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Question
Find the sum invested at `12 1/2` p.a. compound interest on which the interest for the third year exceeds that of the first year by Rs 531.25.
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Solution
Let the sum be P
Interest for first year:
`"P" (1 + 25/(2 xx 100)) - "P"` ..................(i)
Interest for third year:
`"P" (1 + 25/(2 xx 100))^3 - "P" (1 + 25/(2 xx 100))^2` .........(ii)
Subtracting (ii) from (i)
`"P" (1 + 25/(2 xx 100))^3 - "P" (1 + 25/(2 xx 100))^2 - "P" (1 + 25/(2 xx 100)) + "P"` = Rs 531.25
Rs 531. 25 = 1. 423828 P - 1. 265625 P - 1. 125 P + P
Rs 531. 25 = 0. 033203 P
P = Rs 16,000
Hence the sum is Rs 16,000
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Rs.16,000 is invested at 5% compound interest compounded per annum. Use the table, given below, to find the amount in 4 years.
| Year ↓ |
Initial amount (Rs.) |
Interest (Rs.) |
Final amount (Rs.) |
| 1st | 16,000 | 800 | 16,800 |
| 2nd | ........... | ........... | ........... |
| 3rd | ........... | ........... | ........... |
| 4th | ........... | ........... | ........... |
| 5th | ........... | ........... | ........... |
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