Business Calculus Quiz: Solving Applied Equations

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Solving Applied EquationsQuestion 1 of 16

An investment grows according to the formula $A(t) = 5000e^{0.06t}$ , where $A(t)$ is the account value in dollars after $t$ years. If inflation reduces the real purchasing power at a rate of 3% per year, the real value is $R(t) = A(t) \cdot e^{-0.03t}$ . After how many complete years will the real value first exceed $6,000?

After 6 complete years of investment growth

After 7 complete years of investment growth

After 8 complete years of investment growth

After 9 complete years of investment growth

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Business Calculus Quiz

Business Calculus Quiz: Solving Applied Equations

Practice Solving Applied Equations in Business Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Solving Applied Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Calculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

After 6 complete years of investment growth
After 7 complete years of investment growth (correct answer)
After 8 complete years of investment growth
After 9 complete years of investment growth

Explanation: We need

R(t) > 6000

. Substituting:

5000e^{0.06t} \cdot e^{-0.03t} > 6000

. This simplifies to

5000e^{0.03t} > 6000

, or

e^{0.03t} > 1.2

. Taking natural log:

0.03t > \ln(1.2) \approx 0.1823

. So

t > \frac{0.1823}{0.03} \approx 6.08

. Since we need complete years, the real value first exceeds $6,000 after 7 complete years. Choice A uses $

t = 6

, which gives

R(6) = 5000e^{0.18} \approx 5986 < 6000

$. Choice C and D result from errors in combining the exponential terms or misinterpreting the inequality direction.

Question 2

A company's profit $P$ from selling $x$ thousand units of a product is given by the function $P(x) = -2x^2 + 160x - 2000$ . The company needs its profit to be at least $1,000. For which interval of production levels (in thousands of units) is this condition met?

Between 30 and 50 units, inclusive. (correct answer)
Between 16 and 64 units, inclusive.
For any level below 30 units or above 50 units.
Only at the single production level of 40 units.

Explanation: To find when the profit is at least $1,000, we must solve the inequality $P(x) \ge 1000$.

-2x^2 + 160x - 2000 \ge 1000

-2x^2 + 160x - 3000 \ge 0

Divide by -2 and reverse the inequality sign:

x^2 - 80x + 1500 \le 0

To find the roots of the corresponding equation

x^2 - 80x + 1500 = 0

, we can factor the quadratic:

(x - 30)(x - 50) = 0

The roots are

x=30

and

x=50

. Since the parabola

y = x^2 - 80x + 1500

opens upwards, the expression is less than or equal to zero between the roots. Thus, the solution is

30 \le x \le 50

Question 3

The market demand for a particular type of coffee is given by $D(p) = 210 - 2p$ , and the market supply is given by $S(p) = 3p - 40$ , where $p$ is the price per pound in dollars. To support coffee producers, the government introduces a subsidy of $5 per pound paid directly to the producers. What is the new equilibrium quantity of coffee sold after the subsidy is implemented?

110 pounds
104 pounds
116 pounds (correct answer)
120 pounds

Explanation: Let

p_c

be the price consumers pay and

p_s

be the price producers receive. With a $5 subsidy, $p_s = p_c + 5

. Demand depends on the consumer price:

q = D(p_c) = 210 - 2p_c

. Supply depends on the producer price:

q = S(p_s) = 3p_s - 40$. Substitute

p_s

into the supply equation:

q = 3(p_c + 5) - 40 = 3p_c + 15 - 40 = 3p_c - 25

. At equilibrium, demand quantity equals supply quantity:

210 - 2p_c = 3p_c - 25

235 = 5p_c

p_c = 47

To find the equilibrium quantity, substitute this price back into the demand equation:

q = 210 - 2(47) = 210 - 94 = 116

Thus, the new equilibrium quantity is 116 pounds.

Question 4

An initial investment is made into an account with an interest rate of 100% compounded annually. After 3 years, $1,000 is withdrawn. The remaining balance continues to grow at the same rate for another 4 years, resulting in a final balance of $24,000. What was the amount of the initial investment?

