algebra-i-m1-end-of-module-assessment

End-of-Module Assessment Task M1 NYS COMMON CORE MATHEMATICS CURRICULUM ALGEBRA I This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 314 Module 1: Relationships Between Quantities and Reasoning with Equations and Their Graphs 4. Solve for x in each of the equations or inequalities below, and name the property and/or properties used: a. 3 4 x = 9 b. 10 + 3 x = 5 x c. a + x = b d. cx = d e. 1 2 x − g < m f. q + 5 x = 7 x − r

End-of-Module Assessment Task M1 NYS COMMON CORE MATHEMATICS CURRICULUM ALGEBRA I This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 317 Module 1: Relationships Between Quantities and Reasoning with Equations and Their Graphs d. When we replace x by x 2 in the original equation, we get the following new equation: 3 x 2 + 4 = 5 x 2 − 4 . Use the fact that the solution set to the original equation is { 4 } to find the solution set to this new equation. 6. The Zonda Information and Telephone Company (ZI&T) calculates a customer’s total monthly cell phone charge using the formula, C = ( b + rm ) ( 1 + t ) , where C is the total cell phone charge, b is a basic monthly fee, r is the rate per minute, m is the number of minutes used that month, and t is the tax rate. Solve for m , the number of minutes the customer used that month.

End-of-Module Assessment Task M1 NYS COMMON CORE MATHEMATICS CURRICULUM ALGEBRA I This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 320 Module 1: Relationships Between Quantities and Reasoning with Equations and Their Graphs d. For her experiment, Alexus plans to add an anti-bacterial chemical to the Petri dish at 4:00 p.m. that is supposed to kill 99.9% of the bacteria instantaneously. If she started with 100 bacteria at 12:00 noon, how many live bacteria might Alexus expect to find in the Petri dish right after she adds the anti-bacterial chemical? 9. Jack is 27 years older than Susan. In 5 years, he will be 4 times as old as she is. a. Find the present ages of Jack and Susan. b. What calculations would you do to check if your answer is correct?

End-of-Module Assessment Task M1 NYS COMMON CORE MATHEMATICS CURRICULUM ALGEBRA I This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 323 Module 1: Relationships Between Quantities and Reasoning with Equations and Their Graphs 12. The local theater in Jamie’s home town has a maximum capacity of 160 people. Jamie shared with Venus the following graph and said that the shaded region represented all the possible combinations of adult and child tickets that could be sold for one show. a. Venus objected and said there was more than one reason that Jamie’s thinking was flawed. What reasons could Venus be thinking of?

End-of-Module Assessment Task M1 NYS COMMON CORE MATHEMATICS CURRICULUM ALGEBRA I This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 326 Module 1: Relationships Between Quantities and Reasoning with Equations and Their Graphs 3 A-SSE.A.1b Student responds incorrectly or leaves the question blank. Student responds correctly by circling the expression on the left but does not use solid reasoning to explain the answer. Student responds correctly by circling the expression on the left but gives limited explanation or does not use the properties of inequality in the explanation. Student responds correctly by circling the expression on the left and gives a complete explanation that uses the properties of inequality. 4 a–h A-REI.A.1 A-REI.B.3 Student answers incorrectly with no correct steps shown. Student answers incorrectly but has one or more correct steps. Student answers correctly but does not correctly identify the property or properties used. Student answers correctly and correctly identifies the property or properties used. 5 a A-REI.A.1 Student does not answer or demonstrates incorrect reasoning throughout. Student demonstrates only limited reasoning. Student demonstrates solid reasoning but falls short of a complete answer or makes a minor misstatement in the answer. Student answer is complete and demonstrates solid reasoning throughout. b A-REI.A.1 Student does not answer or does not demonstrate understanding of what the question is asking. Student makes more than one misstatement in the definition. Student provides a mostly correct definition with a minor misstatement. Student answers completely and uses a correct definition without error or mis- statement. c A-REI.A.1 Student makes mistakes in both verifications and demonstrates incorrect reasoning or leaves the question blank. Student conducts both verifications but falls short of articulating reasoning to answer the question. Student conducts both verifications and articulates valid reasoning to answer the question but makes a minor error in the verification or a minor misstatement in the explanation. Student conducts both verifications without error and articulates valid reasoning to answer the question.

