Question

a person deposited 80000 in bank at rate of 12p.a interest compounded semi annually for 2 yrs after one year bank revised its policy to pay interest compounded annually at the same rate .what will the percentage difference between interest of second year due to revised policy?​

Answer 1

Answer:

2.96% (2 d.p.) difference

Step-by-step explanation:

Compound Interest Formula

$\large \sf I=P\left(1+\frac{r}{n}\right)^{nt} -P$

where:

• I = total interest
• P = principal amount
• r = interest rate (in decimal form)
• n = number of times interest applied per time period
• t = number of time periods elapsed

Interest earned in Year 1 (Compounded semi-annually)

Given:

• P = 80000
• r = 12% = 0.12
• n = 2 (semi-annually)
• t = 1 year

Substitute the values into the formula:

$\implies \sf I=80000\left(1+\frac{0.12}{2}\right)^{2(1)} -80000$

$\implies \sf I=80000\left(1.06\right)^{2} -80000$

$\implies \sf I=80000\left(1.1236\right) -80000$

$\implies \sf I= 89888-80000$

$\implies \sf I= 9888$

Interest earned in Year 2 (Compounded semi-annually)

Given:

• P = 80000 + 9888
• r = 12% = 0.12
• n = 2 (semi-annually)
• t = 1 year

Substitute the values into the formula:

$\implies \sf I=89888\left(1+\frac{0.12}{2}\right)^{2(1)} -89888$

$\implies \sf I=89888\left(1.06\right)^{2} -89888$

$\implies \sf I=89888\left(1.2636\right) -89888$

$\implies \sf I= 100998.156...-89888$

$\implies \sf I = 11110.16$

Interest earned in Year 2 (compounded annually)

Given:

• P = 80000 + 9888 = 89888
• r = 12% = 0.12
• n = 1 (annually)
• t = 1 year

Substitute the values into the formula:

$\implies \sf I=89888\left(1+\frac{0.12}{1}\right)^{1(1)} -89888$

$\implies \sf I=89888\left(1.12\right) -89888$

$\implies \sf I=100674.56 -89888$

$\implies \sf I=10786.56$

Difference between second year interests

Interest earned in Year 2 (Compounded semi-annually) = 11110.16

Interest earned in Year 2 (compounded annually) = 10786.56

Percentage difference:

$\sf p=\dfrac{|a-b|}{(a+b) \div 2} \times 100$

where:

• a = value 1
• b = value 2

$\sf \implies p=\dfrac{|11110.16-10786.56|}{(11110.16+10786.56) \div 2} \times 100$

$\sf \implies p=\dfrac{323.6}{10948.36} \times 100$

$\sf \implies p=2.95569...$

$\implies \sf p=2.96\% \:\: (2 \:d.p.)$

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