In what time will 4000 at 3% interest produce the same income as 5000 in 5 years at 4%?

A) 5 years

B) 6 years

C) 3 years

D) 4 years

Correct Answer:

Description for Correct answer:
Let time = t years

According to the question,

\( \Large \frac{8000 \times 3 \times t}{100}=\frac{6000 \times 5 \times 4}{100}\)

240 t=1200

t=5 years

Hence required time = 5 years

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We have to work with money every day. While balancing your checkbook or calculating your monthly expenditures on espresso requires only arithmetic, when we start saving, planning for retirement, or need a loan, we need more mathematics.

 

Simple Interest

Discussing interest starts with the principal, or amount your account starts with. This could be a starting investment, or the starting amount of a loan. Interest, in its most simple form, is calculated as a percent of the principal. For example, if you borrowed $100 from a friend and agree to repay it with 5% interest, then the amount of interest you would pay would just be 5% of 100: $100(0.05) = $5. The total amount you would repay would be $105, the original principal plus the interest.

Simple One-time Interest

[latex]\begin{align}&I={{P}_{0}}r\\&A={{P}_{0}}+I={{P}_{0}}+{{P}_{0}}r={{P}_{0}}(1+r)\\\end{align}[/latex]

• I is the interest
• A is the end amount: principal plus interest
• [latex]\begin{align}{{P}_{0}}\\\end{align}[/latex] is the principal (starting amount)
• r is the interest rate (in decimal form. Example: 5% = 0.05)

Examples

A friend asks to borrow $300 and agrees to repay it in 30 days with 3% interest. How much interest will you earn?

Show Solution

[latex]\begin{align}{{P}_{0}}\\\end{align}[/latex] = $300the principalr = 0.033% rateI = $300(0.03) = $9.You will earn $9 interest.

The following video works through this example in detail.

 

One-time simple interest is only common for extremely short-term loans. For longer term loans, it is common for interest to be paid on a daily, monthly, quarterly, or annual basis. In that case, interest would be earned regularly.

For example, bonds are essentially a loan made to the bond issuer (a company or government) by you, the bond holder. In return for the loan, the issuer agrees to pay interest, often annually. Bonds have a maturity date, at which time the issuer pays back the original bond value.

Exercises

Suppose your city is building a new park, and issues bonds to raise the money to build it. You obtain a $1,000 bond that pays 5% interest annually that matures in 5 years. How much interest will you earn?

Show Solution

Each year, you would earn 5% interest: $1000(0.05) = $50 in interest. So over the course of five years, you would earn a total of $250 in interest. When the bond matures, you would receive back the $1,000 you originally paid, leaving you with a total of $1,250.

Further explanation about solving this example can be seen here.

We can generalize this idea of simple interest over time.

Simple Interest over Time

[latex]\begin{align}&I={{P}_{0}}rt\\&A={{P}_{0}}+I={{P}_{0}}+{{P}_{0}}rt={{P}_{0}}(1+rt)\\\end{align}[/latex]

• I is the interest
• A is the end amount: principal plus interest
• [latex]\begin{align}{{P}_{0}}\\\end{align}[/latex] is the principal (starting amount)
• r is the interest rate in decimal form
• t is time

The units of measurement (years, months, etc.) for the time should match the time period for the interest rate.

 

APR – Annual Percentage Rate

Interest rates are usually given as an annual percentage rate (APR) – the total interest that will be paid in the year. If the interest is paid in smaller time increments, the APR will be divided up.

For example, a 6% APR paid monthly would be divided into twelve 0.5% payments.
[latex]6\div{12}=0.5[/latex]

A 4% annual rate paid quarterly would be divided into four 1% payments.
[latex]4\div{4}=1[/latex]

Example

Treasury Notes (T-notes) are bonds issued by the federal government to cover its expenses. Suppose you obtain a $1,000 T-note with a 4% annual rate, paid semi-annually, with a maturity in 4 years. How much interest will you earn?

Show Solution

Since interest is being paid semi-annually (twice a year), the 4% interest will be divided into two 2% payments.

[latex]\begin{align}{{P}_{0}}\\\end{align}[/latex] = $1000the principalr = 0.022% rate per half-yeart = 84 years = 8 half-yearsI = $1000(0.02)(8) = $160. You will earn $160 interest total over the four years.

This video explains the solution.

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Try It

A loan company charges $30 interest for a one month loan of $500. Find the annual interest rate they are charging.

Show Solution

I = $30 of interest
[latex]P_0[/latex] = $500 principal
r = unknown
t = 1 month

Using [latex]I = P_0rt[/latex], we get [latex]30 = 500·r·1[/latex]. Solving, we get r = 0.06, or 6%. Since the time was monthly, this is the monthly interest. The annual rate would be 12 times this: 72% interest.

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Compound Interest

With simple interest, we were assuming that we pocketed the interest when we received it. In a standard bank account, any interest we earn is automatically added to our balance, and we earn interest on that interest in future years. This reinvestment of interest is called compounding.

Suppose that we deposit $1000 in a bank account offering 3% interest, compounded monthly. How will our money grow?

