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Precalculus : Application Problems

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Example Question #1 : Application Problems

If you invest into a savings account which earns an interest rate per year, how much would it take for your deposit to double?

Possible Answers:

$10 \text{ years}$

$2 \text{ years}$

$\frac{\ln 2}{\ln(1+i)}$ $\text{years}$

$\frac{\ln (1+i)}{2^n}$ $\text{years}$

Correct answer:

$\frac{\ln 2}{\ln(1+i)}$ $\text{years}$

Explanation:

The equation of the value for this problem is

$\small R(1+i)^n=2R$ .

We can divide by R to get

$\small \small (1+i)^n=2\,$ .

We want to solve for n in this case, which is the amount of years. If we use the natural log on both sides and properties of logarithms, we get

$\small \ln(1+i)^n=n\ln(1+i)=\ln2$ .

If we solve for n, we get

$\small n = \frac{\ln(2)}{\ln(1+i)}$

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Example Question #1 : Application Problems

If you deposit $\$100$ into a savings account which earns a $5\%$ yearly interest rate, how much is in your account after two years?

Possible Answers:

$\$105$

$\$110.38$

$\$\small 110.25$

$\$110$

Correct answer:

$\$\small 110.25$

Explanation:

Since we are investing for two years with a yearly rate of 5%, we will use the formula to calculate compound interest.

A=P(1+r)^t

where

is the amount of money after time.

is the principal amount (initial amount).

is the interest rate.

is time.

Our amount after two years is:

$\small 100(1.05^2)=100(1.1025)=110.25$

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Example Question #1 : Application Problems

If you deposit into a savings account which compounds interest every month, what is the expression for the amount of money in your account after years if you earn a nominal interest rate of compounded monthly?

Possible Answers:

$\small \small R\bigg(1+\frac{i}{12}\bigg)^{2}$

$\small \small R\big(1+i\big)^{2}$

$\small \small R\big(1+i\big)^{24}$

$\small R\bigg(1+\frac{i}{12}\bigg)^{24}$

Correct answer:

$\small R\bigg(1+\frac{i}{12}\bigg)^{24}$

Explanation:

Since is the nominal interest rate compounded monthly we write the interest term as $\frac{i}{12}$ as it is the effective monthly rate.

We compound for years which is months. Since our interest rate is compounded monthly our time needs to be in the same units thus, months will be the units of time.

Plugging this into the equation for compound interest gives us the expression:

$A=P\left(1+\frac{i}{c}\right)^{ct}$

$A=R\left(1+\frac{i}{12}\right)^{12\cdot 2}$

$A=\small \small R \left(1+\frac{i}{12}\right)^{24}$

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Example Question #4 : Application Problems

John opens a savings account and deposits $\$1000$ into it. This savings account gains $5\%$ interest per year. After years, John withdraws all the money, and deposits it into another savings account with $6\%$ interest per year. years later, John withdraws the money.

How much money does John have after this year period? (Assume compound interest in both accounts)

Possible Answers:

$\$1118.12$

$\$1400.23$

$\$1276.28$

$\$1157.63$

$\$1300.71$

Correct answer:

$\$1300.71$

Explanation:

Plugging our numbers into the formula for compound interest, we have:

1000(1+.05)^3 = 1157.625 .

So John has about $\$1157.63$ after the first three years.

After placing his money into the other savings account, he has

1157.63(1+.06)^2 = 1300.713068 after more years.

So John has accumulated about $\$1300.71$ .

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Example Question #5 : Application Problems

Suppose you took out a loan years ago that gains $6.9\%$ interest. Suppose that you haven't made any payments on it yet, and right now you owe $\$100000$ on the loan. How much was the loan worth when you took it out?

Possible Answers:

$\$89558.26$

$\$51312.47$

$\$38734.83$

$\$35789.47$

None of the other answers.

Correct answer:

$\$51312.47$

Explanation:

The formula for the compund interest is as follows:

A = P(1+i)^t

By substuting known values into the compound interest formula, we have:

$100000 = P(1.069)^{10}$ .

From here, substitute known values.

$\frac{100000}{1.069^{10}} = P$

Divide by $1.069^{10}$

51312.47 = P

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Example Question #6 : Application Problems

How many years does it take for $\$1000$ to grow into $\$5000$ when the $\$1000$ is deposited into a savings account that gains $3\%$ annual compound interest?

Possible Answers:

years

None of the other answers.

years

Correct answer:

None of the other answers.

Explanation:

The correct answer is about 54.45 years.

To find the number of years required, we solve the compound interest formula for .

The formula is as follows:

A = P(1+i)^t

Substitute known values.

