Applied Algebra FX01 Exam (C957)
The MATH 1200 C957 Applied Algebra Exam Practice Questions guide at Ulosca is an in-depth resource aimed at enhancing students' understanding of key algebraic concepts and their practical application in various problem-solving scenarios.
This guide includes 120+ exam-focused practice questions, each paired with detailed rationales that explain the correct answers and address common misconceptions. This ensures students gain a comprehensive understanding of linear equations, polynomial operations, and real-world mathematical applications.
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Free Applied Algebra FX01 Exam (C957) Questions
A company has the following information on last quarter’s sales (month 7 refers to July)
What was the average rate of change over the entire quarter (from July to September)?
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Sales increased an average of $15,000 per month.
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Sales increased an average of $37,000 per month.
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Sales increased an average of $6,285.71 per month.
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Sales increased an average of $10,000 per month.
Explanation
Correct Answer:
Sales increased an average of $15,000 per month.
Explanation:
The average rate of change is calculated as:
Step 1: Calculate the total change in sales from July (Month 7) to September (Month 9):
Total Change in Sales=333−303=30 (in thousands of dollars).
Step 2: Calculate the number of months:
From July to September is 2 months.
Step 3: Calculate the average rate of change:
Average Rate of Change = 302=15 (in thousands of dollars per month).
Thus, the average rate of change is $15,000 per month.
Why the Other Options Are Wrong:
"Sales increased an average of $37,000 per month." This overestimates the rate of change, as the total change in sales is only 30,000, not 37,000.
"Sales increased an average of $6,285.71 per month." This significantly underestimates the rate of change. It does not correspond to the actual total change over the 2 months.
"Sales increased an average of $10,000 per month." This underestimates the rate of change. The correct average increase is $15,000 per month.
An entrepreneur determines that investing in a certain business will return 5 times what was invested. For this situation, E() is a function where E is what was earned and is how much money the entrepreneur invested. What is the correct function notation to represent the earnings if the entrepreneur invested $1,000?
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E(5000) = 1000
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E(200) = 1000
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E(1000) = 200
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E(1000)=5000
Explanation
Correct Answer:
E(1000) = 5000
Explanation:
The entrepreneur earns 5 times the investment. The function can be written as:
E(x) = 5x
where x is the amount invested.
If the entrepreneur invests $1,000
E(1000) = 5 × 1000
= 5000
Thus, the correct representation is E(1000) = 5000.
Why the other options are wrong:
“E(5000) = 1000”: This is incorrect because E(x) represents what is earned, not the amount invested. The earnings for an investment of $5,000 would be E(5000) = 5 × 5000 = 25000, which is not $1,000.
“E(200) = 1000”: This is incorrect because investing $200 would yield E(200) = 5 × 200 =1000. The input here does not match the given scenario of investing $1,000.
“E(1000) = 200”: This is incorrect because the earnings from investing $1,000 are 5 times the investment, which is $5,000, not $200.
A local gerbil population can be estimated with the function where t = 0 represents the gerbil population, P, in the year 2020 and is measured in years. What is P(5) rounded to the nearest integer, and what does it represent?
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P(5) = 204, which means that in 2025, there are approximately 204 gerbils in the population.
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P(5) = 181, which means that in 2025, there are approximately 181 gerbils in the population.
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P(5) = 204, which means that in 2040, there are approximately 5,000 gerbils in the population.
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P(5) = 181, which means that in 2181, there were approximately 5,000 gerbils in the population.
Explanation
Solution:
Step 1: Calculate P(5)
Substitute t = 5 into the function:
P(5) =
Simplify the exponent:
P(5) =
≈ 180.63
Rounded to the nearest integer:
P(5) = 181
Step 2: Interpret P(5)
P(5)=181 means that in the year 2025, the gerbil population is approximately 181.
Correct Answer:
P(5)=181, which means that in 2025, there are approximately 181 gerbils in the population.
Why the Other Options Are Incorrect:
"P(5) = 204, which means that in 2025, there are approximately 204 gerbils in the population." This is incorrect because P(5) = 181, not 204.
