Free Practice Questions for Indices-and-surds in Maths

LCM of 2,3,5 and 7 = 210

$\begin{aligned} 5^{\frac{1}{2}} &=\frac{105}{5^{210}}=\left(5^{105}\right) \frac{1}{210} \\ 4^{\frac{1}{3}} &=4^{70 / 210}=\left(4^{70}\right) \frac{1}{210} \\ 2^{\frac{1}{5}} &=\frac{42}{2^{210}}=\left(2^{42}\right) \frac{1}{210} \\ 3^{\frac{1}{7}} &=\frac{30}{3^{210}}=\left(3^{30}\right) \frac{1}{210} \end{aligned}$

∴ The largest number

$\frac{1}{5^{2}}$

5 is the largest and its order is smallest.

∴ Largest number $=\sqrt{5}$

$\begin{aligned} & \frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}+\frac{\sqrt{3}+1}{\sqrt{3}-1} \\=& \frac{(2+\sqrt{3})^{2}+(2-\sqrt{3})^{2}}{(2-\sqrt{3})(2+\sqrt{3})}+\frac{(\sqrt{3}+1)(\sqrt{3}+1)}{(\sqrt{3})^{2}-(1)^{2}} \\=& \frac{4+3+4 \sqrt{3}+4+3-4 \sqrt{3}}{(4-3)}+\frac{(3+1+2 \sqrt{3})}{3-1} \\=& \frac{14}{1}+\frac{4+2 \sqrt{3}}{2} \\=& 14+(2+\sqrt{3}) \\=& 16+\sqrt{3} \end{aligned}$

$=a+b \sqrt{3}$

$5^{x-2 y}=625$

$x-2 y=4 $.......(i)

$7^{x-3 y}=343$
Similarly $7^{3}=343$,
x − 3y = 3........(ii)
On solving equations (i) and (ii), we get x = 6 and y = 1
x − y = 5

ATQ,

$\frac{\sqrt{192}+\sqrt{320}}{\sqrt{108}+\sqrt{180}}$
$\frac{8 \sqrt{3}+8 \sqrt{5}}{6 \sqrt{3}+6 \sqrt{5}}=\frac{8(\sqrt{3}+\sqrt{5})}{6(\sqrt{3}+\sqrt{5})}=\frac{8}{6}=\frac{4}{3}$

$\Rightarrow 24^{\sqrt[3]{a}}+10^{\sqrt[3]{a}}=26^{\sqrt[3]{a}}$ (given)
⇒ From the observation, it can be seen that 24,10,26 are Pythagoras triplets.
$\Rightarrow 24^{2}+10^{2}=26^{2}$
$\Rightarrow 3 \sqrt{a}=2$
$\Rightarrow a=8$
$\Rightarrow a^{2}-a=64 -8$

$\Rightarrow56$

$(x)^{x^{1.5}}=x^{1.5 x}$
$x^{1.5}=1.5 x$
$\left(x^{1.5}\right)^{2}=(1.5 x)^{2}$
$x^{3}=2.25 x^{2}$
$x=2.25$
$\sqrt{x}=1.5$
$\sqrt{1.5-0.5}=\sqrt{1}=1$

Given, $a=3, b=12, c=10$
$\sqrt{(} 13+a)+\sqrt{(} 112-b)+\sqrt{(} c-1)$
$=\sqrt{(}(13+3)+\sqrt{(} 112-12)+\sqrt{(10-1)}$
$=\sqrt{1} 6+\sqrt{1} 00+\sqrt{9}$
$=4+10+3$
$=17$

$1.66666$$ =1.\overline{6}$
$=1+\frac{6}{9}$
$=1 \frac{2}{3}$
$=\frac{5}{3}$

$(\sqrt{3})^{7} \div(\sqrt{3})^{5}=3^{p}$

$(\sqrt{3})^{2}=3^{p}$

$\mathrm{p}=1$

$\begin{aligned} 22 \sqrt{45} &=22 \times \sqrt{9 \times 5} \\ &=22 \times 3 \sqrt{5} \\ &=66 \sqrt{5} \end{aligned}$
So the multiplying factor is $\sqrt{5}$.

If $\sqrt{72-\sqrt{72-\sqrt{72 \ldots \ldots \infty}}}$
The factor of $72=8 \times 9$
Smaller number is the value of $\mathrm{p}$ i.e. 8
$
\begin{aligned}
\mathrm{p}^{2}-4 &=8^{2}-4 \\
&=60
\end{aligned}
$

$\begin{aligned} \sqrt{729}-\sqrt{\frac{2704}{169}}+\sqrt{\frac{2025}{81}} &=27-\frac{52}{13}+\frac{45}{9} \\ &=27-4+5 \\ &=28 \end{aligned}$

ATQ,
$\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\cdots+\frac{1}{n(n+1)}=\frac{99}{100}$
We know that
$\Rightarrow \frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\cdots+\frac{1}{n(n+1)}=\frac{n}{n+1}$
$\Rightarrow \frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 7}+\cdots+\frac{1}{99 \times 100}=\frac{99}{100}$
$n=99$

ATQ,
$\Rightarrow \quad 15-4 \sqrt{14}$
$\Rightarrow 8+7-4 \sqrt{14}$
$\Rightarrow(\sqrt{8})^{2}+(\sqrt{7})^{2}-2 \times \sqrt{8} \times \sqrt{7} .$
$\Rightarrow\left\{(2 \sqrt{2}){}-(\sqrt{7})\right\}^{2}$
Square root of $(2 \sqrt{2}-\sqrt{7})^{2}$ is
$
\Rightarrow \quad(2 \sqrt{2}-\sqrt{7})
$

$(x+y)\left(x^{2}+y^{2}-x y\right)=x^{3}+y^{3}$
$(\sqrt[3]{3.5}+\sqrt[3]{2.5})\left\{(\sqrt[3]{3.5})^{2}-\sqrt[3]{8.75}+(\sqrt[3]{2.5})^{2}\right\}$
$x=\sqrt[3]{3.5}$
$y=\sqrt[3]{2.5}$
$=(\sqrt[3]{3.5})^{3}+(\sqrt[3]{2.5})^{3}$
$=3.5+2.5$
$=6$

$N=\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}$
$\frac{1}{N}=\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$
$N-\frac{1}{N}=\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}-\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$
$\Rightarrow \frac{(\sqrt{7}+\sqrt{5})(\sqrt{7}+\sqrt{5})-(\sqrt{7}-\sqrt{5})(\sqrt{7}-\sqrt{5})}{(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})}$
$\Rightarrow \frac{(\sqrt{7}+\sqrt{5})^{2}-(\sqrt{7}-\sqrt{5})^{2}}{(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})}$
$=\frac{(12+2 \sqrt{35})-(12-2 \sqrt{35})}{2}$
$=\frac{4 \sqrt{35}}{2}=2 \sqrt{35}$