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.
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Answered by
24
a=13000
p=12500
r=r
t=18months/6=3times
a=P(1+R/100)^t
13000=12500(1+r/100)^3
13000/12500=(100+r)^3
130/125=(100)^3+r^3+3×100×r(100+r)...[a^3+b^3+3ab(a+b)
130/125=1000000+r^3+300r(100+r)
p=12500
r=r
t=18months/6=3times
a=P(1+R/100)^t
13000=12500(1+r/100)^3
13000/12500=(100+r)^3
130/125=(100)^3+r^3+3×100×r(100+r)...[a^3+b^3+3ab(a+b)
130/125=1000000+r^3+300r(100+r)
Answered by
10
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,
(i) Here, After 1 year,
A = 13,000
t = 1/2 years
P= 12,500
Thus, the annual rate = 0.08 = 8%
(ii) Now, P = 125,000
t = 1½ years = 3/2 years
r = 8%
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