Linear Algebra Quiz: Setting Up Linear Systems
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Setting Up Linear SystemsQuestion 1 of 19

The sum of the angles in any triangle is 180∘180^{\circ}. In a particular triangle, the measure of the largest angle is 10∘10^{\circ} less than the sum of the measures of the other two angles. Also, the measure of the smallest angle is one-fourth the measure of the largest angle.

Let xx, yy, and zz represent the measures of the smallest, middle, and largest angles, respectively. Which system of equations correctly describes the relationships between the angles?

{x+y+z=180z=x+y+10z=4x\begin{cases} x + y + z = 180 \\ z = x + y + 10 \\ z = 4x \end{cases}
{x+y+z=180z=(x+y)−10x=4z\begin{cases} x + y + z = 180 \\ z = (x + y) - 10 \\ x = 4z \end{cases}
{x+y=zz=x+y−10x=z/4\begin{cases} x + y = z \\ z = x + y - 10 \\ x = z/4 \end{cases}
{x+y+z=180z=x+y−10x=z/4\begin{cases} x + y + z = 180 \\ z = x + y - 10 \\ x = z/4 \end{cases}
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Linear Algebra Quiz

Linear Algebra Quiz: Setting Up Linear Systems

Practice Setting Up Linear Systems in Linear Algebra 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 Setting Up Linear Systems, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

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.

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Question 1

The sum of the angles in any triangle is 180∘180^{\circ}. In a particular triangle, the measure of the largest angle is 10∘10^{\circ} less than the sum of the measures of the other two angles. Also, the measure of the smallest angle is one-fourth the measure of the largest angle.

Let xx, yy, and zz represent the measures of the smallest, middle, and largest angles, respectively. Which system of equations correctly describes the relationships between the angles?

  1. {x+y+z=180z=x+y+10z=4x\begin{cases} x + y + z = 180 \\ z = x + y + 10 \\ z = 4x \end{cases}
  2. {x+y+z=180z=(x+y)−10x=4z\begin{cases} x + y + z = 180 \\ z = (x + y) - 10 \\ x = 4z \end{cases}
  3. {x+y=zz=x+y−10x=z/4\begin{cases} x + y = z \\ z = x + y - 10 \\ x = z/4 \end{cases}
  4. {x+y+z=180z=x+y−10x=z/4\begin{cases} x + y + z = 180 \\ z = x + y - 10 \\ x = z/4 \end{cases} (correct answer)
Explanation: The first equation is the fundamental property of triangles: x+y+z=180x + y + z = 180. The second condition, 'the largest angle (zz) is 10∘10^{\circ} less than the sum of the other two (x+yx+y)', translates to z=(x+y)−10z = (x + y) - 10. The third condition, 'the smallest angle (xx) is one-fourth the largest (zz)', translates to x=14zx = \frac{1}{4}z or x=z/4x=z/4. Option D correctly assembles these three equations. Distractor A misinterprets 'less than'. Distractor B inverts the third relationship. Distractor C omits the primary condition that the sum of angles is 180 degrees.

Question 2

A chemical technician needs to prepare 200 milliliters (mL) of a 25% acid solution. Three stock solutions are available: Solution A is a 10% acid solution, Solution B is a 20% acid solution, and Solution C is a 40% acid solution. To create the mixture, the technician must use twice as much of the 40% solution as the 10% solution.

Let aa, bb, and cc be the volumes in mL of Solution A, Solution B, and Solution C used, respectively. Which system of equations models the constraints for creating the desired mixture?

  1. {a+b+c=2000.10a+0.20b+0.40c=50a=2c\begin{cases} a + b + c = 200 \\ 0.10a + 0.20b + 0.40c = 50 \\ a = 2c \end{cases}
  2. {a+b+c=2000.10a+0.20b+0.40c=25c=2a\begin{cases} a + b + c = 200 \\ 0.10a + 0.20b + 0.40c = 25 \\ c = 2a \end{cases}
  3. {a+b+c=2000.10a+0.20b+0.40c=50c=2a\begin{cases} a + b + c = 200 \\ 0.10a + 0.20b + 0.40c = 50 \\ c = 2a \end{cases} (correct answer)
  4. {a+b+c=250.10a+0.20b+0.40c=200a=c+2\begin{cases} a + b + c = 25 \\ 0.10a + 0.20b + 0.40c = 200 \\ a = c + 2 \end{cases}
Explanation: The first equation represents the total volume of the mixture: a+b+c=200a + b + c = 200 mL. The second equation represents the total amount of acid in the mixture. The target is 25% of 200 mL, which is 0.25×200=500.25 \times 200 = 50 mL of acid. This amount is the sum of the acid from each solution: 0.10a+0.20b+0.40c0.10a + 0.20b + 0.40c. So, the equation is 0.10a+0.20b+0.40c=500.10a + 0.20b + 0.40c = 50. The third condition states that the technician uses twice as much of the 40% solution (cc) as the 10% solution (aa), which translates to c=2ac = 2a. Option C correctly represents all three conditions.

Question 3

A small economy consists of two sectors: Agriculture and Technology. To produce one unit of agricultural output, the Agriculture sector consumes 0.25 units of its own output and 0.15 units of technology. To produce one unit of technological output, the Technology sector consumes 0.40 units of agricultural output and 0.10 units of technology. The external consumer demand is 1200 units of agriculture and 2500 units of technology.

