14
Q:

# A group of students decided to collect as many paise from each member of group as is the number of members. If the total collection amounts to Rs. 59.29, the number of the member is the group is:

 A) 57 B) 67 C) 77 D) 87

Answer:   C) 77

Explanation:

Money collected = (59.29 x 100) paise = 5929 paise.

${\color{Blue}&space;\therefore&space;}$  Number of members = ${\color{Blue}&space;\sqrt{5929}}$ = 77.

Q:

If $\inline \fn_jvn \small x = 2+2^{\frac{2}{3}}+2^{\frac{1}{3}}$ then find the value of $\inline \fn_jvn \small x^{3}-6x^{2}+6x$ ?

 A) 1 B) 2 C) 3 D) 4

Answer & Explanation Answer: B) 2

Explanation:

Given that $\inline \fn_jvn \small x = 2+2^{\frac{2}{3}}+2^{\frac{1}{3}}$

So x - 2 = $\inline \fn_jvn \small 2^{\frac{2}{3}}+2^{\frac{1}{3}}$

Now cubing on both sides, we get

$\inline \fn_jvn \small (x-2)^{3}=(2^{\frac{2}{3}}+2^{\frac{1}{3}})^{3}$

=> $\inline \fn_jvn \small x^{3}-8-6x^{2}+12x=4+2+6(x-2)$

=> $\inline \fn_jvn \small x^{3}-6x^{2}+6x=-6+8 =2$.

Therefore, $\inline \fn_jvn \small x^{3}-6x^{2}+6x$ = 2.

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3 149
Q:

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

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

Answer & Explanation Answer: B) 4/3

Explanation:

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

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

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

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

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

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3 635
Q:

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

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

Answer & Explanation Answer: C) 14

Explanation:

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

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

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28 908
Q:

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

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

Answer & Explanation Answer: 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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12 901
Q:

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

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

Answer & Explanation Answer: C) a=11,b=-6

Explanation:

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

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

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