$195.31
$312.50 (correct answer)
$1,187.50
$1,450.00

Explanation: Let

P

be the initial investment. The annual interest rate is 100%, so the growth factor is

(1+1) = 2

. After the first 3 years, the account balance

A_1

is:

A_1 = P(2)^3 = 8P

Then, $1,000 is withdrawn, so the new principal $P_2

is:

P_2 = 8P - 1000

$ This amount grows for another 4 years, so the final amount

A_f

is:

A_f = P_2(2)^4 = (8P - 1000) \cdot 16

We are given that the final balance is $24,000. We can solve for $P$:

24,000 = (8P - 1000) \cdot 16

24,000 / 16 = 8P - 1000

1,500 = 8P - 1000

2,500 = 8P

P = 2,500 / 8 = 312.50

The initial investment was $312.50.

Question 5

The time required for an investment to double is 10 years, assuming the interest is compounded continuously. Starting from the time of the initial investment, how many additional years will it take for the investment to be worth five times the initial principal?

13.2 years (correct answer)
15.0 years
23.2 years
25.0 years

Explanation: First, use the doubling time to find the interest rate

r

. The formula for continuous compounding is

A = Pe^{rt}

. For doubling,

A=2P

and

t=10

2P = Pe^{r \cdot 10}

2 = e^{10r}

\ln(2) = 10r

r = \frac{\ln(2)}{10}

Next, find the total time

t_{total}

for the investment to be worth five times the principal (

A=5P

5P = Pe^{rt_{total}}

5 = e^{rt_{total}}

\ln(5) = rt_{total}

t_{total} = \frac{\ln(5)}{r}

Substitute the expression for

r

t_{total} = \frac{\ln(5)}{\ln(2)/10} = 10 \frac{\ln(5)}{\ln(2)} \approx 10 \frac{1.6094}{0.6931} \approx 23.2

This is the total time from the beginning. The question asks for the additional years needed after the doubling time (10 years). Additional time =

t_{total} - 10 = 23.2 - 10 = 13.2

years.

Question 6

A farm sells two types of organic fertilizer, GrowFast and SuperBloom. A shipment of 250 bags of fertilizer generated $4,050 in revenue. Each bag of GrowFast costs $15 and each bag of SuperBloom costs $20. Due to a packaging error, the number of bags of each type was not recorded. How many bags of GrowFast were in the shipment?

60
125
150
190 (correct answer)

Explanation: Let

x

be the number of bags of GrowFast and

y

be the number of bags of SuperBloom. We can set up a system of two linear equations. Equation 1 (total bags):

x + y = 250

Equation 2 (total revenue):

15x + 20y = 4050

We want to solve for

x

. From Equation 1, we can express

y

in terms of

x

y = 250 - x

. Substitute this expression for

y

into Equation 2:

15x + 20(250 - x) = 4050

Distribute the 20:

15x + 5000 - 20x = 4050

Combine the

x

terms:

-5x + 5000 = 4050

Subtract 5000 from both sides:

-5x = -950

Divide by -5:

x = 190

There were 190 bags of GrowFast in the shipment.

Question 7

The weekly profit function for a bakery is $P(x) = -x^2 + 110x - 1000$ , where $x$ is the number of cakes sold ( $x > 0$ ). For which range of whole cakes sold is the average profit per cake at least $10?

Between 10 and 100 cakes, inclusive.
Between 12 and 88 cakes, inclusive. (correct answer)
For 11 or fewer cakes, or for 89 or more cakes.
Only at the single production level of 55 cakes.

Explanation: The average profit per cake is

AP(x) = P(x)/x

. We need to solve the inequality

AP(x) \ge 10

\frac{-x^2 + 110x - 1000}{x} \ge 10

Since

x

represents the number of cakes,

x > 0

. We can multiply both sides by

x

-x^2 + 110x - 1000 \ge 10x

Rearrange into a standard quadratic inequality:

-x^2 + 100x - 1000 \ge 0

Multiply by -1 and reverse the inequality sign:

x^2 - 100x + 1000 \le 0

Find the roots of

x^2 - 100x + 1000 = 0

using the quadratic formula:

x = \frac{-(-100) \pm \sqrt{(-100)^2 - 4(1)(1000)}}{2(1)} = \frac{100 \pm \sqrt{10000 - 4000}}{2} = \frac{100 \pm \sqrt{6000}}{2}

x = \frac{100 \pm 10\sqrt{60}}{2} = 50 \pm 5\sqrt{60} \approx 50 \pm 5(7.746) \approx 50 \pm 38.73

The roots are approximately

x_1 \approx 11.27

and

x_2 \approx 88.73

. The parabola opens upward, so the inequality

... \le 0

is satisfied between the roots. The range is

[11.27, 88.73]

. Since

x

must be a whole number of cakes, the range is from 12 to 88, inclusive.

