The vulgar fraction of 0.3939 is [SSC-CHSL (10+2) Exam, 2015]
(a)$\frac{15}{33}$
(b)$\frac{11}{39}$
(c)$\frac{17}{39}$
(d)$\frac{13}{33}$
Answer
Answer (as printed): D
Explanation
The given expression can be written in this form also $$N=0.3939$$ Multiply equation (i) with 100 on both sides. $$100 N=39.39$$ Subtracting equation (i) from (ii) we get $$\begin{aligned} & \Rightarrow 100 N-N=39 . \overline{39}-0 . \overline{39} \\ & 99 N=39 \\ & \Rightarrow N=\frac{39}{99}=\frac{13}{33} \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.