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## Algebra: Basics Test-8

Congratulations - you have completed Algebra: Basics Test-8. You scored %%SCORE%% out of %%TOTAL%%. You correct answer percentage: %%PERCENTAGE%% . Your performance has been rated as %%RATING%%
 Question 1
$\displaystyle If\,\,\,\frac{\sqrt{7}-2}{\sqrt{7}+2}=p\sqrt{7}+q$
then the value of p is
 A $\displaystyle \frac{5}{3}$ B $\displaystyle -\frac{11}{3}$ C $\displaystyle \frac{11}{3}$ D $\displaystyle \frac{-4\sqrt{7}}{3}$
Question 1 Explanation:
$\displaystyle \begin{array}{l}\frac{\sqrt{7}-2}{\sqrt{7}+2}=\frac{\sqrt{7}-2}{\sqrt{7}+2}\times \frac{\sqrt{7}-2}{\sqrt{7}-2}\\=\frac{{{\left( \sqrt{7}-2 \right)}^{2}}}{7-4}=\frac{7+4-4\sqrt{7}}{3}\\=\frac{11}{3}-\frac{4\sqrt{7}}{3}\\Therefore\,\,\,\,\frac{\sqrt{7}-2}{\sqrt{7}+2}=p\sqrt{7}+q\\\Rightarrow \,\frac{11}{3}-\frac{4}{3}\sqrt{7}=a\sqrt{7}+b\\Clearly,\\p=-\frac{4}{3}\,\,and\,\,\,q=\frac{11}{3}\end{array}$
 Question 2
If (125)p =3125, then the value of p is
 A $\displaystyle 1\frac{1}{5}$ B $\displaystyle 2\frac{3}{5}$ C $\displaystyle 1\frac{2}{3}$ D $\displaystyle 4\frac{5}{7}$
Question 2 Explanation:
$\displaystyle \begin{array}{l}Therefore\,\,\\{{\left( 125 \right)}^{p}}=3125\\\Rightarrow {{\left( {{5}^{3}} \right)}^{p}}={{5}^{5}}\Rightarrow {{5}^{3p}}={{5}^{5}}\\\Rightarrow 3p=5\\\Rightarrow p=\frac{5}{3}\end{array}$
 Question 3
$\displaystyle If\,\,\,{{5}^{\sqrt{k}}}+{{12}^{\sqrt{k}}}={{13}^{\sqrt{k}}},\,\,then\,\,\,k\,\,\,\,is\,\,\,equal\,\,\,to$
 A $\displaystyle 6\frac{5}{4}$ B 4 C $\displaystyle 3\frac{5}{4}$ D 6
Question 3 Explanation:
$\displaystyle \begin{array}{l}{{5}^{\sqrt{k}}}+{{12}^{\sqrt{k}}}={{13}^{\sqrt{k}}}\\We\,\,\,know\,\,\,\,that\,\,\,{{5}^{2}}+{{12}^{2}}={{13}^{2}}\\Therefore\,\,\,\sqrt{k}=2\Rightarrow k={{2}^{2}}=4\end{array}$
 Question 4
If $\displaystyle {{2}^{2p-q}}=16\,\,\,\,and\,\,\,{{2}^{p+q}}=32\,,\,\,\,the\,\,\,\,value\,\,\,of\,\,\,pq\,\,\,is$
 A 8 B 6 C 4 D 2
Question 4 Explanation:
$\displaystyle \begin{array}{l}{{2}^{2p-q}}=16={{2}^{4}}\\\Rightarrow 2p-q=4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,............\left( i \right)\\{{2}^{p+q}}=32={{2}^{5}}\\\Rightarrow p+q=5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...........\left( ii \right)\\on\,\,\,\,adding\,\,\,\,equations\,\,\,\,(i)\,\,\,and\,\,\left( ii \right),\\q=5-p=5-3=2\\Therefore\,\,\,pq=3\times 2=6\end{array}$
 Question 5
If $\displaystyle {{\left( \frac{3}{5} \right)}^{3}}{{\left( \frac{3}{5} \right)}^{-6}}={{\left( \frac{3}{5} \right)}^{2a-1}}$
 A -5 B -4 C -3 D 1
Question 5 Explanation:
$\displaystyle \begin{array}{l}{{\left( \frac{3}{5} \right)}^{3}}{{\left( \frac{3}{5} \right)}^{-6}}={{\left( \frac{3}{5} \right)}^{2a-1}}\\\Rightarrow {{\left( \frac{3}{5} \right)}^{3}}{{\left( \frac{3}{5} \right)}^{-3}}{{\left( \frac{3}{5} \right)}^{-3}}={{\left( \frac{3}{5} \right)}^{2a-1}}\\\Rightarrow {{\left( \frac{3}{6} \right)}^{0}}{{\left( \frac{3}{5} \right)}^{-3}}={{\left( \frac{3}{5} \right)}^{2a-1}}\\\Rightarrow 2a-1=-3\\\Rightarrow 2a=-3+1=-2\\\Rightarrow x=-1\end{array}$
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