"P(5) = 204, which means that in 2040, there are approximately 5,000 gerbils in the population." P(5) is calculated as 181, not 204, and the population is never 5,000 based on the given function.
"P(5) = 181, which means that in 2181, there are approximately 5,000 gerbils in the population." The year t=5 corresponds to 2025, not 2181, and the population is 181, not 5,000.
The graph below shows the number of page views (in thousands) for a website as a function of the number of employees marketing the website.
How many additional page views, on average, are expected for each additional worker when increasing from 10 to 20 employees?
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1,040 page views per worker
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9,590 page views per worker
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959 page views per worker
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366 page views per worker
Explanation
Solution:
where v(20) = 21.99, v(10) = 12.4
= 0.959
Hence on average any additional worker there will be 0.959 thousand page views
Correct answer:
959 page views per worker
Why other options are wrong:
"1,040 page views per worker": This is incorrect because the actual calculation yields 959 page views, not 1,040.
"9,590 page views per worker": This is incorrect because 9,590 page views is the total number of additional views for 10 workers, not per worker.
"366 page views per worker": This is incorrect because the average number of additional page views per worker is 959, not 366.
The number of households adopting solar panels in a suburban community can be modeled with a logistic function. The table below shows data from this model, where H is the number of households with solar panels after m months since a new incentive program was introduced
How should the average rate of change from month 2 to month 6 be interpreted?
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The solar panel adoption is spreading at a decreasing rate of 212 households per month
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The solar panel adoption is spreading at an increasing rate of 212 households per month.
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The solar panel adoption is spreading at a decreasing rate of 180 households per month.
-
The solar panel adoption is spreading at an increasing rate of 180 households per month.
Explanation
Solution:
To calculate the average rate of change, we use the formula:
= 212
Interpretation:
The logistic function shows that the adoption rate slows down as it approaches its maximum potential (asymptote) as fewer households remain without solar panels. Therefore, solar panel adoption is spreading at a decreasing rate of 212 households per month from month 2 to month 6.
Correct Answer:
The solar panel adoption is spreading at a decreasing rate of 212 households per month.
Why the Other Options Are Incorrect:
"The solar panel adoption is spreading at an increasing rate of 212 households per month": While the rate is 212, the logistic function shows decreasing rates over time, not increasing.
"The solar panel adoption is spreading at a decreasing rate of 180 households per month": The value 180 is incorrect; the actual average rate of change is 212.
"The solar panel adoption is spreading at an increasing rate of 180 households per month": Both the rate (180) and the description of increasing are incorrect.
A local business is having a 10-day sale. The company wants to know the number of customers each day and how much profit was made each day.
What can be concluded about the customers during this company's sale?
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The company had more customers the first three days of the sale than the last three days of the sale.
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The company earned more profit on the day it had the fewest customers than on the day it had the most customers.
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The company earned the lowest profits on the same day it had the fewest customers.
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The company had fewer customers the last day of the sale than the first day of the sale.
Explanation
The correct answer:
The company earned more profit on the day it had the fewest customers than on the day it had the most customers.
Solution:
First three days: 20 + 41 + 50 = 111 customers Last three days: 200 + 150 + 140 = 490 customers 111 < 490, so this is false
Fewest customers: Days 1 and 4 tied with 20 customers Most customers: Day 9 with 200 customers Profits on Day 1: $500 Profits on Day 4: $800 Profits on Day 9: $300 $800 > $300, so this is true!
Fewest customers: Days 1 and 4 tied with 20 customers Lowest profit: $105 on Day 8 This is false as the lowest profit day didn't align with the lowest customer day
Last day (Day 10): 150 customers First day (Day 1): 20 customers 150 > 20, so this is false
Why other options are wrong:
“The company had more customers the first three days of the sale than the last three days of the sale” First three days: 20 + 41 + 50 = 111 customers Last three days: 200 + 150 + 140 = 490 customers 111 < 490, so this is false
“The company earned the lowest profits on the same day it had the fewest customers” Fewest customers: Days 1 and 4 tied with 20 customers Lowest profit: $105 on Day 8 This is false as the lowest profit day didn't align with the lowest customer day”
“The company had fewer customers the last day of the sale than the first day of the sale” Last day (Day 10): 150 customers First day (Day 1): 20 customers 150 > 20, so this is false.
