RS Aggarwal 2019,2020 solution class 7 chapter 6 Algebraic Expressions Exercise 6B

Exercise 6B

Page-102

Question 1:

Find the products:

3a² × 8a⁴

Answer 1:

3a² × 8a^{4
$$\begin{gathered} =(3\times 8)\times ({\mathrm{a}}^2\times {\mathrm{a}}^4) \\ =24 \times {\mathrm{a}}^{(2+4)} \\ =24{\mathrm{a}}^6 \end{gathered}$$}

Question 2:

Find the products:

−6x³ × 5x²

Answer 2:

−6x³ × 5x^{2
$$\begin{gathered} =(-6\times 5)\times ({\mathrm{x}}^3\times {\mathrm{x}}^2) \\ =(-30)\times \left({\mathrm{x}}^{(3+2)}\right) \\ =-30{\mathrm{x}}^5 \end{gathered}$$}

Question 3:

Find the products:

(−4ab) × (−3a²bc)

Answer 3:

(−4ab) × (−3a²bc)
$$\begin{gathered} =(-4\times -3)\times (\mathrm{a}\times {\mathrm{a}}^2\times \mathrm{b}\times \mathrm{b}\times \mathrm{c}) \\ =12\times \left({\mathrm{a}}^3{\mathrm{b}}^2\mathrm{c}\right) \\ = 12{\mathrm{a}}^3{\mathrm{b}}^2\mathrm{c} \end{gathered}$$

Question 4:

Find the products:

(2a²b³) × (−3a³b)

Answer 4:

(2a²b³) × (−3a³b)
$$\begin{gathered} =(2\times (-3\left)\right)\times ({\mathrm{a}}^2\times {\mathrm{a}}^3\times {\mathrm{b}}^3\times \mathrm{b}) \\ =(-6)\times ({\mathrm{a}}^{(2+3)}\times {\mathrm{b}}^{(3+1)}) \\ =-6{\mathrm{a}}^5{\mathrm{b}}^4 \end{gathered}$$

Question 5:

Find the products:

$\frac{2}{3}{x}^{2}y\times \frac{3}{5}x{y}^2$

Answer 5:

$$\begin{gathered} =(\frac{2}{3}\times \frac{3}{5})\times ({\mathrm{x}}^2\times \mathrm{x}\times \mathrm{y}\times {\mathrm{y}}^{2)}) \\ =\frac{2}{5}\times {\mathrm{x}}^{(2+1)}\times {\mathrm{y}}^{(1+2)} \\ =\frac{2}{5}{\mathrm{x}}^3{\mathrm{y}}^3 \end{gathered}$$

Question 6:

Find the products:

$\left(\frac{-3}{4}a{b}^3\right)\times \left(\frac{-2}{3}{a}^2{b}^4\right)$

Answer 6:

$$\begin{gathered} =(\frac{-3}{4}\times \frac{-2}{3})\times (\mathrm{a}\times {\mathrm{a}}^2\times {\mathrm{b}}^3\times {\mathrm{b}}^4) \\ =\frac{1}{2}\times {\mathrm{a}}^{(1+2)}\times {\mathrm{b}}^{(3+4)} \\ =\frac{1}{2}{\mathrm{a}}^3{\mathrm{b}}^7 \end{gathered}$$

Question 7:

Find the products:

$\left(\frac{-1}{27}{a}^2{b}^2\right)\times \left(\frac{-9}{2}{a}^{3}b{c}^2\right)$

Answer 7:

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

Question 8:

Find the products:

$\left(\frac{-13}{5}a{b}^{2}c\right)\times \left(\frac{7}{3}{a}^{2}b{c}^2\right)$

