SAT MAth Practice questions – all topics
- Advanced Math Weightage: 35% Questions: 13-15
- Equivalent expressions
- Nonlinear equations in one variable and systems of equations in two variables
- Nonlinear functions
SAT MAth and English – full syllabus practice tests
Question medium
\(z=\frac{x+3}{2y}\)
The given equation relates the distinct positive real numbers x, y, and z. Which equation correctly expresses x in terms of y and z ?
A) x = 2yz + 3
B) x = 2yz − 3
C) \(x=\frac{z}{2y}-3\)
D) \(x=\frac{z-3}{2y}\)
▶️Answer/Explanation
B) x = 2yz − 3
To express \( x \) in terms of \( y \) and \( z \) given the equation \( z = \frac{x + 3}{2y} \), follow these steps:
\[
z = \frac{x + 3}{2y}
\]
Multiply both sides by \( 2y \) to eliminate the denominator:
\[
2yz = x + 3
\]
Isolate \( x \) by subtracting 3 from both sides:
\[
x = 2yz – 3
\]
Thus, the correct equation expressing \( x \) in terms of \( y \) and \( z \) is:
\[
x = 2yz – 3
\]
Question medium
\(L_1=10{(0.89)}^d\)
\(L_2=9.75-0.83d\)
An exponential equation and a linear equation are given. Each equation estimates the luminosity, in units of billions of solar luminosities, for a Type Ia supernova d days after its peak luminosity, for d ≤ 6. The luminosity, in billions of solar luminosities, estimated by the linear equation 3 days after its peak luminosity is how much greater than the luminosity estimated by the exponential equation?
A) 0.21
B) 2.1
C) 21
D) 210
▶️Answer/Explanation
A) 0.21
To find how much greater the luminosity estimated by the linear equation \(L_2\) is than the luminosity estimated by the exponential equation \(L_1\) 3 days after its peak luminosity, we first need to calculate the luminosities from both equations when \(d = 3\).
Calculate \(L_1\) when \(d = 3\):
\[
L_1 = 10 \cdot (0.89)^3
\]
First, compute \((0.89)^3\):
\[
(0.89)^3 \approx 0.704969
\]
Now, multiply by 10:
\[
L_1 \approx 10 \cdot 0.704969 = 7.04969
\]
Calculate \(L_2\) when \(d = 3\):
\[
L_2 = 9.75 – 0.83 \cdot 3
\]
\[
L_2 = 9.75 – 2.49
\]
\[
L_2 = 7.26
\]
3. Find the difference between \(L_2\) and \(L_1\):
\[
\text{Difference} = L_2 – L_1
\]
\[
\text{Difference} = 7.26 – 7.04969
\]
\[
\text{Difference} \approx 0.21031
\]
Thus, the luminosity estimated by the linear equation \(3\) days after its peak luminosity is approximately \(0.21\) billion solar luminosities greater than the luminosity estimated by the exponential equation.
Question medium
In the xy-plane shown, the quadrants are labeled I, II, III, and IV. The graph of \(y = −{(x + h)}^2 + k\) , where h and k are positive constants, is a parabola. In which quadrant is the vertex of this parabola?
A)Quadrant I
B)Quadrant II
C)Quadrant III
D)Quadrant IV
▶️Answer/Explanation
B)Quadrant II
The given equation for the parabola is \( y = -{(x + h)}^2 + k \), where \( h \) and \( k \) are positive constants. This equation represents a parabola that opens downward because the coefficient of the squared term is negative.
Finding the Vertex
The standard form of a downward-opening parabola is \( y = -{(x – h)}^2 + k \), where the vertex is at \((h, k)\). In the given equation, the form is \( y = -{(x + h)}^2 + k \). This can be rewritten as \( y = -{(x – (-h))}^2 + k \), indicating that the vertex is at \((-h, k)\).
Determining the Quadrant
\( h \) is a positive constant, so \(-h\) is negative.
\( k \) is a positive constant.
The vertex \((-h, k)\) is located in the quadrant where \( x \) is negative and \( y \) is positive.
In the coordinate plane, the quadrant where \( x \) is negative and \( y \) is positive is Quadrant II.
Therefore, the vertex of the parabola \( y = -{(x + h)}^2 + k \) is in Quadrant II.
Question Medium
What is the graph of the equation \(y=2(3)^{x}\) ?
▶️Answer/Explanation
Ans: A
For $x=0\Rightarrow =2(3)^{0}=2$
For $x=1\Rightarrow =2(3)^{1}=6$
Question Medium
When a buffet restaurant charges \(\$ 12.00\) per meal, the number of meals it sells per day is 400 . For each \(\$ 0.50\) increase to the price per meal, the number of meals sold per day decreases by 10 . What is the price per meal that results in the greatest sales, in dollars, from meals each day?
A) \(\$ 16.00\)
B) \(\$ 20.00\)
C) \(\$ 24.00\)
D) \(\$ 28.00\)
▶️Answer/Explanation
Ans: A
To determine the price per meal that results in the greatest sales from meals each day, we need to express the total daily sales \( S \) as a function of the price per meal \( p \).
\( p \) as the price per meal in dollars.
\( n \) as the number of meals sold per day.
Given:
At \( p = \$12.00 \), \( n = 400 \) meals are sold.
For each $0.50 increase in price, the number of meals sold decreases by 10.
If \( x \) represents the number of $0.50 increases:
\[ p = 12 + 0.50x \]
\[ n = 400 – 10x \]
The total daily sales \( S \) can be expressed as:
\[ S = p \cdot n \]
\[ S = (12 + 0.50x)(400 – 10x) \]
To find the maximum sales, we need to maximize the function:
\[ S = (12 + 0.50x)(400 – 10x) \]
Expand and simplify the equation:
\[ S = 12(400 – 10x) + 0.50x(400 – 10x) \]
\[ S = 4800 – 120x + 200x – 5x^2 \]
\[ S = 4800 + 80x – 5x^2 \]
This is a quadratic function of the form \( S = -5x^2 + 80x + 4800 \). To find the maximum value, we use the vertex formula \( x = -\frac{b}{2a} \), where \( a = -5 \) and \( b = 80 \).
\[ x = -\frac{80}{2 \cdot -5} \]
\[ x = \frac{80}{10} \]
\[ x = 8 \]
Substitute \( x = 8 \) back into the price equation:
\[ p = 12 + 0.50 \cdot 8 \]
\[ p = 12 + 4 \]
\[ p = 16 \]
Therefore, the price per meal that results in the greatest sales each day is:16