## 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

**[No calc]**** ****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 *= 2*yz *+ 3

*B) x *= 2*yz *− 3

C) \(x=\frac{z}{2y}-3\)

D) \(x=\frac{z-3}{2y}\)

**▶️Answer/Explanation**

*B) x *= 2*yz *− 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

\]

**[Calc]**** ****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.

**[Calc]**** ****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.

**[No- Calc]**** ***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 $x=1\Rightarrow =2(3)^{1}=6$

**[Calc]**** ***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