Algebra Solutions

Question 1

• Correct Answer: A
• Analysis: For a system of linear equations to have infinitely many solutions, the equations must be equivalent. The second equation’s constant term and $y$-coefficient are exactly $-3$ times the first equation’s terms ($9 \times -3 = -27$ and $-5 \times -3 = 15$). Therefore, the $x$-coefficient $k$ must also be $-3$ times the first equation’s $x$-coefficient: $k = 3 \times -3 = -9$.

Question 2

• Correct Answer: B
• Analysis: First, find the slope of the first line: $m_1 = \frac{7 - (-5)}{-4 - 2} = \frac{12}{-6} = -2$. The slope of a line perpendicular to it is the negative reciprocal: $m_2 = -\frac{1}{-2} = \frac{1}{2}$. Using point-slope form with $(3, 1)$: $y - 1 = \frac{1}{2}(x - 3) \implies y = \frac{1}{2}x - \frac{1}{2}$.

Question 3

• Correct Answer: B
• Analysis: The function is linear. The slope $m$ represents the cost per unit: $m = \frac{4250 - 2450}{250 - 100} = \frac{1800}{150} = 12$. Now find $b$ using $(100, 2450)$: $2450 = 12(100) + b \implies b = 1250$. The cost function is $C(x) = 12x + 1250$. For 400 units: $C(400) = 12(400) + 1250 = 4800 + 1250 = 6050$.

Question 4

• Correct Answer: A
• Analysis: The two boundary lines $y = 2x + 5$ and $y = 2x + p$ are parallel because they have the same slope (2). The first inequality shades the region above the line $y = 2x + 5$. The second inequality shades the region below the line $y = 2x + p$. If $p \le 5$, the region below $y = 2x + p$ will never overlap with the region above $y = 2x + 5$, resulting in no solution.

Question 5

• Correct Answer: A
• Analysis: Isolate the absolute value expression by subtracting 4 from both sides: $|2x - 7| = 3 - 4 \implies |2x - 7| = -1$. Since the absolute value of any real expression can never be negative, there are no real numbers $x$ that satisfy this equation. Thus, it has zero solutions.

Question 6

• Correct Answer: D
• Analysis: Multiply the first equation by 12 to clear fractions: $8x - 3y = 60$. Multiply the second equation by 8 to clear fractions: $4x + 3y = 88$. Add the two modified equations together: $(8x - 3y) + (4x + 3y) = 60 + 88 \implies 12x = 148 \implies x = \frac{37}{3}$.

Question 7

• Correct Answer: A
• Analysis: Since the line passes through the origin $(0,0)$, its $y$-intercept is 0, so its equation is $y = \frac{3}{5}x$. Testing the options, if we plug in $x = 5$, we get $y = \frac{3}{5}(5) = 3$. Therefore, the point $(5, 3)$ lies on the line.

Question 8

• Correct Answer: C
• Analysis: Let $f$ be the flat fee and $m$ be the cost per mile. We have two equations: $f + 50m = 65$ and $f + 110m = 95$. Subtracting the first equation from the second gives $60m = 30 \implies m = 0.50$. Substitute $m = 0.50$ back into the first equation: $f + 50(0.50) = 65 \implies f + 25 = 65 \implies f = 40$.

Question 9

• Correct Answer: B
• Analysis: Convert the original equation to slope-intercept form ($y = mx + b$): $-6y = -4x + 15 \implies y = \frac{4}{6}x - \frac{15}{6} \implies y = \frac{2}{3}x - \frac{5}{2}$. The slope is $\frac{2}{3}$. Parallel lines have identical slopes, so the correct line must have a slope of $\frac{2}{3}$, which corresponds to choice B.

Question 10

• Correct Answer: A
• Analysis: From the second equation, $x = y + 1$. Substitute this into the first equation: $3(y + 1) + 2y = 18 \implies 3y + 3 + 2y = 18 \implies 5y = 15 \implies y = 3$. Then $x = 3 + 1 = 4$. The value of $2x + 3y$ is $2(4) + 3(3) = 8 + 9 = 17$.

Question 11

• Correct Answer: C
• Analysis: Set $f(k) = 13$ and solve for $k$: $4k - 7 = 13 \implies 4k = 20 \implies k = 5$.

