RS AGGARWAL CLASS 9 Chapter 1 Number System VERY SHORT ANSWERS

 VERY SHORT ANSWERS

Question 1:

What can you say about the sum of a rational number and an irrational number?

Answer 1:

Sum of a rational number and an irrational number is an irrational number. 
Example: 4 + $\sqrt{5}$ represents sum of rational and an irrational number where 4 is rational and $\sqrt{5}$ is irrational. 

Question 2:

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

Answer 2:

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

Question 3:

The number $\frac{665}{625}$ will terminate after how many decimal places?

Answer 3:

$\frac{665}{625}=\frac{5\times 19\times 7}{5^4}=\frac{19\times 7}{5^3}=\frac{19\times 7\times 2^3}{5^3\times 2^3}=\frac{1064}{1000}=1.064$
So, $\frac{665}{625}$ will terminate after 3 decimal places. 

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Question 4:

Find the value of ${\left(1296\right)}^{0.17}\times {\left(1296\right)}^{0.08}$.

Answer 4:

$$\begin{gathered} {\left(1296\right)}^{0.17}\times {\left(1296\right)}^{0.08} \\ ={\left(1296\right)}^{0.17+0.08} \\ ={\left(1296\right)}^{0.25} \\ ={\left(1296\right)}^{\frac{1}{4}} \\ =\sqrt[4]{1296} \\ =6 \end{gathered}$$

Question 5:

Simplify $6\sqrt{36}+5\sqrt{12}$.

Answer 5:

$$\begin{gathered} 6\sqrt{36}+5\sqrt{12} \\ =6\times 6+5\sqrt{4\times 3} \\ =36+10\sqrt{3} \end{gathered}$$
 

Question 6:

Find an irrational number between 5 and 6.

Answer 6:

A number which is non terminating and non recurring is known as irrational number.

There are infinitely many irrational numbers between 5 and 6.

One of the example is 5.40430045000460000....

Question 7:

Find the value of $\frac{21\sqrt{12}}{10\sqrt{27}}$.

Answer 7:

$$\begin{gathered} \frac{21\sqrt{12}}{10\sqrt{27}}=\frac{21\sqrt{2\times 2\times 3}}{10\sqrt{3\times 3\times 3}} \\ =\frac{21\times 2\sqrt{3}}{10\times 3\sqrt{3}} \\ =\frac{3\times 7\times 2}{5\times 2\times 3} \\ =\frac{7}{5} \end{gathered}$$

Hence, the value of $\frac{21\sqrt{12}}{10\sqrt{27}}$ is $\frac{7}{5}$.

Question 8:

Rationalise $\frac{1}{\sqrt{3}+\sqrt{2}}$.

Answer 8:

$$\begin{gathered} \frac{1}{\sqrt{3}+\sqrt{2}}=\frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} \\ =\frac{\sqrt{3}-\sqrt{2}}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2} \\ =\frac{\sqrt{3}-\sqrt{2}}{3-2} \\ =\frac{\sqrt{3}-\sqrt{2}}{1} \\ =\sqrt{3}-\sqrt{2} \end{gathered}$$

Hence, the rationalised form is $\sqrt{3}-\sqrt{2}$.

Question 9:

Solve for x: ${\left(\frac{2}{5}\right)}^{2x-2}=\frac{32}{3125}$.

Answer 9:

$$\begin{gathered} {\left(\frac{2}{5}\right)}^{2x-2}=\frac{32}{3125} \\ \Rightarrow {\left(\frac{2}{5}\right)}^{2x-2}=\frac{2^5}{5^5} \\ \Rightarrow {\left(\frac{2}{5}\right)}^{2x-2}={\left(\frac{2}{5}\right)}^5 \\ \Rightarrow 2x-2=5 \\ \Rightarrow 2x=5+2 \\ \Rightarrow x=\frac{7}{2} \end{gathered}$$
Hence, $x=\frac{7}{2}$.

Question 10:

Simplify ${\left(32\right)}^{\frac{1}{5}}+{\left(-7\right)}^0+{\left(64\right)}^{\frac{1}{2}}$.

Answer 10:

$$\begin{gathered} {\left(32\right)}^{\frac{1}{5}}+{\left(-7\right)}^0+{\left(64\right)}^{\frac{1}{2}}={\left(2^5\right)}^{\frac{1}{5}}+1+{\left(2^6\right)}^{\frac{1}{2}} \\ =2+1+2^3 \\ =2+1+8 \\ =11 \end{gathered}$$

Hence, ${\left(32\right)}^{\frac{1}{5}}+{\left(-7\right)}^0+{\left(64\right)}^{\frac{1}{2}}$ = 11.

