RS AGGARWAL CLASS 9 Chapter 1 Number System Exercise 1D

 Exercise 1D

Question 1:

Add:
(i) $\left(2\sqrt{3}-5\sqrt{2}\right)\mathrm{and}\left(\sqrt{3}+2\sqrt{2}\right)$
(ii) $\left(2\sqrt{2}+5\sqrt{3}-7\sqrt{5}\right)\mathrm{and}\left(3\sqrt{3}-\sqrt{2}+\sqrt{5}\right)$
(iii) $\left(\frac{2}{3}\sqrt{7}-\frac{1}{2}+6\sqrt{11}\right)\mathrm{and}\left(\frac{1}{3}\sqrt{7}+\frac{3}{2}\sqrt{2}-\sqrt{11}\right)$

Answer 1:

$$\begin{gathered} \left(\mathrm{i}\right) 2\sqrt{3}-5\sqrt{2}+\sqrt{3}+2\sqrt{2} \\ =\left(2\sqrt{3}+\sqrt{3}\right)+\left(2\sqrt{2}-5\sqrt{2}\right) \\ =3\sqrt{3}-3\sqrt{2} \\ \left(\mathrm{ii}\right) 2\sqrt{2}+5\sqrt{3}-7\sqrt{5}+3\sqrt{3}-\sqrt{2}+\sqrt{5} \\ =2\sqrt{2}-\sqrt{2}+5\sqrt{3}+3\sqrt{3}+\sqrt{5}-7\sqrt{5} \\ =\sqrt{2}+8\sqrt{3}-6\sqrt{5} \\ \left(\mathrm{iii}\right) \frac{2}{3}\sqrt{7}-\frac{1}{2}\sqrt{2}+6\sqrt{11}+\frac{1}{3}\sqrt{7}+\frac{3}{2}\sqrt{2}-\sqrt{11} \\ =\frac{2}{3}\sqrt{7}+\frac{1}{3}\sqrt{7}-\sqrt{11}+6\sqrt{11}+\frac{3}{2}\sqrt{2}-\frac{1}{2}\sqrt{2} \\ =\sqrt{7}+5\sqrt{11}+\sqrt{2} \end{gathered}$$

PAGE 28

Question 2:

Multiply:
(i) $3\sqrt{5} \mathrm{by} 2\sqrt{5}$
(ii) $6\sqrt{15} \mathrm{by} 4\sqrt{3}$
(iii) $2\sqrt{6} \mathrm{by} 3\sqrt{3}$
(iv) $3\sqrt{8} \mathrm{by} 3\sqrt{2}$
(v) $\sqrt{10} \mathrm{by} \sqrt{40}$
(vi) $3\sqrt{28} \mathrm{by} 2\sqrt{7}$

Answer 2:

$$\begin{gathered} \left(\mathrm{i}\right) 3\sqrt{5}\times 2\sqrt{5}=3\times 2\times \sqrt{5}\times \sqrt{5}=6\times 5=30 \\ \left(\mathrm{ii}\right) 6\sqrt{15}\times 4\sqrt{3}=6\times 4\times \sqrt{5}\times \sqrt{3}\times \sqrt{3}=24\times 3\times \sqrt{5}=72\sqrt{5} \\ \left(\mathrm{iii}\right) 2\sqrt{6}\times 3\sqrt{3}=2\times 3\times \sqrt{2}\times \sqrt{3}\times \sqrt{3}=6\times 3\times \sqrt{2}=18\sqrt{2} \\ \left(\mathrm{iv}\right) 3\sqrt{8}\times 3\sqrt{2}=3\times 3\times \sqrt{2}\times \sqrt{2}\times \sqrt{2}\times \sqrt{2}=9\times 4=36 \\ \left(\mathrm{v}\right) \sqrt{10}\times \sqrt{40}=\sqrt{2}\times \sqrt{5}\times \sqrt{2}\times \sqrt{2}\times \sqrt{2}\times \sqrt{5}=\sqrt{2}\times \sqrt{2}\times \sqrt{2}\times \sqrt{2}\times \sqrt{5}\times \sqrt{5}=2\times 2\times 5=20 \\ \left(\mathrm{vi}\right) 3\sqrt{28}\times 2\sqrt{7}=6\sqrt{7\times 4}\times \sqrt{7}=6\times 7\times \sqrt{4}=42\times 2=84 \end{gathered}$$

