How to find compound interest - ACT Math

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Question

Michelle makes a bank deposit of $1,500 at 4.2% annual interest, compounded monthly. Approximately how much money will be in Michelle’s bank account in 3 years?

Answer

The formula to use for compounded interest is

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

where P is the principal (original) amount, r is the interest rate (expressed in decimal form), n is the number of times per year the interest compounds, and t is the total number of years the money is left in the bank. In this problem, P=1,500, r=0.042, n=12, and t=3.

By plugging in, we find that Michelle will have about $1,701 at the end of three years.

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Question

Jack has , to invest. If he invests two-thirds of it into a high-yield savings account with an annual interest rate of , compounded quarterly, and the other third in a regular savings account at simple interest, how much does Jack earn after one year?

Answer

First, break the problem into two segments: the amount Jack invests in the high-yield savings, and the amount Jack invests in the simple interest account (10,000 and 5,000 respectively).

Now let's work with the high-yield savings account. $10,000 is invested at an annual rate of 8%, compounded quarterly. We can use the compound interest formula to solve:

$\text{Final Balance}=\text{ Principal} \cdot \bigg(1+\frac{\text{Interest Rate}}{c}\bigg)^{(\text{Time})(c)}$

Plug in the values given:

$=10,000\cdot \bigg(1+\frac{.08}{4}\bigg)^{(1)(4)}$

$=10,000 \cdot (1.02)^{4}$

Therefore, Jack makes $824.32 off his high-yield savings account. Now let's calculate the other interest:

$\text{Interest}=\text{Principal} \cdot \text{ Interest Rate} \cdot \text{ Time}$

$=5,000 \cdot (0.06)\cdot (1)$

Add the two together, and we see that Jack makes a total of, off of his investments.

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Question

Ashley makes a bank deposit of at annual interest, compounded monthly. About how many years will it take her deposit to grow to ?

Answer

The formula for compound interest is

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

where P is the principal (original) amount, r is the interest rate (in decimal form), n is the number of times per year the interest compounds, t is the time in years, and A is the final amount.

In this problem, we are solving for time, t. The given variables from the problem are:

Plugging these into the equation above, we get

$3500 = 2500\left ( 1 + \frac{0.031}{12}\right )^{12t}$

This simplifies to

$1.4 = \left ( 1 + \frac{0.031}{12}\right )^{12t}$

We can solve this by taking the natural log of both sides

$\ln \left (1.4 \right ) = \ln \left (\left ( 1 + \frac{0.031}{12}\right )^{12t} \right )$

$\ln \left (1.4 \right ) = 12t\ln \left ( 1 + \frac{0.031}{12}\right )$

$\frac{\ln \left (1.4 \right ) }{\ln \left (1 + \frac{0.031}{12}\right ) }= 12t$

$\frac{\ln \left (1.4 \right ) }{12\ln \left (1 + \frac{0.031}{12}\right ) }= t$

$t = 10.86 \approx 11$

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Question

Alice wants to invest money such that in years she has . The interest rate is compounded quarterly. How much should she invest?

Answer

The formula for compound interest is

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

where P is the principal (original) amount, r is the interest rate (in decimal form), n is the number of times per year the interest compounds, t is the time in years, and A is the final amount.

In this problem, we are solving for the principal, P. The given variables from the problem are:

Plugging these into the equation above, we get

$3600 = P\left ( 1 + \frac{0.047}{4}\right )^{4\cdot 5}$

Solving for P, we get

$3600 = P\left ( 1.01175)^{20}$

$\frac{3600}{(1.01175)^{20}}} = P$

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Question

A five-year bond is opened with in it and an interest rate of %, compounded annually. This account is allowed to compound for five years. Which of the following most closely approximates the total amount in the account after that period of time?

Answer

Each year, you can calculate your interest by multiplying the principle () by . For one year, this would be:

For two years, it would be:

, which is the same as

Therefore, you can solve for a five year period by doing:

Using your calculator, you can expand the into a series of multiplications. This gives you , which is closest to .

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Question

If a cash deposit account is opened with for a three year period at % interest compounded once annually, which of the following is closest to the positive difference between the interest accrued in the third year and the interest accrued in the second year?

Answer

It is easiest to break this down into steps. For each year, you will multiply by to calculate the new value. Therefore, let's make a chart:

After year 1: ; Total interest:

After year 2: ; Let us round this to ; Total interest:

After year 3: ; Let us round this to ; Total interest:

Thus, the positive difference of the interest from the last period and the interest from the first period is:

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Question

If an account has interest compounded annually at a rate of , what is the balance of the account after years of compounding if the initial balance is ? Round to the nearest cent.

