TSIA2 Mathematics Practice

TSIA2 Mathematics Practice Test — Explanatory Solution Key Page 1 of 3

Part 1: Quantitative & Algebraic Reasoning Solutions (Q1 - Q10)

Q1 (B) — Ratio Proportions

We set up a proportion comparing flour to sugar: \(\frac{3}{2} = \frac{15}{s}\), where \(s\) is the required sugar. Cross-multiplying gives \(3 \times s = 2 \times 15\), which simplifies to \(3s = 30\). Dividing by \(3\), we find \(s = 10\) cups of sugar.

Q2 (B) — Sequential Discount Percentage

Apply the initial \(20\%\) discount: \(\$650 \times (1 - 0.20) = \$650 \times 0.80 = \$520\). Next, apply the extra \(10\%\) discount on this new price: \(\$520 \times (1 - 0.10) = \$520 \times 0.90 = \$468.00\).

Q3 (A) — Distance, Rate, & Time Conversions

Use the time formula \(\text{Time} = \frac{\text{Distance}}{\text{Rate}} = \frac{6 \text{ miles}}{8 \text{ mph}} = 0.75 \text{ hours}\). To convert hours into minutes: \(0.75 \text{ hours} \times 60 \text{ minutes/hour} = 45 \text{ minutes}\).

Q4 (B) — Population Proportion Estimate

The proportion of defective parts in the sample is \(\frac{3}{150} = \frac{1}{50} = 0.02\). To estimate the defective parts in the full batch of \(4,200\), multiply: \(4,200 \times 0.02 = 84\) parts.

Q5 (A) — Mixed Fraction Arithmetic

First, subtract the cut piece from the total length: \(12\frac{3}{4} - 4\frac{1}{3} = \frac{51}{4} - \frac{13}{3} = \frac{153 - 52}{12} = \frac{101}{12} \text{ feet}\). Now, divide this remaining piece into 2 equal segments: \(\frac{101}{12} \times \frac{1}{2} = \frac{101}{24} = 4\frac{5}{24} \text{ feet}\).

Q6 (C) — Distributive Property and Linear Equating

Expand the equation: \(3(2x - 5) - 4(x + 1) = 7 \implies 6x - 15 - 4x - 4 = 7\). Combine like terms: \(2x - 19 = 7\). Add \(19\) to both sides: \(2x = 26 \implies x = 13\).

Q7 (B) — Solving Multi-step Inequalities

Combine variable terms by adding \(5x\) to both sides: \(12 < 2x + 4\). Subtract \(4\) from both sides: \(8 < 2x\). Divide by \(2\) to isolate \(x\): \(4 < x\), which is equivalent to \(x > 4\).

Q8 (A) — System of Linear Equations

Let \(s = \text{students}\) and \(a = \text{adults}\). From the total tickets, \(s + a = 85 \implies s = 85 - a\). Set up the revenue equation: \(4s + 7a = 445\). Substitute \(s\): \(4(85 - a) + 7a = 445 \implies 340 - 4a + 7a = 445 \implies 340 + 3a = 445 \implies 3a = 105 \implies a = 35\).

Q9 (B) — Factoring Trionomial Quadratics (\(a > 1\))

We look to factor \(2x^2 - 7x - 15\). Find two numbers that multiply to \(2 \times (-15) = -30\) and add up to \(-7\). These numbers are \(-10\) and \(3\). Rewrite and group: \(2x^2 - 10x + 3x - 15 = 2x(x - 5) + 3(x - 5) = (2x + 3)(x - 5)\). Thus, \((2x + 3)\) is a factor.

Q10 (D) — Algebraic Translation of Word Problems

Let consecutive integers be \(n\), \(n+1\), and \(n+2\) (largest). The sum is \(3n+3\). Set up equation: \(3n + 3 = 2(n + 2) + 18 \implies 3n + 3 = 2n + 4 + 18 \implies 3n + 3 = 2n + 22\). Subtract \(2n\) and \(3\) from both sides: \(n = 19\). Smallest integer is \(19\).

