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

Calculate percentages: Candidate P received \( \frac{725}{1000} = 0.725 \), or 72.5%. Candidate Q received \( \frac{275}{1000} = 0.275 \), or 27.5%. If 8,000 people vote, Candidate P would receive \( 0.725 \times 8000 = 5800 \) votes, and Candidate Q \( 0.275 \times 8000 = 2200 \) votes. The difference is \( 5800 - 2200 = 3600 \) votes.

Calculate percentages: Candidate L received \( \frac{600}{1000} = 0.6 \), or 60%. Candidate M received \( \frac{400}{1000} = 0.4 \), or 40%. If 7,500 people vote, Candidate L would receive \( 0.6 \times 7500 = 4500 \) votes, and Candidate M \( 0.4 \times 7500 = 3000 \) votes. The difference is \( 4500 - 3000 = 1500 \) votes.

Find the slope using the points, then derive the line's equation. Set \(x = n + 2\) to find \(b = 48\).

To find the doubling time, set \( b(t) = 2 \cdot b(0) \) and solve for \( t \): \( 15,000 \cdot (1.06)^{t/100} = 30,000 \). Use logarithms to isolate \( t \) and convert to hours.

Factor out the greatest common factor, \(6x\), from the expression: \(6x(x^2 + 4x + 5)\). Then factor \(x^2 + 4x + 5\) as \((x + 1)(x + 5)\). The smallest possible value of \(b\) is 1.

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 \(P = \frac{w}{75}\).

Both sides simplify to \( 8z + 4 \), making the equation true for all values of z. Therefore, there are infinitely many solutions.

The vertex of the function \( f(x) = \frac{1}{6}(x - 10)^2 + 4 \) occurs at \( (10, 4) \). This represents the minimum height of the basketball. The vertex form of a quadratic function, \( a(x - h)^2 + k \), indicates the vertex as \( (h, k) \). Since \( h = 10 \) and \( k = 4 \), the minimum height is 4 feet above the court.

To ensure no solution, we set the slopes equal, indicating b = 48 based on intercept calculations.

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

In the function j(x) = -0.5x + 10, the y-intercept (where x = 0) represents the initial amount of water the runner had before starting the marathon. When x = 0, j(0) = 10, indicating that the runner started with 10 liters of water.

To find the doubling time, set \( g(t) = 2 \cdot g(0) \). Thus, \( 20,000 \cdot (1.05)^{t/300} = 40,000 \). Taking logarithms, isolate \( t \) and convert the answer to hours.

Let the number of voters who voted in favor of the bill be w. According to the problem, 3 times as many voters voted against, so the number against is 3w. The report states that 18,000 more people voted against, so the equation is 3w = w + 18,000. Solving for w gives 9,000, and the number against is 3(9,000) = 27,000.

First, find the measure of angle Q in radians: \( \frac{5\pi}{6} + \frac{\pi}{4} = \frac{14\pi}{12} = \frac{7\pi}{6} \). Then, convert to degrees by multiplying by \( \frac{180}{\pi} \): \( \frac{7\pi}{6} \cdot \frac{180}{\pi} = 210 \) degrees.

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.

Substitute \( (1, 5) \) into both inequalities. Both inequalities are satisfied.

Expanding the right side results in \( 10a - 3 \), which is identical to the left side. Therefore, the equation holds for all values of a, implying infinitely many solutions.

We use the Pythagorean theorem: \( c^2 = 16^2 + 30^2 = 256 + 900 = 1156 \). Therefore, \( c = \sqrt{1156} = 34 \). The hypotenuse can be written as \( 2\sqrt{289} \), so \( d = 289 \).

Rearrange and substitute to find that \( y = 10 \).

The values and frequencies are symmetric across the two data sets, which means the mean values are the same.

The vertex of the function \( f(x) = \frac{1}{4}(x - 8)^2 + 6 \) occurs at \( (8, 6) \). This represents the minimum height of the drone. The vertex form of a quadratic function, \( a(x - h)^2 + k \), indicates the vertex as \( (h, k) \). Since \( h = 8 \) and \( k = 6 \), the minimum height is 6 meters above the ground.

