The firts thing we are going to do here is use the simple interest formula: [tex]A=P(1+rt)[/tex]
where
[tex]A[/tex] is the final amount after [tex]t[/tex] years
[tex]P[/tex] is the initial investment
[tex]r[/tex] is the interest rate in decimal form
[tex]t[/tex] is the number of years
With this formula we will find the final amount Azhar's investment after 7 years. We know from our problem that [tex]P=10000[/tex], [tex]r= \frac{2}{100}=0.02 [/tex], and [tex]t=7[/tex]. Lets replace those values in our formula to find [tex]A[/tex]:
[tex]A=10000(1+(0.02)(7))[/tex]
[tex]A=10000(1.14)[/tex]
[tex]A=11400[/tex]
Now, since Sarah is investing in a compound interest account, we are going to use the compound interest formula: [tex]A=P(1+ \frac{r}{n})^{nt} [/tex]
where
[tex]A[/tex] is the final amount after [tex]t[/tex] years
[tex]P[/tex] is the initial investment
[tex]r[/tex] is the interest rate in decimal form
[tex]n[/tex] is the number of times the interest is compounded per year
[tex]t[/tex] is the number of years
Notice that we know from our problem that after 7 years their investments are worth the same amount, so [tex]A=11400[/tex]. We also know that [tex]P=10000[/tex], [tex]r= \frac{x}{100} =0.01x[/tex], and [tex]t=7[/tex]. Since the interest are compounded per year, [tex]n=1[/tex]. Lets replace all the vales in our compound interest formula and solve for [tex]x[/tex] to find our rate:
[tex]11400=10000(1+ \frac{0.01x}{1} )^{(1)(7)} [/tex]
[tex] \frac{11400}{10000} =(1+0.01x) ^{7} [/tex]
[tex](1+0.01x) ^{7} =1.14[/tex]
[tex]1+0.01x= \sqrt[7]{1.14} [/tex]
[tex]0.01x= \sqrt[7]{1.14} -1[/tex]
[tex]x= \frac{ \sqrt[7]{1.14}-1 }{0.01} [/tex]
[tex]x=1.89[/tex]
We can conclude that the interest rate of Sarah's investment is approximately 1.89%, so x=1.89%.