TSIA2 Mathematics Practice Test — Explanatory Solution Key Page 2 of 3

Part 2: Geometric & Spatial Reasoning Solutions (Q11 - Q20)

Q11 (B) — Distance in Coordinate Geometry

Apply distance formula \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) for points \((-3, 5)\) and \((5, -1)\):

\(d = \sqrt{(5 - (-3))^2 + (-1 - 5)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10\)

Q12 (B) — Nested Geometric Path Area

Garden Area = \(12 \times 18 = 216 \text{ sq ft}\). The combined garden plus path has width \(12 + 2(3) = 18 \text{ ft}\) and length \(18 + 2(3) = 24 \text{ ft}\). Total Area = \(18 \times 24 = 432 \text{ sq ft}\). Path Area = Total Area - Garden Area = \(432 - 216 = 216 \text{ sq ft}\).

Q13 (A) — Cylinder Surface Area

Use cylinder surface area formula \(SA = 2\pi rh + 2\pi r^2\). Given \(r = 4\) and \(h = 10\):

\(SA = 2\pi(4)(10) + 2\pi(4)^2 = 80\pi + 32\pi = 112\pi \text{ cm}^2\)

Q14 (B) — Scaling Area of Similar Triangles

The linear scale factor of correspondence is \(k = \frac{DE}{AB} = \frac{9}{6} = 1.5\). The ratio of similar areas is equivalent to the square of the scale factor: \(k^2 = (1.5)^2 = 2.25\). Thus, \(\text{Area of } DEF = 2.25 \times \text{Area of } ABC = 2.25 \times 24 = 54 \text{ square units}\).

Q15 (C) — Completing Square for Circle Standard Form

Group the variables: \((x^2 - 6x) + (y^2 + 8y) = 11\). Complete square by adding \((-6/2)^2 = 9\) and \((8/2)^2 = 16\) to both sides: \((x^2 - 6x + 9) + (y^2 + 8y + 16) = 11 + 9 + 16 \implies (x - 3)^2 + (y + 4)^2 = 36\). Thus, the center is \((3, -4)\) and radius \(r = \sqrt{36} = 6\).

Q16 (C) — Identifying Unknown from Median Values

With an even count of values (\(n=6\)), the median is the average of the 3rd and 4th terms. Since median is \(27\), which is larger than \(25\) and smaller than \(32\), the missing term \(x\) must fall between \(25\) and \(32\), making the sorted order \(12, 18, 25, x, 32, 45\). Thus, \(\frac{25 + x}{2} = 27 \implies 25 + x = 54 \implies x = 29\).

Q17 (C) — Probability of Dependent Events (No Replacement)

Total marbles initially = \(4 + 6 + 5 = 15\). Probability of 1st blue = \(\frac{6}{15} = \frac{2}{5}\). Since we select without replacement, there are \(5\) blue marbles remaining out of \(14\) total. Probability of 2nd blue = \(\frac{5}{14}\). Joint probability = \(\frac{2}{5} \times \frac{5}{14} = \frac{2}{14} = \frac{1}{7}\).

Q18 (B) — Set Theory and Venn Diagram Models

Using principle of inclusion-exclusion, calculate the union of students enrolled in Spanish or Chem: \(|S \cup C| = |S| + |C| - |S \cap C| = 65 + 45 - 20 = 90\). Those enrolled in neither is \(120 - 90 = 30\) students.

Q19 (B) — Interpret Standard Deviation (Spread)

Both classes share the same mean of \(82\%\). Standard deviation measures data variability. A lower standard deviation (\(3.5\%\) for Class A vs. \(8.2\%\) for Class B) indicates that Class A's scores are more uniform, clustered closer to the mean, and show less variance.

Q20 (B) — Combinatorics (Combination Formula)

Because order doesn't matter for a committee of members, we calculate combinations: \(\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56\) unique committees.

TSIA2 Mathematics Practice Test — Explanatory Solution Key Page 3 of 3