1\right)\times 5 \\ =1^6\times 2\times \left(-4\right)\times 5 \\ =1\times 2\times \left(-4\right)\times 5 \\ =-40 \end{gathered}$$

Putting x = 1 in RHS, we get:​

$$\begin{gathered} \text{RHS}=-40x^8 \\ =-40{\left(1\right)}^8 \\ =-40\times 1 \\ =-40 \end{gathered}$$

$\because$ LHS = RHS for x = 1; therefore, the result is correct

Thus, the answer is $-40x^8$.

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:

$$\begin{gathered} \left(-8x^2{y}^6\right)\times \left(-20xy\right) \\ =\left\{\left(-8\right)\times \left(-20\right)\right\}\times \left(x^2\times x\right)\times \left(y^6\times y\right) \\ =\left\{\left(-8\right)\times \left(-20\right)\right\}\times \left(x^{2+1}\right)\times \left(y^{6+1}\right) \\ =-160x^3{y}^7 \end{gathered}$$
$\therefore$ $\left(-8x^2{y}^6\right)\times \left(-20xy\right)=-160x^3{y}^7$

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

$$\begin{gathered} \text{LHS}=\left(-8x^2{y}^6\right)\times \left(-20xy\right) \\ =\left\{-8{\left(2.5\right)}^2{\left(1\right)}^6\right\}\times \left\{-20\left(2.5\right)\left(1\right)\right\} \\ =\left\{-8\left(6.25\right)\left(1\right)\right\}\times \left\{-20\left(2.5\right)\left(1\right)\right\} \\ =\left(-50\right)\times \left(-50\right) \\ =2500 \end{gathered}$$

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

$$\begin{gathered} \text{RHS}=-160x^3{y}^7 \\ =-160{\left(2.5\right)}^3{\left(1\right)}^7 \\ =-160\left(15.625\right)\times 1 \\ =-2500 \end{gathered}$$

Because LHS is equal to RHS, the result is correct.

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

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,

$$\begin{gathered} \left(3.2x^6{y}^3\right)\times \left(2.1x^2{y}^2\right) \\ =\left(3.2\times 2.1\right)\times \left(x^6\times x^2\right)\times \left(y^3\times y^2\right) \\ =6.72x^8{y}^5 \end{gathered}$$
Hence, $\left(3.2x^6{y}^3\right)\times \left(2.1x^2{y}^2\right)=6.72x^8{y}^5$

Now, substitute 1 for x and 0.5  for y in the result.

$$\begin{gathered} 6.72x^8{y}^5 \\ =6.72{\left(1\right)}^8{\left(0.5\right)}^5 \\ =6.72\times 1\times 0.03125 \\ =0.21 \end{gathered}$$

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:

$$\begin{gathered} \left(3.2x^6{y}^3\right)\times \left(2.1x^2{y}^2\right) \\ =\left(3.2\times 2.1\right)\times \left(x^6\times x^2\right)\times \left(y^3\times y^2\right) \\ =\left(3.2\times 2.1\right)\times \left(x^{6+2}\right)\times \left(y^{3+2}\right) \\ =6.72x^8{y}^5 \end{gathered}$$

$\therefore$ $\left(3.2x^6{y}^3\right)\times \left(2.1x^2{y}^2\right)=6.72x^8{y}^5$

Substituting x = 1 and y = 0.5 in the result, we get:

$$\begin{gathered} 6.72x^8{y}^5 \\ =6.72{\left(1\right)}^8{\left(0.5\right)}^5 \\ =6.72\times 1\times 0.03125 \\ =0.21 \end{gathered}$$

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:

$$\begin{gathered} \left(5x^6\right)\times \left(-1.5x^2{y}^3\right)\times \left(-12x{y}^2\right) \\ =\left\{5\times \left(-1.5\right)\times \left(-12\right)\right\}\times \left(x^6\times x^2\times x\right)\times \left(y^3\times y^2\right) \\ =\left\{5\times \left(-1.5\right)\times \left(-12\right)\right\}\times \left(x^{6+2+1}\right)\times \left(y^{3+2}\right) \\ =90x^9{y}^5 \end{gathered}$$

