Question

find the compound interest on 12000 Rs for one and a half years at 10%per annum when compounded half yearly ​

Answer 1

Step-by-step explanation:

Compound interest(C.I)=A−P=14,520−12,000=2520

Hence, the compound interest is Rs. 2,520

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Answer 2

Answer:

Here P=12,000 Rs., n=2 years, R=10

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R )

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R ) n

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R ) n

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R ) n =12000(1+

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R ) n =12000(1+ 100

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R ) n =12000(1+ 10010

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R ) n =12000(1+ 10010

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R ) n =12000(1+ 10010 )

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R ) n =12000(1+ 10010 ) 2

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R ) n =12000(1+ 10010 ) 2

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R ) n =12000(1+ 10010 ) 2 =12000(

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R ) n =12000(1+ 10010 ) 2 =12000( 10

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R ) n =12000(1+ 10010 ) 2 =12000( 1011

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R ) n =12000(1+ 10010 ) 2 =12000( 1011

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R ) n =12000(1+ 10010 ) 2 =12000( 1011 )

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R ) n =12000(1+ 10010 ) 2 =12000( 1011 ) 2

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R ) n =12000(1+ 10010 ) 2 =12000( 1011 ) 2

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R ) n =12000(1+ 10010 ) 2 =12000( 1011 ) 2 =

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R ) n =12000(1+ 10010 ) 2 =12000( 1011 ) 2 = 100

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R ) n =12000(1+ 10010 ) 2 =12000( 1011 ) 2 = 10012000×121

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R ) n =12000(1+ 10010 ) 2 =12000( 1011 ) 2 = 10012000×121

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R ) n =12000(1+ 10010 ) 2 =12000( 1011 ) 2 = 10012000×121

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R ) n =12000(1+ 10010 ) 2 =12000( 1011 ) 2 = 10012000×121 =14,520 Rs.

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R ) n =12000(1+ 10010 ) 2 =12000( 1011 ) 2 = 10012000×121 =14,520 Rs.Compound interest(C.I)=A−P=14,520−12,000=2520

Here P=12,000 Rs., n=2 years, R=10Amount(A)=P(1+ 100R ) n =12000(1+ 10010 ) 2 =12000( 1011 ) 2 = 10012000×121 =14,520 Rs.Compound interest(C.I)=A−P=14,520−12,000=2520Hence, the compound interest is Rs. 2,520

Step-by-step explanation:

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