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
\(\frac{1}{C}=\frac{1}{d}+\frac{1}{e}+\frac{1}{f}\)
An electric circuit contains three capacitors in a particular arrangement. The given equation relates the equivalent capacitance C of the arrangement to d, e, and f, the capacitances of the individual capacitors. Which equation correctly gives C in terms of d, e, and f ?
A. 𝐶 = 𝑑 + 𝑒 + 𝑓
B. \(C=\frac{def}{d+e+f}\)
C. \(C=\frac{d+e+f}{def}\)
D. \(C=\frac{def}{de+df+ef}\)
▶️Answer/Explanation
Ans: D
Given the equation \(\frac{1}{c}=\frac{1}{d}+\frac{1}{e}+\frac{1}{f}\), the equivalent capacitance \(C\) is the reciprocal of the sum of the reciprocals of \(d\), \(e\), and \(f\).
\[
\frac{1}{C} = \frac{1}{d} + \frac{1}{e} + \frac{1}{f}
\]
To find \(C\), we take the reciprocal of both sides:
\[
C = \frac{1}{\frac{1}{d} + \frac{1}{e} + \frac{1}{f}}
\]
To simplify the expression in the denominator, we’ll first find a common denominator:
\[
C = \frac{1}{\frac{ef + df + ed}{def}}
\]
\[
C = \frac{def}{ef + df + ed}
\]
Question Hard
Two numbers, \(a\) and \(b\), are each greater than zero, and 4 times the square root of \(a\) is equal to 9 times the cube root of \(b\). If \(a=\frac{2}{3}\), for what value of \(x\) is \(a^x\) equal to \(b\) ?
▶️Answer/Explanation
7.5,15/2
Given:
\(4\sqrt{a} = 9\sqrt[3]{b}\)
\(a = \frac{2}{3}\)
find \(x\) such that \(a^x = b\).
solving for \(b\) using the given values:
\[ 4\sqrt{\frac{2}{3}} = 9\sqrt[3]{b} \]
\[ \sqrt{\frac{2}{3}} = \frac{9}{4}\sqrt[3]{b}\]
Now, we can isolate \(b\) by squarring both sides:
\[ \frac{2}{3} = \left(\frac{9}{4}\right)^2 b^{\frac{2}{3}} \]
\[ \frac{2}{3} = \frac{81}{16} b^{\frac{2}{3}} \]
\[ b^{\frac{2}{3}} = \frac{2}{3} \times \frac{16}{81} \]
\[ b = \frac{32}{243}^{\frac{3}{2}} \]
\[ b = (\frac{32}{243})^{\frac{3}{2}} \]
\[ b = (\frac{2^5}{3^5})^{\frac{3}{2}} \]
\[ b = ([\frac{2}{3}])^{5\times \frac{3}{2}} \]
\[ b = (\frac{2}{3})^{\frac{15}{2}} \]
Now, we need to find \(x\) such that \(a^x = b\):
\[ \left(\frac{2}{3}\right)^x = b = (\frac{2}{3})^{\frac{15}{2}} \]
\[ x = \frac{15}{2} \]
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
The function f is defined by \(f(x)=\frac{1}{5}x+\frac{9}{10}\) For what value of x does f(x)=1?
▶️Answer/Explanation
Ans: 1/2, .5
To find the value of \(x\) for which \(f(x) = 1\), we set \(f(x)\) equal to \(1\) and solve for \(x\):
\[f(x) = \frac{1}{5}x + \frac{9}{10} = 1\]
To solve for \(x\), we first subtract \( \frac{9}{10} \) from both sides:
\[ \frac{1}{5}x = 1 – \frac{9}{10} \]
\[ \frac{1}{5}x = \frac{10}{10} – \frac{9}{10} \]
\[ \frac{1}{5}x = \frac{1}{10} \]
Now, multiply both sides by \(5\) to isolate \(x\):
\[ x = \frac{1}{10} \times 5 \]
\[ x = \frac{5}{10} \]
\[ x = \frac{1}{2} \]
So, \(f(x) = 1\) when \(x = \frac{1}{2}\).