Answer 8:

$$\begin{gathered} =(\frac{-13}{5}\times \frac{7}{3})\times (\mathrm{a}\times {\mathrm{a}}^2\times {\mathrm{b}}^2\times \mathrm{b}\times \mathrm{c}\times {\mathrm{c}}^2) \\ =\frac{-91}{15}{\mathrm{a}}^{(1+2)}\times {\mathrm{b}}^{(2+1)}\times {\mathrm{c}}^{(1+2)} \\ =\frac{-91}{15}{\mathrm{a}}^3{\mathrm{b}}^3{\mathrm{c}}^3 \end{gathered}$$

Question 9:

Find the products:

$\left(\frac{-18}{5}{x}^{2}z\right)\times \left(\frac{-25}{6}x{z}^{2}y\right)$

Answer 9:

$$\begin{gathered} =(-\frac{18}{5}\times \frac{-25}{6})\times ({\mathrm{x}}^2\times \mathrm{x}\times \mathrm{z}\times {\mathrm{z}}^2\times \mathrm{y}) \\ =15\times {\mathrm{x}}^{(2+1)}\times \mathrm{y}\times {\mathrm{z}}^{(1+2)} \\ =15{\mathrm{x}}^3{\mathrm{yz}}^3 \end{gathered}$$

Question 10:

Find the products:

$\left(\frac{-3}{14}x{y}^4\right)\times \left(\frac{7}{6}{x}^{3}y\right)$

Answer 10:

$$\begin{gathered} =(\frac{-3}{14}\times \frac{7}{6})\times (\mathrm{x}\times {\mathrm{x}}^3\times {\mathrm{y}}^4\times \mathrm{y}) \\ =\frac{-1}{4}{\mathrm{x}}^{(1+3)}\times {\mathrm{y}}^{(4+1)} \\ =\frac{-1}{4}{\mathrm{x}}^4{\mathrm{y}}^5 \end{gathered}$$

Question 11:

Find the products:

$\left(\frac{-7}{5}{x}^{2}y\right)\times \left(\frac{3}{2}x{y}^2\right)\times \left(\frac{-6}{5}{x}^3{y}^3\right)$

Answer 11:

$$\begin{gathered} =(\frac{-7}{5}\times \frac{3}{2}\times \frac{-6}{5})\times ({\mathrm{x}}^2\times \mathrm{x}\times {\mathrm{x}}^3\times \mathrm{y}\times {\mathrm{y}}^2\times {\mathrm{y}}^3) \\ =\frac{63}{25}\times {\mathrm{x}}^{(2+1+3)}\times {\mathrm{y}}^{(1+2+3)} \\ =\frac{63}{25}{\mathrm{x}}^6{\mathrm{y}}^6 \end{gathered}$$

Question 12:

Find the products:

(2a²b) × (−5ab²c) × (−6bc²)

Answer 12:

$$\begin{gathered} =(2\times (-5)\times (-6\left)\right)\times ({\mathrm{a}}^2\times \mathrm{a}\times \mathrm{b}\times {\mathrm{b}}^2\times \mathrm{b}\times \mathrm{c}\times {\mathrm{c}}^2) \\ =60\times {\mathrm{a}}^{(2+1)}\times {\mathrm{b}}^{(1+2+1)}\times {\mathrm{c}}^{(1+2)} \\ =60{\mathrm{a}}^3{\mathrm{b}}^4{\mathrm{c}}^3 \end{gathered}$$

Question 13:

Find the products:

(−4x²) × (−6xy²) × (−3y)

Answer 13:

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

Question 14:

 Find the products:

$\left(\frac{-3}{5}{s}^{2}t\right)\times \left(\frac{15}{7}s{t}^{2}u\right)\times \left(\frac{7}{9}s{u}^2\right)$

Answer 14:

$$\begin{gathered} =(\frac{-3}{5}\times \frac{15}{7}\times \frac{7}{9})\times ({\mathrm{s}}^2\times \mathrm{s}\times \mathrm{s}\times \mathrm{t}\times {\mathrm{t}}^2\times \mathrm{u}\times {\mathrm{u}}^2) \\ =-1\times {\mathrm{s}}^{(2+1+1)}\times {\mathrm{t}}^{(1+2)}\times {\mathrm{u}}^{(1+2)} \\ =-{\mathrm{s}}^4{\mathrm{t}}^3{\mathrm{u}}^3 \end{gathered}$$