Answer

Recall that the equation for compounded interest (with annual compounding) is:

Where is the balance, is the rate of interest, and is the number of years.

Thus, for our data, we need to know:

This is approximately .

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Question

If an account has interest compounded annually at a rate of , what is the balance of the account after years of compounding if the initial balance is ? Round to the nearest cent.

Answer

Recall that the equation for compounded interest (with annual compounding) is:

Where is the balance, is the rate of interest, and is the number of years.

Thus, for our data, we need to know:

This is approximately .

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Question

If an account has interest compounded quarterly at an annual rate of , what is the balance of the account after years of compounding if the initial balance is ? Round to the nearest cent.

Answer

Recall that the equation for compounded interest (with quarterly compounding) is:

$b_n_e_w=b_i_n_i_t_i_a_l(1 + \frac{r}{n})^t^n$

Where is the balance, is the rate of interest, is the number of years, and is the number of times it is compounded per year.

Thus, for our data, we need to know:

$b_n_e_w=12000 (1 + \frac{0.08}{4})^4^^4=12000 (1.02)^1^6$

This is approximately .

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Question

An account is compounded at a given rate of interest annually for years. What is this rate if the beginning balance for the account was and its ending balance ? Round to the nearest hundredth of a percent.

Answer

Recall that the equation for compounded interest (with annual compounding) is:

Where is the balance, is the rate of interest, and is the number of years.

Thus, for our data, we need to know:

Now, let's use for . This gives us:

Using a logarithm, this can be rewritten:

$log_b(\frac{10451.43}{4500})=20$

This can be rewritten:

$\frac{log(2.32254)}{log(b)}=20$

Now, you can solve for :

$\frac{1}{log(b)}=\frac{20}{log(2.32254)}$

or

$log(b)=\frac{log(2.32254)}{20}$

Now, finally you can rewrite this as:

$10^\frac{log(2.32254)}{20}=b$

Thus,

Now, round this to and recall that

Thus, and or

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Question

Michelle makes a bank deposit of $1,500 at 4.2% annual interest, compounded monthly. Approximately how much money will be in Michelle’s bank account in 3 years?

Answer

The formula to use for compounded interest is

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

where P is the principal (original) amount, r is the interest rate (expressed in decimal form), n is the number of times per year the interest compounds, and t is the total number of years the money is left in the bank. In this problem, P=1,500, r=0.042, n=12, and t=3.

By plugging in, we find that Michelle will have about $1,701 at the end of three years.

Compare your answer with the correct one above

Question

Jack has , to invest. If he invests two-thirds of it into a high-yield savings account with an annual interest rate of , compounded quarterly, and the other third in a regular savings account at simple interest, how much does Jack earn after one year?

Answer

First, break the problem into two segments: the amount Jack invests in the high-yield savings, and the amount Jack invests in the simple interest account (10,000 and 5,000 respectively).

Now let's work with the high-yield savings account. $10,000 is invested at an annual rate of 8%, compounded quarterly. We can use the compound interest formula to solve:

$\text{Final Balance}=\text{ Principal} \cdot \bigg(1+\frac{\text{Interest Rate}}{c}\bigg)^{(\text{Time})(c)}$

Plug in the values given:

$=10,000\cdot \bigg(1+\frac{.08}{4}\bigg)^{(1)(4)}$

$=10,000 \cdot (1.02)^{4}$

Therefore, Jack makes $824.32 off his high-yield savings account. Now let's calculate the other interest:

$\text{Interest}=\text{Principal} \cdot \text{ Interest Rate} \cdot \text{ Time}$

$=5,000 \cdot (0.06)\cdot (1)$

Add the two together, and we see that Jack makes a total of, off of his investments.

Compare your answer with the correct one above

Question

Ashley makes a bank deposit of at annual interest, compounded monthly. About how many years will it take her deposit to grow to ?

Answer

The formula for compound interest is

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

where P is the principal (original) amount, r is the interest rate (in decimal form), n is the number of times per year the interest compounds, t is the time in years, and A is the final amount.