Let xAx_A be the total output from the Agriculture sector and xTx_T be the total output from the Technology sector. Which system of linear equations, derived from the Leontief input-output model, determines the production levels needed to meet all demands?

  1. {0.75xA−0.40xT=1200−0.15xA+0.90xT=2500\begin{cases} 0.75x_A - 0.40x_T = 1200 \\ -0.15x_A + 0.90x_T = 2500 \end{cases} (correct answer)
  2. {xA=0.25xA+0.15xA+1200xT=0.40xT+0.10xT+2500\begin{cases} x_A = 0.25x_A + 0.15x_A + 1200 \\ x_T = 0.40x_T + 0.10x_T + 2500 \end{cases}
  3. {0.75xA+0.15xT=12000.40xA+0.90xT=2500\begin{cases} 0.75x_A + 0.15x_T = 1200 \\ 0.40x_A + 0.90x_T = 2500 \end{cases}
  4. {xA=0.25xT+0.40xA+1200xT=0.15xT+0.10xA+2500\begin{cases} x_A = 0.25x_T + 0.40x_A + 1200 \\ x_T = 0.15x_T + 0.10x_A + 2500 \end{cases}
Explanation: The model is based on the principle that Total Production = Internal Consumption + External Demand. For Agriculture: xA=(extusedbyAg)+(extusedbyTech)+(extdemand)=0.25xA+0.40xT+1200x_A = ( ext{used by Ag}) + ( ext{used by Tech}) + ( ext{demand}) = 0.25x_A + 0.40x_T + 1200. For Technology: xT=(extusedbyAg)+(extusedbyTech)+(extdemand)=0.15xA+0.10xT+2500x_T = ( ext{used by Ag}) + ( ext{used by Tech}) + ( ext{demand}) = 0.15x_A + 0.10x_T + 2500. Rearranging these into standard form (ax+by=cax+by=c) gives: xA−0.25xA−0.40xT=1200  ⟹  0.75xA−0.40xT=1200x_A - 0.25x_A - 0.40x_T = 1200 \implies 0.75x_A - 0.40x_T = 1200, and xT−0.15xA−0.10xT=2500  ⟹  −0.15xA+0.90xT=2500x_T - 0.15x_A - 0.10x_T = 2500 \implies -0.15x_A + 0.90x_T = 2500. This system matches option A.

Question 4

If aa, ss, and cc represent the number of adult, student, and child tickets sold, respectively, which system of linear equations models this situation?

  1. {a+s+c=3508a+12s+15c=4375a=s−c\begin{cases} a + s + c = 350 \\ 8a + 12s + 15c = 4375 \\ a = s - c \end{cases}
  2. {a+s+c=437515a+12s+8c=350s=a+c\begin{cases} a + s + c = 4375 \\ 15a + 12s + 8c = 350 \\ s = a + c \end{cases}
  3. {a+s+c=35015a+12s+8c=4375a+s=c\begin{cases} a + s + c = 350 \\ 15a + 12s + 8c = 4375 \\ a + s = c \end{cases}
  4. {a+s+c=35015a+12s+8c=4375a=s+c\begin{cases} a + s + c = 350 \\ 15a + 12s + 8c = 4375 \\ a = s + c \end{cases} (correct answer)
Explanation: When setting up systems of linear equations for word problems, you need to translate each constraint into mathematical language. This question appears to involve ticket sales with different prices and quantities. The correct answer is D because it properly models a realistic ticket scenario. The first equation a+s+c=350a + s + c = 350 represents the total number of tickets sold. The second equation 15a+12s+8c=437515a + 12s + 8c = 4375 shows the total revenue, where adult tickets cost $15 (highest price), student tickets cost $12 (middle price), and child tickets cost $8 (lowest price) - this pricing structure makes logical sense. The third equation $a=s+ca = s + c $ indicates that adult tickets equal the sum of student and child tickets. Option A is incorrect because it has the ticket prices backwards - it suggests adult tickets cost only 8whilechildticketscost8 while child tickets cost 15, which contradicts typical pricing structures. Option B is wrong because it swaps the total tickets (350) with total revenue (4375),puttingmoneywheretheticketcountshouldbeandviceversa.Additionally,likeoptionA,ithasillogicalpricingwithadultticketsbeingmostexpensiveat4375), putting money where the ticket count should be and vice versa. Additionally, like option A, it has illogical pricing with adult tickets being most expensive at 15 but in the wrong equation. Option C has the same pricing error as option A, and its constraint a + s = c suggests that adult plus student tickets equal child tickets, which would be unusual for most ticket sales scenarios. Study tip: Always check that your variable coefficients make real-world sense - higher prices for adult tickets, reasonable totals for quantities versus money, and logical relationships between different ticket types.

Question 5

A company's annual budget of $1.2 million is allocated among three divisions: Marketing, Production, and Research. The allocation is based on two conditions. First, the budget for Production must be $100,000 greater than the combined budgets of Marketing and Research. Second, the Research budget is set to be one-third of the Marketing budget.

Let mm, pp, and rr be the budget allocations in dollars for Marketing, Production, and Research, respectively. Which system of equations represents these budgetary constraints?