End-of-Module Assessment Task M1 NYS COMMON CORE MATHEMATICS CURRICULUM ALGEBRA I This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 329 Module 1: Relationships Between Quantities and Reasoning with Equations and Their Graphs 1 1 a A-REI.C.6 Student is unable to demonstrate the understanding that two equations with ( 3 , 4 ) as a solution are needed. Student provides two equations that have ( 3 , 4 ) as a solution (or attempts to provide such equations) but makes one or more errors. Student may provide an equation with a negative slope. Student shows one minor error in the answer but attempts to provide two equations both containing ( 3 , 4 ) as a solution and both with positive slope. Student provides two equations both containing ( 3 , 4 ) as a solution and both with positive slope. b A-REI.C.6 Student is unable to identify the multiple correctly. Student identifies the multiple as 3 . N/A Student correctly identifies the multiple as 2 . c A-REI.C.6 Student is unable to demonstrate even a partial understanding of how to find the solution to the system. Student shows some reasoning required to find the solution but makes multiple errors. Student makes a minor error in finding the solution point. Student successfully identifies the solution point as ( 3 , 4 ) . d A-REI.C.5 A-REI.C.6 A-REI.D.10 Student is unable to answer or to support the answer with any solid reasoning. Student concludes yes or no but is only able to express limited reasoning in support of the answer. Student correctly explains that all the systems have the solution point ( 3 , 4 ) but incorrectly assumes this is true for all cases of m . Student correctly explains that while in most cases this is true, if m = 1 , the two lines are coinciding lines, resulting in a solution set consisting of all the points on the line. 1 2 a MP.2 A-REI.D.12 Student is unable to articulate any sound reasons. Student is only able to articulate one sound reason. Student provides two sound reasons but makes minor errors in the expression of reasoning. Student is able to articulate at least two valid reasons. Valid reasons include the following: the graph assumes x could be less than zero, the graph assumes y could be less than zero, the graph assumes a and b could be non-whole numbers, the graph assumes 160 children could attend with no adults.

End-of-Module Assessment Task M1 NYS COMMON CORE MATHEMATICS CURRICULUM ALGEBRA I This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 332 Module 1: Relationships Between Quantities and Reasoning with Equations and Their Graphs b. If c = 42 − d is true, then which is greater: c or d or are you not able to tell? Explain how you know your choice is correct. There is no way to tell. We only know that the sum of c and d is 42. If d were 10, c would be 32 and, therefore, greater than d. But if d were 40, c would be 2 and, therefore, less than d. 3. If a < 0 and c > b , circle the expression that is greater: a ( b − c ) or a ( c − b ) Use the properties of inequalities to explain your choice. 4. Solve for x in each of the equations or inequalities below and name the property and/or properties used: a. 3 4 x = 9 x = 9 ∙ ( 4 3 ) Multiplication p roperty of equality x = 12 Since c ¿ b, it follows that c − ¿ b ¿ 0. so (c − ¿ b) is positive. And since a is negative, the product of a ∙ (c − ¿ b) ¿ a ∙ (b − ¿ c). Since c ¿ b , it follows that 0 ¿ b − ¿ c, and since a ¿ 0, a is negative, and the product of two negatives will be a positive.

End-of-Module Assessment Task M1 NYS COMMON CORE MATHEMATICS CURRICULUM ALGEBRA I This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 333 Module 1: Relationships Between Quantities and Reasoning with Equations and Their Graphs b. 10 + 3 x = 5 x 10 = 2x Addition property of equality 5 = x Multiplication property of equality c. a + x = b x = b − a Addition property of equality d. cx = d x = d c , c ≠ 0 Multiplication property of equality e. 1 2 x − g < m 1 2 x < m + g Addition property of equality x < 2 ⋅ ( m + g ) Multiplication property of equality f. q + 5 x = 7 x − r q + r = 2x Addition property of equality ( q+r ) 2 = x Multiplication property of equality g. 3 4 ( x + 2 ) = 6 ( x + 12 ) 3 ∙ (x + 2) = 24 ∙ (x + 12) Multiplication property of equality 3x + 6 = 24x + 288 Distributive property