$\therefore$ $\left(5x^6\right)\times \left(-1.5x^2{y}^3\right)\times \left(-12x{y}^2\right)=90x^9{y}^5$

Substituting x = 1 and y = 0.5 in the result, we get:

$$\begin{gathered} 90x^9{y}^5 \\ =90{\left(1\right)}^9{\left(0.5\right)}^5 \\ =90\times 1\times 0.03125 \\ =2.8125 \end{gathered}$$

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:

$$\begin{gathered} \left(2.3a^5{b}^2\right)\times \left(1.2a^2{b}^2\right) \\ =\left(2.3\times 1.2\right)\times \left(a^5\times a^2\right)\times \left(b^2\times b^2\right) \\ =\left(2.3\times 1.2\right)\times \left(a^{5+2}\right)\times \left(b^{2+2}\right) \\ =2.76a^7{b}^4 \end{gathered}$$

$\therefore$ $\left(2.3a^5{b}^2\right)\times \left(1.2a^2{b}^2\right)=2.76a^7{b}^4$

Substituting a =1 and b = 0.5 in the result, we get:

$$\begin{gathered} 2.76a^7{b}^4 \\ =2.76{\left(1\right)}^7{\left(0.5\right)}^4 \\ =2.76\times 1\times 0.0625 \\ =0.1725 \end{gathered}$$

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:

$$\begin{gathered} \left(-8x^2{y}^6\right)\times \left(-20xy\right) \\ =\left\{\left(-8\right)\times \left(-20\right)\right\}\times \left(x^2\times x\right)\times \left(y^6\times y\right) \\ =\left\{\left(-8\right)\times \left(-20\right)\right\}\times \left(x^{2+1}\right)\times \left(y^{6+1}\right) \\ =160x^3{y}^7 \end{gathered}$$

$\therefore$ $\left(-8x^2{y}^6\right)\times \left(-20xy\right)=160x^3{y}^7$

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

$$\begin{gathered} 160x^3{y}^7 \\ =160{\left(2.5\right)}^3{\left(1\right)}^7 \\ =160\times 15.625 \\ =2500 \end{gathered}$$

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:

$$\begin{gathered} \left(-x{y}^3\right)\times \left(y{x}^3\right)\times \left(xy\right) \\ =\left(-1\right)\times \left(x\times x^3\times x\right)\times \left(y^3\times y\times y\right) \\ =\left(-1\right)\times \left(x^{1+3+1}\right)\times \left(y^{3+1+1}\right) \\ =-x^5{y}^5 \end{gathered}$$

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

$$\begin{gathered} \text{LHS }=\left(-x{y}^3\right)\times \left(y{x}^3\right)\times \left(xy\right) \\ =\left\{\left(-1\right)\times 1\times 2^3\right\}\times \left(2\times 1^3\right)\times \left(1\times 2\right) \\ =\left\{\left(-1\right)\times 1\times 8\right\}\times \left(2\times 1\right)\times 2 \\ =\left(-8\right)\times 2\times 2 \\ =-32 \end{gathered}$$

Substituting x = 1 and y = 2 in RHS, we get:​

$$\begin{gathered} \text{RHS}=-x^5{y}^5 \\ =\left(-1\right){\left(1\right)}^5{\left(2\right)}^5 \\ =\left(-1\right)\times 1\times 32 \\ =-32 \end{gathered}$$

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:
$\left(\frac{1}{8}{x}^2{y}^4\right)\times \left(\frac{1}{4}{x}^4{y}^2\right)\times \left(xy\right)\times 5$