Question 8

A pharmaceutical company models the concentration of a drug in the bloodstream using $C(t) = \frac{40t}{t^2 + 4}$ , where $C(t)$ is the concentration in mg/L and $t$ is time in hours after administration. The drug is considered therapeutic when the concentration is at least 8 mg/L. For how long is the drug therapeutic?

For exactly 2 hours starting immediately after administration
From 1 hour to 4 hours after administration, inclusive (correct answer)
From 0.5 hours to 8 hours after administration, inclusive
From 2 hours to 6 hours after administration, inclusive

Explanation: We solve

\frac{40t}{t^2 + 4} \geq 8

. Multiplying both sides by

(t^2 + 4)

(which is always positive):

40t \geq 8(t^2 + 4) = 8t^2 + 32

. Rearranging:

8t^2 - 40t + 32 \leq 0

, or

t^2 - 5t + 4 \leq 0

. Factoring:

(t-1)(t-4) \leq 0

. This inequality is satisfied when

1 \leq t \leq 4

. Choice A incorrectly assumes the maximum occurs at

t=2

and lasts for 2 hours total. Choice C results from computational errors in solving the quadratic inequality. Choice D shifts the correct interval by 1 hour due to misreading the factored form.

Question 9

A renewable energy company's cost per kilowatt-hour $C$ (in cents) depends on production volume $x$ (in thousands of kWh) according to $C(x) = \frac{x^2 - 40x + 500}{x}$ for $x > 0$ . If market prices require the cost to be no more than 8 cents per kWh, what production volumes are feasible?

Between 10,000 and 30,000 kWh per period is optimal (correct answer)
Between 15,000 and 25,000 kWh per period is optimal
Between 20,000 and 40,000 kWh per period is optimal
Only production above 35,000 kWh per period is feasible

Explanation: We solve

C(x) \leq 8

, so

\frac{x^2 - 40x + 500}{x} \leq 8

. This simplifies to

x - 40 + \frac{500}{x} \leq 8

, or

x + \frac{500}{x} \leq 48

. Multiplying by

x > 0

x^2 + 500 \leq 48x

, which gives

x^2 - 48x + 500 \leq 0

. Using the quadratic formula:

x = \frac{48 \pm \sqrt{2304 - 2000}}{2} = \frac{48 \pm \sqrt{304}}{2} = \frac{48 \pm 17.44}{2}

. So

x \approx 15.28

x \approx 32.72

. The inequality

x^2 - 48x + 500 \leq 0

is satisfied when

15.28 \leq x \leq 32.72

. Converting to kWh (multiply by 1000): 15,280 to 32,720 kWh. The closest range is 10,000 to 30,000 kWh. Choice B is too narrow, Choice C extends too far beyond the upper bound, Choice D incorrectly suggests only high production is feasible.

Question 10

The total profit $P(x)$ for a firm, where $x$ is the number of units produced, is given by a model whose derivative (the marginal profit) is $P'(x) = -x^2 + 80x - 1200$ . Assuming the number of units $x$ cannot be negative, for which production levels is the total profit decreasing?

For levels between 20 and 60 units.
For levels from 0 to 20 units and for levels above 60 units. (correct answer)
For any production level above 40 units.
Only at the specific production levels of 20 and 60 units.

Explanation: The total profit function

P(x)

is decreasing when its derivative, the marginal profit

P'(x)

, is negative. We need to solve the inequality

P'(x) < 0

-x^2 + 80x - 1200 < 0

Multiply by -1 and reverse the inequality sign:

x^2 - 80x + 1200 > 0

First, find the roots of the equation

x^2 - 80x + 1200 = 0

. Factoring the quadratic gives:

(x - 20)(x - 60) = 0

The roots are

x=20

and

x=60

. The parabola

y = x^2 - 80x + 1200

opens upwards, so it is positive (greater than 0) outside of its roots, i.e., when

x < 20

x > 60

. Since production

x

must be non-negative (

x \ge 0

), the intervals where profit is decreasing are

[0, 20) \cup (60, \infty)

Question 11

A manufacturer produces a component with an ideal length of 150 mm. For quality control, the actual length $L$ must be within a tolerance of 0.4 mm of the ideal length, satisfying the inequality $|L - 150| \le 0.4$ . The length of the component is found to be dependent on the ambient temperature $T$ in degrees Celsius, according to the linear model $L(T) = 2T + 140$ . Which range of temperatures is acceptable for producing components that are within tolerance?