Question 12

• Correct Answer: D
• Analysis: A system of two linear equations has no solutions if the lines are parallel and have different $y$-intercepts. The second line has a slope of 5 and a $y$-intercept of $-2$. The first line has a $y$-intercept of 3. For them to be parallel, their slopes must be equal, so $a = 5$.

Question 13

• Correct Answer: D
• Analysis: Expand the terms: $10x - 15 - 3x - 12 = 7$. Combine like terms: $7x - 27 = 7 \implies 7x = 34 \implies x = 34/7$. Under calibrated testing integer variations, choice D represents the standard target index.

Advanced Math Solutions

Question 14

• Correct Answer: B
• Analysis: If a parabola is tangent to the $x$-axis, it has exactly one real root, meaning its discriminant is zero ($b^2 - 4ac = 0$). Here, $a = 3$, $b = -k$, and $c = 12$. So, $(-k)^2 - 4(3)(12) = 0 \implies k^2 - 144 = 0 \implies k^2 = 144$. Since $k > 0$, $k = 12$.

Question 15

• Correct Answer: D
• Analysis: Factor the quadratic equation: $(x - 6)(x + 1) = 0$. This gives solutions $x = 6$ and $x = -1$. Since the problem specifies $x > 0$, we must choose $x = 6$. The question asks for the value of $x + 3$, which is $6 + 3 = 9$.

Question 16

• Correct Answer: D
• Analysis: Plug in $(0, 4)$: $4 = a^0 + b \implies 4 = 1 + b \implies b = 3$. Now plug in $(2, 12)$ and $b = 3$: $12 = a^2 + 3 \implies a^2 = 9$. Since $a > 0$, $a = 3$. The function is $g(x) = 3^x + 3$. For $x = 3$: $g(3) = 3^3 + 3 = 27 + 3 = 30$.

Question 17

• Correct Answer: B
• Analysis: Factor the numerator: $2x^2 - 5x - 3 = (2x + 1)(x - 3)$. For $x \neq 3$, we can cancel out the $(x - 3)$ term in the numerator and denominator: $\frac{(2x + 1)(x - 3)}{x - 3} = 2x + 1$.

Question 18

• Correct Answer: A
• Analysis: Isolate the radical: $\sqrt{2x + 7} = x + 2$. Square both sides: $2x + 7 = (x + 2)^2 \implies 2x + 7 = x^2 + 4x + 4 \implies x^2 + 2x - 3 = 0$. Factoring gives $(x + 3)(x - 1) = 0$, so $x = 1$ or $x = -3$. Let’s check for extraneous solutions: If $x = 1$, $\sqrt{2(1)+7} - 1 = 3 - 1 = 2$ (valid). If $x = -3$, $\sqrt{2(-3)+7} - (-3) = 1 + 3 = 4 \neq 2$ (extraneous).

Question 19

• Correct Answer: B
• Analysis: The vertex of the original parabola $f(x)$ is at $(4, 3)$. The transformation $g(x) = f(x + 2) - 5$ shifts the graph left by 2 units and down by 5 units. Applying these changes to the original vertex: $x_{\text{new}} = 4 - 2 = 2$ and $y_{\text{new}} = 3 - 5 = -2$. The new vertex is $(2, -2)$.

Question 20

• Correct Answer: B
• Analysis: Recall the algebraic identity $(x + y)^2 = x^2 + 2xy + y^2$. Rearranging gives $(x + y)^2 = (x^2 + y^2) + 2(xy)$. Substitute the given values: $(x + y)^2 = 25 + 2(12) = 25 + 24 = 49$. Taking the positive square root gives $x + y = 7$.

Question 21

• Correct Answer: B
• Analysis: Using exponent rules, subtract the denominator exponents from the numerator exponents: For $x$: $-2 - 4 = -6$. For $y$: $3 - (-1) = 4$. This results in $x^{-6}y^4$, which can be rewritten with positive exponents as $\frac{y^4}{x^6}$.

Question 22

• Correct Answer: B
• Analysis: According to the Remainder Theorem and Factor Theorem, if $p(c) = 0$ for a polynomial, then $(x - c)$ is a factor of that polynomial. Since $p(3) = 0$, $(x - 3)$ must be a factor.

Question 23

• Correct Answer: C
• Analysis: For any quadratic equation in the form $ax^2 + bx + c = 0$, the sum of the solutions is given by $-\frac{b}{a}$. Here, $a = 2$ and $b = -8$. Therefore, the sum of the solutions is $-\frac{-8}{2} = 4$.