The sum of the solutions is calculated separately for each factor. Using the given total sum of \( \frac{25}{2} \), solve for the value of \( p \).

With \( h(4) = 810 \), substitute into \( h(x) = 10 \cdot c^x \): \( 10 \cdot c^4 = 810 \Rightarrow c^4 = 81 \Rightarrow c = 3 \). Now find \( h(-1) \): \( h(-1) = 10 \cdot 3^{-1} = 10 \cdot \frac{1}{3} = \frac{10}{3} \).

To find \( p \), use the equation: \( 39 = \frac{p}{100} \times 65 \Rightarrow p = \frac{39 \times 100}{65} = 60 \). Then, calculate \( 60 \% \) of \( 39 \): \( \frac{60}{100} \times 39 = 23.4 \).

In the equation 4𝑛 + 25𝑚 = 1,150, the coefficient 4 corresponds to the bird sanctuary, meaning 𝑛 represents the number of nests per hectare in the bird sanctuary.

The change in degrees Fahrenheit is determined by multiplying the temperature increase in kelvins by \( \frac{9}{5} \), since the \( x \) term in the function is being multiplied by this factor. Therefore, the increase in Fahrenheit temperature is \( \frac{9}{5} \times 2.10 = 3.78 \).

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 \( g(\frac{1}{2}) \): \( g(\frac{1}{2}) = 25 \cdot 5^{\frac{1}{2}} = 25 \cdot \sqrt{5} = 25 \sqrt{5} \).

To find \( a \), set \( q(x) = 0 \): \( -7x + 21 = 0 Rightarrow x = 3 \), so \( a = 3 \). For \( b \), evaluate \( q(0) = 21 \). Thus, \( a + b = 3 + 21 = 24 \).

Sum the solutions for each factor individually, then equate it to the given total of 15 to determine \( p \).

To find \( f(0) \), substitute \( 0 \) for \( x \) in the function: \( f(0) = 300(0.2)^0 \). Since any non-zero number raised to the power of \( 0 \) is \( 1 \), the expression becomes \( 300 imes 1 = 300 \).

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 equation 5𝑏 + 40𝑐 = 2,450, the coefficient 5 corresponds to the sports field, meaning 𝑏 represents the number of benches per acre in the sports field.

In the function k(x) = -8x + 64, the y-intercept (where x = 0) represents the initial amount of concrete the worker had before building any foundations. When x = 0, k(0) = 64, indicating that the worker started with 64 cubic feet of concrete.

Rewrite \(f(x)\) in vertex form as \(-3(x - 2)^2 + 7\). The maximum of \(f(x)\) is at \(x = 2\), so \(g(x) = f(x - 6)\) shifts this maximum to \(x = 8\).

For a quadratic in the form \( y = ax^2 + bx + c \), the minimum or maximum value of \( y \) occurs at \( x = \frac{-b}{2a} \). Here, \( a = 1 \) and \( b = -10 \), so \( x = \frac{10}{2} = 5 \).

Calculate the slope with the given points and determine the equation of the line. Substitute \(x = p - 6\) to find \(b = -41\).

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.

To find the surface area of the cube, we need to determine its side length. The formula for the volume of a cube is \( V = s^3 \). Given \( V = 474552 \), solve for \( s \): \( s = \sqrt[3]{474552} \approx 78 \). The formula for surface area is \( 6s^2 \). Thus, the surface area is \( 6(78)^2 = 36,504 \) square units.

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

Since the decay factor is 0.70 annually, this translates to a 30% decrease per year, so \(q = 30\).

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.

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

In similar triangles, corresponding angles have the same trigonometric values. Thus, \( \sin(J) = \sin(G) = \frac{24}{25} \).

Using \( P = 40,000 \) and \( C = 5,000 \), set up the equation \( 40,000 = 5,000(2)^{5r} \). Dividing both sides by 5,000 gives \( 8 = (2)^{5r} \). Taking the logarithm base 2 of both sides yields \( 3 = 5r \). Therefore, \( r = \frac{3}{5} \).