mr lalit invested 5000 at a certain rate of interest compounded annually for 2 years at the end of first year it amounts to 5325 calculate (i) rate (ii) the amount at the end of second year to the nearest rupee.. ans with full solution
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140
P=5000
Amount at the end of first year=5325
S.I=A-P
=5325-5000
=325
1)RATE=SI×100÷P×T
=325×100÷5000×1
=325÷50
=6.5%p.a
2)amount at the end of the second year= a(1+r/100)^2
= 5000(1+6.5/100)^2
=5000×1065/1000×1065/1000
=1065×1065/200
=213×213/8
=45399/8
=5671
Amount at the end of first year=5325
S.I=A-P
=5325-5000
=325
1)RATE=SI×100÷P×T
=325×100÷5000×1
=325÷50
=6.5%p.a
2)amount at the end of the second year= a(1+r/100)^2
= 5000(1+6.5/100)^2
=5000×1065/1000×1065/1000
=1065×1065/200
=213×213/8
=45399/8
=5671
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32
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