Question

What interest rate per annum is a fund paying its clients if the third to the last payment of a depositor became Php 1,040.40 from an initial savings payment of Php 1,000.00?

Answer 1

Answer: 1.3%

Step-by-step explanation:

To find rate of interest (r) per annum for a fund deposit by depositor .

Initial saving payment  P= $1,000

Amount after three years A =$1,040.40

we know that

$A=P(1+\frac{r}{100} )^n\\\\\Rightarrow\ 1040=1000(1+\frac{r}{100} )^3\\\Rightarrow\ \frac{1040}{1000} =(1+\frac{r}{100} )^3\\\Rightarrow\ 1.040=(1+\frac{r}{100} )^3\\\text{by taking log on both sides we get,}\\log(1.040)=log((1+\frac{r}{100} )^3)\\\Rightarrow\ 0.039=3log(1+\frac{r}{100})\\\Rightarrow\ 0.013=log(1+\frac{r}{100})\\\Rightarrow\ e^0.013=1+\frac{r}{100}\\\Rightarrow\ 1.013=(1+\frac{r}{100})\\\Rightarrow\ 1.013-1=\frac{r}{100}$

r=0.013×100=1.3%

Answer 2

The interest rate per annum if a fund paying its clients is the third to the last payment of a depositor became Php 1,040.40 from an initial savings payment of Php 1,000.00 is 1.32%

To get the interest rate, we will use the formula for calculating the compound interest as shown:

$A=P(1+\frac{r}{n} )^{nt}$

P is the amount of investment

n is the time of compounding

t is the time

A is the amount after t years

Given the following parameters

P = 1,000.00

A = Php 1,040.40

n = 1

Substitute the parameters into the formula

$1040.4=1000(1+\frac{r}{1} )^{3}\\\frac{1040.4}{1000}=(1+r)^3\\1.0404 = (1+r)^3\\ln(1.0404) = 3ln(1+r)\\ 0.0396=3ln(1+r)\\ln(1+r)=0.0132\\1+r = e^{0.0132}\\1+r =1.0132\\r=1.0132-1\\r=0.0132\\r=1.32\%$

Hence the interest rate per annum if a fund paying its clients is the third to the last payment of a depositor became Php 1,040.40 from an initial savings payment of Php 1,000.00 is 1.32%.

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