Question 15:

 Find the products:

$\left(\frac{-2}{7}{u}^{4}v\right)\times \left(\frac{-14}{5}u{v}^3\right)\times \left(\frac{-3}{4}{u}^2{v}^3\right)$

Answer 15:

$$\begin{gathered} =(\frac{-2}{7}\times \frac{-14}{5}\times \frac{-3}{4})\times ({\mathrm{u}}^4\times \mathrm{u}\times {\mathrm{u}}^2\times \mathrm{v}\times {\mathrm{v}}^3\times {\mathrm{v}}^3) \\ =\frac{-3}{5}\times {\mathrm{u}}^{(4+1+2)}\times {\mathrm{v}}^{(1+3+3)} \\ =\frac{-3}{5}{\mathrm{u}}^7{\mathrm{v}}^7 \end{gathered}$$

Question 16:

 Find the products:

(ab²) × (−b²c) × (−a²c³) × (−3abc)

Answer 16:

$$\begin{gathered} =(-3\times -1\times -1)\times (\mathrm{a}\times {\mathrm{a}}^2\times \mathrm{a}\times {\mathrm{b}}^2\times {\mathrm{b}}^2\times \mathrm{b}\times \mathrm{c}\times {\mathrm{c}}^3\times \mathrm{c} \\ =-3\times {\mathrm{a}}^{(1+2+1)}\times {\mathrm{b}}^{(2+2+1)}\times {\mathrm{c}}^{(1+4+1)} \\ =-3{\mathrm{a}}^4{\mathrm{b}}^5{\mathrm{c}}^5 \end{gathered}$$

Question 17:

 Find the products:

$\left(\frac{4}{3}{x}^{2}yz\right)\times \left(\frac{1}{3}{y}^{2}zx\right)\times \left(-6xy{z}^2\right)$

Answer 17:

$$\begin{gathered} =(\frac{4}{3}\times \frac{1}{3}\times (-6\left)\right)\times ({\mathrm{x}}^2\times \mathrm{x}\times \mathrm{x}\times \mathrm{y}\times {\mathrm{y}}^2\times \mathrm{y}\times \mathrm{z}\times \mathrm{z}\times {\mathrm{z}}^2) \\ =\frac{-8}{3}\times {\mathrm{x}}^{(2+1+1)}\times {\mathrm{y}}^{(1+2+1)}\times {\mathrm{z}}^{(1+1+2)} \\ =\frac{-8}{3}{\mathrm{x}}^4{\mathrm{y}}^4{\mathrm{z}}^4 \end{gathered}$$

Question 18:

Multiply $-\frac{2}{3}{a}^{2}b \mathrm{by}\frac{6}{5}{a}^3{b}^2$ and verify your result for a = 2 and b = 3.

Answer 18:

$$\begin{gathered} \frac{-2}{3}{\mathrm{a}}^2\mathrm{b}\times \frac{6}{5}{\mathrm{a}}^3{\mathrm{b}}^2 \\ =(\frac{-2}{3}\times \frac{6}{5})\times ({\mathrm{a}}^2\times {\mathrm{a}}^3\times \mathrm{b}\times {\mathrm{b}}^2) \\ =\frac{-4}{5}\times {\mathrm{a}}^{(2+3)}\times {\mathrm{b}}^{(1+2)} \\ =\frac{-4}{5}{\mathrm{a}}^5{\mathrm{b}}^3 \end{gathered}$$


