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6.Interest
easy
The least number of completed years in which a sum of money put out at $20 \%$ $C.I.$ will be more than doubled is
A
$3$
B
$4$
C
$5$
D
$6$
Solution
(b) $x\left(1+\frac{20}{100}\right)^{n}>2 x$ or $\left(\frac{6}{5}\right)^{n}>2$
Now, $\quad\left(\frac{6}{5} \times \frac{6}{5} \times \frac{6}{5} \times \frac{6}{5}\right)>2 \quad \therefore \quad n=4$ years.
Standard 13
Quantitative Aptitude