5000 = 1000(1+.03)^t

Divide by 1000 .

5 = (1.03)^t

Take the natural log of both sides.

$\ln5 = \ln ((1.05)^t)$

Use the log power rule.

$\ln5 = t\ln1.03$

Divide by $\ln1.03$ .

$\frac{\ln5}{\ln1.03} = t$

54.44869 = t use a calculator to simplify.

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Example Question #7 : Application Problems

The amount of phosphorus present in a sample at a given time is given by the following equation:

$Q(t)=Q_0e^{-0.03t}$

Where is in days, Q(t) is the amount of phosphorus after days, and Q_0 is the initial amount of phosphorus at the beginning of the first day. What percent of the initial amount of phosphorus is left after days of decay?

Possible Answers:

$47\%$

$11\%$

$83\%$

$56\%$

$23\%$

Correct answer:

$47\%$

Explanation:

The problem asks us for the percent that the amount of phosphorus after t days is of the original amount of phosphorus. If we think of this percentage with respect to the variables present in the equation, we can see that the following fraction expresses the amount of phosphorus after t days as a percentage of the initial amount of phosphorus:

$\frac{Q(t)}{Q_0}$

So if this is the fraction we want to solve for, we should divide both sides of the equation by Q_0 to obtain this fraction on the left side of the equation:

$Q(t)=Q_0e^{-0.03t}$

$\frac{Q(t)}{Q_0}=\frac{Q_0e^{-0.03t}}{Q_0}$

$\frac{Q(t)}{Q_0}=e^{-0.03t}$

We now have the fraction we want to solve for in terms of just one variable, , for which we plug in 25 days to find the percentage of phosphorus left of the initial amount after 25 days:

$\frac{Q(25)}{Q_0}=e^{-0.03(25)}=0.47$

So after 25 days of decay, the amount of phosphorus is 47% of the initial amount.

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Example Question #1 : Application Problems

The exponential decay of an element is given by the following function:

$Q(t)=Q_0e^{-0.04223t}$

Where Q(t) is the amount of the element left after days, and Q_0 is the initial amount of the element. If there are kg of the element left after days, what was the initial amount of the element?

Possible Answers:

127.1 kg

63.7 kg

45.3 kg

82.4 kg

106.3 kg

Correct answer:

106.3 kg

Explanation:

The problem asks us for the initial amount of the element, so first let's solve our equation for Q_0 :

$Q(t)=Q_0e^{-0.04223t}$

$Q_0=Q(t)e^{0.04223t}$

The problem tells us that 25 days has passed, which gives us , and it also tells us the amount left after 25 days, which gives us Q(25) . Now that we have our equation for Q_0 , we can plug in the given values to find the initial amount of the element:

$Q_0=Q(25)e^{0.04223(25)}=(37)e^{0.04223(25)}=106.3$ kg

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Example Question #2 : Application Problems

The exponential decay of an element is given by the function:

$Q(t) = Q_oe^{-0.42t}$

In this function, Q(t) is the amount of the element left after days, and Q_o is the initial amount of the element. If $25\:kg$ of the element is left after seven days, how much of the element was there to begin with, rounded to the nearest kilogram?

Possible Answers:

$140\:kg$

$180\:kg$

$122\:kg$

$512\:kg$

$473\:kg$

Correct answer:

$473\:kg$

Explanation:

To find the initial amount, you must rearrange the equation to solve for Q_o :

$Q(t) = Q_oe^{-0.42t}$

Divide both sides by $e^{-0.42t}$ :

$Q_o = \frac{Q(t)}{e^{-0.42t}}$

Substituting in the values from the problem gives

$Q_o = \frac{25}{e^{-0.42(7)}} \approx 473\:kg$

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Example Question #1 : Application Problems

The exponential decay of an element is given by the function

$Q(t) = Q_o e^{-0.9t}$

In this function, Q(t) is the amount left after days, and Q_o is the initial amount of the element. What percent of the element is left after ten days, rounded to the nearest whole percent?

Possible Answers:

$0\%$

$47\%$

$26\%$

$52\%$

$10\%$

Correct answer:

$0\%$

Explanation:

To find the final percentage of the element left, we must rearrange the equation to solve for $e^{-0.9t}$ :

$\frac{Q(t)}{Q_o} = e^{-0.9t}$

Now, using the ten days as , we can solve for the percent of the element left after ten days:

$\frac{Q(10)}{Q_o} = e^{-0.9*10} \approx 0.0001 = 0.01\% \approx 0\%$

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Precalculus : Application Problems

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Application Problems

Example Question #1 : Application Problems

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Example Question #4 : Application Problems

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