The 3% interest is an annual percentage rate (APR) – the total interest to be paid during the year. Since interest is being paid monthly, each month, we will earn [latex]\frac{3%}{12}[/latex]= 0.25% per month.

In the first month,

• P0 = $1000
• r = 0.0025 (0.25%)
• I = $1000 (0.0025) = $2.50
• A = $1000 + $2.50 = $1002.50

In the first month, we will earn $2.50 in interest, raising our account balance to $1002.50.

 

In the second month,

• P0 = $1002.50
• I = $1002.50 (0.0025) = $2.51 (rounded)
• A = $1002.50 + $2.51 = $1005.01

Notice that in the second month we earned more interest than we did in the first month. This is because we earned interest not only on the original $1000 we deposited, but we also earned interest on the $2.50 of interest we earned the first month. This is the key advantage that compounding interest gives us.

Calculating out a few more months gives the following:

MonthStarting balanceInterest earnedEnding Balance11000.002.501002.5021002.502.511005.0131005.012.511007.5241007.522.521010.0451010.042.531012.5761012.572.531015.1071015.102.541017.6481017.642.541020.1891020.182.551022.73101022.732.561025.29111025.292.561027.85121027.852.571030.42

We want to simplify the process for calculating compounding, because creating a table like the one above is time consuming. Luckily, math is good at giving you ways to take shortcuts. To find an equation to represent this, if Pm represents the amount of money after m months, then we could write the recursive equation:

P0 = $1000

Pm = (1+0.0025)Pm-1

You probably recognize this as the recursive form of exponential growth. If not, we go through the steps to build an explicit equation for the growth in the next example.

Example

Build an explicit equation for the growth of $1000 deposited in a bank account offering 3% interest, compounded monthly.

Show Solution

• P0 = $1000
• P­1 = 1.0025P­0 = 1.0025 (1000)
• P­2 = 1.0025P­1 = 1.0025 (1.0025 (1000)) = 1.0025 2(1000)
• P­3 = 1.0025P­2 = 1.0025 (1.00252(1000)) = 1.00253(1000)
• P­4 = 1.0025P­3 = 1.0025 (1.00253(1000)) = 1.00254(1000)

Observing a pattern, we could conclude

• Pm = (1.0025)m($1000)

Notice that the $1000 in the equation was P0, the starting amount. We found 1.0025 by adding one to the growth rate divided by 12, since we were compounding 12 times per year.

 

Generalizing our result, we could write

[latex]{{P}_{m}}={{P}_{0}}{{\left(1+\frac{r}{k}\right)}^{m}}[/latex]

 

In this formula:

• m is the number of compounding periods (months in our example)
• r is the annual interest rate
• k is the number of compounds per year.

View this video for a walkthrough of the concept of compound interest.

While this formula works fine, it is more common to use a formula that involves the number of years, rather than the number of compounding periods. If N is the number of years, then m = N k. Making this change gives us the standard formula for compound interest.

Compound Interest

[latex]P_{N}=P_{0}\left(1+\frac{r}{k}\right)^{Nk}[/latex]

• PN is the balance in the account after N years.
• P0 is the starting balance of the account (also called initial deposit, or principal)
• r is the annual interest rate in decimal form
• k is the number of compounding periods in one year
  • If the compounding is done annually (once a year), k = 1.
  • If the compounding is done quarterly, k = 4.
  • If the compounding is done monthly, k = 12.
  • If the compounding is done daily, k = 365.

The most important thing to remember about using this formula is that it assumes that we put money in the account once and let it sit there earning interest. 

In the next example, we show how to use the compound interest formula to find the balance on a certificate of deposit after 20 years.

Example

A certificate of deposit (CD) is a savings instrument that many banks offer. It usually gives a higher interest rate, but you cannot access your investment for a specified length of time. Suppose you deposit $3000 in a CD paying 6% interest, compounded monthly. How much will you have in the account after 20 years?

Show Solution

In this example,

P0 = $3000the initial depositr = 0.066% annual ratek = 1212 months in 1 yearN = 20 since we’re looking for how much we’ll have after 20 years

So [latex]{{P}_{20}}=3000{{\left(1+\frac{0.06}{12}\right)}^{20\times12}}=\$9930.61[/latex] (round your answer to the nearest penny)

A video walkthrough of this example problem is available below.

Let us compare the amount of money earned from compounding against the amount you would earn from simple interest

YearsSimple Interest ($15 per month)6% compounded monthly = 0.5% each month.5$3900$4046.5510$4800$5458.1915$5700$7362.2820$6600$9930.6125$7500$13394.9130$8400$18067.7335$9300$24370.65

As you can see, over a long period of time, compounding makes a large difference in the account balance. You may recognize this as the difference between linear growth and exponential growth.

Try It

 

Evaluating exponents on the calculator

When we need to calculate something like [latex]5^3[/latex] it is easy enough to just multiply [latex]5\cdot{5}\cdot{5}=125[/latex].  But when we need to calculate something like [latex]1.005^{240}[/latex], it would be very tedious to calculate this by multiplying [latex]1.005[/latex] by itself [latex]240[/latex] times!  So to make things easier, we can harness the power of our scientific calculators.