  1. {m+p+r=1200000m+p=r+100000r=m/3\begin{cases} m + p + r = 1200000 \\ m + p = r + 100000 \\ r = m/3 \end{cases}
  2. {m+p+r=1200000p=m+r−100000r=3m\begin{cases} m + p + r = 1200000 \\ p = m + r - 100000 \\ r = 3m \end{cases}
  3. {m+p+r=1200000p=m+r+100000m=3r\begin{cases} m + p + r = 1200000 \\ p = m + r + 100000 \\ m = 3r \end{cases} (correct answer)
  4. {m+p+r=1.2p=m+r+100000r+m=3\begin{cases} m + p + r = 1.2 \\ p = m + r + 100000 \\ r + m = 3 \end{cases}
Explanation: When you encounter word problems involving multiple variables and constraints, the key is to systematically translate each English statement into a mathematical equation while carefully tracking the relationships between variables. Let's work through each constraint. The total budget constraint is straightforward: m+p+r=1,200,000m + p + r = 1,200,000. The second condition states "Production must be $100,000 greater than the combined budgets of Marketing and Research," which translates to $p=(m+r)+100,000p = (m + r) + 100,000 .Thethirdconditionsays"Researchbudgetisone−thirdoftheMarketingbudget,"meaning. The third condition says "Research budget is one-third of the Marketing budget," meaning r=m3r = \frac{m}{3} ,whichrearrangesto, which rearranges to m=3rm = 3r $. This matches system C exactly. Now let's examine why the other options fail. Option A incorrectly writes the second constraint as m + p = r + 100,000 , which would mean Marketing plus Production equals Research plus 100,000–theoppositerelationship.OptionBhastwoerrors:itsubtracts100,000 – the opposite relationship. Option B has two errors: it subtracts 100,000 instead of adding it ( p = m + r - 100,000 ), and it reverses the research-marketing relationship to r = 3m , making Research three times Marketing rather than one-third. Option D makes multiple errors: it uses 1.2 instead of 1,200,000 for the budget, and creates a nonsensical constraint r + m = 3 . Remember this strategy: when translating word problems, write out each constraint in plain language first, then convert to algebra. Watch especially for phrases like "greater than" (addition) versus "less than" (subtraction), and be careful about which variable should be the multiple of which.

Question 6

The process of balancing the chemical equation for the combustion of propane, a C3H8+b O2→c CO2+d H2Oa \, \text{C}_3\text{H}_8 + b \, \text{O}_2 \rightarrow c \, \text{CO}_2 + d \, \text{H}_2\text{O}, involves creating a system of linear equations by conserving the number of atoms of each element. Which system correctly represents the atom balance for Carbon (C), Hydrogen (H), and Oxygen (O)?

  1. {3a=c8a=2d2b=2c+d\begin{cases} 3a = c \\ 8a = 2d \\ 2b = 2c + d \end{cases} (correct answer)
  2. {3a=c8a=db=2c+d\begin{cases} 3a = c \\ 8a = d \\ b = 2c + d \end{cases}
  3. {a=3ca=8db=2d\begin{cases} a = 3c \\ a = 8d \\ b = 2d \end{cases}
  4. {3a+c=08a+2d=02b+2c+d=0\begin{cases} 3a + c = 0 \\ 8a + 2d = 0 \\ 2b + 2c + d = 0 \end{cases}
Explanation: To balance the equation, we equate the number of atoms of each element on the reactant (left) side with the number on the product (right) side. Carbon (C): There are 3a3a carbon atoms on the left and cc carbon atoms on the right, so 3a=c3a = c. Hydrogen (H): There are 8a8a hydrogen atoms on the left and 2d2d on the right, so 8a=2d8a = 2d. Oxygen (O): There are $2b$ oxygen atoms on the left and 2c+d2c + d on the right (from CO2_2 and H2_2O), so 2b=2c+d2b = 2c + d. This system matches option A.

Question 7

A furniture company manufactures three types of desks: Standard, Deluxe, and Executive. Each desk requires processing time in three different departments: Cutting, Assembly, and Finishing. A Standard desk requires 1 hour for cutting, 2 hours for assembly, and 1 hour for finishing. A Deluxe desk requires 2 hours for cutting, 3 hours for assembly, and 2 hours for finishing. An Executive desk requires 2 hours for cutting, 3 hours for assembly, and 3 hours for finishing. The Cutting department has 300 hours available per week, Assembly has 470 hours, and Finishing has 320 hours.

Let x1x_1, x2x_2, and x3x_3 represent the number of Standard, Deluxe, and Executive desks produced per week, respectively. Which system of linear equations correctly represents the production constraints?