Question 3:

Divide:
(i) $16\sqrt{6} \mathrm{by} 4\sqrt{2}$
(ii) $12\sqrt{5} \mathrm{by} 4\sqrt{3}$
(iii) $18\sqrt{21} \mathrm{by} 6\sqrt{7}$

Answer 3:

$$\begin{gathered} \left(\mathrm{i}\right) \frac{16\sqrt{6}}{4\sqrt{2}}=\frac{16\sqrt{2}\sqrt{3}}{4\sqrt{2}}=4\sqrt{3} \\ \left(\mathrm{ii}\right) \frac{12\sqrt{15}}{4\sqrt{3}}=\frac{12\sqrt{5}\times \sqrt{3}}{4\sqrt{3}}=3\sqrt{5} \\ \left(\mathrm{iii}\right) \frac{18\sqrt{21}}{6\sqrt{7}}=\frac{18\sqrt{7}\sqrt{3}}{6\sqrt{7}}=3\sqrt{3} \end{gathered}$$

Question 4:

Simplify
(i) $\left(3-\sqrt{11}\right) \left(3+\sqrt{11}\right)$
(ii) $\left(-3+\sqrt{5}\right) \left(-3-\sqrt{5}\right)$
(iii) ${\left(3-\sqrt{3}\right)}^2$
(iv) ${\left(\sqrt{5}-\sqrt{3}\right)}^2$
(v) $\left(5+\sqrt{7}\right) \left(2+\sqrt{5}\right)$
(vi) $\left(\sqrt{5}-\sqrt{2}\right) \left(\sqrt{2}-\sqrt{3}\right)$

Answer 4:

(i) $\left(3-\sqrt{11}\right) \left(3+\sqrt{11}\right)$

$$\begin{gathered} =3^2-{\left(\sqrt{11}\right)}^2 [\left(a-b\right)\left(a+b\right)=a^2-b^2] \\ =9-11 \\ =-2 \end{gathered}$$

(ii) $\left(-3+\sqrt{5}\right) \left(-3-\sqrt{5}\right)$

$$\begin{gathered} ={\left(-3\right)}^2-{\left(\sqrt{5}\right)}^2 [\left(a+b\right)\left(a-b\right)=a^2-b^2] \\ =9-5 \\ =4 \end{gathered}$$

(iii) ${\left(3-\sqrt{3}\right)}^2$

$$\begin{gathered} =3^2+{\left(\sqrt{3}\right)}^2-2\times 3\times \sqrt{3} [{\left(a-b\right)}^2=a^2+b^2-2ab] \\ =9+3-6\sqrt{3} \\ =12-6\sqrt{3} \end{gathered}$$

(iv) ${\left(\sqrt{5}-\sqrt{3}\right)}^2$

$$\begin{gathered} ={\left(\sqrt{5}\right)}^2+{\left(\sqrt{3}\right)}^2-2\times \sqrt{5}\sqrt{3} [{\left(a-b\right)}^2=a^2+b^2-2ab] \\ =5+3-2\sqrt{15}=8-2\sqrt{15} \end{gathered}$$
$$\begin{gathered} =\sqrt{5}\times \sqrt{2}-\sqrt{5}\times \sqrt{3}-\sqrt{2}\times \sqrt{2}+\sqrt{2}\times \sqrt{3} \\ =\sqrt{10}-\sqrt{15}-2+\sqrt{6} \end{gathered}$$

