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 Easy
What value of \(p\) satisfies the equation \(5p+180=250\) ?
A 14
B 65
C 86
D 250
▶️Answer/Explanation
Ans: A
Choice A is correct. Subtracting \(180\) from both sides of the given equation yields \(5p=70\) . Dividing both sides of this equation by \(5\) yields \(p=14\) . Therefore, the value of that satises the equation \(5p+180=250\) is \(14\).
Choice B is incorrect. This value of \(p\) satises the equation \(5p+180=505\).
Choice C is incorrect. This value of \(p\) satises the equation \(5p+180=610\).
Choice D is incorrect. This value of \(p\) satises the equation \(5p+180=1430\).
Question Easy
The perimeter of an isosceles triangle is \(83\) inches. Each of the two congruent sides of the triangle has a length of \(24\) inches. What is the length, in inches, of the third side?
▶️Answer/Explanation
Ans: \(35\)
Rationale
The correct answer is \(35\) . It’s given that the perimeter of an isosceles triangle is \(83\) inches and that each of the two congruent sides has a length of \(24\) inches. The perimeter of a triangle is the sum of the lengths of its three sides. The equation \(24+24+x=83\) can be used to represent this situation, where \(x\) is the length, in inches, of the third side. Combining like terms on the left-hand side of this equation yields \(48+x=83\) . Subtracting \(48\) from both sides of this equation yields \(x=35\). Therefore, the length, in inches, of the third side is \(35\).
Question Easy
If 2(x − 4) = x , what value of x makes the equation true?
A) \(\frac{4}{3}\)
B) \(\frac{8}{3}\)
C) 4
D) 8
▶️Answer/Explanation
D) 8
We are given the equation \( 2(x – 4) = x \) and need to find the value of \( x \) that makes the equation true.
Distributing the 2 on the left-hand side:
\[
2(x – 4) = 2x – 8
\]
So, the equation becomes:
\[
2x – 8 = x
\]
Subtract \( x \) from both sides to isolate \( x \):
\[
2x – 8 – x = x – x
\]
\[
x – 8 = 0
\]
Add 8 to both sides to solve for \( x \):
\[
x – 8 + 8 = 0 + 8
\]
\[
x = 8
\]
Question Easy
Which equation has the same solution as 8x = 2x + 12?
A) 10x = −12
B) 10x = 12
C) 6x = −12
D) 6x = 12
▶️Answer/Explanation
D) 6x = 12
To find the equation that has the same solution as \(8x = 2x + 12\):
Solve the original equation for \(x\):
\[
8x – 2x = 12
\]
\[
6x = 12
\]
\[
x = 2
\]
Check each option to see which one also has \(x = 2\) as a solution:
Option A: \(10x = -12\)
\[
x = -\frac{12}{10} = -\frac{6}{5}
\]
This does not match \(x = 2\).
Option B: \(10x = 12\)
\[
x = \frac{12}{10} = \frac{6}{5}
\]
This does not match \(x = 2\).
Option C: \(6x = -12\)
\[
x = -2
\]
This does not match \(x = 2\).
Option D: \(6x = 12\)
\[
x = 2
\]
This matches the solution of the original equation.
Question Easy
Line segment \(A C\) has a length of 120 and contains point \(B\). If \(A B=5 x+20\) and \(B C=6 x-10\), which equation shows the relationship between the lengths of line segments \(A B, B C\), and \(A C\) ?
A) \(5 x+20=120\)
B) \(6 x-10=120\)
C) \((5 x+20)-(6 x-10)=120\)
D) \((5 x+20)+(6 x-10)=120\)
▶️Answer/Explanation
D
We need to find the equation that represents the relationship between the lengths of the line segments \(AB\), \(BC\), and \(AC\).Since \(AC = AB + BC\):
\[ AC = 120 \]
\[ AB + BC = 120 \]
\[ (5x + 20) + (6x – 10) = 120 \]
Simplifying:
\[ 5x + 20 + 6x – 10 = 120 \]
\[ 11x + 10 = 120 \]
Therefore, the correct equation is:
\[ (5x + 20) + (6x – 10) = 120 \]
So the answer is:
\[ \boxed{D} \]