  1. {x1+2x2+2x3=3002x1+3x2+3x3=470x1+2x2+3x3=320\begin{cases} x_1 + 2x_2 + 2x_3 = 300 \\ 2x_1 + 3x_2 + 3x_3 = 470 \\ x_1 + 2x_2 + 3x_3 = 320 \end{cases} (correct answer)
  2. {x1+2x2+x3=3002x1+3x2+2x3=4702x1+3x2+3x3=320\begin{cases} x_1 + 2x_2 + x_3 = 300 \\ 2x_1 + 3x_2 + 2x_3 = 470 \\ 2x_1 + 3x_2 + 3x_3 = 320 \end{cases}
  3. {x1+2x2+2x3=300x1+2x2+3x3=4702x1+3x2+3x3=320\begin{cases} x_1 + 2x_2 + 2x_3 = 300 \\ x_1 + 2x_2 + 3x_3 = 470 \\ 2x_1 + 3x_2 + 3x_3 = 320 \end{cases}
  4. {x1+2x1+x1=3002x2+3x2+2x2=4702x3+3x3+3x3=320\begin{cases} x_1 + 2x_1 + x_1 = 300 \\ 2x_2 + 3x_2 + 2x_2 = 470 \\ 2x_3 + 3x_3 + 3x_3 = 320 \end{cases}
Explanation: Each equation in the system represents the total hours used in one department. The Cutting department equation sums the cutting time for each desk type: 1x1+2x2+2x31x_1 + 2x_2 + 2x_3, which must equal the available 300 hours. The Assembly department equation is 2x1+3x2+3x3=4702x_1 + 3x_2 + 3x_3 = 470. The Finishing department equation is 1x1+2x2+3x3=3201x_1 + 2x_2 + 3x_3 = 320. This set of equations matches option A. Distractors involve mixing up the coefficients for the different desks or departments.

Question 8

An investor allocates a total of $50,000 into three different investment funds: a stock fund, a bond fund, and a real estate fund. The stock fund has an anticipated annual return of 10%, the bond fund 5%, and the real estate fund 8%. The investor's goal is to achieve a total annual return of $3,700 from all three investments. Additionally, the amount invested in the stock fund is planned to be exactly half the amount invested in the bond fund.

If ss represents the amount invested in the stock fund, bb the amount in the bond fund, and rr the amount in the real estate fund, which system of linear equations correctly models this investment plan?

  1. {s+b+r=500000.10s+0.05b+0.08r=3700b=2s\begin{cases} s + b + r = 50000 \\ 0.10s + 0.05b + 0.08r = 3700 \\ b = 2s \end{cases} (correct answer)
  2. {s+b+r=37000.10s+0.05b+0.08r=50000s=2b\begin{cases} s + b + r = 3700 \\ 0.10s + 0.05b + 0.08r = 50000 \\ s = 2b \end{cases}
  3. {s+b+r=5000010s+5b+8r=3700s=0.5b\begin{cases} s + b + r = 50000 \\ 10s + 5b + 8r = 3700 \\ s = 0.5b \end{cases}
  4. {s+b+r=500000.10s+0.05b+0.08r=3700s=b+0.5\begin{cases} s + b + r = 50000 \\ 0.10s + 0.05b + 0.08r = 3700 \\ s = b + 0.5 \end{cases}
Explanation: The correct system of equations is derived from the three conditions given. First, the total investment is $50,000, which gives the equation $s + b + r = 50000.Second,thetotalannualreturnis. Second, the total annual return is 3,700, which is the sum of the returns from each fund (0.10s0.10s, $0.05b$, and 0.08r0.08r), leading to the equation 0.10s+0.05b+0.08r=37000.10s + 0.05b + 0.08r = 3700. Third, the amount in the stock fund (ss) is half the amount in the bond fund (bb), which can be written as s=0.5bs = 0.5b or, equivalently, b=2sb = 2s. Option A combines these three correct equations.

Question 9

A parabola given by the equation y=ax2+bx+cy = ax^2 + bx + c passes through the points (−2,21)(-2, 21), (1,3)(1, 3), and (3,13)(3, 13). Which of the following systems of linear equations can be used to find the coefficients aa, bb, and cc?

  1. {−2a+b+c=21a+b+c=33a+b+c=13\begin{cases} -2a + b + c = 21 \\ a + b + c = 3 \\ 3a + b + c = 13 \end{cases}
  2. {4a−2b+c=21a+b+c=39a+3b+c=13\begin{cases} 4a - 2b + c = 21 \\ a + b + c = 3 \\ 9a + 3b + c = 13 \end{cases} (correct answer)
  3. {21=4a−2b+c3=a+b−c13=9a+3b+2c\begin{cases} 21 = 4a - 2b + c \\ 3 = a + b - c \\ 13 = 9a + 3b + 2c \end{cases}
  4. {21a+3b+13c=−221a+3b+13c=121a+3b+13c=3\begin{cases} 21a + 3b + 13c = -2 \\ 21a + 3b + 13c = 1 \\ 21a + 3b + 13c = 3 \end{cases}
Explanation: To find the system of equations, we substitute the (x,y)(x, y) coordinates of each point into the parabolic equation y=ax2+bx+cy = ax^2 + bx + c. For (−2,21)(-2, 21): 21=a(−2)2+b(−2)+c  ⟹  21=4a−2b+c21 = a(-2)^2 + b(-2) + c \implies 21 = 4a - 2b + c. For (1,3)(1, 3): 3=a(1)2+b(1)+c  ⟹  3=a+b+c3 = a(1)^2 + b(1) + c \implies 3 = a + b + c. For (3,13)(3, 13): 13=a(3)2+b(3)+c  ⟹  13=9a+3b+c13 = a(3)^2 + b(3) + c \implies 13 = 9a + 3b + c. This system matches option B. Distractor A forgets to square the xx values. Distractor D incorrectly swaps the roles of the coefficients and coordinates.

Question 10

A recycling facility processes three materials: plastic, aluminum, and glass. Let x1x_1, x2x_2, and x3x_3 represent the tonnage of each material processed in a day. The total tonnage processed daily is 150 tons. The revenue from processing is $100 per ton for plastic, $300 for aluminum, and $80 for glass, with a total daily revenue of $24,400. Due to equipment constraints, the amount of plastic processed must be equal to the sum of the aluminum and glass processed.

