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 Medium
Some values of \(\mathrm{x}\) and the corresponding values of \(f(x)\) are given in the table shown.
If there is a linear relationship between \(\mathrm{x}\) and \(f(\mathrm{x})\), which of the following equations gives this relationship?
A) \(f(x)=\frac{1}{2} x+\frac{1}{2}\)
B) \(f(x)=\frac{1}{2} x-\frac{1}{2}\)
C) \(f(x)=\frac{1}{6} x+\frac{5}{6}\)
D) \(f(x)=\frac{1}{6} x+\frac{2}{3}\)
▶️Answer/Explanation
Ans:D
To determine the equation representing the linear relationship between \(x\) and \(f(x)\), we can use the point-slope form of a linear equation:
\[f(x) – f(x_1) = m(x – x_1)\]
where \(m\) is the slope of the line and \((x_1, f(x_1))\) is a point on the line.
Let’s choose the point \((2, 1)\) as it is the first point in the table.
Substituting the values into the point-slope form:
\[f(x) – 1 = m(x – 2)\]
Now, let’s calculate the slope \(m\) using another point, for example, \((5, 1.5)\). The slope \(m\) is given by:
\[m = \frac{f(x_2) – f(x_1)}{x_2 – x_1} = \frac{1.5 – 1}{5 – 2} = \frac{0.5}{3} = \frac{1}{6}\]
Now, substituting \(m = \frac{1}{6}\) and \((x_1, f(x_1)) = (2, 1)\) into the equation:
\[f(x) – 1 = \frac{1}{6}(x – 2)\]
\[f(x) – 1 = \frac{1}{6}x – \frac{1}{3}\]
\[f(x) = \frac{1}{6}x – \frac{1}{3} + 1\]
\[f(x) = \frac{1}{6}x + \frac{2}{3}\]
So, the equation representing the linear relationship between \(x\) and \(f(x)\) is:
D) \(f(x) = \frac{1}{6}x + \frac{2}{3}\)
Question medium
Which equation has no solution?
A. 4(𝑥 + 1) = 𝑥 + 4
B. 4(𝑥 + 1) = 𝑥 + 1
C. 4(𝑥 + 1) = 4𝑥 + 4
D. 4(𝑥 + 1) = 4x
▶️Answer/Explanation
▶️Answer/Explanation
Ans: D
We need to determine which equation has no solution. To do this, we’ll solve each equation and check if it leads to a contradiction.
A. \(4(x+1)=x+4\) becomes \(4x + 4 = x + 4 \), which simplifies to \(3x = 0 \), leading to \( x = 0 \). So, this equation has a solution.
B. \(4(x+1)=x+1\) simplifies to \(4x + 4 = x + 1 \), which leads to \(3x = -3 \), giving \( x = -1 \). So, this equation has a solution.
C. \(4(x+1)=4x+4\) becomes \(4x + 4 = 4x + 4 \), which simplifies to \(0 = 0 \), indicating all \(x\) are solutions. So, this equation has infinite solutions.
D. \(4(x+1)=4x\) simplifies to \(4x + 4 = 4x \), which leads to \(4 = 0 \), showing a contradiction. Hence, this equation has no solution.
Therefore, the answer is D.
Question Medium
Line \(k\) is defined by \(y=2 x+14\). Line \(j\) is perpendicular to line \(k\) in the \(x y\)-plane. What is the slope of line \(j\) ?
A) \(-\frac{1}{2}\)
B) \(\frac{1}{14}\)
C) \(\frac{1}{2}\)
D) 2
▶️Answer/Explanation
A
Line \(k\) is defined by \(y = 2x + 14\).
To find the slope of line \(j\), which is perpendicular to line \(k\), we know that the product of the slopes of perpendicular lines is -1.
The slope of line \(k\) is 2. So, the slope of line \(j\) will be the negative reciprocal of 2.
\[ \text{Slope of } j = -\frac{1}{\text{Slope of } k} = -\frac{1}{2} \]
Therefore, the answer is:
\[ \boxed{A) \, -\frac{1}{2}} \]
Question medium
Line k is shown in the xy-plane. Line j (not shown) is perpendicular to line k. What is the slope of line j?
▶️Answer/Explanation
Ans: 5/2, 2.5
Since line K is passing through points (-5,0) and (0,-2).
\[
\text { slope of } k=m_k=\frac{y_2-y_1}{x_2-x_1}
\]
Substitute the given points \((-5,0)\) and \((0,-2)\) :
\[
m_k=\frac{-2-0}{0-(-5)}=\frac{-2}{5}
\]
Since line \(j\) is perpendicular to line \(k\), the slope of line \(j\left(m_j\right)\) is the negative reciprocal of the slope of line \(k\).
\[
m_j=-\frac{1}{m_k}=-\frac{1}{\left(\frac{-2}{5}\right)}=\frac{5}{2}
\]
Thus, the slope of line \(j\) is:
\[
m_j=\frac{5}{2}
\]
Question Medium
The equation of line \(k\) is \(y=7 x+2\). What is the slope of a line that is parallel to line \(k\) in the \(x y\)-plane?
▶️Answer/Explanation
7
The equation of line \(k\) is given by:
\[ y = 7x + 2 \]
The slope-intercept form of a line is \(y = mx + b\), where \(m\) represents the slope. Here, the slope \(m\) of line \(k\) is 7. For a line to be parallel to line \(k\) in the \(xy\)-plane, it must have the same slope.
Thus, the slope of a line parallel to line \(k\) is:
\[ \boxed{7} \]