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50 anya invests $6000 in an account that pays compound interest at a ra…

Question

50 anya invests $6000 in an account that pays compound interest at a rate of r% per year. at the end of 8 years, the account has earned $621.70 in interest. calculate the value of r. r = .........................

Explanation:

Step1: Recall compound interest formula

The compound interest formula is $A = P(1 + \frac{r}{100})^n$, where $A$ is the amount, $P$ is the principal, $r$ is the rate of interest per annum, and $n$ is the number of years. The interest earned $I$ is $A - P$. So we have $I=P(1 + \frac{r}{100})^n - P$.

Given $P = 6000$, $n = 8$, $I=621.70$. Then $A=P + I=6000 + 621.70 = 6621.70$.

Step2: Substitute values into the formula

Substitute $A = 6621.70$, $P = 6000$, $n = 8$ into $A = P(1+\frac{r}{100})^n$:

$$6621.70=6000(1 + \frac{r}{100})^8$$

Divide both sides by 6000:

$$\frac{6621.70}{6000}=(1 + \frac{r}{100})^8$$

Calculate $\frac{6621.70}{6000}=1.10361667$

So we have $(1 + \frac{r}{100})^8 = 1.10361667$

Step3: Take the 8th root of both sides

Take the 8th root of both sides. Let $x = 1+\frac{r}{100}$, then $x^8=1.10361667$. So $x = 1.10361667^{\frac{1}{8}}$

Calculate $1.10361667^{\frac{1}{8}}$. Using a calculator, $1.10361667^{\frac{1}{8}}\approx1.0125$

Step4: Solve for r

Since $x = 1+\frac{r}{100}=1.0125$

Subtract 1 from both sides:

$\frac{r}{100}=1.0125 - 1=0.0125$

Multiply both sides by 100:

$r = 0.0125\times100 = 1.25$

Answer:

$1.25$