TSIA2 Mathematics Assessment Comprehensive Study Notes

Unit Conversions and Proportional Relationships

  • Question 1: Converting Kilograms to Pounds

    • Problem: If there are 2.2 pounds2.2\,pounds in 1 kilogram1\,kilogram, how many pounds are there in x kilogramsx\,kilograms?

    • Choices:

      • A. x2.2x^{2.2}

      • B. 2.2x2.2x

      • C. 2.2+x2.2 + x

      • D. x2.2\frac{x}{2.2}

    • Correct Answer: Choice B.

    • Rationale: The conversion factor is 2.2 lbs/kg2.2\,lbs/kg. To find the total pounds for xx kilograms, multiply the generic variable by the unit rate: 2.2×x=2.2x2.2 \times x = 2.2x.

  • Question 5: Rate, Time, and Distance Calculation

    • Problem: Running at an average rate of 6 miles6\,miles per hour, how many minutes would it take Kyle to run 3 miles3\,miles?

    • Choices:

      • A. 1818

      • B. 3030

      • C. 4040

      • D. 4545

    • Correct Answer: Choice B.

    • Rationale: There are 60 minutes60\,minutes in one hour. A rate of 6 miles6\,miles in 60 minutes60\,minutes simplifies to 660=110\frac{6}{60} = \frac{1}{10} miles per minute. This means it takes Kyle 10 minutes10\,minutes to run 1 mile1\,mile. To run 3 miles3\,miles, the time required is 3×10=30 minutes3 \times 10 = 30\,minutes.

Algebraic Operations and Equation Solving

  • Question 4: Solving First-Degree Equations

    • Problem: If 7p−4=87p - 4 = 8, what is the value of pp?

    • Choices:

      • A. 47\frac{4}{7}

      • B. 712\frac{7}{12}

      • C. 127\frac{12}{7}

      • D. 74\frac{7}{4}

    • Correct Answer: Choice C.

    • Rationale: To isolate pp, first add 44 to both sides of the equation: 7p=127p = 12. Then, divide by 77 to find p=127p = \frac{12}{7}.

  • Question 9: Solving for a Variable to Evaluate an Expression

    • Problem: If 5c−2=3c5c - 2 = 3c, then what is the value of 24c24c?

    • Choices:

      • A. 66

      • B. 88

      • C. 1616

      • D. 2424

    • Correct Answer: Choice D.

    • Rationale: Solve for cc by subtracting 3c3c from both sides: 2c−2=02c - 2 = 0. Add 22 to both sides: 2c=22c = 2. Dividing by 22 yields c=1c = 1. To find the final answer, evaluate 24c24c as 24×1=2424 \times 1 = 24.

  • Question 14: Representing Consecutive Odd Integers

    • Problem: If nn is the least of two consecutive odd integers, which expression represents the sum of the two integers?

    • Choices:

      • A. n+1n + 1

      • B. n+2n + 2

      • C. 2n+12n + 1

      • D. 2n+22n + 2

    • Correct Answer: Choice D.

    • Rationale: Consecutive odd integers are separated by a value of 22. If the smaller integer is nn, the next consecutive odd integer is n+2n + 2. The sum is n+(n+2)=2n+2n + (n + 2) = 2n + 2.

Graphical Analysis: Bar Graphs and Dot Plots

  • Question 2: Interpreting Store Customer Data

    • Problem: A bar graph shows customer counts (in hundreds): Mon (22), Tue (33), Wed (55), Thu (44). If Friday's customers represented a one-fifth increase from Thursday, how many customers shopped on Friday?

    • Choices:

      • A. 480480

      • B. 500500

      • C. 525525

      • D. 600600

    • Correct Answer: Choice A.

    • Rationale: Thursday had 400400 customers (44 units of 100100). A one-fifth increase of 400400 is calculated as 15×400=80\frac{1}{5} \times 400 = 80. Adding this increase to the original Thursday total: 400+80=480400 + 80 = 480.

  • Question 6: Probability and Ratios from a Dot Plot

    • Problem: A dot plot tracks pets per family for 2020 families. What fraction of families have more than two pets?

    • Choices:

      • A. 320\frac{3}{20}

      • B. 15\frac{1}{5}

      • C. 14\frac{1}{4}

      • D. 920\frac{9}{20}

    • Correct Answer: Choice B.

    • Rationale: Families with "more than two pets" include those with 33 pets and 44 pets. The plot shows 33 families with three pets and 11 family with four pets, totaling 44 families. As a fraction of the total (2020), this is 420\frac{4}{20}, which simplifies to 15\frac{1}{5}.

Word Problems and Systems of Equations

  • Question 3: Multi-Year Predictive Sales Modeling

    • Problem: Last year, a bakery sold ww loaves. This year, they sold three more than twice last year's amount. Next year, they plan to sell twice this year's amount. How many loaves are expected next year?

    • Choices:

      • A. 2w2w

      • B. 2w+32w + 3

      • C. 4w+34w + 3

      • D. 4w+64w + 6

    • Correct Answer: Choice D.