Question 11:

Evaluate ${\left(\frac{81}{49}\right)}^{\frac{-3}{2}}$.

Answer 11:

$$\begin{gathered} {\left(\frac{81}{49}\right)}^{\frac{-3}{2}}={\left(\frac{49}{81}\right)}^{\frac{3}{2}} \\ ={\left(\frac{7^2}{9^2}\right)}^{\frac{3}{2}} \\ ={\left[{\left(\frac{7}{9}\right)}^2\right]}^{\frac{3}{2}} \\ ={\left(\frac{7}{9}\right)}^3 \\ =\frac{7^3}{9^3} \\ =\frac{343}{729} \end{gathered}$$

Hence, ${\left(\frac{81}{49}\right)}^{\frac{-3}{2}}=\frac{343}{729}$.

Question 12:

Simplify $\sqrt[4]{81x^8{y}^4{z}^{16}}$.

Answer 12:

$$\begin{gathered} \sqrt[4]{81x^8{y}^4{z}^{16}} \\ ={\left(3^4\times x^8\times y^4\times z^{16}\right)}^{\frac{1}{4}} \\ ={\left(3^4\right)}^{\frac{1}{4}}\times {\left(x^8\right)}^{\frac{1}{4}}\times {\left(y^4\right)}^{\frac{1}{4}}\times {\left(z^{16}\right)}^{\frac{1}{4}} \left[{\left(x\times y\times z\times ...\right)}^a=x^a\times y^a\times z^a\times ...\right] \end{gathered}$$
$$\begin{gathered} =\left(3^{4\times \frac{1}{4}}\right)\times \left(x^{8\times \frac{1}{4}}\right)\times \left(y^{4\times \frac{1}{4}}\right)\times \left(z^{16\times \frac{1}{4}}\right) \left[{\left(x^a\right)}^b=x^{ab}\right] \\ =3\times x^2\times y\times z^4 \\ =3x^{2}y{z}^4 \end{gathered}$$
 

Question 13:

If a = 1, b = 2 then find the value of (a^b + b^a)^–1.

Answer 13:

For a = 1and b = 2,
$$\begin{gathered} {\left(a^b+b^a\right)}^{-1} \\ ={\left(1^2+2^1\right)}^{-1} \\ ={\left(1+2\right)}^{-1} \end{gathered}$$
$$\begin{gathered} =3^{-1} \\ =\frac{1}{3} \left(x^{-a}=\frac{1}{x^a}\right) \end{gathered}$$
Thus, the value of (a^b + b^a)^–1 when a = 1 and b = 2 is $\frac{1}{3}$.

Question 14:

Simplify ${\left(\frac{3125}{243}\right)}^{\frac{4}{5}}$.

Answer 14:

$$\begin{gathered} {\left(\frac{3125}{243}\right)}^{\frac{4}{5}} \\ ={\left(\frac{5\times 5\times 5\times 5\times 5}{3\times 3\times 3\times 3\times 3}\right)}^{\frac{4}{5}} \\ ={\left(\frac{5^5}{3^5}\right)}^{\frac{4}{5}} \\ ={\left[{\left(\frac{5}{3}\right)}^5\right]}^{\frac{4}{5}} \left[{\left(\frac{x}{y}\right)}^a=\frac{x^a}{y^a}\right] \end{gathered}$$
$$\begin{gathered} ={\left(\frac{5}{3}\right)}^{5\times \frac{4}{5}} \left[{\left(x^a\right)}^b=x^{ab}\right] \\ ={\left(\frac{5}{3}\right)}^4 \\ =\frac{5\times 5\times 5\times 5}{3\times 3\times 3\times 3} \\ =\frac{625}{81} \end{gathered}$$

Question 15:

Give an example of two irrational numbers whose sum as well as product is rational.

Answer 15:

Let the two irrational numbers be $2+\sqrt{3}$ and $2-\sqrt{3}$.
Sum of these irrational numbers $=\left(2+\sqrt{3}\right)+\left(2-\sqrt{3}\right)=4$, which is rational
Product of these irrational numbers $=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=2^2-{\left(\sqrt{3}\right)}^2=4-3=1$, which is rational
 

Question 16:

Is the product of a rational and an irrational number always irrational? Give an example.

Answer 16:

Yes, the product of a rational and an irrational number is always an irrational number.
Example: 
2 is a rational number and $\sqrt{3}$ is an irrational number.
Now, $2\times \sqrt{3}=2\sqrt{3}$, which is an irrational number.