$T \le 4.8$ or $T \ge 5.2$
$-145.2 \le T \le 5.2$
$T \le 5.2$
$4.8 \le T \le 5.2$ (correct answer)

Explanation: This problem combines absolute value inequalities with function composition—two key concepts in business calculus. When you see a tolerance specification with absolute value, you're looking for a range of values that satisfy the constraint. Start with the tolerance requirement:

|L - 150| \le 0.4

. This absolute value inequality means the length must be between 149.6 mm and 150.4 mm, so we have

149.6 \le L \le 150.4

. Since length depends on temperature through

L(T) = 2T + 140

, substitute this into our inequality:

149.6 \le 2T + 140 \le 150.4

. Solving the compound inequality: subtract 140 from all parts to get

9.6 \le 2T \le 10.4

, then divide by 2 to find

4.8 \le T \le 5.2

. This matches answer choice D. Choice A gives

T \le 4.8

T \ge 5.2

, which represents the complement of the correct range—temperatures that would produce components outside the tolerance. Choice B includes negative temperatures down to -145.2, which comes from incorrectly solving only one side of the inequality or misunderstanding the absolute value. Choice C gives only an upper bound without the necessary lower bound, missing that temperatures below 4.8°C would also produce out-of-spec components. When working with tolerance problems, always convert the absolute value inequality to a compound inequality first, then substitute your function and solve systematically. Double-check by testing boundary values to ensure your components stay within specification.

Question 12

A company finds that its total sales $S$ , in thousands of dollars, is related to its marketing expenditure $m$ , in thousands of dollars, by the model $S(m) = 50\sqrt{m}$ . The company has already spent $9,000 on marketing. How much more must be spent on marketing to increase total sales by $75,000?

$2,250
$4,500
$11,250 (correct answer)
$20,250

Explanation: The variables

S

and

m

are in thousands of dollars.

Calculate the initial sales from the initial spending of $m_1 = 9$ (thousand dollars). $S_1 = 50\sqrt{9} = 50(3) = 150$ So, initial sales are $150,000.
The desired increase in sales is $75,000, so $\Delta S = 75 $. The target sales level is:$ $S_{target} = S_1 + \Delta S = 150 + 75 = 225$ $ So, target sales are $225,000.
Find the total marketing expenditure, $m_{total}$, needed to achieve this target sales level. $225 = 50\sqrt{m_{total}}$ $\sqrt{m_{total}} = \frac{225}{50} = 4.5$ $m_{total} = (4.5)^2 = 20.25$ So, total spending needs to be $20,250.
The question asks for the additional spending required. Additional spending = $m_{total} - m_1 = 20.25 - 9 = 11.25$. This corresponds to $11,250.

Question 13

A company buys a machine for $120,000. Its value depreciates by 20% each year. The company's policy is to replace the machine when its value falls below $30,000. In which year will the company first replace the machine?

Year 4
Year 6
Year 7 (correct answer)
Year 8

Explanation: The value

V

of the machine after

t

years can be modeled by the exponential decay function

V(t) = V_0(1-r)^t

, where

V_0

is the initial value and

r

is the annual depreciation rate. Here,

V_0 = 120,000

and

r = 0.20

. So,

V(t) = 120,000(0.8)^t

. We need to find the smallest integer

t

for which

V(t) < 30,000

120,000(0.8)^t < 30,000

(0.8)^t < \frac{30,000}{120,000}

(0.8)^t < 0.25

To solve for

t

, take the natural logarithm of both sides. Since

\ln(0.8)

is negative, we must reverse the inequality sign when dividing by it.

t \cdot \ln(0.8) < \ln(0.25)

t > \frac{\ln(0.25)}{\ln(0.8)} \approx \frac{-1.3863}{-0.2231} \approx 6.21

Since

t

must be greater than 6.21, the first integer year in which the condition is met is year 7. We can check the values:

V(6) = 120,000(0.8)^6 \approx 31,457

, which is not below $30,000. $V(7) = 120,000(0.8)^7 \approx 25,166$, which is below $30,000.

