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

Question

2 months ago

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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.

Mathematics

2 answers:

AnnZ [12.3K] · Answer 1 · 2026-03-16T06:19:59+03:00

Answer:

2.1%

Step-by-step explanation:

We know that

The formula for compound interest is represented as

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

In this context,

A represents the Final Investment Value

P denotes the Principal amount to invest

r indicates the interest rate in decimal form

t denotes the Number of Time Periods

n signifies the frequency of interest compounding per year

In this scenario, we have

For the initial 2 years

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

Substituting in the equation outlined above

$A=6,000(1+\frac{0.015}{1})^{1*2}$

$A=6,000(1.015)^{2}$

$A=\$6,181.35$

Determine the interest rate for the third year

Given that

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

Input values

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

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

$6,311.16=6,181.35(1+r)$

$r=(6,311.16/6,181.35)-1$

$r=0.021$

Convert to percentage

$r=0.021*100=2.1\%$

AnnZ [12.3K] · Answer 2 · 2026-03-16T06:23:10+03:00

Answer:

2.1%

Step-by-step explanation:

The compound interest formula can be expressed as:

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

Starting with a Principal amount of $6000 and applying an interest rate of 1.5% for the first two years:

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

We calculate compound interest $A_2$ for one year at rate i, resulting in $6311.16:

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

Thus, the interest rate for the third year is 2.1%