7

# $\inline \fn_jvn {\color{Black} \frac{3+\sqrt{6}}{5\sqrt{3}-2\sqrt{12}-\sqrt{32}+\sqrt{50}}=?}$

 A) 3 B) 3sqrt{2} C) 6 D) None of these

Explanation:

$\inline \fn_jvn {\color{Blue}Given\; exp.= \frac{3+\sqrt{6}}{5\sqrt{3}-4\sqrt{3}-4\sqrt{2}+5\sqrt{2}}=\frac{3+\sqrt{6}}{\sqrt{3}+\sqrt{2}}}$

$\inline \fn_jvn {\color{Blue}= \frac{3+\sqrt{6}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{3\sqrt{3}-3\sqrt{2}+3\sqrt{2}-2\sqrt{3}}{3-2}=\sqrt{3}}$

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• Related Questions

$\inline \fn_jvn {\color{Black}If\; a= \frac{\sqrt{5}+1}{\sqrt{5}-1}\; and\; b=\frac{\sqrt{5}-1}{\sqrt{5}+1},the\; value\; of\; \left [ \frac{a^{2}+ab+b^{2}}{a^{2}-ab+b^{2}} \right ]is:}$

 A) 3/4 B) 4/3 C) 3/5 D) 5/3

Explanation:

$\inline \fn_jvn {\color{Blue}a=\frac{(\sqrt{5}+1)}{(\sqrt{5}-1)}\times \frac{(\sqrt{5}+1)}{(\sqrt{5}+1)}=\frac{(\sqrt{5}+1)^2}{(5-1)}=\frac{5+1+2\sqrt{5}}{4}=\frac{3+\sqrt{5}}{2}}$

$\inline \fn_jvn {\color{Blue}b=\frac{(\sqrt{5}-1)}{(\sqrt{5}+1)}\times \frac{(\sqrt{5}-1)}{(\sqrt{5}-1)}=\frac{(\sqrt{5}-1)^2}{(5-1)}=\frac{5+1-2\sqrt{5}}{4}=\frac{3-\sqrt{5}}{2}}$

$\inline \fn_jvn {\color{Blue}\therefore\; \; a^{2}+b^{2}=\frac{(3+\sqrt{5})^2}{4}+\frac{(3-\sqrt{5})^2}{4}=\frac{(3+\sqrt{5})^2+(3-\sqrt{5})^2}{4}=\frac{2(9+5)}{4}=7}$

$\inline \fn_jvn {\color{Blue}Also,ab=\frac{(3+\sqrt{2})}{2}\times \frac{(3-\sqrt{2})}{2}=\frac{9-5}{4}=1}$

$\inline \fn_jvn {\color{Blue}\therefore \; \; \frac{a^{2}+ab+b^{2}}{a^{2}-ab+b^{2}}=\frac{(a^{2}+b^{2})+ab}{(a^{2}+b^{2})-ab}=\frac{7+1}{7-1}=\frac{8}{6}=\frac{4}{3}}$

Subject: Square Roots and Cube Roots - Quantitative Aptitude - Arithmetic Ability

3

$\inline \fn_jvn {\color{Black}If \; x=(7-4\sqrt{3}),then\; the \; value\; of\; \left ( x+\frac{1}{x} \right )is:}$

 A) 3sqrt{3} B) 8sqrt{3} C) 14 D) 14+8sqrt{3}

Explanation:

$\inline \fn_jvn {\color{Blue} x+\frac{1}{x} =(7-4\sqrt{3})+\frac{1}{(7-4\sqrt{3})}\times \frac{(7+4\sqrt{3})}{(7+4\sqrt{3})}=(7-4\sqrt{3})+\frac{(7+4\sqrt{3})}{(49-48)}}$

$\inline \fn_jvn {\color{Blue} =(7-4\sqrt{3})+(7+4\sqrt{3})=14}$

Subject: Square Roots and Cube Roots - Quantitative Aptitude - Arithmetic Ability

22

$\inline \fn_jvn {\color{Black}If\; \frac{5+2\sqrt{3}}{7+4\sqrt{3}}=a+b\sqrt{3},then:}$

 A) a=-11,b=-6 B) a=-11,b=6 C) a=11,b=-6 D) a=11,b=6

Explanation:

$\inline \fn_jvn {\color{Blue} a+b\sqrt{3}=\frac{(5+2\sqrt{3})}{(7+4\sqrt{3})}\times \frac{(7-4\sqrt{3})}{(7-4\sqrt{3})}=\frac{35-20\sqrt{3}+14\sqrt{3}-24}{(7)^{2}-(4\sqrt{3})^{2}}=\frac{11-6\sqrt{3}}{49-48}=11-6\sqrt{3}}$

$\inline \fn_jvn {\color{Blue} \therefore a=11,b=-6}$

Subject: Square Roots and Cube Roots - Quantitative Aptitude - Arithmetic Ability

3

$\inline \fn_jvn {\color{Black}\left [ \frac{3\sqrt{2}}{\sqrt{6}-\sqrt{3}}-\frac{4\sqrt{3}}{\sqrt{6}-\sqrt{2}} -\frac{6}{\sqrt{8}-\sqrt{12}}\right ]=?}$

 A) sqtr{3}-sqrt{2} B) sqtr{3}+sqrt{2} C) 5sqrt{3} D) 1

Explanation:

$\inline \fn_jvn {\color{Blue}Given \; exp.= \frac{3\sqrt{2}}{\sqrt{6}-\sqrt{3}}\times \frac{\sqrt{6}+\sqrt{3}}{\sqrt{6}+\sqrt{3}}-\frac{4\sqrt{3}}{\sqrt{6}-\sqrt{2}}\times \frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}}-\frac{6}{2(\sqrt{2}-\sqrt{3})}}$

$\inline \fn_jvn {\color{Blue}= \frac{3\sqrt{2}(\sqrt{6}+\sqrt{3})}{6-3}-\frac{4\sqrt{3}(\sqrt{6}+\sqrt{2})}{6-2}+\frac{3}{(\sqrt{3}-\sqrt{2})}\times \frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}+\sqrt{2})}}$

$\inline \fn_jvn {\color{Blue}=\sqrt{2}(\sqrt{6}+\sqrt{3})-\sqrt{3}(\sqrt{6}+\sqrt{2})+3(\sqrt{3}+\sqrt{2}) }$

$\inline \fn_jvn {\color{Blue}=\sqrt{12}+\sqrt{6} -\sqrt{18}-\sqrt{6}+3\sqrt{3}+3\sqrt{2}}$

$\inline \fn_jvn {\color{Blue}=2\sqrt{3}-3\sqrt{2}+3\sqrt{3}+3\sqrt{2}=5\sqrt{3}}$

Subject: Square Roots and Cube Roots - Quantitative Aptitude - Arithmetic Ability

1

$\inline \fn_jvn {\color{Black} If\; 3a=4b=6c\; and\; a+b+c=27\sqrt{29},then\sqrt{a^{2}+b^{2}+c^{2}}\; is:}$

 A) 81 B) 83 C) 87 D) None of these

Explanation:

$\inline \fn_jvn {\color{Blue} 4b=6c \Rightarrow b=\frac{3}{2}c \; \; and\; \; 3a=4b\Rightarrow a=\frac{4}{3}b=\frac{4}{3}\left ( \frac{3}{2}c \right )=2c}$

$\inline \fn_jvn {\color{Blue} a+b+c=27\sqrt{29}\Rightarrow 2c+\frac{3}{2}c+c=27\sqrt{29}\Rightarrow \frac{9}{2}c=27\sqrt{29}\Rightarrow c=6\sqrt{29}}$

$\inline \fn_jvn {\color{Blue} \therefore \sqrt{a^{2}+b^{2}+c^{2}}=\sqrt{(a+b+c)^{2}-2(ab+bc+ca)}}$

$\inline \fn_jvn {\color{Blue} =\sqrt{(27\sqrt{29})^{2}-2\left ( 2c\times \frac{3}{2} c+\frac{3}{2}c\times c+c\times 2c\right )}}$

$\inline \fn_jvn {\color{Blue} =\sqrt{(729\times 29)-2\left ( 3c^{2}+\frac{3}{2}c^{2} +2c^{2}\right )}=\sqrt{(729\times 29)-2\times \frac{13}{2}c^{2}}}$

$\inline \fn_jvn {\color{Blue} =\sqrt{(729\times 29)-13\times (6\sqrt{29})^2}=\sqrt{29(729-468)}}$

$\inline \fn_jvn {\color{Blue} =\sqrt{(29\times 261)}=\sqrt{29\times 29\times 9}=29\times 3=87}$