Question

the rate of interest at which an qamount of 1800 on compounded interest become 1984.50 in two years​

Answer 1

Answer:

A = P (1+r100)2

Therefor, 1984.50 = 1800(1+r100)2

39692×1800=(1+r100)2

4412×200=(1+r100)2

1+r100=2120

r100=2120−1

r=120×100 = 5%

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