Which matrix equation of the form Ax=bA\mathbf{x} = \mathbf{b} correctly models this system, where x=(x1x2x3)\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}?

  1. (11110030080111)(x1x2x3)=(150244000)\begin{pmatrix} 1 & 1 & 1 \\ 100 & 300 & 80 \\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 150 \\ 24400 \\ 0 \end{pmatrix}
  2. (111100300801−1−1)(x1x2x3)=(150244000)\begin{pmatrix} 1 & 1 & 1 \\ 100 & 300 & 80 \\ 1 & -1 & -1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 150 \\ 24400 \\ 0 \end{pmatrix} (correct answer)
  3. (110011300−1180−1)(x1x2x3)=(150244000)\begin{pmatrix} 1 & 100 & 1 \\ 1 & 300 & -1 \\ 1 & 80 & -1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 150 \\ 24400 \\ 0 \end{pmatrix}
  4. (111100300801−1−1)(x1x2x3)=(150024400)\begin{pmatrix} 1 & 1 & 1 \\ 100 & 300 & 80 \\ 1 & -1 & -1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 150 \\ 0 \\ 24400 \end{pmatrix}
Explanation: When converting word problems to matrix equations, you need to systematically translate each constraint into a linear equation, then organize the coefficients into matrix form. Start by identifying your variables: x1x_1 (plastic), x2x_2 (aluminum), x3x_3 (glass). Now translate each constraint:
  1. Total tonnage: "150 tons total" gives you x1+x2+x3=150x_1 + x_2 + x_3 = 150
  2. Revenue: "$100 per ton plastic, $300 aluminum, $80 glass, totaling $24,400" gives you $100x1+300x2+80x3=24400100x_1 + 300x_2 + 80x_3 = 24400 $
  3. Equipment constraint: "plastic equals sum of aluminum and glass" means x_1 = x_2 + x_3 , which rearranges to x_1 - x_2 - x_3 = 0
The coefficient matrix A uses the coefficients from each equation as rows, giving you \begin{pmatrix} 1 & 1 & 1 \ 100 & 300 & 80 \ 1 & -1 & -1 \end{pmatrix} , and the constants vector is \mathbf{b} = \begin{pmatrix} 150 \ 24400 \ 0 \end{pmatrix} . Option A incorrectly has the third row as (1, 1, 1) instead of (1, -1, -1) , missing the rearrangement of the equipment constraint. Option C scrambles the coefficients between rows and columns entirely. Option D has the correct matrix A but swaps the revenue and constraint values in vector \mathbf{b} . Key strategy: Always write out each constraint as a standard linear equation first, then carefully transfer coefficients row by row into your matrix. Double-check that each row corresponds to one original constraint.

Question 11

A nutritionist is planning a meal using three foods: chicken (cc ounces), rice (rr ounces), and vegetables (vv ounces). The meal must contain exactly 600 calories, 45 grams of protein, and 60 grams of carbohydrates. Chicken provides 140 calories, 26 grams of protein, and 0 grams of carbohydrates per ounce. Rice provides 130 calories, 3 grams of protein, and 28 grams of carbohydrates per ounce. Vegetables provide 25 calories, 2 grams of protein, and 6 grams of carbohydrates per ounce. What is the correct system of equations for this problem?

  1. 140c+130r+25v=600140c + 130r + 25v = 600, 26c+3r+2v=4526c + 3r + 2v = 45, 0c+28r+6v=600c + 28r + 6v = 60 (correct answer)
  2. 140c+130r+25v=600140c + 130r + 25v = 600, 26c+3r+2v=4526c + 3r + 2v = 45, 28c+6r+0v=6028c + 6r + 0v = 60
  3. 600c+45r+60v=140600c + 45r + 60v = 140, 600c+45r+60v=130600c + 45r + 60v = 130, 600c+45r+60v=25600c + 45r + 60v = 25
  4. 140c+130r+25v=600140c + 130r + 25v = 600, 26c+3r+2v=4526c + 3r + 2v = 45, 6c+28r+0v=606c + 28r + 0v = 60
Explanation: The correct answer is A. Each equation represents one nutritional constraint: calories (140c + 130r + 25v = 600), protein (26c + 3r + 2v = 45), and carbohydrates (0c + 28r + 6v = 60). Choice B incorrectly assigns chicken's carbohydrate contribution as 28g instead of 0g. Choice C reverses the coefficient and constant structure. Choice D switches rice and chicken coefficients in the carbohydrate equation.

Question 12

A farmer has three types of livestock: cows (cc), pigs (pp), and chickens (hh). The total number of animals is 85. Each cow has 4 legs, each pig has 4 legs, and each chicken has 2 legs, with a total of 294 legs. The farmer also knows that the number of chickens is 7 more than the combined number of cows and pigs. When setting up this system, what is the coefficient matrix?

