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A = P(1+r100)2
I = A - P
Hence I=P(1+r100)2−P315=P(1+r100)2−P315=P(1+10100)2−P315=P(1110)2−P315=121P100−P315=P(121100−1)315=21P100
P=315×10021=105×1007=₹1500
Therefore, the initial amount = ₹ 1500
Standard VIII Mathematics
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