When a =2 and b =3, we get:

$$\begin{gathered} \frac{-2}{3}{\mathrm{a}}^2\mathrm{b} = \frac{-2}{3}\times 2^2\times 3 = -8 \\ \frac{6}{5}{\mathrm{a}}^3{\mathrm{b}}^2= \frac{6}{5}\times 2^3\times 3^2 = \frac{432}{5} \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.\hspace{0.17em}=\frac{-2}{3}{\mathrm{a}}^2\mathrm{b} \times \frac{6}{5}{\mathrm{a}}^3{\mathrm{b}}^2 = -8\times \frac{432}{5}=\frac{-3456}{5} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}. = \frac{-4}{5}{\mathrm{a}}^5{\mathrm{b}}^3=\frac{-4}{5}\times 2^5\times 3^3 = \frac{-3456}{5} \end{gathered}$$

L.H.S. = R.H.S.

Hence, the result is verified.

Question 19:

Multiply $-\frac{8}{21}{x}^2{y}^3 \mathrm{by}-\frac{7}{16}x{y}^2$ and verify your result for x = 3 and y = 2.

Answer 19:

$$\begin{gathered} \frac{-8}{21}{\mathrm{x}}^2{\mathrm{y}}^3 \times \frac{-7}{16}{\mathrm{xy}}^2 = \left(\frac{-8}{21}\times \frac{-7}{16}\right)\left({\mathrm{x}}^{2+1}\right)\left({\mathrm{y}}^{3+2}\right) = \frac{1}{6}\times {\mathrm{x}}^3\times {\mathrm{y}}^{5 } \\ \mathrm{When} x=3 \mathrm{and} y=2, \mathrm{we} \mathrm{get}: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.\hspace{0.17em}= \frac{-8}{21}{\mathrm{x}}^2{\mathrm{y}}^3 \times \frac{-7}{16}{\mathrm{xy}}^2 = \frac{-192}{7}\times \frac{-21}{4}= 144 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}. = \frac{1}{6}{\mathrm{x}}^3{\mathrm{y}}^5= \frac{1}{6}\times 3^3\times 2^{5 }= 144 \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.\hspace{0.17em}= \mathrm{R}.\mathrm{H}.\mathrm{S}. \\ \therefore \frac{-8}{21}{\mathrm{x}}^2{\mathrm{y}}^3 \times \frac{-7}{16}{\mathrm{xy}}^2 = \frac{1}{6}{\mathrm{x}}^3{\mathrm{y}}^5 \end{gathered}$$

Question 20:

Find the value of (2.3a⁵b²) × (1.2a²b²), when a = 1 and b = 0.5.

Answer 20:

$$\begin{gathered} =(2.3\times 1.2)\times ({\mathrm{a}}^5\times {\mathrm{a}}^2\times {\mathrm{b}}^2\times {\mathrm{b}}^2) \\ =2.76\times {\mathrm{a}}^{(5+2)}\times {\mathrm{b}}^{(2+2)} \\ =2.76{\mathrm{a}}^7{\mathrm{b}}^4 \\ \mathrm{When} \mathrm{a} = 1 \mathrm{and} \mathrm{b}= 0.5, \mathrm{we} \mathrm{get}: \\ 2.76{\mathrm{a}}^7{\mathrm{b}}^4 = 2.76\times 1^7\times 0.5^4 = 0.1725 \end{gathered}$$

Question 21:

Find the value of (−8u²v⁶) × (−20uv) for u = 2.5 and v = 1.

Answer 21:

$$\begin{gathered} =(-8\times (-20\left)\right)\times ({\mathrm{u}}^2\times \mathrm{u}\times {\mathrm{v}}^6\times \mathrm{v}) \\ = 160\times {\mathrm{u}}^{(2+1)}\times {\mathrm{v}}^{(6+1)} \\ =160{\mathrm{u}}^3{\mathrm{v}}^7 \\ 160{\mathrm{u}}^3{\mathrm{v}}^7 = 160\times 2.5^3\times 1^7 = 2500 \end{gathered}$$

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Question 22:

Find the product and verify the result for a = 1, b = 2 and c = 3.