In this problem, we are solving for time, t. The given variables from the problem are:

Plugging these into the equation above, we get

$3500 = 2500\left ( 1 + \frac{0.031}{12}\right )^{12t}$

This simplifies to

$1.4 = \left ( 1 + \frac{0.031}{12}\right )^{12t}$

We can solve this by taking the natural log of both sides

$\ln \left (1.4 \right ) = \ln \left (\left ( 1 + \frac{0.031}{12}\right )^{12t} \right )$

$\ln \left (1.4 \right ) = 12t\ln \left ( 1 + \frac{0.031}{12}\right )$

$\frac{\ln \left (1.4 \right ) }{\ln \left (1 + \frac{0.031}{12}\right ) }= 12t$

$\frac{\ln \left (1.4 \right ) }{12\ln \left (1 + \frac{0.031}{12}\right ) }= t$

$t = 10.86 \approx 11$

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Question

Alice wants to invest money such that in years she has . The interest rate is compounded quarterly. How much should she invest?

Answer

The formula for compound interest is

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

where P is the principal (original) amount, r is the interest rate (in decimal form), n is the number of times per year the interest compounds, t is the time in years, and A is the final amount.

In this problem, we are solving for the principal, P. The given variables from the problem are:

Plugging these into the equation above, we get

$3600 = P\left ( 1 + \frac{0.047}{4}\right )^{4\cdot 5}$

Solving for P, we get

$3600 = P\left ( 1.01175)^{20}$

$\frac{3600}{(1.01175)^{20}}} = P$

Compare your answer with the correct one above

Question

A five-year bond is opened with in it and an interest rate of %, compounded annually. This account is allowed to compound for five years. Which of the following most closely approximates the total amount in the account after that period of time?

Answer

Each year, you can calculate your interest by multiplying the principle () by . For one year, this would be:

For two years, it would be:

, which is the same as

Therefore, you can solve for a five year period by doing:

Using your calculator, you can expand the into a series of multiplications. This gives you , which is closest to .

Compare your answer with the correct one above

Question

If a cash deposit account is opened with for a three year period at % interest compounded once annually, which of the following is closest to the positive difference between the interest accrued in the third year and the interest accrued in the second year?

Answer

It is easiest to break this down into steps. For each year, you will multiply by to calculate the new value. Therefore, let's make a chart:

After year 1: ; Total interest:

After year 2: ; Let us round this to ; Total interest:

After year 3: ; Let us round this to ; Total interest:

Thus, the positive difference of the interest from the last period and the interest from the first period is:

Compare your answer with the correct one above

Question

If an account has interest compounded annually at a rate of , what is the balance of the account after years of compounding if the initial balance is ? Round to the nearest cent.

Answer

Recall that the equation for compounded interest (with annual compounding) is:

Where is the balance, is the rate of interest, and is the number of years.

Thus, for our data, we need to know:

This is approximately .

Compare your answer with the correct one above

Question

If an account has interest compounded annually at a rate of , what is the balance of the account after years of compounding if the initial balance is ? Round to the nearest cent.

Answer

Recall that the equation for compounded interest (with annual compounding) is:

Where is the balance, is the rate of interest, and is the number of years.

Thus, for our data, we need to know:

This is approximately .

Compare your answer with the correct one above

Question

If an account has interest compounded quarterly at an annual rate of , what is the balance of the account after years of compounding if the initial balance is ? Round to the nearest cent.

Answer

Recall that the equation for compounded interest (with quarterly compounding) is:

$b_n_e_w=b_i_n_i_t_i_a_l(1 + \frac{r}{n})^t^n$

Where is the balance, is the rate of interest, is the number of years, and is the number of times it is compounded per year.

Thus, for our data, we need to know:

$b_n_e_w=12000 (1 + \frac{0.08}{4})^4^^4=12000 (1.02)^1^6$

This is approximately .

Compare your answer with the correct one above

Question

An account is compounded at a given rate of interest annually for years. What is this rate if the beginning balance for the account was and its ending balance ? Round to the nearest hundredth of a percent.

Answer

Recall that the equation for compounded interest (with annual compounding) is:

Where is the balance, is the rate of interest, and is the number of years.

Thus, for our data, we need to know:

Now, let's use for . This gives us:

Using a logarithm, this can be rewritten:

$log_b(\frac{10451.43}{4500})=20$

This can be rewritten:

$\frac{log(2.32254)}{log(b)}=20$

Now, you can solve for :

$\frac{1}{log(b)}=\frac{20}{log(2.32254)}$

or

$log(b)=\frac{log(2.32254)}{20}$

Now, finally you can rewrite this as:

$10^\frac{log(2.32254)}{20}=b$

Thus,

Now, round this to and recall that

Thus, and or

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