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
There are infinitely many solutions to which of the following equations?
I. \(4(x+1)+2=4 x+6\)
II. \(4(x+1)+1=4 x+7\)
A) I only
B) II only
C) I and II
D) Neither I nor II
▶️Answer/Explanation
Ans:A
We need to determine which of the given equations has infinitely many solutions.
Equation I:
\[
4(x+1)+2=4x+6
\]
First, simplify the left side:
\[
4x + 4 + 2 = 4x + 6 \\
4x + 6 = 4x + 6
\]
This equation is always true for any value of \( x \), which means there are infinitely many solutions for this equation.
Equation II:
\[
4(x+1)+1=4x+7
\]
First, simplify the left side:
\[
4x + 4 + 1 = 4x + 7 \\
4x + 5 = 4x + 7
\]
This equation simplifies to:
\[
4x + 5 = 4x + 7 \implies 5 = 7
\]
This is a contradiction, so there are no solutions for this equation.
Thus, the correct answer is:
\[
\boxed{\text{I only}}
\]
Question medium
How many solutions does the equation
3x − 8 = x + 2(x − 4) have?
A) Zero
B) Exactly one
C) Exactly two
D) Infinitely many
▶️Answer/Explanation
D) Infinitely many
To determine how many solutions the equation \(3x – 8 = x + 2(x – 4)\) has, follow these steps:
Distribute the terms on the right-hand side:
\[
3x – 8 = x + 2x – 8
\]
\[
3x – 8 = 3x – 8
\]
Simplify the equation:
\[
3x – 8 = 3x – 8
\]
This is an identity, meaning it holds true for all \(x\). Therefore, the equation has infinitely many solutions.
Question medium
x(x − 12) − 12(x − 12) = 0
How many distinct real solutions does the given equation have?
A) Zero
B) Exactly one
C) Exactly two
D) Infinitely many
▶️Answer/Explanation
B) Exactly one
To find the number of distinct real solutions for the equation \(x(x – 12) – 12(x – 12) = 0\), let’s first simplify it:
Distribute and combine like terms:
\[ x(x – 12) – 12(x – 12) = 0 \]
\[ x^2 – 12x – 12x + 144 = 0 \]
\[ x^2 – 24x + 144 = 0 \]
Now, we can see that this is a quadratic equation in the form \(ax^2 + bx + c = 0\), where \(a = 1\), \(b = -24\), and \(c = 144\).
We can use the discriminant (\(b^2 – 4ac\)) to determine the number of distinct real solutions:
\[ \text{Discriminant} = (-24)^2 – 4(1)(144) = 576 – 576 = 0 \]
Since the discriminant is equal to 0, the quadratic equation has exactly one real solution.
Question Medium
The function f is defined by f(x) = -2x + 8. The x-intercept of the graph of y = f(x) in the xy-plane is (x, 0). What is the value of x ?
▶️Answer/Explanation
4
To find the \(x\)-intercept of the graph of \(y = f(x)\) where \(f(x) = -2x + 8\), we set \(y = 0\) and solve for \(x\).
1. Set \(y = 0\):
\[
0 = -2x + 8
\]
2. Solve for \(x\):
\[
2x = 8
\]
\[
x = 4
\]
Question Medium
When a ball is thrown straight down with an initial speed of’32 feet per second, the ball hits the ground and bounces up. The equation \(y = \frac{1}{2} x + 8\) represents the relationship between the ball’s initial height x, in feet, and the maximum height y, in feet, that the ball reaches after bouncing once.
If the initial height is 24 feet, what is the maximum height, in feet, the ball reaches after bouncing once ?
A) 12
B) 16
C) 20
D) 32
▶️Answer/Explanation
C) 20
Given the equation \( y = \frac{1}{2} x + 8 \) that represents the relationship between the ball’s initial height \( x \), in feet, and the maximum height \( y \), in feet, that the ball reaches after bouncing once, we need to find the maximum height when the initial height is 24 feet.
1. Substitute \( x = 24 \) into the equation:
\[
y = \frac{1}{2} \times 24 + 8
\]
\[
y = 12 + 8
\]
\[
y = 20
\]
Thus, the maximum height the ball reaches after bouncing once is:
\[ \boxed{20} \]