SAT Math FREE Sample Practice Questions

1- C
When 5 times the number \(x\) is added to 10, the result is
\(10 + 5x\). Since this result is equal to 35, the equation \(10 + 5x = 35\) is true.
Subtracting 10 from each side of \(10 + 5x = 35\) gives \(5x=25\),
and then dividing both sides by 5 gives \(x=5\).
Therefore, 3 times x added to 6, or
\(6 + 3x\), is equal to \(6 + 3(5) = 21.\)

2- B
Since \(f(x)\) is linear function with a negative slop, then when \(x=-2, f(x)\) is maximum and when \(x = 3, f(x)\) is minimum. Then the ratio of the minimum value to the maximum value of the function is:
\(\frac{f(3)}{f(-2)}=\frac{-3(3)+1}{-3(-2)+1}=\frac{-8}{7}=-\frac{8}{7}\)

3- D
The description \(8+2x\) is 16 more than 20 can be written as the equation \(8+2x=16+20\), which is equivalent to \(8+2x=36\).
Subtracting 8 from each side of \(8+2x=36\) gives \(2x=28\).
Since \(6x\) is 3 times \(2x\), multiplying both sides of \(2x=28\) by 3 gives \(6x = 84\)

4- D
Of the 30 employees, there are 5 females under age 45 and 6 males age 45 or older. Therefore, the probability that the person selected will be either a female under age 45 or a male age 45 or older is: \(\frac{5}{30} + \frac{6}{30} = \frac{11}{30}\)

5- A
Substituting \(x\) for \(y\) in first equation.
\(5x+2y=3\)
\(5x+2(x)=3\)
\(7x=3\)
Divide both side of \(7x = 3\) by 3 gives \(x = \frac{3}{7}\)

6- D
Substituting 6 for\( x\) and 14 for \(y\) in \(y = nx+2\) gives \(14=(n)(6)+2\),
which gives \(n=2\). Hence, \(y=2x+2\).
Therefore, when \(x = 10\), the value of \(y\) is
\(y=(2)(10)+2 = 22\)

7- A
Subtracting \(2x\) and adding 5 to both sides of \(2x – 5 ≥ 3x – 1\) gives \(-4 ≥ x\).
Therefore, x is a solution to \(2x – 5 ≥ 3x – 1\) if and only if \(x\) is less than or equal to \(-4\) and \(x\) is NOT a solution to \(2x – 5 ≥ 3x – 1\) if and only if \(x\) is greater than \(-4\).
Of the choices given, only \(-2\) is greater than \(-4\) and, therefore, cannot be a value of \(x\).

8- C
The sum of the two polynomials is \((4x^2+6x-3)+(3x^2-5x+8)\)
This can be rewritten by combining like terms:
\((4x^2+6x-3)+(3x^2-5x+8)=(4x^2+3x^2 )+(6x-5x)+(-3+8)=
7x^2+x+5\)

9- C
Plugin the values of \(x\) in the choices provided. The points are \((1,-1)\),\((2,-3)\),and \((3,-5)\)
For \((1,-1)\) check the options provided:
A.\(g(x)=2x+1→-1=2(1)+1→-1=3\) This is NOT true.
B. \(g(x)=2x-1→-1=2(1)-1=1 \) This is NOT true.
C.\(g(x)=-2x+1→-1=2(-1)+1→-1=-1\) This is true.
D.\(g(x)=x+2→-1=1+2→-1=3\) This is NOT true.
From the choices provided, only choice C is correct.

10- B
To solve the equation for \(y\), multiply both sides of the equation by the reciprocal of \(\frac{6}{5}\), which is \(\frac{5}{6}\),
this gives \(\frac{5}{6}× \frac{6}{5}y=\frac{3}{2} ×\frac{5}{6}\), which simplifies to
\(y=\frac{15}{12}=\frac{5}{4}\)

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