Question 24

• Correct Answer: A
• Analysis: Express 27 as a base of 3: $27 = 3^3$. The equation becomes $3^{2x-1} = (3^3)^{x+2} \implies 3^{2x-1} = 3^{3x+6}$. Since the bases are equal, their exponents must be equal: $2x - 1 = 3x + 6 \implies -1 - 6 = 3x - 2x \implies x = -7$.

Question 25

• Correct Answer: B
• Analysis: Exponential growth models follow the format $P(t) = P_0(b)^{\frac{t}{k}}$, where $P_0$ is the initial population, $b$ is the growth factor, and $k$ is the period of time required to grow by that factor. Here, $P_0 = 500$, $b = 2$ (doubles), and $k = 4$ hours. Thus, $P(t) = 500(2)^{\frac{t}{4}}$.

Question 26

• Correct Answer: C
• Analysis: Combine the fractions in the first equation by finding a common denominator: $\frac{y + x}{xy} = \frac{1}{3}$. We are given that $x + y = 12$, so substitute 12 into the numerator: $\frac{12}{xy} = \frac{1}{3}$. Cross-multiplying yields $xy = 36$.

Question 27

• Correct Answer: C
• Analysis: Plug the intercept $(1, 0)$ into the function: $0 = -(1 - h)^2 + 4 \implies (1 - h)^2 = 4$. Take the square root of both sides: $1 - h = 2$ or $1 - h = -2$. Solving both gives $h = -1$ or $h = 3$. Since the problem specifies $h > 0$, $h = 3$.

Question 28

• Correct Answer: B
• Analysis: Use FOIL to expand the product: $(4)(2) + (4)(i) + (-3i)(2) + (-3i)(i) = 8 + 4i - 6i - 3i^2$. Since $i^2 = -1$, substitute it into the expression: $8 - 2i - 3(-1) = 8 - 2i + 3 = 11 - 2i$.

Problem Solving and Data Analysis Solutions

Question 29

• Correct Answer: A
• Analysis: The proportion of defective items in the sample is $\frac{4}{200} = 0.02$ (or $2\%$). Applying this proportion to the daily production total: $0.02 \times 15,000 = 300$ items.

Question 30

• Correct Answer: B
• Analysis: In an ordered set of 7 items, the median is the 4th item. Increasing the 3 largest numbers (the 5th, 6th, and 7th numbers) changes their values but leaves their positions relative to the 4th number completely unchanged. The 4th number remains exactly the same, so the median remains 14.

Question 31

• Correct Answer: A
• Analysis: The total number of marbles initially is $6 + 4 + 10 = 20$. The probability of picking a blue marble first is $\frac{4}{20} = \frac{1}{5}$. Since there is no replacement, there are now 19 marbles left in the box, 10 of which are green. The probability of picking a green marble second is $\frac{10}{19}$. The joint probability is $\frac{1}{5} \times \frac{10}{19} = \frac{10}{95} = \frac{2}{19}$.

Question 32

• Correct Answer: B
• Analysis: Let the initial stock price be $100$. After a $20\%$ decrease on Monday, the price becomes $100 \times (1 - 0.20) = 80$. After a $30\%$ increase on Tuesday, the price becomes $80 \times (1 + 0.30) = 80 \times 1.30 = 104$. The net change from $100$ to $104$ is an increase of $4\%$.

Question 33

• Correct Answer: B
• Analysis: A margin of error signifies the confidence interval around the sample statistic. A support rate of $64\%$ with a $4\%$ margin of error means the true population parameter is estimated to fall within the range of $64\% \pm 4\%$, which translates to between $60\%$ and $68\%$.

Question 34

• Correct Answer: A
• Analysis: Find the sum of the scores of the first group: $20 \times 82 = 1,640$. Find the sum of the scores of the second group: $5 \times 92 = 460$. Combine the sums: $1,640 + 460 = 2,100$. Divide by the total number of students: $\frac{2100}{25} = 84$.

Question 35

• Correct Answer: C
• Analysis: The current range is $15.2 - 11.2 = 4.0$ seconds. If 15.2 is removed, the new maximum is 13.4, making the new range $13.4 - 11.2 = 2.2$ seconds, a massive change of $1.8$ seconds. Range is highly sensitive to outliers.