The following table shows the number of bacteria in a petri dish after t hours:
What is the average rate of change in this time period?
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-90 bacteria per hour
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bacteria per hour
-
bacteria per hour
-
90 bacteria per hour
Explanation
Correct Answer:
90 bacteria per hour
Explanation:
The average rate of change is calculated using the formula:
= 90 bacteria per hour
Since the bacteria count is increasing, the rate of change is positive.
Why other options are wrong:
“−90 bacteria per hour” This is incorrect because the bacteria count is increasing, not decreasing. A negative rate of change would imply the bacteria count is reducing, which is not the case here.
“-190 bacteria per hour” This is incorrect because the average rate of change is not a fraction. It is a whole number representing the increase in bacteria per hour.
“90 bacteria per hour” This is correct because the bacteria count increases by 90 bacteria every hour between 5 and 10 hours.
Computers frequently use algorithms to make large files smaller, a process called compression. When given the size of the file (in gigabytes), f, the amount of time in seconds, S, for the standard algorithm to compress a file is given by the equation S(f) = 0.5f3. Its graph is shown below as the solid curve. The time to use a new algorithm for this same process is given by the function N(f) = 0.5f2 +2, which is shown below in the dashed curve.
Based on this graph, which algorithm is quicker for a file that is 5 gigabytes in size?
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The new algorithm, N, is better since N(5) < S(5).
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The standard algorithm, S, is better since N(5) > S(5).
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The new algorithm, N, is better since S(5) < N(5).
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The standard algorithm, S, is better since S(5) < N(5).
Explanation
Solution
S(f) = 0.5f3
S(5) = 0.5 * 53
= 62.5 seconds
N(f) = 0.5f2 +2
N(5) = 0.5 * 52 + 2
= 14.5 seconds
Correct answer:
The new algorithm, N, is better since N(5) < S(5).
Why other options are wrong:
“The standard algorithm, S, is better since N(5) > S(5).” N(5) < S(5) hence the option is wrong
“The new algorithm, N, is better since S(5) < N(5)” S(5) > N(5) hence the option is wrong.
“The standard algorithm, S, is better since S(5) < N(5)” S(5) > N(5) hence the option is wrong.
A local business wants to model the relationship between the number of repeat customers in a three-month period and its quarterly revenue (in thousands of dollars). The company fit its data with the following linear model, which has r² value of 0.91.
Does the proposed function appropriately fit the data?
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No, the data should be modeled with a quadratic function.
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Yes, there is a moderate positive correlation.
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No, the value is not close enough to 1.
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Yes, there is a strong positive correlation.
Explanation
Correct answer:
Yes, there is a strong positive correlation.
Why others are wrong:
“No, the data should be modeled with a quadratic function.” This is wrong since the already fitted function has an r squared of 0.91 which is very strong.
“Yes, there is a moderate positive correlation.” The correlation is not moderate since the square root of 0.91 is 0.95 which is a strong correlation.
“No, the r² value is not close enough to 1.” For a linear relationship the value does not necessarily have to be close to 1. The observation just needs to follow the straight line.
The manager of a technical support department at a company is concerned because many of the employees in the department missed their target of reviewing 200 email accounts within the past 12 months. The manager gathered data on the number of combined sick and vacation hours 10 of the employees took and the number of email accounts they fell short on. The graph below shows the data the manager collected, along with the best-fit line, which has r² value of 0.76.
The manager wants to try to ensure that next year the employees on the team do not exceed 20 incomplete reviews. The manager concludes that if the sick and vacation policy is revised to limit employees to no more than 160 combined hours of sick and vacation time per year, then the employees shouldn't fall more than 20 accounts short of their target next year. Is this a valid conclusion and why?
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Yes, it is a valid conclusion because the model appropriately fits the data, and the model predicts that employees who take a combined 160 hours of sick and vacation time will fall less than 20 accounts short of their target.
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No, it is not a valid conclusion because (127,17) is an outlier, which causes predictions made with the best-fit line to be inaccurate.
