A sum of ₹ 12,500 is deposited for 1½ years, compounded half yearly. It amounts to ₹ 13,000/- at the end of first half year. Find: (i) The rate of interest (ii) The final amount. Give your answer correct to the nearest rupee.
Answers
Answer:
i) The rate of interest = 8 %
(ii) Final amount after 3/2 years = 14061 rupee (Approx)
Step-by-step explanation:
Since, An amount P is deposited in annual rate of r (in decimals) for t years and compounded half years.
Then after t years the final amount is,
A = P(1+\frac{r}{2})^{2t}A=P(1+2r)2t
(i) Here, After 1 year,
A = 13,000
t = 1/2 years
P= 12,500
Thus, 12500(1+\frac{r}{2})^{1}=13,00012500(1+2r)1=13,000
125(1+\frac{r}{2})^{1}=130125(1+2r)1=130
25(1+\frac{r}{2})^{1}=2625(1+2r)1=26
(1+\frac{r}{2})^{1}=1.04(1+2r)1=1.04
\frac{r}{2}=0.042r=0.04
r=0.08r=0.08
Thus, the annual rate = 0.08 = 8%
(ii) Now, P = 125,000
t = 1½ years = 3/2 years
r = 8%
Thus, A = 12500(1+\frac{0.08}{2})^{2\times \frac{3}{2}}A=12500(1+20.08)2×23
A = 12500(1+\frac{0.08}{2})^{3}A=12500(1+20.08)3
A = 12500\times 1.124864=14060.8\approx 14061A=12500×1.124864=14060.8≈14061