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
Questions Hard
\[
-2|x-5|=-4 x
\]
What are all possible solutions to the given equation?
A) -5
B) \(\frac{5}{3}\)
C) -5 and \(\frac{5}{3}\)
D) 5 and \(-\frac{5}{3}\)
▶️Answer/Explanation
Ans: B
To solve the equation:
\[ -2|x-5| = -4x \]
First, divide both sides by \(-2\):
\[ |x-5| = 2x \]
We consider two cases for the absolute value equation:
Case 1: \(x-5 = 2x\)
\[
x – 5 = 2x
\]
\[
-5 = x
\]
So, \(x = -5\).
Case 2: \(x-5 = -2x\)
\[
x – 5 = -2x
\]
\[
x + 2x = 5
\]
\[
3x = 5
\]
\[
x = \frac{5}{3}
\]
We must check if both solutions satisfy the original equation.
For \(x = -5\):
\[
-2| -5 – 5 | = -2| -10 | = -2 \times 10 = -20
\]
\[
-4 \times -5 = 20
\]
Clearly, \(-20 \neq 20\), so \(x = -5\) is not a valid solution.
For \(x = \frac{5}{3}\):
\[
-2 \left| \frac{5}{3} – 5 \right| = -2 \left| \frac{5}{3} – \frac{15}{3} \right| = -2 \left| \frac{-10}{3} \right| = -2 \times \frac{10}{3} = -\frac{20}{3}
\]
\[
-4 \times \frac{5}{3} = -\frac{20}{3}
\]
Both sides are equal, so \(x = \frac{5}{3}\) is a valid solution.
Thus, the correct answer is:
\[
\boxed{\frac{5}{3}}
\]
Question Hard
$
2.1(h+3)=3 h+2.1
$
What value of \(h\) is the solution to the given equation?
▶️Answer/Explanation
Ans: 4.66,14 / 3,4.67
To solve the equation \(2.1(h+3) = 3h + 2.1\), we can start by distributing \(2.1\) on the left side:
\[2.1h + 2.1(3) = 3h + 2.1\]
\[2.1h + 6.3 = 3h + 2.1\]
Now, let’s isolate \(h\) by moving all terms with \(h\) to one side of the equation:
\[2.1h – 3h = 2.1 – 6.3\]
\[-0.9h = -4.2\]
Divide both sides by \(-0.9\) to solve for \(h\):
\[h = \frac{-4.2}{-0.9} = 4.67\]
So, the value of \(h\) that is the solution to the given equation is \(h = 4.67\).
Question Hard
If \(\frac{x-3}{7}=\frac{x-3}{9}\) , the value of 𝑥 − 3 is between which of the following pairs of values?
A. -7 and -9
B. -1 and 1
C. 2.5 and 3.5
D. 6.75 and 9.25
▶️Answer/Explanation
Ans: B
To find the value of \(x-3\), let’s first cross multiply the equation \(\frac{x-3}{7} = \frac{x-3}{9}\):
\[ 9(x-3) = 7(x-3) \]
Expanding both sides:
\[ 9x – 27 = 7x – 21 \]
Now, let’s solve for \(x\):
\[ 9x – 7x = -21 + 27 \]
\[ 2x = 6 \]
\[ x = 3 \]
Now that we know \(x = 3\), we can substitute it back into the equation to find \(x – 3\):
\[ x – 3 = 3 – 3 = 0 \]
So, the value of \(x – 3\) is \(0\). Therefore, the correct answer is option B, \(-1\) and \(1\).
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
2 | x – 9 |= 20
What is the sum of the solutions to the given equation?
▶️Answer/Explanation
18
To solve the equation \( 2|x-9|=20 \), we’ll isolate \( |x-9| \) first.
Dividing both sides by 2:
\[ |x – 9| = 10 \]
This equation means that \( x – 9 \) could be either 10 or -10, since the absolute value of a number is its distance from zero on the number line.
So, we have:
\[ x – 9 = 10 \]
\[ \text{or} \]
\[ x – 9 = -10 \]
Solving each equation:
1. \( x – 9 = 10 \)
Adding 9 to both sides:
\[ x = 19 \]
2. \( x – 9 = -10 \)
Adding 9 to both sides:
\[ x = -1 \]
The sum of the solutions is \( 19 + (-1) = 18 \).