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:

$$\begin{gathered} \left(\frac{1}{8}{x}^2{y}^4\right)\times \left(\frac{1}{4}{x}^4{y}^2\right)\times \left(xy\right)\times 5 \\ =\left(\frac{1}{8}\times \frac{1}{4}\times 5\right)\times \left(x^2\times x^4\times x\right)\times \left(y^4\times y^2\times y\right) \\ =\left(\frac{1}{8}\times \frac{1}{4}\times 5\right)\times \left(x^{2+4+1}\right)\times \left(y^{4+2+1}\right) \\ =\frac{5}{32}{x}^7{y}^7 \end{gathered}$$

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

$$\begin{gathered} \text{LHS}=\left(\frac{1}{8}{x}^2{y}^4\right)\times \left(\frac{1}{4}{x}^4{y}^2\right)\times \left(xy\right)\times 5 \\ =\left\{\frac{1}{8}\times {\left(1\right)}^2\times {\left(2\right)}^4\right\}\times \left\{\frac{1}{4}\times {\left(1\right)}^4\times {\left(2\right)}^2\right\}\times \left(1\times 2\right)\times 5 \\ =\left(\frac{1}{8}\times 1\times 16\right)\times \left(\frac{1}{4}\times 1\times 4\right)\times \left(1\times 2\right)\times 5 \\ =2\times 1\times 2\times 5 \\ =20 \end{gathered}$$

Substituting x = 1 and y = 2 in RHS, we get:​

$$\begin{gathered} \text{RHS}=\frac{5}{32}{x}^7{y}^7 \\ =\frac{5}{32}{\left(1\right)}^7{\left(2\right)}^7 \\ =\frac{5}{\cancel{32}}\times 1\times {\cancel{128}}^4 \\ =20 \end{gathered}$$

Because LHS is equal to RHS, the result is correct.

Thus, the answer is $\frac{5}{32}{x}^7{y}^7$.

Question 29:

Express each of the following product as a monomials and verify the result for x = 1, y = 2:
$\left(\frac{2}{5}{a}^{2}b\right)\times \left(-15b^{2}ac\right)\times \left(-\frac{1}{2}{c}^2\right)$

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:

$$\begin{gathered} \left(\frac{2}{5}{a}^{2}b\right)\times \left(-15b^{2}ac\right)\times \left(-\frac{1}{2}{c}^2\right) \\ =\left\{\frac{2}{5}\times \left(-15\right)\times \left(-\frac{1}{2}\right)\right\}\times \left(a^2\times a\right)\times \left(b\times b^2\right)\times \left(c\times c^2\right) \\ =\left\{\frac{2}{5}\times \left(-15\right)\times \left(-\frac{1}{2}\right)\right\}\times \left(a^{2+1}\right)\times \left(b^{1+2}\right)\times \left(c^{1+2}\right) \\ =3a^3{b}^3{c}^3 \end{gathered}$$

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

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

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

Question 30:

Express each of the following product as a monomials and verify the result for x = 1, y = 2:
$\left(-\frac{4}{7}{a}^{2}b\right)\times \left(-\frac{2}{3}{b}^{2}c\right)\times \left(-\frac{7}{6}{c}^{2}a\right)$

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:

$$\begin{gathered} \left(-\frac{4}{7}{a}^{2}b\right)\times \left(-\frac{2}{3}{b}^{2}c\right)\times \left(-\frac{7}{6}{c}^{2}a\right) \\ =\left\{\left(-\frac{4}{7}\right)\times \left(-\frac{2}{3}\right)\times \left(-\frac{7}{6}\right)\right\}\times \left(a^2\times a\right)\times \left(b\times b^2\right)\times \left(c\times c^2\right) \\ =\left\{\left(-\frac{4}{7}\right)\times \left(-\frac{2}{3}\right)\times \left(-\frac{7}{6}\right)\right\}\times \left(a^{2+1}\right)\times \left(b^{1+2}\right)\times \left(c^{1+2}\right) \\ =-\frac{4}{9}{a}^3{b}^3{c}^3 \end{gathered}$$