End-of-Module Assessment Task M1 NYS COMMON CORE MATHEMATICS CURRICULUM ALGEBRA I This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 334 Module 1: Relationships Between Quantities and Reasoning with Equations and Their Graphs − 282 21 = x Addition property of equality and multiplication − 94 7 = x Property of equality − 94 7 = x h. 3 ( 5 – 5 x ) > 5 x 15 − 15x > 5x Distributive property 15 > 20x Addition property of inequality 3 4 > x Multiplication property of equality 5. The equation, 3 x + 4 = 5 x − 4 , has the solution set { 4 } . a. Explain why the equation, ( 3 x + 4 ) + 4 = ( 5 x − 4 ) + 4 , also has the solution set { 4 } . Since the new equation can be created by applying the addition property of equality, the solution set does not change. OR Each side of this equation is 4 more than the sides of the original equation. Whatever value(s) make 3x + ¿ 4 ¿ 5x − ¿ 4 true would also make 4 more than 3x + ¿ 4 equal to 4 more than 5x − ¿ 4. b. In part (a), the expression ( 3 x + 4 )+ 4 is equivalent to the expression 3 x + 8 . What is the definition of equivalent algebraic expressions? Describe why changing an expression on one side of an equation to an equivalent expression leaves the solution set unchanged? Algebraic expressions are equivalent if (possibly repeated) use of the distributive, associative, and commutative properties and/or the properties of rational exponents can be applied to one expression to convert it to the other expression. When two expressions are equivalent, assigning the same value to x in both expressions will give an equivalent numerical expression, which then evaluates to the same number. Therefore, changing the expression to something equivalent will not change the truth value of the equation once values are assigned to x.

End-of-Module Assessment Task M1 NYS COMMON CORE MATHEMATICS CURRICULUM ALGEBRA I This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 335 Module 1: Relationships Between Quantities and Reasoning with Equations and Their Graphs c. When we square both sides of the original equation, we get the following new equation: ( 3 x + 4 ) 2 = ( 5 x − 4 ) 2 . Show that 4 is still a solution to the new equation. Show that 0 is also a solution to the new equation but is not a solution to the original equation. Write a sentence that describes how the solution set to an equation may change when both sides of the equation are squared. ( 3 ∙ 4 + 4 ) 2 = ( 5 ∙ 4 − 4 ) 2 gives 16 2 = 16 6 , which is true. ( 3 ∙ 0 + 4 ) 2 = ( 5 ∙ 0 − 4 ) 2 gives 4 2 = ( − 4 ) 2 , which is true. But, ( 3 ∙ 0 + 4 ) = ( 5 ∙ 0 − 4 ) gives 4 = − 4 , which is false. When both sides are squared, you might introduce new numbers to the solution set because statements like 4 = − 4 are false, but statements like 4 2 = ( − 4 ) 2 are true. d. When we replace x by x 2 in the original equation, we get the following new equation: 3 x 2 + 4 = 5 x 2 − 4 . Use the fact that the solution set to the original equation is {4} to find the solution set to this new equation. Since the original equation 3x + ¿ 4 ¿ 5x − ¿ 4 was true when x ¿ 4, the new equation 3x 2 + 4 = 5x 2 − 4 should be true when x 2 = 4 . And, x 2 = 4 when x = 2 , so the solution set to the new equation is { − 2,2 } .