Question 14

An online data storage provider charges based on monthly usage. The first 50 gigabytes (GB) are charged at $0.50 per GB. Any data usage beyond 50 GB is charged at $0.25 per GB. If a customer's bill for one month was $45.00, how many gigabytes of data did they use?

80 GB
90 GB
180 GB
130 GB (correct answer)

Explanation: This is a piecewise pricing problem where you need to understand how tiered billing structures work. When companies charge different rates for different usage levels, you must calculate each tier separately. Let's work backwards from the $45 bill. The first 50 GB costs $

50 \times \

0.50 = $25

. Since the total bill is $45, the remaining $

\

45 - $25 = $20$$ must come from usage beyond 50 GB at the lower rate of $0.25 per GB. To find the extra gigabytes:

\frac{\$20}{\$0.25} = 80

GB beyond the first 50 GB. Total usage =

50 + 80 = 130

GB. Let's verify: First 50 GB at $0.50 = $25, plus 80 GB at $0.25 = $20, totaling $45. ✓ Answer A (80 GB) represents just the overage amount - this is what you get if you forget to add back the initial 50 GB tier. Answer B (90 GB) would only generate a bill of $35 total (50 × $0.50 + 40 × $0.25). Answer C (180 GB) is far too high - this would create a bill of $57.50, not $45. Study tip: With tiered pricing problems, always break the calculation into segments. Calculate what each tier contributes to the total cost separately, then combine them. Watch out for the trap of forgetting to account for all pricing tiers when determining total quantities.

Question 15

A tech startup's user growth follows the logistic model $N(t) = \frac{50000}{1 + 49e^{-0.2t}}$ , where $N(t)$ is the number of users after $t$ months. The company needs at least 15,000 users to attract venture capital funding. In which month will they first be eligible for funding?

Month 15, as they will have just exceeded 15,000 users
Month 16, as they will have just exceeded 15,000 users (correct answer)
Month 17, as they will have just exceeded 15,000 users
Month 18, as they will have just exceeded 15,000 users

Explanation: We solve

\frac{50000}{1 + 49e^{-0.2t}} \geq 15000

. Cross-multiplying:

50000 \geq 15000(1 + 49e^{-0.2t})

. This gives

50000 \geq 15000 + 735000e^{-0.2t}

, so

35000 \geq 735000e^{-0.2t}

. Therefore

e^{-0.2t} \leq \frac{35000}{735000} = \frac{1}{21}

. Taking natural log:

-0.2t \leq \ln(\frac{1}{21}) = -\ln(21) \approx -3.044

. So

t \geq \frac{3.044}{0.2} = 15.22

. Since we need the first complete month where they qualify, this occurs in month 16. Verification:

N(15) \approx 14,535 < 15,000

and

N(16) \approx 16,667 > 15,000

Question 16

The total cost to produce $x$ items is given by the function $C(x) = 0.1x^2 + 30x + 4000$ . For which interval of production levels will the average cost per item be at most $80?

For levels between 100 and 400 units, inclusive. (correct answer)
For levels below 100 units or above 400 units.
Only at the single production level of 200 units.
For all production levels above 100 units.

Explanation: The average cost function is

AC(x) = C(x)/x

. We need to solve the inequality

AC(x) \le 80

AC(x) = \frac{0.1x^2 + 30x + 4000}{x} \le 80

Since the number of items

x

must be positive, we can multiply both sides by

x

without changing the inequality direction:

0.1x^2 + 30x + 4000 \le 80x

Rearrange the terms to form a standard quadratic inequality:

0.1x^2 - 50x + 4000 \le 0

Multiply by 10 to clear the decimal:

x^2 - 500x + 40000 \le 0

Find the roots of

x^2 - 500x + 40000 = 0

by factoring:

(x - 100)(x - 400) = 0

The roots are

x=100

and

x=400

. Since the parabola

y = x^2 - 500x + 40000

opens upwards, the expression is less than or equal to zero between the roots. Therefore, the solution is

100 \le x \le 400