Question : If $7\mathrm{b}-\frac{1}{4 \mathrm{b}}=7$, what is the value of $16 \mathrm{b}^2+\frac{1}{49 \mathrm{b}^2}$?
Option 1: $\frac{80}{49}$
Option 2: $\frac{104}{7}$
Option 3: $\frac{120}{7}$
Option 4: $\frac{7}{2}$
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Correct Answer: $\frac{120}{7}$
Solution : $7\mathrm{b}-\frac{1}{4\mathrm{b}}=7$ Multiplying both sides by $\frac{4}{7}$. $4\mathrm{b}-\frac{1}{7\mathrm{b}}=4$ Squaring both sides, we get $(4\mathrm{b}-\frac{1}{7\mathrm{b}})^2=4^2$ ⇒ $16\mathrm{b^2}+\frac{1}{49\mathrm{b^2}}-\frac{8}{7 }=16$ ⇒ $16\mathrm{b}^2+\frac{1}{49\mathrm{b}^2}=\frac{120}{7}$ Hence, the correct answer is $\frac{120}{7}$.
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Question : If $\mathrm{p}=7+4 \sqrt{3}$, then what is the value of $\frac{\mathrm{p}^6+\mathrm{p}^4+\mathrm{p}^2+1}{\mathrm{p}^3}$?
Option 1: 2617
Option 2: 2167
Option 3: 2716
Option 4: 2176
Question : If $\mathrm{y}+\frac{1}{\mathrm{y}}=3$, then what is the value of $\frac{1}{\mathrm{y}^3}+\mathrm{y}^3+2$ ?
Option 1: 24
Option 2: 18
Option 3: 20
Option 4: 29
Question : If $\mathrm{k}+\frac{1}{\mathrm{k}}=4$, then what is the value of $\mathrm{k}^4+\frac{1}{\mathrm{k}^4}$ ?
Option 1: 410
Option 2: 192
Option 3: 212
Option 4: 194
Question : If $\mathrm{p}=\frac{\sqrt{2}+1}{\sqrt{2}-1}$ and $\mathrm{q}=\frac{\sqrt{2}-1}{\sqrt{2}+1}$ then, find the value of $\frac{\mathrm{p}^2}{\mathrm{q}}+\frac{\mathrm{q}^2}{\mathrm{p}}$.
Option 1: 200
Option 2: 196
Option 3: 198
Option 4: 188
Question : $\triangle \mathrm{XYZ} \sim \triangle \mathrm{GST}$ and $\mathrm{XY}: \mathrm{GS}=2: 3, \mathrm{XV}$ is the median to the side $\mathrm{YZ}$, and $\mathrm{GD}$ is the median to the side ST. The value of $\left(\frac{\mathrm{YV}}{\mathrm{SD}}\right)^2$ is:
Option 1: $\frac{4}{9}$
Option 2: $\frac{3}{5}$
Option 3: $\frac{1}{4}$
Option 4: $\frac{2}{3}$
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