(v) $\left(5+\sqrt{7}\right) \left(2+\sqrt{5}\right)$

$$\begin{gathered} \left(5+\sqrt{7}\right)\left(2+\sqrt{5}\right) \\ =5\times 2+5\times \sqrt{5}+\sqrt{7}\times 2+\sqrt{7}\times \sqrt{5} \\ =10+5\sqrt{5}+2\sqrt{7}+\sqrt{35} \end{gathered}$$

(vi) $\left(\sqrt{5}-\sqrt{2}\right) \left(\sqrt{2}-\sqrt{3}\right)$

$$\begin{gathered} \left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{2}-\sqrt{3}\right) \\ =\sqrt{5}\times \sqrt{2}-\sqrt{5}\times \sqrt{3}-\sqrt{2}\times \sqrt{2}+\sqrt{2}\times \sqrt{3} \\ =\sqrt{10}-\sqrt{15}-2+\sqrt{6} \end{gathered}$$

Question 5:

Simplify $\left(3+\sqrt{3}\right) {\left(2+\sqrt{2}\right)}^2$.

Answer 5:

$$\begin{gathered} \left(3+\sqrt{3}\right) {\left(2+\sqrt{2}\right)}^2 \\ =\left(3+\sqrt{3}\right) \left[2^2+{\left(\sqrt{2}\right)}^2+2\times 2\sqrt{2}\right] \\ =\left(3+\sqrt{3}\right) \left[4+2+4\sqrt{2}\right] \\ =\left(3+\sqrt{3}\right) \left[6+4\sqrt{2}\right] \end{gathered}$$

$$\begin{gathered} =3\times 6+3\times 4\sqrt{2}+\sqrt{3}\times 6+\sqrt{3}\times 4\sqrt{2} \\ =18+12\sqrt{2}+6\sqrt{3}+4\sqrt{6} \end{gathered}$$

Question 6:

Examine whether the following numbers are rational or irrational:

(i) $\left(5-\sqrt{5}\right) \left(5+\sqrt{5}\right)$

(ii) ${\left(\sqrt{3}+2\right)}^2$

(iii) $\frac{2\sqrt{13}}{3\sqrt{52}-4\sqrt{117}}$

(iv) $\sqrt{8}+4\sqrt{32}-6\sqrt{2}$

Answer 6:

(i) $\left(5-\sqrt{5}\right) \left(5+\sqrt{5}\right)$

$$\begin{gathered} =5^2-{\left(\sqrt{5}\right)}^2 \left[\left(a-b\right)\left(a+b\right)=a^2-b^2\right] \\ =25-5 \\ =20, \mathrm{which} \mathrm{is} \mathrm{an} \mathrm{integer} \end{gathered}$$

Hence, $\left(5-\sqrt{5}\right) \left(5+\sqrt{5}\right)$ is rational.

(ii) ${\left(\sqrt{3}+2\right)}^2$

$$\begin{gathered} ={\left(\sqrt{3}\right)}^2+2^2+2\times \sqrt{3}\times 2 \left[{\left(a+b\right)}^2=a^2+b^2+2ab\right] \\ =3+4+4\sqrt{3} \\ =7+4\sqrt{3} \end{gathered}$$

Since, the sum and product of rational numbers and an irrational number is always an irrational.

$\Rightarrow$$7+4\sqrt{3}$ is irrational.

Hence, ${\left(\sqrt{3}+2\right)}^2$​ is irrational.

(iii) $\frac{2\sqrt{13}}{3\sqrt{52}-4\sqrt{117}}$

$$\begin{gathered} =\frac{2\sqrt{13}}{3\sqrt{13\times 4}-4\sqrt{13\times 9}} \\ =\frac{2\sqrt{13}}{\sqrt{13}\left(3\sqrt{4}-4\sqrt{9}\right)} \\ =\frac{2}{\left(3\times 2-4\times 2\right)} \end{gathered}$$

$$\begin{gathered} =\frac{2}{6-8} \\ =\frac{2}{-2} \\ =-1, \mathrm{which} \mathrm{is} \mathrm{an} \mathrm{integer} \end{gathered}$$

Hence, $\frac{2\sqrt{13}}{3\sqrt{52}-4\sqrt{117}}$ is rational.

