Question

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.

Answer 1

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,

$A = P(1+\frac{r}{2})^{2t}$

(i) Here, After 1 year,

A = 13,000

t =  1/2 years

P= 12,500

Thus, $12500(1+\frac{r}{2})^{1}=13,000$

$125(1+\frac{r}{2})^{1}=130$

$25(1+\frac{r}{2})^{1}=26$

$(1+\frac{r}{2})^{1}=1.04$

$\frac{r}{2}=0.04$

$r=0.08$

Thus, the annual rate = 0.08 = 8%

(ii) Now, P = 125,000

t = 1½ years = 3/2 years

r = 8%

Thus, $A = 12500(1+\frac{0.08}{2})^{2\times \frac{3}{2}}$

$A = 12500(1+\frac{0.08}{2})^{3}$

$A = 12500\times 1.124864=14060.8\approx 14061$

Answer 2

Answer:1

13000=12500(1+r/100×2)^1/2×2

13000÷12500=(200+r/200)^1

26÷25=(200+r/200)^1

26÷25×200-200=r

8%=r

Step-by-step explanation:

In compounded half yearly

Rate becomes half and times doubles