a man borrows rs 16000 at 15% p.a compounded annually. if he repays Rs4500 at the end of each year,how much loan is outstanding against him at the beginning of the third year?
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Answer:
begining of the third year 27000
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he pay in 2 years 4500 ×2 =₹9000
amount after 2 year = p(1+R/100)^2 = ₹21160
balance at the beginning of 3rd year = ₹21160-₹9000= ₹ 12160
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