Question

find the difference between the compound interest on 120000 for 1 year at 20% p.nic.in compounded half yearly and quarterly​

Answer 1

Answer:

Difference = 660.75

Step-by-step explanation:

P = initial amount in Rupees = 120000

R = rate of interest per cent per year = 20

T = time period in years = 1

m = compounding interval

A = final amount at the end of period T

CI = Compound interest = A - P

$A = P*(1 + \frac{\frac{R}{m}}{100} )^{mT}$

A = P*(1 + (R/m)/100))^(mT) ............(i)

Case 1:

m = 2 (half-yearly compounding means two intervals in a year)

mT = 2*1 = 2

R/m = 20/2 = 10

=> A = 120000*(1 + 10/100)^2

=> A = 120000*(1.1)^2

=> A = 120000*1.21

=> A = 145200

CI = A - P

CI = 145200 - 120000 = 25200 ..........(i)

Case 2:

m = 4 (quarterly compounding means four intervals in a year)

mT = 4*1 = 4

R/m = 20/4 = 5

=> A = 120000*(1 + 5/100)^4

=> A = 120000*(1.05)^4

=> A = 120000*(1.215506)

=> A = 145860.75

CI = A - P

CI = 145860.75 - 120000 = 25860.75 ..........(ii)

Difference in CI in the two cases

= 25860.75 - 25200

= 660.75

Answer 2

Answer:

P = initial amount in Rupees = 120000

R = rate of interest per cent per year = 20

T = time period in years = 1

m = compounding interval

A = final amount at the end of period T

CI = Compound interest = A - P

A = P*(1 + R/m/100)mT

A = P*(1 + (R/m)/100))^(mT) ............(i)

Case 1:

m = 2 (half-yearly compounding means two intervals in a year)

mT = 2*1 = 2

R/m = 20/2 = 10

=> A = 120000*(1 + 10/100)^2

=> A = 120000*(1.1)^2

=> A = 120000*1.21

=> A = 145200

CI = A - P

CI = 145200 - 120000 = 25200 ..........(i)

Case 2:

m = 4 (quarterly compounding means four intervals in a year)

mT = 4*1 = 4

R/m = 20/4 = 5

=> A = 120000*(1 + 5/100)^4

=> A = 120000*(1.05)^4

=> A = 120000*(1.215506)

=> A = 145860.75

CI = A - P

CI = 145860.75 - 120000 = 25860.75 ..........(ii)

Difference in CI in the two cases

= 25860.75 - 25200

= 660.75

Hope this is helpful for you.