Answers
Answered by
1
Answer:
\Large A = P \left(1+\frac{r}{100}\right)^{n} => 7986 = 6000 \left(1+\frac{10}{100}\right)^{n} \)
=> \( \Large \frac{7986}{6000}=\frac{11}{10}^{n} => \frac{1331}{1000}= \left(\frac{11}{10}\right)^{n} => n = 3 years \)
Similar questions