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 Hard
\(\frac{x^{2}+x}{x+5}\)
The given expression can be rewritten as \(A+\frac{20}{x+5}=\) where A is a polynomial. Which of the following represents A?
A. 𝑥 − 4
B. 𝑥 + 4
C. \(x^{2}+x\)
D. \(x^{2}+x-20\)
▶️Answer/Explanation
Ans: A
To rewrite the given expression \(\frac{x^2 + x}{x + 5}\) in the form \(A + \frac{20}{x + 5}\), we need to perform polynomial division.
\[
\frac{x^2 + x}{x + 5}
\]
Divide \(x^2\) by \(x\):
\[
x^2 \div x = x
\]
Multiply \(x\) by \(x + 5\):
\[
x(x + 5) = x^2 + 5x
\]
Subtract \(x^2 + 5x\) from \(x^2 + x\):
\[
(x^2 + x) – (x^2 + 5x) = x – 5x = -4x
\]
\[
\frac{-4x}{x + 5} = -4
\]
Now our quotient is \(x – 4\) and the remainder is:
\[
-4x + 20
\]
\[
\frac{x^2 + x}{x + 5} = x – 4 + \frac{20}{x + 5}
\]
Therefore, the expression can be rewritten as:
\[
x – 4 + \frac{20}{x + 5}
\]
So, \(A = x – 4\).
[No calc] Question Hard
Which of the following could be the graph of the equation y=\(\frac{-4x+16}{x+2}\)?
▶️Answer/Explanation
Ans: D
To find the \(y\)-intercept of the equation \(y=\frac{-4 x+16}{x+2}\), we substitute \(x=0\) into the equation since the \(y\)-intercept occurs where \(x=0\).
So, when \(x=0\) :
\[
\begin{aligned}
& y=\frac{-4(0)+16}{0+2} \\
& y=\frac{16}{2} \\
& y=8
\end{aligned}
\]
Therefore, the \(y\)-intercept of the graph is at the point \((0,8)\). Which is only correctly fit in option – D
[Calc] Question Hard
If ax − 3 is a factor of \(6x^3 + 27x^2 − 54x\) , where a is a positive constant, what is the value of a ?
▶️Answer/Explanation
2
To factorize the polynomial \(6x^3 + 27x^2 – 54x\),
The GCF of \(6x^3\), \(27x^2\), and \(-54x\) is \(3x\).
\[
6x^3 + 27x^2 – 54x = 3x(2x^2 + 9x – 18)
\]
To factor \(2x^2 + 9x – 18\), we look for two numbers that multiply to \(2 \cdot (-18) = -36\) and add to \(9\).
These numbers are \(12\) and \(-3\):
\[
2x^2 + 9x – 18 = 2x^2 + 12x – 3x – 18
\]
\[
2x^2 + 12x – 3x – 18 = 2x(x + 6) – 3(x + 6)
\]
\[
2x(x + 6) – 3(x + 6) = (2x – 3)(x + 6)
\]
Thus, the factorized form of \(6x^3 + 27x^2 – 54x\) is:
\[
\boxed{3x(2x – 3)(x + 6)}
\]
Now comparing with ax − 3 .
$a=2$
[Calc] Question Hard
What is the graph of the equation \(y = {(x + 2)}^2 – 4\)?
▶️Answer/Explanation
D
\(y = {(x + 2)}^2 – 4\) If we put $x=-2$ then value of $y = {(-2 + 2)}^2 – 4\Rightarrow -4$
and For $x=0$ then value of $y = {(0 + 2)}^2 – 4\Rightarrow 0$
[No- Calc] Questions Hard
\(\frac{1}{2x}+5=kx+7\)
In the given equation, k is a constant. The equation has no solution. What is the value of k ?
▶️Answer/Explanation
Ans: 1/2, .5
To have no solution, the coefficients of \(x\) on both sides of the equation should be equal, but the constants should be different. Therefore, we set the coefficients equal to each other:
\[kx = \frac{1}{2}x \]
\[k = \frac{1}{2} \]
So, the value of \(k\) is \(\frac{1}{2}\).
Question Hard
\[
\frac{(x-4)(x+2)}{(x-4)}=0
\]
Which value is a solution to the given equation?
A) 4
B) 2
C) 0
D) -2
▶️Answer/Explanation
D
Given the equation:
\[ \frac{(x-4)(x+2)}{(x-4)} = 0 \]
First, simplify the equation:
\[ x + 2 = 0 \]
\[ x = -2 \]
We must also consider the domain restriction from the denominator \( (x-4) \):
\[ x \neq 4 \]
Since \( x = -2 \) does not violate this restriction, it is a valid solution.
So the answer is:
\[ \boxed{D} \]
Question Hard
The table shows several values of x and their corresponding values of f(x). The function f is defined by f(x) = mx + b, where m and b are constants. What is the value of b ?
▶️Answer/Explanation
Ans: 16
To find the value of \(b\), we can use the fact that \(f(x) = mx + b\).
Let’s choose the point \((2, 106)\).
Substituting \(x = 2\) and \(f(x) = 106\) into the equation, we get:
\[106 = m \cdot 2 + b\]
Now, we need to find the slope \(m\) of the function. use another point from the table, say \((3, 151)\), to find \(m\).
Substituting \(x = 3\) and \(f(x) = 151\) into the equation, we get:
\[151 = m \cdot 3 + b\]
Now, we have two equations:
\[106 = 2m + b\]
\[151 = 3m + b\]
We can subtract the first equation from the second to eliminate \(b\):
\[151 – 106 = 3m – 2m\]
\[45 = m\]
Now, we can substitute \(m = 45\) into one of the equations to solve for \(b\).
\[106 = 2 \cdot 45 + b\]
\[106 = 90 + b\]
\[b = 106 – 90\]
\[b = 16\]
So, the value of \(b\) is \(\mathbf{16}\).
Question Hard
$
|x+1|=5
$
What positive value of \(\mathrm{x}\) satisfies the given equation?
▶️Answer/Explanation
Ans:4
To solve \(|x + 1| = 5\), we consider the definition of absolute value. This gives us two equations:
\[x + 1 = 5\]
\[x + 1 = -5\]
Solving these:
\[x + 1 = 5 \implies x = 4\]
\[x + 1 = -5 \implies x = -6\]
The positive value of \(x\) that satisfies the equation is:
\[x = 4\]
Question Hard
$$
(x-3)^4=0
$$
What value of $x$ makes the equation above true?
▶️Answer/Explanation
3