    • Rationale:

      1. Last year: ww

      2. This year: 2w+32w + 3

      3. Next year: 2(2w+3)=4w+62(2w + 3) = 4w + 6

  • Question 8: System of Linear Equations (Pizza and Soda)

    • Problem: Richard bought 33 slices of cheese pizza and 22 sodas for $8.75\$8.75. Jordan bought 22 slices of cheese pizza and 44 sodas for $8.50\$8.50. What is the cost of 11 slice of pizza and 33 sodas?

    • Choices:

      • A. $3.25\$3.25

      • B. $5.25\$5.25

      • C. $7.75\$7.75

      • D. $17.25\$17.25

    • Correct Answer: Choice B.

    • Rationale:

      1. Define variables: c=cost of pizza slicec = \text{cost of pizza slice}, s=cost of sodas = \text{cost of soda}.

      2. Equation 1: 3c+2s=8.753c + 2s = 8.75

      3. Equation 2: 2c+4s=8.502c + 4s = 8.50

      4. Multiply Eq 1 by 22: 6c+4s=17.506c + 4s = 17.50

      5. Subtract Eq 2 from result: (6c+4s)−(2c+4s)=17.50−8.50→4c=9.00→c=2.25(6c + 4s) - (2c + 4s) = 17.50 - 8.50 \rightarrow 4c = 9.00 \rightarrow c = 2.25.

      6. Substitute cc into Eq 2: 2(2.25)+4s=8.50→4.50+4s=8.50→4s=4.00→s=1.002(2.25) + 4s = 8.50 \rightarrow 4.50 + 4s = 8.50 \rightarrow 4s = 4.00 \rightarrow s = 1.00.

      7. Target calculation: 1c+3s=2.25+3(1.00)=5.251c + 3s = 2.25 + 3(1.00) = 5.25.

  • Question 13: Area and Unit Cost

    • Problem: Carpeting costs $2.50\$2.50 per square foot. How much will it cost to carpet a rectangular floor that is 10 feet10\,feet by 12 feet12\,feet?

    • Choices:

      • A. $112.00\$112.00

      • B. $120.00\$120.00

      • C. $250.00\$250.00

      • D. $300.00\$300.00

    • Correct Answer: Choice D.

    • Rationale: Area of rectangular floor = 10 ft×12 ft=120 ft210\,ft \times 12\,ft = 120\,ft^2. Total cost = Area ×\times unit cost = 120 ft2×$2.50/ft2=$300.00120\,ft^2 \times \$2.50/ft^2 = \$300.00.

Polynomials and Exponents

  • Question 12: Algebraic Equivalence in Factoring

    • Problem: Which of the following is NOT equivalent to (3x−12)(x+4)(3x - 12)(x + 4)?

    • Choices:

      • A. 3(x2−8x+16)3(x^2 - 8x + 16)

      • B. 3(x2−16)3(x^2 - 16)

      • C. 3x2−483x^2 - 48

      • D. 3x(x+4)−12(x+4)3x(x + 4) - 12(x + 4)

    • Correct Answer: Choice A.

    • Rationale: Expanding (3x−12)(x+4)(3x - 12)(x + 4) gives 3x2+12x−12x−48=3x2−483x^2 + 12x - 12x - 48 = 3x^2 - 48. Factoring out 33 gives 3(x2−16)3(x^2 - 16). Choice D is simply the distributive expansion. Choice A is 3(x−4)23(x - 4)^2, which expands to 3x2−24x+483x^2 - 24x + 48, thus not equivalent.

  • Question 15: Simplification of Exponential Expressions

    • Problem: Simplify (x−5yy3)−1(\frac{x^{-5}y}{y^3})^{-1}.

    • Choices:

      • A. x5y2x^5y^2

      • B. y2x5\frac{y^2}{x^5}

      • C. x5y2x^5y^2

      • D. x5y−3x^5y^{-3}

    • Correct Answer: Choice C.

    • Rationale: Inside the parentheses, simplify yy terms: yy3=y1−3=y−2\frac{y}{y^3} = y^{1-3} = y^{-2}. The expression is now (x−5y−2)−1(x^{-5}y^{-2})^{-1}. Apply the power of −1-1 to each factor: (x−5)−1(y−2)−1=x5y2(x^{-5})^{-1}(y^{-2})^{-1} = x^5y^2.

Geometric Calculations and Coordinate Geometry

  • Question 7: Volume of a Cylinder

    • Problem: Volume formula is V=πr2hV = \pi r^2 h. If r=2br = 2b and h=5b+3h = 5b + 3, what is the volume in terms of bb?

    • Choices:

      • A. 10πb2+6πb10\pi b^2 + 6\pi b

      • B. 20πb3+12πb220\pi b^3 + 12\pi b^2

      • C. 20π2b3+12π2b220\pi^2 b^3 + 12\pi^2 b^2

      • D. 50πb3+20πb2+90πb50\pi b^3 + 20\pi b^2 + 90\pi b

    • Correct Answer: Choice B.