Most scientific calculators have a button for exponents.  It is typically either labeled like:

^ ,   [latex]y^x[/latex] ,   or [latex]x^y[/latex] .

To evaluate [latex]1.005^{240}[/latex] we’d type [latex]1.005[/latex] ^ [latex]240[/latex], or [latex]1.005 \space{y^{x}}\space 240[/latex].  Try it out – you should get something around 3.3102044758.

Example

You know that you will need $40,000 for your child’s education in 18 years. If your account earns 4% compounded quarterly, how much would you need to deposit now to reach your goal?

Show Solution

In this example, we’re looking for P0.

r = 0.044%k = 44 quarters in 1 yearN = 18Since we know the balance in 18 yearsP18 = $40,000The amount we have in 18 years

In this case, we’re going to have to set up the equation, and solve for P0.

[latex]\begin{align}&40000={{P}_{0}}{{\left(1+\frac{0.04}{4}\right)}^{18\times4}}\\&40000={{P}_{0}}(2.0471)\\&{{P}_{0}}=\frac{40000}{2.0471}=\$19539.84 \\\end{align}[/latex]

So you would need to deposit $19,539.84 now to have $40,000 in 18 years.

Try It

 

Rounding

It is important to be very careful about rounding when calculating things with exponents. In general, you want to keep as many decimals during calculations as you can. Be sure to keep at least 3 significant digits (numbers after any leading zeros). Rounding 0.00012345 to 0.000123 will usually give you a “close enough” answer, but keeping more digits is always better.

Example

To see why not over-rounding is so important, suppose you were investing $1000 at 5% interest compounded monthly for 30 years.

P0 = $1000the initial depositr = 0.055%k = 1212 months in 1 yearN = 30since we’re looking for the amount after 30 years

If we first compute r/k, we find 0.05/12 = 0.00416666666667

Here is the effect of rounding this to different values:

 

r/k rounded to:

Gives P­30­ to be:Error0.004$4208.59$259.150.0042$4521.45$53.710.00417$4473.09$5.350.004167$4468.28$0.540.0041667$4467.80$0.06no rounding$4467.74

If you’re working in a bank, of course you wouldn’t round at all. For our purposes, the answer we got by rounding to 0.00417, three significant digits, is close enough – $5 off of $4500 isn’t too bad. Certainly keeping that fourth decimal place wouldn’t have hurt.

View the following for a demonstration of this example.

 

Using your calculator

In many cases, you can avoid rounding completely by how you enter things in your calculator. For example, in the example above, we needed to calculate [latex]{{P}_{30}}=1000{{\left(1+\frac{0.05}{12}\right)}^{12\times30}}[/latex]

We can quickly calculate 12×30 = 360, giving [latex]{{P}_{30}}=1000{{\left(1+\frac{0.05}{12}\right)}^{360}}[/latex].

Now we can use the calculator.

Type thisCalculator shows0.05 ÷ 12 = .0.00416666666667+ 1 = .1.00416666666667yx 360 = .4.46774431400613× 1000 = .4467.74431400613

Using your calculator continued

The previous steps were assuming you have a “one operation at a time” calculator; a more advanced calculator will often allow you to type in the entire expression to be evaluated. If you have a calculator like this, you will probably just need to enter:

1000 ×  ( 1 + 0.05 ÷ 12 ) yx 360 =

Solving For Time

Note: This section assumes you’ve covered solving exponential equations using logarithms, either in prior classes or in the growth models chapter.

Often we are interested in how long it will take to accumulate money or how long we’d need to extend a loan to bring payments down to a reasonable level.

Examples

If you invest $2000 at 6% compounded monthly, how long will it take the account to double in value?

Show Solution

This is a compound interest problem, since we are depositing money once and allowing it to grow. In this problem,

P0 = $2000the initial depositr = 0.066% annual ratek = 1212 months in 1 year

So our general equation is [latex]{{P}_{N}}=2000{{\left(1+\frac{0.06}{12}\right)}^{N\times12}}[/latex]. We also know that we want our ending amount to be double of $2000, which is $4000, so we’re looking for N so that PN = 4000. To solve this, we set our equation for PN equal to 4000.

[latex]4000=2000{{\left(1+\frac{0.06}{12}\right)}^{N\times12}}[/latex]              Divide both sides by 2000

[latex]2={{\left(1.005\right)}^{12N}}[/latex]                                   To solve for the exponent, take the log of both sides

[latex]\log\left(2\right)=\log\left({{\left(1.005\right)}^{12N}}\right)[/latex]             Use the exponent property of logs on the right side

[latex]\log\left(2\right)=12N\log\left(1.005\right)[/latex]                     Now we can divide both sides by 12log(1.005)

[latex]\frac{\log\left(2\right)}{12\log\left(1.005\right)}=N[/latex]                              Approximating this to a decimal

N = 11.581

It will take about 11.581 years for the account to double in value. Note that your answer may come out slightly differently if you had evaluated the logs to decimals and rounded during your calculations, but your answer should be close. For example if you rounded log(2) to 0.301 and log(1.005) to 0.00217, then your final answer would have been about 11.577 years.