$\left(\frac{2}{5}{a}^{2}b\right)\times \left(-15b^{2}ac\right)\times \left(-\frac{1}{2}{c}^2\right)$

Answer 22:

$$\begin{gathered} =(\frac{2}{5}\times -15\times \frac{-1}{2})\times ({\mathrm{a}}^2\times \mathrm{a}\times \mathrm{b}\times {\mathrm{b}}^2\times \mathrm{c}\times {\mathrm{c}}^2) \\ =3\times {\mathrm{a}}^{(2+1)}\times {\mathrm{b}}^{(1+2)}\times {\mathrm{c}}^{(1+2)} \\ =3{\mathrm{a}}^3{\mathrm{b}}^3{\mathrm{c}}^3 \\ \mathrm{When} \mathrm{a} = 1, \mathrm{b} = 2 \mathrm{and} \mathrm{c} = 3, \mathrm{we} \mathrm{get}: \\ \frac{2}{5}{\mathrm{a}}^2\mathrm{b} = \frac{2}{5}\times 1^2\times 2= \frac{4}{5} \\ -15{\mathrm{b}}^2\mathrm{ac} = -15\times 2^2\times 1\times 3= -180 \\ \frac{-1}{2}{\mathrm{c}}^2 = \frac{-1}{2}\times 3^2 = \frac{-9}{2} \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.\hspace{0.17em}= \frac{2}{5}{\mathrm{a}}^2\mathrm{b}\times -15{\mathrm{b}}^2\mathrm{ac}\times \frac{-1}{2}{\mathrm{c}}^2 = \frac{4}{5}\times -180\times \frac{-9}{2} = 648 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.\hspace{0.17em}= 3{\mathrm{a}}^3{\mathrm{b}}^3{\mathrm{c}}^3 = 3\times 1^3\times 2^3\times 3^3= 648 \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.\hspace{0.17em}= \mathrm{R}.\mathrm{H}.\mathrm{S}. \\ \therefore \frac{2}{5}{\mathrm{a}}^2\mathrm{b}\times -15{\mathrm{b}}^2\mathrm{ac}\times \frac{-1}{2}{\mathrm{c}}^2 =3{\mathrm{a}}^3{\mathrm{b}}^3{\mathrm{c}}^3 \end{gathered}$$

Question 23:

Find the product and verify the result for a = 1, b = 2 and c = 3.

$\left(\frac{1}{4}abc\right)\times \left(-6b^{2}c\right)\times \left(-\frac{1}{3}{c}^3\right)$

Answer 23:

$$\begin{gathered} =(\frac{1}{4}\times -6\times -\frac{1}{3})\times (\mathrm{a}\times \mathrm{b}\times {\mathrm{b}}^2\times \mathrm{c}\times \mathrm{c}\times {\mathrm{c}}^3) \\ =\frac{1}{2}\times \mathrm{a}\times {\mathrm{b}}^{(1+2)}\times {\mathrm{c}}^{(1+1+3)} \\ =\frac{1}{2}{\mathrm{ab}}^3{\mathrm{c}}^5 \\ \mathrm{When} \mathrm{a} =1, \mathrm{b} = 2 \mathrm{and} \mathrm{c} = 3, \mathrm{we} \mathrm{get}: \\ \frac{1}{4}\mathrm{abc} = \frac{1}{4}\times 1\times 2\times 3 = \frac{3}{2} \\ -6{\mathrm{b}}^2\mathrm{c} = -6 \times 2^2\times 3 = -72 \\ -\frac{1}{3}{\mathrm{c}}^3= \frac{-1}{3}\times 3^3=-9 \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.\hspace{0.17em}= \frac{1}{4}\mathrm{abc} \times -6{\mathrm{b}}^2\mathrm{c} \times -\frac{1}{3}{\mathrm{c}}^3 = \frac{3}{2}\times -72\times -9 = 972 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}. = \frac{1}{2}{\mathrm{ab}}^3{\mathrm{c}}^5 = \frac{1}{2}\times 1\times 2^3\times 3^5= 972 \\ \mathrm{L}.\mathrm{H}.\mathrm{S}. = \mathrm{R}.\mathrm{H}.\mathrm{S}. \\ \therefore \frac{1}{4}\mathrm{abc} \times -6{\mathrm{b}}^2\mathrm{c} \times -\frac{1}{3}{\mathrm{c}}^3 = \frac{1}{2}{\mathrm{ab}}^3{\mathrm{c}}^5 \end{gathered}$$

