This is the seventh practice test adapted for the new Digital SAT format. To reflect the adaptive nature of the exam, the questions are categorized into the four official testing domains: Algebra, Advanced Math, Problem-Solving and Data Analysis, and Geometry and Trigonometry.
Focus: Linear equations, systems of linear equations, and inequalities.
What is the solution to the equation \(3(x - 4) + 2 = 5x - 6\)?
First, distribute the 3 on the left side of the equation:
\[ 3x - 12 + 2 = 5x - 6 \]
Combine the constant terms on the left side:
\[ 3x - 10 = 5x - 6 \]
Subtract \(3x\) from both sides to group the variables on the right:
\[ -10 = 2x - 6 \]
Add \(6\) to both sides to isolate the variable term:
\[ -4 = 2x \]
Divide by \(2\) to find the final value of \(x\):
\[ x = -2 \]
The system of equations is given by \(2x + 3y = 12\) and \(x - y = 1\). What is the value of \(x + y\)?
Use substitution to solve the system. From the second equation, isolate \(x\):
\[ x = y + 1 \]
Substitute this expression for \(x\) into the first equation:
\[ 2(y + 1) + 3y = 12 \]
Distribute and combine like terms:
\[ 2y + 2 + 3y = 12 \Rightarrow 5y + 2 = 12 \]
\[ 5y = 10 \Rightarrow y = 2 \]
Substitute \(y = 2\) back into our expression for \(x\):
\[ x = 2 + 1 = 3 \]
The question asks for the sum of \(x\) and \(y\):
\[ x + y = 3 + 2 = 5 \]
A linear function \(f\) is defined by \(f(x) = -3x + c\), where \(c\) is a constant. If \(f(2) = 5\), what is the value of \(f(-1)\)?
First, use the given point \((2, 5)\) to find the value of the constant \(c\):
\[ f(2) = -3(2) + c = 5 \]
\[ -6 + c = 5 \Rightarrow c = 11 \]
Now rewrite the complete function rule: \[ f(x) = -3x + 11 \]
Finally, evaluate the function for \(x = -1\):
\[ f(-1) = -3(-1) + 11 = 3 + 11 = 14 \]
Focus: Quadratics, higher-order polynomials, and non-linear equations.
If \(x > 1\) and \(\dfrac{x^4 \cdot x^5}{x^{-2}} = x^k\), what is the value of \(k\)?
Apply exponent rules to simplify the left side. When multiplying terms with the same base, add their exponents:
\[ x^4 \cdot x^5 = x^{4+5} = x^9 \]
When dividing, subtract the exponent in the denominator from the numerator:
\[ \dfrac{x^9}{x^{-2}} = x^{9 - (-2)} = x^{9 + 2} = x^{11} \]
Since the bases are identical, we conclude that \(k = 11\).
What is the minimum value of the function \(g(x) = 2x^2 - 12x + 23\)?
Because the leading coefficient is positive, the function is a parabola that opens upward. Its minimum value occurs at the vertex.
Step 1: Find the x-coordinate of the vertex
\[ x = -\dfrac{b}{2a} = -\dfrac{-12}{2(2)} = \dfrac{12}{4} = 3 \]
Step 2: Find the function value at the vertex
Substitute \(x = 3\) into the function:
\[ g(3) = 2(3)^2 - 12(3) + 23 \]
\[ g(3) = 18 - 36 + 23 = 5 \]
The minimum value of the function is 5.
The polynomial \(P(x) = 2x^3 - kx^2 + 4x - 5\) is divided by \(x - 1\). If the remainder is 2, what is the value of the constant \(k\)?
According to the Polynomial Remainder Theorem, if you divide \(P(x)\) by \((x - 1)\), the remainder is equal to \(P(1)\).
\[ P(1) = 2(1)^3 - k(1)^2 + 4(1) - 5 = 2 \]
\[ 2 - k + 4 - 5 = 2 \Rightarrow 1 - k = 2 \]
Isolate \(k\): \[ -k = 1 \Rightarrow k = -1 \]
During a sale, a store offers a 25% discount. If the discounted price of a jacket is $60, what was the original price?
Let \(x\) be the original price. A 25% discount means the customer pays 75% of the original price:
\[ 0.75x = 60 \]
Divide by 0.75 to solve for \(x\):
\[ x = \dfrac{60}{0.75} = 80 \]
The original price was $80.
A bag contains red, blue, and green marbles in the ratio of 2:3:5. If there are exactly 15 blue marbles, what is the total number of marbles?
Let the parts of the ratio be \(2x, 3x,\) and \(5x\). We are given that blue marbles (\(3x\)) equals 15.
\[ 3x = 15 \Rightarrow x = 5 \]
Total marbles = \(2x + 3x + 5x = 10x\). Substitute \(x = 5\):
\[ \text{Total} = 10(5) = 50 \]
A printer prints 400 pages in 5 minutes. At this constant rate, how many minutes will it take to print 2000 pages?
Find the rate per minute: \[ \text{Rate} = 400 / 5 = 80 \text{ pages/min} \]
Time required: \[ \text{Time} = 2000 / 80 = 25 \text{ minutes} \]
(Shortcut: Since 2000 is 5 times 400, the time is 5 times the original 5 minutes, which is 25).
Triangle \(ABC\) is similar to triangle \(DEF\). Side \(AB = 4\) and corresponding side \(DE = 12\). If the area of \(ABC\) is 10, what is the area of \(DEF\)?
Calculate the linear scale factor: \[ k = 12 / 4 = 3 \]
When the scale factor of sides is \(k\), the scale factor of the area is \(k^2\):
\[ \text{Area Factor} = 3^2 = 9 \]
Area of \(DEF = 10 \times 9 = 90\).
The equation of a circle is \(x^2 + y^2 - 10x + 6y + 18 = 0\). What is the radius of the circle?
Complete the square for the \(x\) and \(y\) groups:
\[ (x^2 - 10x + 25) + (y^2 + 6y + 9) = -18 + 25 + 9 \]
\[ (x - 5)^2 + (y + 3)^2 = 16 \]
Since \(r^2 = 16\), the radius \(r = 4\).
In a right triangle with acute angle \(\theta\), \(\tan(\theta) = 3/4\). What is the value of \(\cos(\theta)\)?
Using SOH CAH TOA, \(\tan(\theta) = \text{Opposite} / \text{Adjacent} = 3 / 4\). This identifies a 3-4-5 right triangle where the hypotenuse is 5.
\[ \cos(\theta) = \text{Adjacent} / \text{Hypotenuse} = 4 / 5 = 0.8 \]