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 Hard
19
$
\sqrt{14-2 x}=x-7
$
What value of \(\mathrm{x}\) satisfies the given equation?
▶️Answer/Explanation
Ans:7
To solve the equation \(\sqrt{14 – 2x} = x – 7\) for \(x\):
Square both sides to eliminate the square root:
\[
(\sqrt{14 – 2x})^2 = (x – 7)^2
\]
\[
14 – 2x = (x – 7)^2
\]
Expand the right-hand side:
\[
14 – 2x = x^2 – 14x + 49
\]
Rearrange the equation to set it to zero:
\[
x^2 – 14x + 49 + 2x – 14 = 0
\]
\[
x^2 – 12x + 35 = 0
\]
Factor the quadratic equation:
\[
(x – 5)(x – 7) = 0
\]
Solve for \(x\):
\[
x – 5 = 0 \quad \text{or} \quad x – 7 = 0
\]
\[
x = 5 \quad \text{or} \quad x = 7
\]
Check both solutions in the original equation:
For \(x = 5\):
\[
\sqrt{14 – 2(5)} = 5 – 7
\]
\[
\sqrt{4} = -2 \quad (\text{Not true, so } x = 5 \text{ is not a solution})
\]
For \(x = 7\):
\[
\sqrt{14 – 2(7)} = 7 – 7
\]
\[
\sqrt{0} = 0 \quad (\text{True, so } x = 7 \text{ is a solution})
\]
The value of \(x\) that satisfies the equation is \(7\).
Question Hard
Which of the following expressions is equivalent to \({(2\sqrt{x}-\sqrt{y})}^{\frac{2}{5}}\) , where x>y and y>0?
A) \({(4x-y)}^5\)
B) \(\sqrt[5]{4x-y}\)
C) \({(4x-4 \sqrt{xy}+y)}^{\frac{1}{5}}\)
D) \(\sqrt[5]{4x-4xy+y}\)
▶️Answer/Explanation
C) \({(4x-4\sqrt{xy}+y)}^{\frac{1}{5}}\)
We need to determine the expression equivalent to \((2 \sqrt{x} – \sqrt{y})^{\frac{2}{5}}\).
1. Simplify the expression inside the parentheses:
Given: \((2 \sqrt{x} – \sqrt{y})^{\frac{2}{5}}\)
2. Examine each option:
Option A: \((4 x – y)^5\)
\[
(4x – y)^{\frac{1}{5}} \neq (2\sqrt{x} – \sqrt{y})^{\frac{2}{5}}
\]
Option B: \(\sqrt[5]{4 x – y}\)
\[
(4x – y)^{\frac{1}{5}} \neq (2\sqrt{x} – \sqrt{y})^{\frac{2}{5}}
\]
Option C: \((4 x – 4 \sqrt{x y} + y)^{\frac{1}{5}}\)
\[
(4x – 4\sqrt{xy} + y)^{\frac{1}{5}} = (2\sqrt{x} – \sqrt{y})^{\frac{2}{5}}
\]
Option D: \(\sqrt[5]{4 x – 4xy + y}\)
\[
(4x – 4xy + y)^{\frac{1}{5}} \neq (2\sqrt{x} – \sqrt{y})^{\frac{2}{5}}
\]
Thus, the correct expression is:
\[ \boxed{(4 x – 4 \sqrt{x y} + y)^{\frac{1}{5}}} \]
Question Hard
Which of the following is equivalent to \((\sqrt{32})(\sqrt[5]{64})\) ?
A. \(6\left(\sqrt[7]{2^5}\right)\)
B. \(6\left(\sqrt[10]{2^7}\right)\)
c. \(8\left(\sqrt[7]{2^5}\right)\)
D. \(8\left(\sqrt[10]{2^7}\right)\)
▶️Answer/Explanation
Ans:D
To simplify \((\sqrt{32})(\sqrt[5]{64})\), we can first rewrite the numbers under the square roots as powers of 2 :
\[
\begin{aligned}
& \sqrt{32}=\sqrt{2^5}=2^{\frac{5}{2}} \\
& \sqrt[5]{64}=\sqrt[5]{2^6}=2^{\frac{6}{5}}
\end{aligned}
\]
Now, when we multiply these together, we add the exponents:
\[
(\sqrt{32})(\sqrt[5]{64})=2^{\frac{5}{2}} \cdot 2^{\frac{6}{5}}=2^{\frac{5}{2}+\frac{6}{5}}=2^{\frac{25}{10}+\frac{12}{10}}=2^{\frac{37}{10}}
\]
A. \(6\left(\sqrt[7]{2^5}\right)\)
\[6\left(\sqrt[7]{2^5}\right) = 6 \times 2^{\frac{5}{7}}\]
This expression cannot be simplified to \(2^{\frac{37}{10}}\).
B. \(6\left(\sqrt[10]{2^7}\right)\)
\[6\left(\sqrt[10]{2^7}\right) = 6 \times 2^{\frac{7}{10}}\]
This expression cannot be simplified to \(2^{\frac{37}{10}}\).
C. \(8\left(\sqrt[7]{2^5}\right)\)
\[8\left(\sqrt[7]{2^5}\right) = 8 \times 2^{\frac{5}{7}}\]
This expression cannot be simplified to \(2^{\frac{37}{10}}\).
D. \(8\left(\sqrt[10]{2^7}\right)\)
\[8\left(\sqrt[10]{2^7}\right) = 8 \times 2^{\frac{7}{10}}\]
This expression can be simplified to \(2^{\frac{37}{10}}\), since \(8 \times 2^{\frac{7}{10}} = 2^{\frac{3}{10}} \times 2^{\frac{7}{10}} = 2^{\frac{3}{10} + \frac{7}{10}} = 2^{\frac{37}{10}}\).
So, after solving each option, we see that only option D simplifies to \(2^{\frac{37}{10}}\).
Question Hard
Which of the following expressions is equivalent to $(\sqrt{2 q}+\sqrt{2 r})^{\frac{2}{3}}$, where $q>0$ and $r<0$ ?
A) $(2 q+2 r)^3$
B) $\sqrt[3]{2 q+2 r}$
C) $\sqrt[3]{2 q+2 \sqrt{q r}+2 r}$
D) $\sqrt[3]{2 q+4 \sqrt{q r}+2 r}$
▶️Answer/Explanation
D
Question
In 480 BC, the population of the Persian Empire was approximately 49.4 million. The population of the Persian Empire was 44% of the world population at that time. Which of the following is the best estimate of the world population in 480 BC?
- 21.7 million
- 89.1 million
- 93.4 million
- 112.3 million
▶️Answer/Explanation
D