Question

max invests $6000 in a savings account for 3 years. the account pays compound interest at a rate of 1.5% per year for the first 2 years. The compound interest rate changes for the third year. At the end of 3 years, there is a total of $6311.16 in the account. Work out the compound interest rate for the third year.

Answer 1

Answer:

2.1%

Step-by-step explanation:

The formula for compound interest is given as:

$A=P(1+I)^n\\\\P-Principal \\A-amount\\i-compound \ interest \ rate$

Given the Principal amount as $6000, and the rate in the first two years as 1.5%:

$A=P(1+i)^n\\\\A_2=6000(1+0.015)^2\\\\A_2=6181.35$

We compound $A_2$ for 1 year at rate i to obtain $6311.16:

$A=P(1+i)^n, n=1, i=i, P=6181.35, A=6311.16\\\\6311.16=6181.35(1+i)^1\\\\\frac{6311.16}{6181.35}=(1+i)\\\\i=\frac{6311.16}{6181.35}-1\\\\i=0.02100$

Hence, the compound interest rate in the third year is 2.1%

Answer 2

Answer:

2.1%

Step-by-step explanation:

we know that    

The compound interest formula is equal to  

$A=P(1+\frac{r}{n})^{nt}$  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

in this problem we have  

For the first 2 years

$t=2\ years\\ P=\ 6,000\\ r=1.5\%=1.5/100=0.015\\n=1$  

substitute in the formula above

$A=6,000(1+\frac{0.015}{1})^{1*2}$  

$A=6,000(1.015)^{2}$  

$A=\ 6,181.35$  

Find the interest rate for the third year

we have

$t=1\ years\\ P=\ 6,181.35\\ r=?\\n=1\\A=\ 6,311.16$  

substitute

$6,311.16=6,181.35(1+\frac{r}{1})^{1*1}$  

$6,311.16=6,181.35(1+r)^{1}$  

$6,311.16=6,181.35(1+r)$  

$r=(6,311.16/6,181.35)-1$  

$r=0.021$

Convert to percentage

$r=0.021*100=2.1\%$