A man borrows 15000 at 12% per annum, compounded annually. If he repays 4400 at the
end of each year, find the amount outstanding against him at the beginning of the third year.
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Given:-
- Principal = 15000
- Rate = 12%
- Amount = 4400
To find:-
- The amount outstanding against him at the beginning of the third year.
Solutions:-
Simple interest = Principal × rate × time / 100
=> 15000 × 12 × 1 / 100
=> 1500 × 12 / 100
=> 150 × 12
=> 1800
Amount at the end of first year => p + 1
=> 15000 + 1800
=> 16800
Amount left after repaying
=> 16800 - 4400
=> 12400
Principal of second year:-
=> 12400 × 12 × 1 /100
=> 12400 × 12 / 100
=> 124 × 12
=> 1488
Amount with interest
=> 12400 - 1488
=> 13888
Amount after repaying
=> 13888 - 4400
=> 9488
Hence, the third year is 9488.
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