Question

If $47.2506$ $=4A+\frac{7}{B}+2C+\frac{5}{D}+6E,$ then the value of $5A+3B+6C+D+3E$ is:

• A. $63.6003$
• B. $53.603$
• C. $153.6003$
• D. $213.0003$

Answer 1

The correct option is C $153.6003$

Given $47.2506=4A+\frac{7}{B}+2C+\frac{5}{D}+6E..............\left(1\right)$

Now write $47.2506=40+7+0.2+0.05+\mathrm{0.0006...........................}\left(2\right)$

comparing (1) and (2) we get

$4A=10\Rightarrow A=10$

$\frac{7}{B}=7\Rightarrow B=1and2C=0.2\Rightarrow C=0.1$

$\frac{5}{D}=0.05\Rightarrow D=\frac{5}{0.05}$$=100$

$6E=0.0006\Rightarrow E=0.0001$

So velue of $5A+3B+6C+D+3E$=$(5\times 10)+(3\times 1)+(6\times 0.1)+100+(3\times 0.0001)$

$=50+3+0.6+100+0.0003=153.6003$