SAT MAth Practice questions – all topics
- Algebra Weightage: 35% Questions: 13-15
- Linear equations in one variable
- Linear equations in two variables
- Linear functions
- Systems of two linear equations in two variables
- Linear inequalities in one or two variables
SAT MAth and English – full syllabus practice tests
Question Hard
Line \(m\) is shown in the \(x y\)-plane, and the point with coordinates \((0.25, r)\) is on line \(m\). What is the value of \(r\) ?
▶️Answer/Explanation
Ans:5/3
To find the equation of line \(m\), we can first determine its slope using the given points \((0,2)\) and \((1.5,0)\). Then, we can use the slope-intercept form of a line (\(y = mx + b\)), where \(m\) is the slope and \(b\) is the y-intercept.
Calculate the slope (\(m\)):
\[m = \frac{{y_2 – y_1}}{{x_2 – x_1}}\]
\[m = \frac{{0 – 2}}{{1.5 – 0}}\]
\[m = \frac{{-2}}{{1.5}}\]
\[m = -\frac{4}{3}\]
Use the slope-intercept form to find the equation of the line:
Given that the y-intercept is 2, we can substitute \(m\) and \(b\) into the equation:
\[y = -\frac{4}{3}x + 2\]
So, the equation of line \(m\) is \(y = -\frac{4}{3}x + 2\).
Now, we’re given that the point \((0.25, r)\) is on line \(m\). We can substitute \(x = 0.25\) into the equation of line \(m\) to find \(r\):
\[r = -\frac{4}{3}(0.25) + 2\]
\[r = -\frac{1}{3} + 2\]
\[r = \frac{5}{3}\]
Therefore, the coordinates of the point on line \(m\) are \((0.25, \frac{5}{3})\).
Question Hard
For the linear equation \(y=m x+b\), where \(m\) and \(b\) are positive constants, which of the following tables gives three values of \(x\) and their corresponding values of \(y\) ?
▶️Answer/Explanation
Ans:B
Given the linear equation \( y = mx + b \), we need to match the given values of \( x \) with their corresponding values of \( y \) from the options.
1. For \( x = -2 \):
\[
y = m(-2) + b = -2m + b
\]
2. For \( x = 1 \):
\[
y = m(1) + b = m + b
\]
3. For \( x = \frac{-b}{m} \):
\[
y = m \left( \frac{-b}{m} \right) + b = -b + b = 0
\]
Option A:
For \( x = -2 \): \( y = -2m \) (incorrect, should be \( -2m + b \))
For \( x = 1 \): \( y = m \) (incorrect, should be \( m + b \))
For \( x = \frac{b}{m} \): \( y = 0 \) (incorrect \( x \) value)
Option B:
For \( x = -2 \): \( y = -2m + b \) (correct)
For \( x = 1 \): \( y = m + b \) (correct)
For \( x = \frac{-b}{m} \): \( y = 0 \) (correct)
Option C:
For \( x = -2 \): \( y = -2m + b \) (correct)
For \( x = 1 \): \( y = m + b \) (correct)
For \( x = b \): \( y = 0 \) (incorrect \( x \) value)
Option D:
For \( x = -2 \): \( y = -2m \) (incorrect, should be \( -2m + b \))
For \( x = 1 \): \( y = m \) (incorrect, should be \( m + b \))
For \( x = b \): \( y = 0 \) (incorrect \( x \) value)
Question Hard
\[
-9 x+24 q x=36
\]
In the given equation, \(q\) is a constant. The equation has no solution. What is the value of \(q\) ?
▶️Answer/Explanation
$.375, 3 / 8$
If the equation \( -9x + 24qx = 36 \) has no solution, it means that the coefficients of \(x\) on both sides of the equation are equal, but their constant terms are not equal.
Given:
Coefficient of \(x\) on the left side: \(-9\)
Coefficient of \(x\) on the right side: \(-24q\)
For the equation to have no solution, these coefficients must be equal to each other. Therefore, we have:
\[ -9 =- 24q \]
Now, let’s solve for \(q\):
\[ q = \frac{9}{24} \]
\[ q = \frac{3}{8} \]
So, the value of \(q\) is \( \boxed{\frac{3}{8}} \).
Question Hard
When the quadratic function \(f\) is graphed in
the \(x y\)-plane, where \(y=f(x)\), its vertex is \((-2,5)\). One of the \(x\)-intercepts of this graph is \(\left(-\frac{7}{3}, 0\right)\). What is the other \(x\)-intercept of the graph?
A. \(\left(-\frac{13}{3}, 0\right)\)
B. \(\left(-\frac{5}{3}, 0\right)\)
C. \(\left(\frac{1}{3}, 0\right)\)
D. \(\left(\frac{7}{3}, 0\right)\)
▶️Answer/Explanation
Ans:B
Question Hard
y= 3x + 12
One of the two linear equations in a system is given. The system has no solution. Which equation could be the second equation in the system ?
A) y=3(x+3)
B) y= 3(x+ 4)
C) y=4(x+3)
D) y= 4(x +4)
▶️Answer/Explanation
A) y=3(x+3)
Given the equation \(y = 3x + 12\), we need to find an equation for the second line in the system such that the system has no solution. This occurs when the lines are parallel, meaning they have the same slope but different \(y\)-intercepts.
1. Identify the slope of the given line:
The slope of \(y = 3x + 12\) is 3.
2. Find an equation with the same slope:
We need another equation with a slope of 3 but a different \(y\)-intercept.
Check the options:
\(y = 3(x + 3)\):
\[
y = 3x + 9 \quad \text{(slope = 3, intercept = 9)}
\]
\(y = 3(x + 4)\):
\[
y = 3x + 12 \quad \text{(slope = 3, intercept = 12)}
\]
\(y = 4(x + 3)\):
\[
y = 4x + 12 \quad \text{(slope = 4)}
\]
\(y = 4(x + 4)\):
\[
y = 4x + 16 \quad \text{(slope = 4)}
\]
Since the second equation must have the same slope but a different intercept, the correct answer is:\[ \boxed{\text{A}} \]