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

$\therefore$ 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:
$\left(\frac{4}{9}ab{c}^3\right)\times \left(-\frac{27}{5}{a}^3{b}^2\right)\times \left(-8b^{3}c\right)$

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:

$$\begin{gathered} \left(\frac{4}{9}ab{c}^3\right)\times \left(-\frac{27}{5}{a}^3{b}^2\right)\times \left(-8b^{3}c\right) \\ =\left\{\left(\frac{4}{9}\right)\times \left(-\frac{27}{5}\right)\times \left(-8\right)\right\}\times \left(a\times a^3\right)\times \left(b\times b^2\times b^3\right)\times \left(c^3\times c\right) \\ =\left\{\left(\frac{4}{9}\right)\times \left(-\frac{27}{5}\right)\times \left(-8\right)\right\}\times \left(a^{1+3}\right)\times \left(b^{1+2+3}\right)\times \left(c^{3+1}\right) \\ =\frac{96}{5}{a}^4{b}^6{c}^4 \end{gathered}$$

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

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

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

Question 32:

Evaluate each of the following when x = 2, y = −1.
$\left(2xy\right)\times \left(\frac{x^{2}y}{4}\right)\times \left(x^2\right)\times \left(y^2\right)$

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:

$$\begin{gathered} \left(2xy\right)\times \left(\frac{x^{2}y}{4}\right)\times \left(x^2\right)\times \left(y^2\right) \\ =\left(2\times \frac{1}{4}\right)\times \left(x\times x^2\times x^2\right)\times \left(y\times y\times y^2\right) \\ =\left(2\times \frac{1}{4}\right)\times \left(x^{1+2+2}\right)\times \left(y^{1+1+2}\right) \\ =\frac{1}{2}{x}^5{y}^4 \end{gathered}$$

$\therefore$ $\left(2xy\right)\times \left(\frac{x^{2}y}{4}\right)\times \left(x^2\right)\times \left(y^2\right)=\frac{1}{2}{x}^5{y}^4$

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

$$\begin{gathered} \frac{1}{2}{x}^5{y}^4 \\ =\frac{1}{2}{\left(2\right)}^5{\left(-1\right)}^4 \\ =\frac{1}{2}\times 32\times 1 \\ =16 \end{gathered}$$

Thus, the answer is $16$.

Question 33:

Evaluate each of the following when x = 2, y = −1.
$\left(\frac{3}{5}{x}^{2}y\right)\times \left(-\frac{15}{4}x{y}^2\right)\times \left(\frac{7}{9}{x}^2{y}^2\right)$

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:

$$\begin{gathered} \left(\frac{3}{5}{x}^{2}y\right)\times \left(-\frac{15}{4}x{y}^2\right)\times \left(\frac{7}{9}{x}^2{y}^2\right) \\ =\left\{\frac{3}{5}\times \left(-\frac{15}{4}\right)\times \frac{7}{9}\right\}\times \left(x^2\times x\times x^2\right)\times \left(y\times y^2\times y^2\right) \\ =\left\{\frac{3}{5}\times \left(-\frac{15}{4}\right)\times \frac{7}{9}\right\}\times \left(x^{2+1+2}\right)\times \left(y^{1+2+2}\right) \\ =-\frac{7}{4}{x}^5{y}^5 \end{gathered}$$

$\therefore$ $\left(\frac{3}{5}{x}^{2}y\right)\times \left(-\frac{15}{4}x{y}^2\right)\times \left(\frac{7}{9}{x}^2{y}^2\right)=-\frac{7}{4}{x}^5{y}^5$.

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

$$\begin{gathered} -\frac{7}{4}{x}^5{y}^5 \\ =-\frac{7}{4}{\left(2\right)}^5{\left(-1\right)}^5 \\ =\left(-\frac{7}{4}\right)\times 32\times \left(-1\right) \\ =56 \end{gathered}$$

Thus, the answer is 56.