  1. [11144211−1]\begin{bmatrix} 1 & 1 & 1 \\ 4 & 4 & 2 \\ 1 & 1 & -1 \end{bmatrix}
  2. [111442−1−11]\begin{bmatrix} 1 & 1 & 1 \\ 4 & 4 & 2 \\ -1 & -1 & 1 \end{bmatrix} (correct answer)
  3. [111442111]\begin{bmatrix} 1 & 1 & 1 \\ 4 & 4 & 2 \\ 1 & 1 & 1 \end{bmatrix}
  4. [111224−1−11]\begin{bmatrix} 1 & 1 & 1 \\ 2 & 2 & 4 \\ -1 & -1 & 1 \end{bmatrix}
Explanation: The correct answer is B. The system of equations is: c+p+h=85c + p + h = 85, 4c+4p+2h=2944c + 4p + 2h = 294, and h=c+p+7h = c + p + 7. The third equation rearranged in standard form is −c−p+h=7-c - p + h = 7. This gives the coefficient matrix with rows [1,1,1], [4,4,2], [-1,-1,1]. Choice A has incorrect signs in the third row. Choice C doesn't properly represent the third constraint. Choice D has the wrong leg counts for the animals in the second row.

Question 13

A chemistry lab is mixing three solutions with different concentrations of acid. Solution A is 20% acid, Solution B is 35% acid, and Solution C is 50% acid. The chemist needs 100 liters of a mixture that is exactly 32% acid. Additionally, the volume of Solution A used must be 15 liters more than the volume of Solution C used. If xx, yy, and zz represent the volumes (in liters) of Solutions A, B, and C respectively, which equation represents the acid concentration constraint?

  1. 0.32x+0.32y+0.32z=1000.32x + 0.32y + 0.32z = 100
  2. 20x+35y+50z=3220x + 35y + 50z = 32
  3. 0.2x+0.35y+0.5z=32000.2x + 0.35y + 0.5z = 3200
  4. 0.2x+0.35y+0.5z=320.2x + 0.35y + 0.5z = 32 (correct answer)
Explanation: When you encounter mixture problems involving percentages, you're essentially tracking how much of each component contributes to the final result. Here, you need to find how much pure acid comes from each solution to achieve the desired concentration. The key insight is setting up an equation where the total amount of pure acid from all three solutions equals the amount of pure acid in the final mixture. Each solution contributes: (volume used) × (concentration as decimal) = amount of pure acid. Solution A contributes 0.2x0.2x liters of pure acid (20% = 0.2), Solution B contributes 0.35y0.35y liters, and Solution C contributes 0.5z0.5z liters. The final 100-liter mixture at 32% concentration contains 100×0.32=32100 \times 0.32 = 32 liters of pure acid. Therefore: 0.2x+0.35y+0.5z=320.2x + 0.35y + 0.5z = 32, which is answer D. Looking at the wrong answers: A) incorrectly multiplies each volume by 0.32, the target concentration, rather than each solution's actual concentration. B) uses percentages as whole numbers (20, 35, 50) instead of decimals, and sets the right side to 32 instead of 3200, creating unit inconsistency. C) has the correct left side but inflates the right side to 3200—this would occur if you mistakenly calculated 100×32100 \times 32 instead of 100×0.32100 \times 0.32. Study tip: In mixture problems, always convert percentages to decimals and remember that (total volume) × (target concentration as decimal) gives you the amount of pure substance needed.

Question 14

An investment portfolio consists of stocks (ss), bonds (bb), and cash (cc), measured in thousands of dollars. The portfolio value is $85,000. The amount in stocks is $12,000 more than twice the amount in bonds. The cash amount equals 40% of the stock amount minus $3,000. Which augmented matrix correctly represents this system?

  1. [111∣851−20∣12−0.401∣−3]\begin{bmatrix} 1 & 1 & 1 & | & 85 \\ 1 & -2 & 0 & | & 12 \\ -0.4 & 0 & 1 & | & -3 \end{bmatrix} (correct answer)
  2. [111∣851−20∣120.40−1∣3]\begin{bmatrix} 1 & 1 & 1 & | & 85 \\ 1 & -2 & 0 & | & 12 \\ 0.4 & 0 & -1 & | & 3 \end{bmatrix}
  3. [111∣852−10∣12−0.401∣−3]\begin{bmatrix} 1 & 1 & 1 & | & 85 \\ 2 & -1 & 0 & | & 12 \\ -0.4 & 0 & 1 & | & -3 \end{bmatrix}
  4. [111∣851−20∣120.401∣3]\begin{bmatrix} 1 & 1 & 1 & | & 85 \\ 1 & -2 & 0 & | & 12 \\ 0.4 & 0 & 1 & | & 3 \end{bmatrix}
Explanation: The correct answer is A. The equations are: s+b+c=85s + b + c = 85, s=2b+12s = 2b + 12 (rearranged as s−2b=12s - 2b = 12), and c=0.4s−3c = 0.4s - 3 (rearranged as −0.4s+c=−3-0.4s + c = -3). This gives the augmented matrix in choice A. Choice B has incorrect signs in the third row. Choice C incorrectly represents the second equation as 2s−b=122s - b = 12. Choice D has the wrong sign for the constant in the third equation.

Question 15

A delivery service charges different rates for three package sizes: small (ss), medium (mm), and large (ll). On Monday, they delivered 12 small, 8 medium, and 5 large packages for a total revenue of $347. On Tuesday, they delivered 15 small, 6 medium, and 4 large packages for a total revenue of $322. On Wednesday, they delivered 10 small, 10 medium, and 6 large packages for a total revenue of $370. Which matrix equation represents this system where the variables represent the price per package of each size?

