What will be the simple interest on rupees 12000 at 10% per annum for 3 years?

Solution : p=12000Rs
r=12
n=10
`A=P(1+r/100)^n`
`A=12000(1+12/100)^10`
`A=12000(1.12)^10`
taking log both side
`logA=log12000+10log1.12`
`logA=4.0792+10*(0.0493)`
`logA=4.0792+0.493`
`=4.5722`
`A=antilog(4.5722)=37350`
Amount after 10 years=37350
Compound interest=37350-12000=25350.

Principal for the first year = Rs 12000

Rate of interest = 10% p.a.

Interest for the first year = Rs (12000 × 10 × 1) / 100

= Rs 1200

Amount at the end of first year = Rs 12000 + Rs 1200

= 13200

Principal for the second year = Rs 13200

Interest for the second year = Rs (13200 × 10 × 1) / 100

= Rs 1320

Amount at the end of second year = Rs 13200 + Rs 1320

= Rs 14520

Principal for the third year = Rs 14520

Interest for the third year = Rs (14520 × 10 × 1) / 100

= Rs 1452

Amount at the end of third year = Rs 14520 + Rs 1452

= Rs 15972

Hence,

Compound interest for 3 year = Final amount – (original) Principal

= Rs 15972 – Rs 12000

= Rs 3972

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