Question 24:

Find the product and verify the result for a = 1, b = 2 and c = 3.

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

Answer 24:

$$\begin{gathered} =(\frac{4}{9}\times \frac{-27}{5}\times -8)\times (\mathrm{a}\times {\mathrm{a}}^3\times \mathrm{b}\times {\mathrm{b}}^2\times {\mathrm{b}}^3\times {\mathrm{c}}^3\times \mathrm{c}) \\ =\frac{96}{5}\times {\mathrm{a}}^{(1+3)}\times {\mathrm{b}}^{(1+2+3)}\times {\mathrm{c}}^{(3+1)} \\ =\frac{96}{5}{\mathrm{a}}^4{\mathrm{b}}^6{\mathrm{c}}^4 \\ \mathrm{When} \mathrm{a}=1, \mathrm{b}=2 \mathrm{and} \mathrm{c}=3: \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.: (\frac{4}{9}\times \frac{-27}{5}\times -8)\times (\mathrm{a}\times {\mathrm{a}}^3\times \mathrm{b}\times {\mathrm{b}}^2\times {\mathrm{b}}^3\times {\mathrm{c}}^3\times \mathrm{c}) \\ = (\frac{4}{9}\times \frac{-27}{5}\times -8)\times \left(1\times 1^3\times 2\times 2^2\times 2^3\times 3^3\times 3\right) \\ = \frac{497664}{5} \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.: \frac{96}{5}{\mathrm{a}}^4{\mathrm{b}}^6{\mathrm{c}}^4 = \frac{96}{5}\left(1^4\times 2^6\times 3^4\right) = \frac{497664}{5} \\ \mathrm{L}.\mathrm{H}.\mathrm{S}. = \mathrm{R}.\mathrm{H}.\mathrm{S}. \\ \mathrm{Hence}, \mathrm{verified}. \end{gathered}$$

Question 25:

Find the product and verify the result for a = 1, b = 2 and c = 3.

$\left(\frac{-4}{7}{a}^{2}b\right)\times \left(\frac{-2}{3}{b}^{2}c\right)\times \left(\frac{-7}{6}{c}^{2}a\right)$

Answer 25:

$$\begin{gathered} =(\frac{-4}{7}\times \frac{-2}{3}\times \frac{-7}{6})\times ({\mathrm{a}}^2\times \mathrm{a}\times \mathrm{b}\times {\mathrm{b}}^2\times \mathrm{c}\times {\mathrm{c}}^2) \\ =-\frac{4}{9}{\mathrm{a}}^{(2+1)}\times {\mathrm{b}}^{(1+2)}\times {\mathrm{c}}^{(1+2)} \\ =\frac{-4}{9}{\mathrm{a}}^3{\mathrm{b}}^3{\mathrm{c}}^3 \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.: (\frac{-4}{7}\times \frac{-2}{3}\times \frac{-7}{6})\times (1^2\times 1\times 2\times 2^2\times 3\times 3^2) \\ = -96 \\ \mathrm{R}.\mathrm{H}.\mathrm{S}.: \frac{-4}{9}\times 1^3\times 2^3\times 3^3 = -96 \\ \mathrm{L}.\mathrm{H}.\mathrm{S}. = \mathrm{R}.\mathrm{H}.\mathrm{S}. \\ \mathrm{Hence}, \mathrm{verified}. \end{gathered}$$