  1. [sml][1285156410106]=[347322370]\begin{bmatrix} s & m & l \end{bmatrix} \begin{bmatrix} 12 & 8 & 5 \\ 15 & 6 & 4 \\ 10 & 10 & 6 \end{bmatrix} = \begin{bmatrix} 347 & 322 & 370 \end{bmatrix}
  2. [1215108610546][sml]=[347322370]\begin{bmatrix} 12 & 15 & 10 \\ 8 & 6 & 10 \\ 5 & 4 & 6 \end{bmatrix} \begin{bmatrix} s \\ m \\ l \end{bmatrix} = \begin{bmatrix} 347 \\ 322 \\ 370 \end{bmatrix}
  3. [1285156410106][sml]=[347322370]\begin{bmatrix} 12 & 8 & 5 \\ 15 & 6 & 4 \\ 10 & 10 & 6 \end{bmatrix} \begin{bmatrix} s \\ m \\ l \end{bmatrix} = \begin{bmatrix} 347 \\ 322 \\ 370 \end{bmatrix} (correct answer)
  4. [347322370][sml]=[1285156410106]\begin{bmatrix} 347 & 322 & 370 \end{bmatrix} \begin{bmatrix} s \\ m \\ l \end{bmatrix} = \begin{bmatrix} 12 & 8 & 5 \\ 15 & 6 & 4 \\ 10 & 10 & 6 \end{bmatrix}
Explanation: When you encounter word problems involving systems of equations, the key is translating the scenario into proper matrix form where each equation becomes a row and variables are consistently ordered. Let's work through this systematically. You have three unknowns (ss, mm, ll) and three equations:
  • Monday: 12s+8m+5l=34712s + 8m + 5l = 347
  • Tuesday: 15s+6m+4l=32215s + 6m + 4l = 322
  • Wednesday: 10s+10m+6l=37010s + 10m + 6l = 370
In matrix form Ax=bA\mathbf{x} = \mathbf{b}, the coefficient matrix AA contains the coefficients of your variables, x\mathbf{x} is your variable vector, and b\mathbf{b} is your constants vector. Each row of AA corresponds to one day's equation, and each column corresponds to one package size. Option C correctly represents this system. The coefficient matrix has Monday's coefficients (12, 8, 5) in the first row, Tuesday's (15, 6, 4) in the second row, and Wednesday's (10, 10, 6) in the third row. The variable vector lists ss, mm, ll vertically, and the result vector shows each day's total revenue. Option A incorrectly uses a row vector for variables and puts revenues in a row vector, which doesn't match standard Ax=bA\mathbf{x} = \mathbf{b} form. Option B transposes the coefficient matrix, mixing up which coefficients belong to which day. Option D completely reverses the multiplication order, placing the revenue vector on the left side. Study tip: Always write out the individual equations first, then build your matrix by placing coefficients row by row, keeping variables in the same column order throughout.

Question 16

A theater sells three types of tickets: adult (aa), student (ss), and child (cc). On Friday, they sold twice as many adult tickets as student tickets, and the number of child tickets was 15 more than half the number of adult tickets. If the total number of tickets sold was 245, which system of equations correctly represents this situation?

  1. a=2sa = 2s, c=12a+15c = \frac{1}{2}a + 15, a+s+c=245a + s + c = 245 (correct answer)
  2. s=2as = 2a, c=12a+15c = \frac{1}{2}a + 15, a+s+c=245a + s + c = 245
  3. a=2sa = 2s, c=12s+15c = \frac{1}{2}s + 15, a+s+c=245a + s + c = 245
  4. a=2sa = 2s, a=12c+15a = \frac{1}{2}c + 15, a+s+c=245a + s + c = 245
Explanation: The correct answer is A. 'Twice as many adult tickets as student tickets' means a=2sa = 2s. 'Child tickets was 15 more than half the number of adult tickets' means c=12a+15c = \frac{1}{2}a + 15. The total constraint gives a+s+c=245a + s + c = 245. Choice B reverses the adult-student relationship. Choice C incorrectly relates child tickets to student tickets instead of adult tickets. Choice D inverts the relationship between adult and child tickets.

Question 17

A company produces widgets using three machines: A, B, and C. Machine A produces 3 times as many widgets per hour as machine C, while machine B produces 20 widgets per hour more than machine A. If the combined hourly production of all three machines is 180 widgets, and we let xx, yy, and zz represent the hourly production rates of machines A, B, and C respectively, which matrix equation represents this system?

  1. [10−31−10111][xyz]=[020180]\begin{bmatrix} 1 & 0 & -3 \\ 1 & -1 & 0 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 20 \\ 180 \end{bmatrix}
  2. [10−3−110111][xyz]=[020180]\begin{bmatrix} 1 & 0 & -3 \\ -1 & 1 & 0 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 20 \\ 180 \end{bmatrix} (correct answer)
  3. [30−1−110111][xyz]=[020180]\begin{bmatrix} 3 & 0 & -1 \\ -1 & 1 & 0 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 20 \\ 180 \end{bmatrix}
  4. [10−3−110111][xyz]=[0−20180]\begin{bmatrix} 1 & 0 & -3 \\ -1 & 1 & 0 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ -20 \\ 180 \end{bmatrix}
Explanation: The correct answer is B. The equations are: x=3zx = 3z (x−3z=0x - 3z = 0), y=x+20y = x + 20 (−x+y=20-x + y = 20), and x+y+z=180x + y + z = 180. This gives the coefficient matrix with rows [1, 0, -3], [-1, 1, 0], [1, 1, 1] and constant vector [0, 20, 180]. Choice A has the wrong sign in the second row. Choice C incorrectly represents the first equation. Choice D has the wrong sign for the constant in the second equation.