Question 17:

Give an example of a number x such that x² is an irrational number and x³ is a rational number.

Answer 17:

The cube roots of natural numbers which are not perfect cubes are all irrational numbers.
Let $x=\sqrt[3]{2}=2^{\frac{1}{3}}$.
Now,
$x^2={\left(2^{\frac{1}{3}}\right)}^2=2^{\frac{2}{3}}={\left(2^2\right)}^{\frac{1}{3}}=4^{\frac{1}{3}}$, which is an irrational number
Also,
$x^3={\left(2^{\frac{1}{3}}\right)}^3=2^{3\times \frac{1}{3}}=2$, which is a rational number

Question 18:

Write the reciprocal of $\left(2+\sqrt{3}\right)$.

Answer 18:

The reciprocal of $\left(2+\sqrt{3}\right)$

$$\begin{gathered} =\frac{1}{\left(2+\sqrt{3}\right)} \\ =\frac{1}{\left(2+\sqrt{3}\right)}\times \frac{\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)} \\ =\frac{\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)} \\ =\frac{\left(2-\sqrt{3}\right)}{2^2-{\left(\sqrt{3}\right)}^2} \left[\left(a+b\right)\left(a-b\right)=a^2-b^2\right] \end{gathered}$$

$$\begin{gathered} =\frac{\left(2-\sqrt{3}\right)}{4-3} \\ =\frac{\left(2-\sqrt{3}\right)}{1} \\ =\left(2-\sqrt{3}\right) \end{gathered}$$

Question 19:

If $\sqrt{10}=3.162$, find the value of $\frac{1}{\sqrt{10}}$.

Answer 19:

The value of $\frac{1}{\sqrt{10}}=\frac{1}{\sqrt{10}}\times \frac{\sqrt{10}}{\sqrt{10}}=\frac{\sqrt{10}}{10}=\frac{3.162}{10}=0.3162$

Question 20:

Simplify ${\left(2\sqrt{5}+3\sqrt{2}\right)}^2$.

Answer 20:

$$\begin{gathered} {\left(2\sqrt{5}+3\sqrt{2}\right)}^2 \\ ={\left(2\sqrt{5}\right)}^2+{\left(3\sqrt{2}\right)}^2+2\left(2\sqrt{5}\right)\left(3\sqrt{2}\right) \left[{\left(a+b\right)}^2=a^2+b^2+2ab\right] \\ =20+18+12\sqrt{10} \\ =38+12\sqrt{10} \end{gathered}$$

Question 21:

If 10^x = 64, find the value of ${10}^{\left(\frac{x}{2}+1\right)}$.

Answer 21:

We have,

${10}^x=64$

Taking square root from both sides, we get

$$\begin{gathered} \sqrt{{10}^x}=\sqrt{64} \\ \Rightarrow {\left({10}^x\right)}^{\frac{1}{2}}=8 \\ \Rightarrow {10}^{\left(\frac{x}{2}\right)}=8 \end{gathered}$$

Multiplying both sides by 10, we get

$$\begin{gathered} {10}^{\left(\frac{x}{2}\right)}\times 10=8\times 10 \\ \therefore {10}^{\left(\frac{x}{2}+1\right)}=80 \end{gathered}$$

Question 22:

Evaluate $\frac{2^n+2^{n-1}}{2^{n+1}-2^n}$.

Answer 22:

$$\begin{gathered} \frac{2^n+2^{n-1}}{2^{n+1}-2^n} \\ =\frac{2^n\left(1+2^{-1}\right)}{2^n\left(2-1\right)} \\ =\frac{\left(1+{\displaystyle \frac{1}{2}}\right)}{1} \\ =\frac{2+1}{2} \\ =\frac{3}{2} \end{gathered}$$

Question 23:

Simplify ${\left[{\left\{{\left(256\right)}^{-\frac{1}{2}}\right\}}^{\frac{1}{4}}\right]}^2$.

Answer 23:

$$\begin{gathered} {\left[{\left\{{\left(256\right)}^{-\frac{1}{2}}\right\}}^{\frac{1}{4}}\right]}^2 \\ ={\left\{{\left(256\right)}^{-\frac{1}{2}}\right\}}^{\frac{1}{2}} \\ ={\left(256\right)}^{-\frac{1}{4}} \\ ={\left(4^4\right)}^{-\frac{1}{4}} \\ =4^{-1} \\ =\frac{1}{4} \end{gathered}$$