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## Algebra: Basics Test-5

Congratulations - you have completed Algebra: Basics Test-5.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%%
 Question 1
$\displaystyle f\,\,\,\frac{1}{4}\times \frac{2}{6}\times \frac{3}{8}\times \frac{4}{10}\times \frac{5}{12}\times ..........\times \frac{31}{64}=\frac{1}{{{2}^{t}}},$
the value of t is
 A 38 B 40 C 36 D 58
Question 1 Explanation:
$\displaystyle \begin{array}{l}=\,\,\,\frac{1}{4}\times \frac{2}{6}\times \frac{3}{8}\times \frac{4}{10}\times \frac{5}{12}\times ..........\times \frac{31}{64}=\frac{1}{{{2}^{p}}}\\\Rightarrow \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times .....to\,\,30\,\,terms\,\,\times \frac{1}{64}=\frac{1}{{{2}^{p}}}\\\Rightarrow \frac{1}{{{2}^{30}}}\times \frac{1}{{{2}^{6}}}=\frac{1}{{{2}^{p}}}\\\Rightarrow \frac{1}{{{2}^{36}}}=\frac{1}{{{2}^{p}}}\\\Rightarrow p=36\end{array}$
 Question 2
The value of is $\displaystyle \frac{{{\left( 243 \right)}^{\frac{k}{5}}}{{.3}^{2k+1}}}{{{9}^{k}}{{.3}^{k-1}}}$
 A 11 B 9 C 7 D 3n
Question 2 Explanation:
$\displaystyle \begin{array}{l}=\frac{{{\left( 243 \right)}^{\frac{k}{5}}}\times {{3}^{2k+1}}}{{{9}^{k}}\times {{3}^{k-1}}}\\=\frac{{{\left( {{3}^{5}} \right)}^{\frac{k}{5}}}\times {{3}^{2k+1}}}{{{\left( {{3}^{2}} \right)}^{k}}\times {{3}^{k-1}}}\\=\frac{{{3}^{k}}\times {{3}^{2k+1}}}{{{3}^{2k}}\times {{3}^{k-1}}}\\=\frac{{{3}^{k+2k+1}}}{{{3}^{2k+k-1}}}\\=\frac{{{3}^{3k+1}}}{{{3}^{3k-1}}}\\={{3}^{3k+1-3k+1}}\\={{3}^{2}}\\=9\end{array}$
 Question 3
$\displaystyle If\,\,a=0.5\,\,\,and\,\,b=0.2,\,\,\,then\,\,\,\,value\,\,\,of\,\,\,\sqrt{0.6}\,\,\times \,\,{{\left( 3b \right)}^{a}}\,\,\,is\,\,\,equal\,\,\,to$
 A 0.9 B 0.6 C 0.5 D 1.3
Question 3 Explanation:
$\displaystyle \begin{array}{l}a=0.5\,\,\,and\,\,b=0.2\left( Given \right)\\Therefore\,\,\sqrt{0.6}\times {{\left( 3b \right)}^{a}}\\=\sqrt{0.6}\times {{\left( 3\times 0.2 \right)}^{0.5}}\\=\sqrt{0.6}\times {{\left( 0.6 \right)}^{\frac{1}{2}}}\\=\sqrt{0.6\times .06}=0.6\end{array}$
 Question 4
$\displaystyle if\,\,\,{{s}^{s\sqrt{s}}}={{\left( s\sqrt{s} \right)}^{s}},\,\,then\,\,s\,\,equals$
 A $\displaystyle \frac{2}{9}$ B $\displaystyle \frac{1}{3}$ C $\displaystyle \frac{9}{4}$ D $\displaystyle \frac{3}{2}$
Question 4 Explanation:
$\displaystyle \begin{array}{l}{{s}^{s\sqrt{s}}}={{\left( s\sqrt{s} \right)}^{s}}\\\Rightarrow {{s}^{s.s\frac{1}{2}}}={{\left( s\times {{s}^{\frac{1}{2}}} \right)}^{s}}\\\Rightarrow {{s}^{1+\frac{1}{2}}}={{\left( {{s}^{1+\frac{1}{2}}} \right)}^{s}}\\{{s}^{{{s}^{\frac{3}{2}}}}}={{\left( {{s}^{\frac{3}{2}}} \right)}^{s}}={{s}^{\frac{3s}{2}}}\\\Rightarrow {{s}^{\frac{3}{2}}}=\frac{3s}{2}\Rightarrow {{s}^{\frac{3}{2}}}-\frac{3s}{2}=0\\\Rightarrow s\left( {{s}^{\frac{1}{2}}}-\frac{3}{2} \right)=0\\\Rightarrow s=0\,\,\,or\,\,\,{{s}^{\frac{1}{2}}}=\frac{3}{2}\\\Rightarrow s={{\left( \frac{3}{2} \right)}^{2}}=\frac{9}{4}\\s=0\,\,given\,\,\,in\det er\min ate\,\,\,\,value.\\Therefore\,\,\,s=\frac{9}{4}\end{array}$
 Question 5
If f =7, g=5 and h=3, then the value of f2+ g2+ h2-fg-gh-hf is:
 A 14 B 12 C -12 D 13
Question 5 Explanation:
$\displaystyle \begin{array}{l}{{f}^{2}}+{{g}^{2}}+{{h}^{2}}-fg-gh-hf\\=\frac{1}{2}\left[ {{\left( f-g \right)}^{2}}+{{\left( g-h \right)}^{2}}+{{\left( h-f \right)}^{2}} \right]\\=\frac{1}{2}\left[ {{\left( 7-5 \right)}^{2}}+{{\left( 5-3 \right)}^{2}}+{{\left( 3-7 \right)}^{2}} \right]\\=\frac{1}{2}\left( 4+4+16 \right)\\=\frac{1}{2}\times 24=12\end{array}$
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