Q24 / 60 · KVPY 2012 (SA) · Mathematics

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

  • Correct answer
    A
    $1 / 2$
  • B
    1
  • C
    2
  • D
    Not determinable

Solution

$\begin{array}{l} \frac{a}{3}=\frac{a+b}{4}=\frac{a+b+c}{5}=\frac{a+b+c+d}{6}=k \\ a=3 k, b=k, c=k, d=k \\ \text { Hence:- } \frac{a}{b+2 c+3 d}=\frac{3 k}{k+2 k+3 k}=\frac{1}{2} \end{array}$

Hence, the correct answer is option: 'A'.

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