Simplification-Test 6 (Based on Surds and Indices/Percentages)

Answer: Option E

Explanation

$\begin{array}{l}\text{Here, }\\ \Rightarrow {\left(3^{?}\right)}^{?}\text{= 19683}\\ \\ \text{Putting x at the place of (?)}\\ \\ {\text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3}}^{x\times x}{\text{=(27)}}^{\text{3}}\text{ [}\because {\text{19683=27}}^{\text{3}}\text{]}\\ \\ \Rightarrow {\text{3}}^{x^{\text{2}}}{\text{= 3}}^{\text{9}}\Rightarrow x^{\text{2}}\text{= 9}\\ \\ \Rightarrow x\text{\hspace{0.17em}=}\sqrt{\text{9}}\text{ =3}\end{array}$

Hence, Option E is correct.

Answer: Option C

Explanation

$\begin{array}{l}\text{Here, }\\ \Rightarrow {{\left(2\right)}^{3\times }}^{1.3}\times {\left(2^2\right)}^{x0.6}\times {\left(2\right)}^{4\times 0.2}=2^{?}\\ \\ \Rightarrow 2^{3.9+1.2+0.8}=2^{?}\\ \\ [\because m^x\times m^y\times m^z=m^{x+y+z}]\\ \\ \Rightarrow 2^{5.9}=2^{?}\\ \\ \Rightarrow ?=5.9\end{array}$

Hence, option C is correct.

Answer: Option B

Explanation

Here,

$\begin{array}{l}\Rightarrow 7^{2.3}\times {\left(7\right)}^{2\times 4.7}\times {(7\times 9)}^{3.4}\times {\left(9^2\right)}^{5.85}={63}^{?}\\ \\ \Rightarrow {\left(7\right)}^{\text{2.3+9.4 +3.4}}\times {\left(\text{9}\right)}^{3.4+11.70}={63}^{?}\\ \\ \Rightarrow 7^{\text{1.51}}\times {\text{9}}^{\text{15.1}}\text{=}{\left(63\right)}^{\text{?}}\\ \\ \Rightarrow {\left(63\right)}^{15.1}={\left(63\right)}^{?}\\ \\ \Rightarrow ?=15.1\end{array}$

Hence, option B is correct.

Answer: Option E

Explanation

$\begin{array}{l}\text{Here,\hspace{0.17em}}\\ \Rightarrow {\text{27}}^{\text{1.5}}\times {27}^{3.5}{\text{=27}}^{\text{?}}\\ \\ [\because m^x\times m^y=m^{x+y}]\\ \\ \Rightarrow {27}^5={27}^{?}\\ \\ \Rightarrow ?=5\end{array}$

Hence, Option E is correct.

Answer: Option D

Explanation

$\begin{array}{l}\text{Here,}\\ \\ \Rightarrow \sqrt{\text{(2704}}\text{ + 78)}÷\text{13+ ?\% of 70 = 31}\\ \\ \Rightarrow \text{(52+ 78)}÷\text{13+}\frac{70x}{100}\text{= 31}\\ \\ \Rightarrow \text{130}÷\text{13+}\frac{70x}{100}\text{= 31}\\ \\ \Rightarrow \text{10+}\frac{7}{10}x\text{= 31}\\ \\ \Rightarrow \frac{7}{10}x\text{\hspace{0.17em}= 31}-\text{10}\\ \\ \Rightarrow x=\frac{\text{21}\times \text{10}}{7}\\ \\ \text{= 30.}\end{array}$

Hence, Option D is correct.