SAT Randomized Questions - 1 Full Math Practice Test - Answers and Detailed Explanations at the END

To find \( p \): \( 50 = \frac{p}{100} \times 80 \implies p = \frac{50 \times 100}{80} = 62.5 \). Calculate \( 62.5 \% \) of \( 50 \): \( \frac{62.5}{100} \times 50 = 31.25 \).

Since the function decreases by 60%, it retains 40% of its value at each step. Thus, the multiplier for the decay is 0.40. Therefore, the correct equation is \( p(x) = 50(0.40)^x \).

Since the triangles are similar, corresponding angles have equal trigonometric ratios, so \( \sin(U) = \sin(R) = \frac{40}{41} \).

Since \( g(2) = 625 \), substitute into \( g(x) = 25 \cdot b^x \) to solve for \( b \): \( 25 \cdot b^2 = 625 \Rightarrow b^2 = 25 \Rightarrow b = 5 \). Now find \( g(\frac{1}{2}) \): \( g(\frac{1}{2}) = 25 \cdot 5^{\frac{1}{2}} = 25 \cdot \sqrt{5} = 25 \sqrt{5} \).

Let the number of people who voted in favor of the policy be x. According to the problem, 5 times as many people voted against it, so the number against is 5x. The article states that 40,000 more people voted against it, so the equation is 5x = x + 40,000. Solving for x gives 10,000.

To cover the walls twice, double the total area \(w\) to get \(2w\). Since each gallon covers 150 square feet, divide \(2w\) by 150 to get the equation \(P = \frac{w}{75}\).

Using the quadratic formula, \(x = \frac{8 \pm \sqrt{64 - 28}}{2} = 4 \pm \sqrt{9}\). So, the solution can be written as \(4 + \sqrt{9}\) and the correct value of \(k\) is 9.

Using the slope \( \frac{1}{3} \) and point \( (-3, 5) \) in the point-slope formula, the resulting equation is \( y = \frac{1}{3}x + 6 \).

To find \( p \): \( 120 = \frac{p}{100} \times 200 \implies p = \frac{120 \times 100}{200} = 60 \). Calculate \( 60 \% \) of \( 120 \): \( \frac{60}{100} \times 120 = 72 \).

The lines have the same slope but different intercepts, indicating that they are parallel and do not intersect.

To isolate \(2p + 9q\), multiply both sides by \((2p + 9q)\), resulting in \(g = w(2p + 9q)\). Dividing by \(w\) gives \(2p + 9q = \frac{g}{w}\).

In the function g(x) = -3x + 50, the y-intercept (where x = 0) represents the initial amount of dough the baker had before making any loaves. When x = 0, g(0) = 50, indicating that the baker started with 50 pounds of dough.

The function \(g(x) = 1.5x\) represents a proportional increase with respect to \(x\). As \(x\) increases, \(g(x)\) also increases linearly. Therefore, the correct answer is 'Increasing Linear.'

Since \( m(3) = 4096 \), substitute into \( m(x) = 16 \cdot q^x \): \( 16 \cdot q^3 = 4096 \Rightarrow q^3 = 256 \Rightarrow q = 4 \). Now find \( m(\frac{1}{4}) \): \( m(\frac{1}{4}) = 16 \cdot 4^{\frac{1}{4}} = 16 \cdot \sqrt[4]{4} = 16 \cdot \sqrt{2} = 16 \sqrt{2} \).

Rewriting the equation: \( x^2 + x - 56 = 4x^2 - 28x \). Rearranging terms, we get \( -3x^2 + 29x - 56 = 0 \). Solving using the quadratic formula, the sum of the solutions is given by \( -b/a \), where \( a = -3 \) and \( b = 29 \), which yields \( 29/3 \).

The system has no solution when the slopes are the same but the lines have different intercepts. Solving this condition gives n = 45.

Since \(\frac{r}{s} = 6\), we get \(r = 6s\). Substitute into \(\frac{48r}{ks} = 6\) to obtain \(\frac{48(6s)}{ks} = 6\). Simplify to \(\frac{288s}{ks} = 6\), cancel \(s\) and solve for \(k = 48\).

Solving by elimination or substitution, we find that \( y = 14 \).

We use the Pythagorean theorem: \( c^2 = 7^2 + 24^2 = 49 + 576 = 625 \). Therefore, \( c = \sqrt{625} = 25 \). Since the length is already in the form \( \sqrt{625} [/latex}, the value of [latex] d \) is 625.

Since \(g(x) = f(x+4) = 7(2)^{x+4} = 7 \cdot 16 \cdot (2)^x = 112(2)^x\), the correct answer is g(x) = 112(2)^x.

Rewrite \(f(x)\) as \((x - 5)^2\). The minimum of \(f(x)\) is at \(x = 5\), so \(g(x) = f(x - 2)\) shifts this minimum to \(x = 7\).

The percentage can be calculated using the formula: \( \frac{62.5}{250} \times 100 = 25\% \).

Dividing the equation by 16, completing the square, and simplifying gives n = 8.

The minimum value of \( y \) occurs at \( x = \frac{-b}{2a} \). Here, \( a = 1 \) and \( b = -8 \), so \( x = \frac{8}{2} = 4 \).

