In what time will Rs. 3300 becomes Rs. 3399 at 6% per annum interest compounded half-yearly?

16 months
21 year
3 $1 \frac{1}{2}$ $1\frac{1}{2}$ years
43 months

Answer:- 1

Explanation:-

Solution:

\begin{aligned} P = R s . 3300 \\ A = R s . 3399 \\ R = 6 % per annum \\ Let the time be n years . \\ Compound interest is taken half - yearly . \\ A = P \times {[1 + (\frac{R}{2} \times 100)]}^{2 n} \\ 3399 = 3300 {(1 + \frac{3}{100})}^{2 n} \\ {(1.03)}^{2 n} = \frac{3399}{3300} \\ {(1.03)}^{2 n} = {(1.03)}^{1} \\ T h u s, 2 n = 1 y e a r \\ n = \frac{1}{2} year = 6 months \end{aligned}

$\eqalign{ & P = Rs.\,3300 \cr & A = Rs.\,3399 \cr & R = 6\% \,{\text{per}}\,{\text{annum}} \cr & {\text{Let}}\,{\text{the}}\,{\text{time}}\,{\text{be}}\,{\text{n}}\,{\text{years}}{\text{.}} \cr & {\text{Compound}}\,{\text{interest}}\,{\text{is}}\,{\text{taken}}\,{\text{half - yearly}}. \cr & A = P \times {\left[ {1 + \left( {\frac{R}{2} \times 100} \right)} \right]^{2n}} \cr & 3399 = 3300{\left( {1 + \frac{3}{{100}}} \right)^{2n}} \cr & {\left( {1.03} \right)^{2n}} = \frac{{3399}}{{3300}} \cr & {\left( {1.03} \right)^{2n}} = {\left( {1.03} \right)^1} \cr & Thus,\,2n = 1\,year \cr & n = \frac{1}{2}{\text{year}} = 6\,{\text{months}} \cr}$