### Number Systems - Test Papers

CBSE Test Paper 01

CH-1 Number Systems

1. $\sqrt{12}×\sqrt{15}$ =

1. 5

2. $5\sqrt{6}$

3. $6\sqrt{5}$

4. 6

2. Which of the following is equal to ‘x’?

1. $\sqrt[12]{{\left({x}^{4}\right)}^{\frac{1}{3}}}$

2. ${x}^{\frac{12}{7}}×{x}^{\frac{7}{12}}$

3. ${\left(\sqrt{{x}^{3}}\right)}^{\frac{2}{3}}$

4. ${x}^{\frac{12}{7}}-{x}^{\frac{5}{7}}$

3. An irrational number between 5 and 6 is

1. none of these

2. $\sqrt{5+6}$

3. $\sqrt{5-6}$

4. $\sqrt{5×6}$

4. $2\sqrt{3}+\sqrt{3}$ is equal to

1. $2\sqrt{6}$

2. $3\sqrt{6}$

3. 3

4. $3\sqrt{3}$

5. If $x=3+\sqrt{8}$ , then the value of $\left({x}^{2}+\frac{1}{{x}^{2}}\right)$ is

1. 32

2. 34

3. 6

4. 12

6. Fill in the blanks:

The two rational number which are their own multiplicative inverses are ________.

7. Fill in the blanks:

The value of $\frac{2}{\sqrt{5}-\sqrt{3}}$ is ________.

8. Find:${125}^{\frac{-1}{3}}$

9. Identify whether $\sqrt{45}$ is rational number or irrational number.

10. Write the decimal form and state its kind of decimal expansion.$\frac{36}{100}$

11. Simplify the expression: $\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)$

12. Recall, $\pi$ is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, $\pi =\frac{c}{d}$ This seems to contradict the fact that is irrational. How will you resolve this contradiction?

13. Find three different irrational numbers between the rational numbers  $\frac{5}{9}$ and $\frac{9}{11}$.

14. Rationalize a denominator of the following: $\frac{\sqrt{3}-1}{\sqrt{3}+1}$

15. Prove that: $\frac{{3}^{-3}×{6}^{2}×\sqrt{98}}{{5}^{2}×\sqrt[3]{1/25}×\left(15{\right)}^{-4/3}×{3}^{1/3}}$ = 28$\sqrt{2}$

CBSE Test Paper 01
CH-1 Number Systems

Solution

1. (c) $6\sqrt{5}$
Explanation:
$\sqrt{12}=\sqrt{3×{2}^{2}}=2\sqrt{3}$
and $\sqrt{15}=\sqrt{5}×\sqrt{3}$
so, $\sqrt{12}×\sqrt{15}=2\sqrt{3}×\sqrt{3}×\sqrt{5}$
$=2×3\sqrt{5}$$=6\sqrt{5}$
2. (c) ${\left(\sqrt{{x}^{3}}\right)}^{\frac{2}{3}}$
Explanation:
$\left(\sqrt{{x}^{3}}{\right)}^{\frac{2}{3}}$
$={\left({x}^{\frac{3}{2}}\right)}^{\frac{2}{3}}$

=x
3. (d) $\sqrt{5×6}$
Explanation:
We know that,
If a and b are two distinct positive rational numbers such that ab is not a perfect square of a rational number,then

$\sqrt{ab}$ is an irrational number lying between a and b,

Here also we have 5 and 6 two distinct rational numbers and 5×6=30 is not a perfect square,
So irrational number between 5 and 6 =$\sqrt{5×6}$

4. (d) $3\sqrt{3}$
Explanation:
$2\sqrt{3}+\sqrt{3}$
$=\sqrt{3}\left(2+1\right)$
$=3\sqrt{3}$
5. (b) 34
Explanation:
given $x=\left(3+\sqrt{8}\right)$
$\frac{1}{x}=\frac{1}{\left(3+\sqrt{8}\right)}=\frac{1}{\left(3+\sqrt{8}\right)}×\frac{\left(3-\sqrt{8}\right)}{\left(3-\sqrt{8}\right)}$
$=\frac{\left(3-\sqrt{8}\right)}{\left({3}^{2}-\left(\sqrt{8}{\right)}^{2}\right)}=\frac{\left(3+\sqrt{8}\right)}{\left(9-8\right)}=\left(3-\sqrt{8}\right)$
$\left(x+\frac{1}{x}\right)=\left(3+\sqrt{8}\right)+\left(3-\sqrt{8}\right)=6$
$⇒{\left(x+\frac{1}{x}\right)}^{2}={6}^{2}=36$
$⇒\left({x}^{2}+\frac{1}{{x}^{2}}\right)+2×x×\frac{1}{x}=36$
$⇒\left({x}^{2}+\frac{1}{{x}^{2}}\right)+2=36$
$⇒\left({x}^{2}+\frac{1}{{x}^{2}}\right)=36-2=34$
6. 1, -1

7. $\sqrt{5}+\sqrt{3}$

8. ${125}^{\frac{-1}{3}}$
We know that ${a}^{-n}=\frac{1}{{a}^{n}}$
We conclude that ${125}^{\frac{-1}{3}}$ can also be written as $\frac{1}{{125}^{\frac{1}{3}}}$,or ${\left(\frac{1}{125}\right)}^{\frac{1}{3}}$
We know that
We know that ${\left(\frac{1}{125}\right)}^{\frac{1}{3}}$ can also be written as
$\sqrt[3]{\left(\frac{1}{125}\right)}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\sqrt[3]{\left(\frac{1}{5}×\frac{1}{5}×\frac{1}{5}\right)}=\frac{1}{5}$.
Therefore, the value of ${125}^{\frac{-1}{3}}$ will be $\frac{1}{5}$.

9. We have,
$\sqrt{45}$ = $\sqrt{9×5}$ = 3$\sqrt{5}$
Since 3 is a rational number and $\sqrt{5}$ is an irrational number. Therefore, the product 3$\sqrt{5}$ = $\sqrt{45}$ is an irrational number.

10. $\frac{36}{100}$ = 0.36
The decimal expansion is terminating.

11. $\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)$