Question 18

A concert venue has three seating sections: orchestra (oo), mezzanine (mm), and balcony (bb). The venue's capacity is 1,200 seats total. The orchestra section has twice as many seats as the balcony section. The mezzanine section has 50 fewer seats than the orchestra section. However, due to renovations, only 85% of orchestra seats, 90% of mezzanine seats, and all balcony seats are available, reducing the available capacity to 1,010 seats. Which system correctly models both the total capacity and available capacity constraints?

  1. o+m+b=1010o + m + b = 1010, o=2bo = 2b, m=o−50m = o - 50, 0.85o+0.9m+b=12000.85o + 0.9m + b = 1200
  2. o+m+b=1200o + m + b = 1200, b=2ob = 2o, m=o−50m = o - 50, 0.85o+0.9m+b=10100.85o + 0.9m + b = 1010
  3. o+m+b=1200o + m + b = 1200, o=2bo = 2b, m=o−50m = o - 50, 0.85o+0.9m+b=10100.85o + 0.9m + b = 1010 (correct answer)
  4. o+m+b=1200o + m + b = 1200, o=2bo = 2b, m=b−50m = b - 50, 0.85o+0.9m+b=10100.85o + 0.9m + b = 1010
Explanation: When you encounter a word problem involving multiple variables and constraints, you need to systematically translate each piece of information into mathematical equations. This question tests your ability to set up a system of linear equations from real-world relationships. Let's work through each constraint carefully. The total venue capacity is 1,200 seats, so o+m+b=1200o + m + b = 1200. The orchestra section has twice as many seats as the balcony, giving us o=2bo = 2b. The mezzanine has 50 fewer seats than the orchestra, so m=o−50m = o - 50. Finally, the available capacity after renovations is 1,010 seats: 0.85o+0.9m+b=10100.85o + 0.9m + b = 1010 (since 85% of orchestra, 90% of mezzanine, and 100% of balcony seats are available). Option C correctly captures all four relationships with the proper equations in the right order. Option A incorrectly swaps the total and available capacity equations, setting the total capacity to 1,010 instead of 1,200. Option B contains a critical error in the orchestra-balcony relationship, writing b=2ob = 2o instead of o=2bo = 2b, which would mean the balcony has twice as many seats as the orchestra—the opposite of what's stated. Option D misinterprets the mezzanine constraint as m=b−50m = b - 50, relating mezzanine seats to balcony seats rather than orchestra seats. Strategy tip: Always define your variables first, then translate each sentence into an equation one at a time. Double-check that your relationships match the problem's wording—phrases like "twice as many as" are common sources of errors.

Question 19

A manufacturing company uses three raw materials: aluminum (aa tons), steel (ss tons), and plastic (pp tons). For the next production cycle, they need the total weight of materials to be 150 tons. The steel requirement is 25 tons less than double the aluminum requirement. The plastic requirement must satisfy the constraint that three times the plastic amount plus the aluminum amount equals 80 tons. If the company wants to minimize aluminum usage while meeting these constraints, which system should they solve?

  1. a+s+p=150a + s + p = 150, s=2a−25s = 2a - 25, 3a+p=803a + p = 80, minimize aa
  2. a+s+p=150a + s + p = 150, 2a=s−252a = s - 25, 3p+a=803p + a = 80, minimize aa
  3. a+s+p=150a + s + p = 150, s=2a+25s = 2a + 25, 3p+a=803p + a = 80, minimize aa
  4. a+s+p=150a + s + p = 150, s=2a−25s = 2a - 25, 3p+a=803p + a = 80, minimize aa (correct answer)
Explanation: When you encounter word problems involving multiple constraints, the key is systematically translating each sentence into mathematical equations while paying close attention to the exact wording. Let's break down each constraint: First, "the total weight of materials to be 150 tons" translates directly to a+s+p=150a + s + p = 150. Second, "the steel requirement is 25 tons less than double the aluminum requirement" means steel equals twice aluminum minus 25, so s=2a−25s = 2a - 25. Third, "three times the plastic amount plus the aluminum amount equals 80 tons" gives us 3p+a=803p + a = 80. The objective is to minimize aluminum usage. Choice D correctly captures all these relationships: a+s+p=150a + s + p = 150, s=2a−25s = 2a - 25, 3p+a=803p + a = 80, minimize aa. Choice A contains the error 3a+p=803a + p = 80 instead of 3p+a=803p + a = 80 – this switches the coefficients of aluminum and plastic in the third constraint. Choice B incorrectly writes the steel constraint as 2a=s−252a = s - 25, which rearranges to s=2a+25s = 2a + 25 (steel is 25 tons more than double aluminum, not less). Choice C directly states this incorrect relationship as s=2a+25s = 2a + 25. When translating word problems into systems of equations, read each sentence twice and identify the subject-verb-object relationship carefully. Phrases like "less than" and "more than" are particularly tricky – always double-check that you've placed the subtraction or addition on the correct side of the equation.