Question

If $63.9805=6A+\frac{3}{B}+9C+\frac{8}{D}+5E$, then what is the value of $4A+7B+6C+D+3E$?

• A. $47.603$
• B. $4.7603$
• C. $147.6003$
• D. $47.6003$

Answer 1

The correct option is C $147.6003$

$63.9805=6A+\frac{3}{B}+9C+\frac{8}{D}+5E$

$\Rightarrow$ $63.9805=6\times 10+3\times 1+9\times \frac{1}{10}+8\times \frac{1}{100}+5\times \frac{1}{10000}$

$\therefore$ $A=10,B=1,C=\frac{1}{10},D=100,E=\frac{1}{10000}$

$\Rightarrow$ $4A+7B+6C+D+3E$

$\Rightarrow$ $4\times 10+7\times 1+6\times \frac{1}{10}+100+3\times \frac{1}{10000}$

$\Rightarrow$ $4+7+0.6+100+0.0003$

$\Rightarrow$ $\mathrm{147.6003.}$

$\therefore$ $4A+7B+6C+D+3E=147.6003$