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Question 63
A sum of money becomes Rs. 1,08,000/- after 3 years and becomes Rs. 1,92,000/- after 9 years on same rate of compound interest. Then, that sum of money is (in rupees)
Solution
Let the rate of interest be R and $$1+\frac{R}{100}=m$$
Hence, 108000 = P*$$m^3$$ and 192000 = P*$$m^9$$
Dividing the second by first, we get $$m^6=\frac{16}{9}$$
Hence, $$m^3=\frac{4}{3}$$ , substituting in the first equation, we get P = 81,000
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