Since \(g(x) = f(x+3) = 3(6)^{x+3} = 3 \cdot 216 \cdot (6)^x = 648(6)^x\), the correct answer is g(x) = 648(6)^x.

To find the number of votes Angel Cruz is expected to win by, first determine the percentage of votes each candidate received. Angel received \( \frac{483}{803} \approx 0.601 \), or 60.1%. Terry received \( \frac{320}{803} \approx 0.399 \), or 39.9%. If 6,424 people vote, Angel would receive \( 0.601 \times 6424 \approx 3864 \) votes and Terry would receive \( 0.399 \times 6424 \approx 2560 \) votes. The difference is \( 3864 - 2560 = 1304 \) votes.

The function \(q(x) = 1.3x\) represents a proportional increase, showing a steady, linear growth with respect to \(x\). Therefore, 'Increasing Linear' is correct.

The vertex of the function \( f(x) = \frac{1}{8}(x - 6)^2 + 5 \) occurs at \( (6, 5) \). This represents the minimum height of the kite. The vertex form of a quadratic function, \( a(x - h)^2 + k \), indicates the vertex as \( (h, k) \). Since \( h = 6 \) and \( k = 5 \), the minimum height is 5 feet above the ground.

The percentage can be calculated using the formula: \( \frac{120}{400} \times 100 = 30\% \).

Expanding the equation, we obtain \( x^2 + 2x - 10 = 3x^2 - 15x \). Rearranging terms, we get \( -2x^2 + 17x - 10 = 0 \). Using the quadratic formula, the sum of the solutions is given by \( -b/a \), where \( a = -2 \) and \( b = 17 \), which yields \( 17/2 \).

To isolate \(7m + 5n\), multiply both sides by \((7m + 5n)\) to get \(k = p(7m + 5n)\), then divide by \(p\) to find \(7m + 5n = \frac{k}{p}\).

Using the quadratic formula, \(x = \frac{12 \pm \sqrt{144 - 44}}{2} = 6 \pm \sqrt{25}\). So, the solution can be written as \(6 + \sqrt{25}\) and the correct value of \(k\) is 25.

To cover the walls twice, double the total area \(w\) to get \(2w\). Since each gallon covers 300 square feet, divide \(2w\) by 300 to get \(P = \frac{w}{150}\).

In the equation 10𝑎 + 30𝑏 = 1,800, the coefficient 10 corresponds to the orchard, meaning 𝑎 represents the number of apple trees per acre in the orchard.

Let the number of small candles be represented by \( x \), and the number of large candles by \( y \). The total number of candles is at least 200, so we have the inequality:

\( x + y geq 200 \).

The budget constraint is that the total cost must not exceed $2,200. Therefore, we have the equation:

\( 4.90x + 11.60y leq 2,200 \).

We want to maximize the number of large candles, \( y \). First, solve for \( x \) in terms of \( y \) using the inequality:

\( x = 200 - y \).

Substitute this into the budget constraint:

\( 4.90(200 - y) + 11.60y leq 2,200 \)

Simplify:

\( 980 - 4.90y + 11.60y leq 2,200 \)

Combine like terms:

\( 980 + 6.70y leq 2,200 \)

Subtract 980 from both sides:

\( 6.70y leq 1,220 \)

Solve for \( y \):

\( y leq rac{1,220}{6.70} approx 182.09 \)

Since the number of candles must be an integer, the maximum number of large candles is 182.

Data set A is skewed toward lower values, while data set B is skewed toward higher values. Therefore, the mean of Data set A is less than the mean of Data set B.

To find the surface area, find the side length first. The volume of the cube is \( V = s^3 \). Given \( V = 729 \), \( s = \sqrt[3]{729} = 9 \). The surface area formula is \( 6s^2 \). Therefore, the surface area is \( 6(9)^2 = 486 \) square units.

Convert the area to square miles by dividing by \(1760^2\): \(1,549,000 / 3,097,600 = 0.5\) square miles.

Rewrite \(f(x)\) in vertex form as \(2(x + 2)^2 - 2\). The minimum of \(f(x)\) is at \(x = -2\), so \(g(x) = f(x + 1)\) shifts this to \(x = -3\).

Since \(g(x) = f(x+2) = 8(5)^{x+2} = 8 \cdot 25 \cdot (5)^x = 200(5)^x\), the correct answer is g(x) = 200(5)^x.

The half-life is 3 years, so the remaining quantity after \( x \) minutes is \( Q = 5 \cdot \left(\frac{1}{2}\right)^{\frac{x}{3 \cdot 365 \cdot 24 \cdot 60}} \).

Calculate the slope using the two points and find the equation of the line. Substitute \(x = m - 4\) to find the y-intercept \(b = 45\).

The population grows by 2% annually, so the growth factor is 1.02. Since \( x \) is in seconds, we convert it to years by dividing by \( 365 \cdot 24 \cdot 60 \cdot 60 \). The equation is \( p = 50,000 \cdot 1.02^{\frac{x}{365 \cdot 24 \cdot 60 \cdot 60}} \).

To find the doubling time, solve \( j(t) = 2 \cdot j(0) \), leading to \( 100,000 \cdot (1.03)^{t/350} = 200,000 \). Taking logarithms of both sides allows solving for \( t \) and then converting to hours.