    • Rationale: Substitute variables into the formula: V=π(2b)2(5b+3)V = \pi (2b)^2 (5b + 3). Simplify the squared term: V=π(4b2)(5b+3)V = \pi (4b^2)(5b + 3). Distribute 4πb24\pi b^2: V=20πb3+12πb2V = 20\pi b^3 + 12\pi b^2.

  • Question 10: Comparison of Slopes

    • Problem: Slope of y=mx−4y = mx - 4 is less than the slope of y=x−4y = x - 4. What must be true about mm?

    • Choices:

      • A. m=−1m = -1

      • B. m=1m = 1

      • C. m<1m < 1

      • D. m>1m > 1

    • Correct Answer: Choice C.

    • Rationale: In slope-intercept form (y=mx+by = mx + b), the coefficient of xx represents the slope. The slope of the second line is 11. If the slope of the first line is less than that of the second, then m<1m < 1.

  • Question 17: Finding the y-intercept

    • Problem: Determine the y-intercept of the graph of y=6x−(x+3)2y = 6x - (x + 3)^2.

    • Choices:

      • A. −9-9

      • B. −12-12

      • C. 33

      • D. 99

    • Correct Answer: Choice A.

    • Rationale: The y-intercept is the value of yy when x=0x = 0. Substitute 00 for xx: y=6(0)−(0+3)2=0−(3)2=−9y = 6(0) - (0 + 3)^2 = 0 - (3)^2 = -9.

  • Question 18: Solving for Variable via Triangle Area

    • Problem: A triangle has base xx, height x+1x + 1, and area 2121. Solve for xx.

    • Choices:

      • A. 33

      • B. 66

      • C. 77

      • D. 1111

    • Correct Answer: Choice B.

    • Rationale: Area 12bh=21\frac{1}{2}bh = 21. Substitute: 12(x)(x+1)=21\frac{1}{2}(x)(x+1) = 21. Multiply by 22: x2+x=42x^2 + x = 42. Set to zero: x2+x−42=0x^2 + x - 42 = 0. Factor: (x+7)(x−6)=0(x + 7)(x - 6) = 0. Since height cannot be negative, x=6x = 6.

Statistics and Probability

  • Question 11: Weighted Averages

    • Problem: A history class has 1212 tenth graders (avg. 7777) and 99 eleventh graders (avg. 9191). Calculate the class average.

    • Choices:

      • A. 8282

      • B. 8383

      • C. 8484

      • D. 8585

    • Correct Answer: Choice B.

    • Rationale: Find the total sum of all scores: (12×77)+(9×91)=924+819=1,743(12 \times 77) + (9 \times 91) = 924 + 819 = 1,743. Divide by the total number of students (12+9=2112 + 9 = 21): 1,74321=83\frac{1,743}{21} = 83.

  • Question 16: Probability of Combined Coin Values

    • Problem: Reyna has 55 coins (10 cents each) and 44 coins (25 cents each). If she picks two at random, what is the probability they are worth at least 35 cents?

    • Choices:

      • A. 518\frac{5}{18}

      • C. 1318\frac{13}{18}

      • D. 7172\frac{7}{172}

    • Correct Answer: Choice C.

    • Rationale: The only combination worth less than 35 cents is picking two 10-cent coins. Total coins = 99.

      1. Prob(first coin is 10c) = 59\frac{5}{9}.

      2. Prob(second coin is 10c | first was 10c) = 48=12\frac{4}{8} = \frac{1}{2}.

      3. Prob(two 10c coins) = 59×12=518\frac{5}{9} \times \frac{1}{2} = \frac{5}{18}.

      4. Complementary outcome (at least 35c) = 1−518=13181 - \frac{5}{18} = \frac{13}{18}.

Functions and Growth Models

  • Question 19: Determining if a Value is Real (Function Domain)

    • Problem: For which value of xx is f(x)=4−x2f(x) = \sqrt{4 - x^2} NOT defined as a real number?

    • Choices:

      • A. −2-2

      • B. 00

      • C. 22

      • D. 44

    • Correct Answer: Choice D.

    • Rationale: A square root function yields non-real results if the radicand is negative. Test x=4x = 4: 4−42=4−16=−124 - 4^2 = 4 - 16 = -12. Since −12\sqrt{-12} is not a real number, the function is undefined at x=4x=4. Other options (−2,0,2-2, 0, 2) result in radicands of 0,4,00, 4, 0, which are real.

  • Question 20: Modeling Exponential Growth

    • Problem: A population of 100100 doubles every nine years. Which expression gives the population after tt years?

    • Choices:

      • A. 2×1009t2 \times 100^{9t}

      • B. t2×1009t^2 \times \frac{100}{9}

      • C. 100×29t100 \times 2^{9t}

      • D. 100×2t9100 \times 2^{\frac{t}{9}}

    • Correct Answer: Choice D.

    • Rationale: The initial population is 100100. The growth factor is 22 (doubling). The doubling occurs over intervals of 99. The number of periods is t9\frac{t}{9}. Thus, the population at time tt is 100×2t9100 \times 2^{\frac{t}{9}}.