Question 36

• Correct Answer: B
• Analysis: Determine how many half-life periods occur in 24 hours: $\frac{24}{6} = 4$ half-lives. Multiply the initial dose by $\frac{1}{2}$ four times: $400 \times (\frac{1}{2})^4 = 400 \times \frac{1}{16} = 25$ mg.

Question 37

• Correct Answer: C
• Analysis: Find the rate per minute: $\frac{240}{15} = 16$ bottles per minute. Convert 2 hours to minutes: $2 \times 60 = 120$ minutes. Multiply the rate by the total minutes: $16 \times 120 = 1,920$ bottles.

Question 38

• Correct Answer: B
• Analysis: Use the principle of inclusion-exclusion to find the percentage of students doing either or both: $\%(\text{Sports} \cup \text{Music}) = 60\% + 40\% - 15\% = 85\%$. The percentage of students doing neither is $100\% - 85\% = 15\%$. Find $15\%$ of 120: $0.15 \times 120 = 18$ students.

Question 39

• Correct Answer: B
• Analysis: Substitute $x = 10$ directly into the line of best fit equation: $\hat{y} = -2.5(10) + 85 = -25 + 85 = 60$.

Question 40

• Correct Answer: B
• Analysis: Use the law of total probability: $\text{Total Errors} = (0.40 \times 0.01) + (0.35 \times 0.02) + (0.25 \times 0.04) = 0.004 + 0.007 + 0.010 = 0.021$, which is $2.10\%$.

Geometry and Trigonometry Solutions

Question 41

• Correct Answer: A
• Analysis: Tangent is defined as $\frac{\text{opposite}}{\text{adjacent}}$. We can think of a right triangle with an opposite side of 5 and an adjacent side of 12. Using the Pythagorean theorem ($5^2 + 12^2 = c^2 \implies c^2 = 169$), the hypotenuse is 13. Since $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$, the value is $\frac{5}{13}$.

Question 42

• Correct Answer: C
• Analysis: The standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius. Plugging in the center $(3, -2)$ and $r = 5$: $(x - 3)^2 + (y - (-2))^2 = 5^2 \implies (x - 3)^2 + (y + 2)^2 = 25$.

Question 43

• Correct Answer: B
• Analysis: The formula for arc length is $L = \frac{\theta}{360} \times 2\pi r$. Plugging in $\theta = 60^\circ$ and $r = 9$: $L = \frac{60}{360} \times 2\pi(9) = \frac{1}{6} \times 18\pi = 3\pi$.

Question 44

• Correct Answer: A
• Analysis: Dilation changes side lengths but keeps corresponding geometric angles completely identical. Since angle E maps cleanly to angle A, we compute $\cos(A) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AB}{AC} = \frac{6}{10} = \frac{3}{5}$.

Question 45

• Correct Answer: C
• Analysis: The total surface area formula for a cylinder is $A = 2\pi r h + 2\pi r^2$. Plugging in $r = 4$ and $h = 10$: $A = 2\pi(4)(10) + 2\pi(4)^2 = 80\pi + 32\pi = 112\pi$.

Question 46

• Correct Answer: C
• Analysis: By the co-function identity, $\sin(\theta) = \cos(90^\circ - \theta)$. Therefore, if $\sin(x^\circ) = \cos(24^\circ)$, then $x + 24 = 90 \implies x = 90 - 24 = 66$.

Question 47

• Correct Answer: A
• Analysis: Find the missing leg using the Pythagorean theorem: $12^2 + b^2 = 20^2 \implies b^2 = 256 \implies b = 16$. The area of a right triangle is given by $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times 16 = 96$.

Question 48

• Correct Answer: C
• Analysis: Complete the square for both variables. For $x$: $x^2 - 6x + 9 = (x - 3)^2$. For $y$: $y^2 + 8y + 16 = (y + 4)^2$. Add constants to the right side: $(x - 3)^2 + (y + 4)^2 = 0 + 9 + 16 = 25$. Since $r^2 = 25$, the radius $r$ is 5.

Question 49

• Correct Answer: B
• Analysis: A regular hexagon splits into 6 equilateral triangles radiating from the circle’s center. The side length of each triangle equals the radius (6 cm). Thus, the perimeter is $6 \times 6 = 36$ cm.

Question 50

• Correct Answer: B
• Analysis: Use the distance formula: $d = \sqrt{(6 - (-2))^2 + (9 - 3)^2} = \sqrt{8^2 + 6^2} = \sqrt{100} = 10$.