(iv) $\sqrt{8}+4\sqrt{32}-6\sqrt{2}$

$$\begin{gathered} =2\sqrt{2}+4\times 4\sqrt{2}-6\sqrt{2} \\ =2\sqrt{2}+16\sqrt{2}-6\sqrt{2} \\ =12\sqrt{2} \end{gathered}$$

Since, the product of a rational number and an irrational number is always an irrational.

Hence, $\sqrt{8}+4\sqrt{32}-6\sqrt{2}$ is rational.

Question 7:

On her birthday Reema distributed chocolates in an orphanage. The total number of chocolates she distributed is given by $\left(5+\sqrt{11}\right) \left(5-\sqrt{11}\right)$.
(i) Find the number of chocolates distributed by her.
(ii) Write the moral values depicted here by Reema.

Answer 7:

(i) As, $\left(5+\sqrt{11}\right) \left(5-\sqrt{11}\right)$

$$\begin{gathered} =5^2-{\left(\sqrt{11}\right)}^2 \left[\left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =25-11 \\ =14 \end{gathered}$$

Hence, the number of chocolates distributed by Reema is 14.

(ii) The moral values depicted here by Reema is helpfulness and caring.

Disclaimer: The moral values may vary from person to person.

Question 8:

Simplify

(i) $3\sqrt{45}-\sqrt{125}+\sqrt{200}-\sqrt{50}$

(ii) $\frac{2\sqrt{30}}{\sqrt{6}}-\frac{3\sqrt{140}}{\sqrt{28}}+\frac{\sqrt{55}}{\sqrt{99}}$

(iii) $\sqrt{72}+\sqrt{800}-\sqrt{18}$

Answer 8:

(i) $3\sqrt{45}-\sqrt{125}+\sqrt{200}-\sqrt{50}$

$$\begin{gathered} =3\sqrt{9\times 5}-\sqrt{25\times 5}+\sqrt{100\times 2}-\sqrt{25\times 2} \\ =3\times 3\sqrt{5}-5\sqrt{5}+10\sqrt{2}-5\sqrt{2} \\ =9\sqrt{5}-5\sqrt{5}+5\sqrt{2} \\ =4\sqrt{5}+5\sqrt{2} \end{gathered}$$

(ii) $\frac{2\sqrt{30}}{\sqrt{6}}-\frac{3\sqrt{140}}{\sqrt{28}}+\frac{\sqrt{55}}{\sqrt{99}}$

$$\begin{gathered} =\frac{2\sqrt{6\times 5}}{\sqrt{6}}-\frac{3\sqrt{28\times 5}}{\sqrt{28}}+\frac{\sqrt{5\times 11}}{\sqrt{9\times 11}} \\ =\frac{2\sqrt{6}\times \sqrt{5}}{\sqrt{6}}-\frac{3\sqrt{28}\times \sqrt{5}}{\sqrt{28}}+\frac{\sqrt{5}\times \sqrt{11}}{\sqrt{9}\times \sqrt{11}} \\ =2\sqrt{5}-3\sqrt{5}+\frac{\sqrt{5}}{3} \end{gathered}$$

$$\begin{gathered} =-\sqrt{5}+\frac{\sqrt{5}}{3} \\ =\frac{-3\sqrt{5}+\sqrt{5}}{3} \\ =\frac{-2\sqrt{5}}{3} \end{gathered}$$

(iii) $\sqrt{72}+\sqrt{800}-\sqrt{18}$

$$\begin{gathered} =\sqrt{36\times 2}+\sqrt{400\times 2}-\sqrt{9\times 2} \\ =6\sqrt{2}+20\sqrt{2}-3\sqrt{2} \\ =23\sqrt{2} \end{gathered}$$