Question

If a, b, c, d are positive real number such that $\frac{a}{3}=\frac{a+b}{4}=\frac{a+b+c}{5}=\frac{a+b+c+d}{6}$, then $\frac{a}{b+2c+3d}$ is

• A. $\frac{1}{2}$
• B. 1
• C. 2
• D. not determinable

Answer 1

The correct option is A $\frac{1}{2}$

$a=3k$
$b=k$
$c=5k-4k=k$
$d=6k-5k=k$
$\frac{a}{b+2c+3d}=\frac{3k}{k+2k+3k}=\frac{1}{2}$