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: $\frac{7}{2}$
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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