End-of-Module Assessment Task M1 NYS COMMON CORE MATHEMATICS CURRICULUM ALGEBRA I This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 336 Module 1: Relationships Between Quantities and Reasoning with Equations and Their Graphs 6. The Zonda Information and Telephone Company calculates a customer’s total monthly cell phone charge using the formula, C = ( b + rm ) ( 1 + t ) , where C is the total cell phone charge, b is a basic monthly fee, r is the rate per minute, m is the number of minutes used that month, and t is the tax rate. Solve for m , the number of minutes the customer used that month. C = b + bt + rm + rmt C − b − bt = m ∙ ( r + rt ) C − b − bt r + rt = m t ≠ − 1 r ≠ 0 7. Students and adults purchased tickets for a recent basketball playoff game. All tickets were sold at the ticket booth—season passes, discounts, etc., were not allowed. Student tickets cost $ 5 each, and adult tickets cost $ 10 each. A total of $ 4,500 was collected. 700 tickets were sold. a. Write a system of equations that can be used to find the number of student tickets, s , and the number of adult tickets, a , that were sold at the playoff game. 5s + 10a = 4500 s + a = 700 b. Assuming that the number of students and adults attending would not change, how much more money could have been collected at the playoff game if the ticket booth charged students and adults the same price of $ 10 per ticket? 700 × $10 = $7000 $7000 − $4500 = $2500 more

End-of-Module Assessment Task M1 NYS COMMON CORE MATHEMATICS CURRICULUM ALGEBRA I This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 337 Module 1: Relationships Between Quantities and Reasoning with Equations and Their Graphs c. Assuming that the number of students and adults attending would not change, how much more money could have been collected at the playoff game if the student price was kept at $ 5 per ticket and adults were charged $ 15 per ticket instead of $ 10 ? First solve for a and s 5s + 10a = 4500 -5s − 5a = − 3500 5a = 1000 a = 200 s = 500 $5 ∙ (500) + ¿ $15 ∙ (200) ¿ $5500 $1,000 more OR $5 more per adult ticket (200 ∙ $5 ¿ $1000more) 8. Alexus is modeling the growth of bacteria for an experiment in science. She assumes that there are B bacteria in a Petri dish at 12:00 noon. In reality, each bacterium in the Petri dish subdivides into two new bacteria approximately every 20 minutes. However, for the purposes of the model, Alexus assumes that each bacterium subdivides into two new bacteria exactly every 20 minutes. a. Create a table that shows the total number of bacteria in the Petri dish at 1 3 hour intervals for 2 hours starting with time 0 to represent 12:00 noon. Time Number of Bacteria 0 B 1/3 hour 2B 2/3 hour 4B 1 hour 8B 1 1/3 hour 16B 1 2/3 hour 32B 2 hour 64B b. Write an equation that describes the relationship between total number of bacteria T and time h in hours, assuming there are B bacteria in the Petri dish at h = 0 . T = B ∙ ( 2 ) 3h or T = B ∙ 8 h c. If Alexus starts with 100 bacteria in the Petri dish, draw a graph that displays the total number of bacteria with respect to time from 12:00 noon ( h = 0 ) to 4:00 p.m. ( h = 4 ). Label points on your graph at time h = 0 , 1 , 2 , 3 , 4 .

End-of-Module Assessment Task M1 NYS COMMON CORE MATHEMATICS CURRICULUM ALGEBRA I This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 338 Module 1: Relationships Between Quantities and Reasoning with Equations and Their Graphs d. For her experiment, Alexus plans to add an anti-bacterial chemical to the Petri dish at 4:00 p.m. that is supposed to kill 99.9% of the bacteria instantaneously. If she started with 100 bacteria at 12:00 noon, how many live bacteria might Alexus expect to find in the Petri dish right after she adds the anti-bacterial chemical? ( 1 − 0.999 ) ∙ 409 600 = 409.6 about 410 live bacteria 9. Jack is 27 years older than Susan. In 5 years time, he will be 4 times as old as she is. a. Find the present ages of Jack and Susan. J = S + 27 J + 5 = 4 ∙ ( S + 5 ) S + 27 + 5 = 4S + 20 S + 32 = 4S + 20 12 = 3S S = 4 J = 4 + 27 J = 31 Jack is 31 and Susan is 4. b. What calculations would you do to check if your answer is correct? Is Jack’s age – Susan’s age = 27? Add 5 years to Jack’s and Susan’s ages, and see if that makes Jack 4 times as old as Susan.

