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
Question Medium
The expression \(\frac{4 x^{11}}{12 x^5}\) is equivalent to \(\frac{1}{3} x^b\), where \(b\) is a constant and \(x>0\). What is the value of \(b\) ?
▶️Answer/Explanation
Ans: 6
To simplify the expression \(\frac{4x^{11}}{12x^5}\) and find the value of \(b\), follow these steps:
1. Simplify the coefficients:
\[
\frac{4}{12} = \frac{1}{3}
\]
2. Simplify the exponents using the property \(a^m / a^n = a^{m-n}\):
\[
\frac{x^{11}}{x^5} = x^{11-5} = x^6
\]
So the expression becomes:
\[
\frac{4x^{11}}{12x^5} = \frac{1}{3}x^6
\]
Therefore, \(b = 6\), and the value of \(b\) is:
\[
\boxed{6}
\]
Question Medium
Which expression is equivalent to \(\sqrt[4]{a^9}\) ?
A) \(a^{\frac{9}{4}}\)
B) \(a^{\frac{4}{9}}\)
C) \(a^{13}\)
D) \(a^{36}\)
▶️Answer/Explanation
Ans: A
We need to simplify the expression \(\sqrt[4]{a^9}\).
Using the property of exponents that \(\sqrt[n]{a^m} = a^{\frac{m}{n}}\), we can write:
\[
\sqrt[4]{a^9} = a^{\frac{9}{4}}
\]
So, the correct answer is:
\[
\boxed{A}
\]
Question medium
Which expression is equivalent to \(\left(x^2+4\right)^2+(x-2)(x+2)\) ?
A. \(x^4+x^2+20\)
B. \(x^4+5 x^2+16\)
C. \(x^4+9 x^2\)
D. \(x^4+9 x^2+12\)
▶️Answer/Explanation
Ans:D
To simplify the expression \((x^2+4)^2+(x-2)(x+2)\), let’s first expand \((x^2+4)^2\) using the binomial theorem:
\[
(x^2+4)^2 = (x^2+4)(x^2+4) = x^4 + 4x^2 + 4x^2 + 16 = x^4 + 8x^2 + 16
\]
Now, let’s distribute \((x-2)(x+2)\):
\[
(x-2)(x+2) = x^2 + 2x – 2x – 4 = x^2 – 4
\]
Now, add the two expressions together:
\[
(x^2 + 8x^2 + 16) + (x^2 – 4) = x^4 + 9x^2 + 12
\]
So, the expression \(\left(x^2+4\right)^2+(x-2)(x+2)\) is equivalent to \(x^4 + 9x^2 + 12\). Therefore, the correct answer is option D.
Question medium
The expression \(\frac{32 x^6}{4 x^3}\) is equivalent to \(c x^d\), where \(c\) and \(d\) are constants and \(x>0\). What is the value of \(c+d\) ?
▶️Answer/Explanation
Ans:11
\[
\frac{32 x^6}{4 x^3} = \frac{32}{4} \cdot \frac{x^6}{x^3} = 8 \cdot x^{6-3} = 8x^3
\]
So, \(c = 8\) and \(d = 3\), and \(c + d = 8 + 3 = 11\).
Question medium
$
q=s(r-1)^2
$
The given equation relates the positive numbers \(q, r\), and \(s\). Which equation gives \(r\) in terms of \(q\) and \(s\), when \(r>1\) ?
A. \(r=1+\sqrt{\frac{q}{s}}\)
B. \(r=1+\frac{\sqrt{q}}{s}\)
c. \(r=-1-\sqrt{\frac{q}{s}}\)
D. \(r=-1-\frac{\sqrt{q}}{s}\)
▶️Answer/Explanation
Ans:A
To solve for \(r\) in terms of \(q\) and \(s\) in the equation \(q = s(r-1)^2\), we’ll first isolate \(r\).
\[q = s(r-1)^2\]
Divide both sides by \(s\):
\[\frac{q}{s} = (r-1)^2\]
Now, take the square root of both sides:
\[\sqrt{\frac{q}{s}} = |r-1|\]
Since \(r > 1\), we can drop the absolute value:
\[r – 1 = \sqrt{\frac{q}{s}}\]
Add 1 to both sides:
\[r = 1 + \sqrt{\frac{q}{s}}\]
So, the equation that gives \(r\) in terms of \(q\) and \(s\) when \(r > 1\) is:
A. \(r = 1 + \sqrt{\frac{q}{s}}\)
Question Medium
\(r^q= t^s\)
The given equation relates the distinct positive real numbers q, r, sand t. Which equation correctly expresses tin terms of q, r, and s ?
A) \(t=r^{\frac{q}{s}}\)
B) \(t=r^{\frac{s}{q}}\)
C) \(t=r^{q-s}\)
D) \(t=\frac{r^q}{s}\)
▶️Answer/Explanation
A) \(t=r^{\frac{q}{s}}\)
Given the equation \(r^q = t^s\), we need to express \(t\) in terms of \(q\), \(r\), and \(s\).
1. Rewrite the equation:
\[
r^q = t^s
\]
2. Take the \(s\)th root on both sides to solve for \(t\):
\[
t = \left( r^q \right)^{\frac{1}{s}}
\]
3. Simplify the exponent:
\[
t = r^{\frac{q}{s}}
\]
Thus, the correct expression for \(t\) in terms of \(q\), \(r\), and \(s\) is:
\[ \boxed{t = r^{\frac{q}{s}}} \]