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Yes, it is a valid conclusion because the model appropriately fits the data, and the model guarantees that employees who fall 20 or more reviews short of their target will miss more than 160 hours of combined sick and vacation time.
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No, it is not a valid conclusion because there were no employees who took 160 hours of combined sick and vacation time.
Explanation
Correct answer:
Yes, it is a valid conclusion because the model appropriately fits the data, and the model predicts that employees who take a combined 160 hours of sick and vacation time will fall less than 20 accounts short of their target.
Explanation
The correct conclusion is that the model appropriately fits the data, and it predicts that employees who take a combined 160 hours of sick and vacation time will fall less than 20 accounts short of their target. This is a valid conclusion because the linear regression model gives an estimated trend based on the data, and the predicted outcome of fewer than 20 incomplete reviews for employees with 160 hours of sick and vacation time is consistent with the pattern observed in the graph.
Why others are wrong:
“No, it is not a valid conclusion because (127,17) is an outlier, which causes predictions made with the best-fit line to be inaccurate.” From the graph there are no outliers.
“Yes, it is a valid conclusion because the model appropriately fits the data, and the model guarantees that employees who fall 20 or more reviews short of their target will miss more than 160 hours of combined sick and vacation time” is wrong because a model does not guarantee outcomes; it only makes predictions based on trends in the data. Predictions are estimates, not certainties, and the relationship is not causal.
“No, it is not a valid conclusion because there were no employees who took 160 hours of combined sick and vacation time.” From the graph there were employees who took more than 160 hours of combined sick and vacation time.
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Comprehensive Guide for MATH 1200 C957 Applied Algebra Exam Questions
Chapter 1: Foundations of Algebra
1.1 Numbers and Operations
Types of Numbers:
Natural numbers (ℕ): 1, 2, 3, ...
Whole numbers (ℕ_0): 0, 1, 2, ...
Integers (ℤ): The integer line contains these values: -3 -2 -1 0 1 2 3 ...
Rational numbers (ℚ): A number shows up as p divided by q when q does not equal zero.
Irrational numbers: Decimals that continue senselessly when written in digits without technical names like π and √2.
Properties of Operations:
Commutative: a + b = b + a; a × b = b × a.
Associative: When you combine numbers they stay in this order a then b and finally c. Also when you multiply numbers they follow the same order a times b then c.
Distributive: a(b + c) = ab + ac.
Order of Operations (PEMDAS):
Parentheses
Exponents
We do Division or Multiplication starting from the left to right end.
Put additions and subtractions in their normal reading sequence starting from the left.
Terms: An expression part consists of numbers and variables that mathematically join by using plus or minus symbols (such as 3x + 5 has two parts 3x and 5).
Like Terms: An expression shows the same variables with the same power together (like 3x and 4x).
Simplification:
Combine like terms.
Distribute expressions when needed.
1.3 Solving Linear Equations
General form: ax + b = 0.
Steps:
Work through both sides by spreading out factors and uniting like terms.
Move numeric values and numbers to isolate the variable in its equation.
Verify the solution.
You can solve rules using the same methods as you use for equations.
Key difference: When working with negative factors in equations you must switch the greater than and less than sign.
Graphing solutions:
Open circle: We use a different symbol for every type of inequality besides these three core operators (≠, >, <).
Closed circle: usage of signs ≤ and ≥ marks the end of a range in inequality systems.
Chapter 2: Functions and Graphs
2.1 Functions
Definition: A relation gives a single output result y for every input value x.
Notation: f(x) = y.
Domain: A function takes all possible numbers for input.
Range: All values that come out of the system appear here.
2.2 Linear Functions
General form: y = mx + b.
m: slope (∆y/∆x).
b: The line crosses the y-axis at its y-intercept.
Graphing:
Plot the y-intercept.
The slope points you to another location on the graph.
2.3 Slope-Intercept Form
Slope (m): Slope/Run equals the output difference between the two points divided by their input difference.
Equation of a line: y = mx + b.
2.4 Systems of Linear Equations
Methods to solve:
Graphing: Draw both lines on the graph and the point where they cross reveals the answer.