End-of-Module Assessment Task M1 NYS COMMON CORE MATHEMATICS CURRICULUM ALGEBRA I This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 339 Module 1: Relationships Between Quantities and Reasoning with Equations and Their Graphs 10. a. Find the product: ( x 2 − x + 1 ) ( 2 x 2 + 3 x + 2 ) 2x 4 + 3 x 3 + 2 x 2 − 2x 3 − 3x 2 − 2x + 2x 2 + 3x + 2 2x 4 + x 3 + x 2 + x + 2 b. Use the results of part (a) to factor 21,112 as a product of a two-digit number and a three-digit number. ( 100 − 10 + 1 ) ∙ ( 200 + 30 + 2 ) ( 91 ) ∙ ( 232 ) 11. Consider the following system of equations with the solution x = 3 , y = 4 . Equation A1: y = x + 1 Equation A2: y =− 2 x + 10 a. Write a unique system of two linear equations with the same solution set. This time make both linear equations have positive slope. 1. y = 4 3 x

End-of-Module Assessment Task M1 NYS COMMON CORE MATHEMATICS CURRICULUM ALGEBRA I This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 340 Module 1: Relationships Between Quantities and Reasoning with Equations and Their Graphs Equation B1: _______ Equation B2: ________ b. The following system of equations was obtained from the original system by adding a multiple of equation A2 to equation A1. Equation C1: y = x + 1 Equation C2: 3 y =− 3 x + 21 What multiple of A2 was added to A1? 2 times A2 was added to A1. c. What is the solution to the system given in part (b)? (3,4) d. For any real number m , the line y = m ( x − 3 ) + 4 passes through the point ( 3,4 ) . Is it certain then that the system of equations: Equation D1: y = x + 1 Equation D2: y = m ( x − 3 ) + 4 has only the solution x = 3 , y = 4 ? Explain. No. If m = 1, then the two lines have the same slope. Both lines pass through the point (3,4), and the lines are parallel; therefore, they coincide. There are infinite solutions. The solution set is all the points on the line. Any other nonzero value of m would create a system with the only solution of (3,4). 2.

End-of-Module Assessment Task M1 NYS COMMON CORE MATHEMATICS CURRICULUM ALGEBRA I This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 341 Module 1: Relationships Between Quantities and Reasoning with Equations and Their Graphs

End-of-Module Assessment Task M1 NYS COMMON CORE MATHEMATICS CURRICULUM ALGEBRA I This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 342 Module 1: Relationships Between Quantities and Reasoning with Equations and Their Graphs 12. The local theater in Jamie’s home town has a maximum capacity of 160 people. Jamie shared with Venus the following graph and said that the shaded region represented all the possible combinations of adult and child tickets that could be sold for one show. a. Venus objected and said there was more than one reason that Jamie’s thinking was flawed. What reasons could Venus be thinking of? 1. The graph implies that the number of tickets sold could be a fractional amount, but really it only makes sense to sell whole number tickets. x and y must be whole numbers. 2. The graph also shows that negative ticket amounts could be sold, which does not make sense.

End-of-Module Assessment Task M1 NYS COMMON CORE MATHEMATICS CURRICULUM ALGEBRA I This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG I-M1-TE-1.3.0-07.2015 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 343 Module 1: Relationships Between Quantities and Reasoning with Equations and Their Graphs b. Use equations, inequalities, graphs, and/or words to describe for Jamie the set of all possible combinations of adult and child tickets that could be sold for one show. The system would be { a + c ≤ 160 a ≥ 0 c ≥ 0 where a and c are whole numbers. c. The theater charges $ 9 for each adult ticket and $ 6 for each child ticket. The theater sold 144 tickets for the first showing of the new release. The total money collected from ticket sales for that show was $ 1,164 . Write a system of equations that could be used to find the number of child tickets and the number of adult tickets sold, and solve the system algebraically. Summarize your findings using the context of the problem. a: the number of adult tickets sold (must be a whole number) c: the number of child tickets sold (must be a whole number) { 9a + 6c = 1164 a + c = 144 9a + 6c = 1164 -6a − 6c = -8643a = 300a = 100, c = 44 In all, 100 adult tickets and 44 child tickets were sold.