Substitution: You find one equation's direct formula for its variable then put it into the remaining equation.
Elimination: Join two or more equations to get rid of one variable.
Chapter 3: Polynomials.
Chapter 3: Polynomials.
3.1 Definitions
Monomial: A single term (e.g., 3x).
Binomial: Two terms (e.g., x + 2).
Trinomial: An equation made of three mathematical terms.
Polynomial: A sum of monomials.
3.2 Operations with Polynomials
Addition/Subtraction: Combine like terms.
Multiplication: Distribute each term.
Division: Perform division one term at a time using standard or synthetic processing.
3.3 Factoring
Common techniques:
Pulling out the highest common divisor from all terms.
Difference of squares: a² − b² = (a + b)(a − b).
Trinomials: ax² + bx + c.
3.To find solutions of polynomial equations needs our focus
Put the polynomial equation equal to zero.
Set the equation to zero and find the value of the variable.
Check your answer by putting the new values into the initial equation.
Chapter 4: This chapter studies Rational Expressions and Rational Equations
4.1 Processing Rational Fractions
Divide out shared parts between top and bottom sections.
Restrictions: Denominator ≠ 0.
4.We Work on Rational Expression Systems
Addition/Subtraction: Use a common denominator.
Multiplication: For multiplication work with numerator and denominator values together.
Division: Divide by the conversion factor of the divisor.
4.3 Solving Rational Equations
Steps:
Find a common denominator.
You can remove the denominator by multiplying each term of the equation.
Solve the resulting equation.
Look for solutions that do not exist in reality.
Chapter 5: Radicals and Exponents
Product Rule: a^m × a^n = a^(m+n).
Quotient Rule: a^m / a^n = a^(m−n).
Power Rule: (a^m)^n = a^(m×n).
Negative Exponents: a^(−n) = 1/a^n.
Zero Exponent: a^0 = 1 (a ≠ 0).
5.2 Simplifying Radicals
Product Property: √(ab) = √a √× b.
Quotient Property: √(a/b) = (√a)/(√b).
Rationalizing the Denominator: Cancelling radicals from the bottom of expressions.
Isolate the radical.
Square both sides.
Find solutions then verify any extra values.
Chapter 6: Quadratic Equations
6.1 Standard Form Models a Quadratic Equation
General form: ax² + bx + c = 0.
6.2 Solving Quadratic Equations
Factoring: Solve (x + p)(x + q) = 0.
Completing the Square:
Rewrite as x² + bx = −c.
Add (b/2)² to both sides.
Solve for x.
x = [−b ± √(b² − 4ac)] / 2a.
Discriminant (Δ): Shows how roots will appear.
Δ > 0: Two distinct real roots.
Δ = 0: One real root.
Δ < 0: Two complex roots.
6.Technology helps draw graphed curves based on quadratic functions.
Vertex form: y = a(x − h)² + k.
Vertex: (h, k).
Axis of symmetry: x = h.
Parabola opens:
Upward if a > 0.
Downward if a < 0.
Frequently Asked Question
This course covers a variety of algebraic concepts, including solving linear equations and inequalities, quadratic equations, graphing functions, systems of equations, exponents, polynomials, and real-world applications of algebra.
Yes, all practice questions and study materials are meticulously updated and aligned with the 2025 standards to ensure students are prepared for current academic and professional requirements.
The course includes practical scenarios such as calculating interest, analyzing business data, optimizing resources, and modeling real-world situations with algebraic equations and graphs.
Yes, each practice question comes with step-by-step rationales to explain the correct answers and address common mistakes, helping students develop a deeper understanding of the material.
You can prepare by reviewing the practice questions, studying the provided rationales, working on real-world application problems, and using the exam-focused blueprints included in the course materials.
Yes, a graphing calculator is highly recommended, as it is essential for solving complex problems, graphing functions, and visualizing algebraic concepts.
The course offers video tutorials, interactive problem-solving exercises, concise topic summaries, and access to an online community of students and tutors for support.
Progress is tracked through quizzes, assignments, and practice exams included in the course. These tools provide detailed feedback on areas of strength and those that need improvement.