5 SAT Math Questions That 95% of Our Users Get Wrong
Tip
Want to test yourself for real? You can scroll down to see the answers, or you can try to solve these live in our free practice test. It simulates the real SAT timer and interface, giving you a much better sense of whether you can beat the odds when it counts.
Question 1
In the \(xy\)-plane, the graph of \(4x^2 - 24bx + 4y^2 + 32by = 200b^2\) is a circle with center \((12,-16)\).
Answer Preview
<h3>The Solution: Step-by-Step</h3>
<p><strong>1. Simplify the Equation</strong><br>
First, notice that every term is divisible by 4. To put this into the standard circle format, we need our \( x^2 \) and \( y^2 \) coefficients to be 1.</p>
<p>\[ x^2 - 6bx + y^2 + 8by = 50b^2 \]</p>
<p><strong>2. Complete the Square</strong><br>
We need to turn the \( x \) terms and \( y \) terms into perfect squares.<br>
For \( x \): Half of \( -6b \) is \( -3b \). Square it to get \( 9b^2 \).<br>
For \( y \): Half of \( 8b \) is \( 4b \). Square it to get \( 16b^2 \).<br>
Add these values to <strong>both sides</strong> of the equation to keep it balanced:</p>
<p>\[ (x^2 - 6bx + 9b^2) + (y^2 + 8by + 16b^2) = 50b^2 + 9b^2 + 16b^2 \]</p>
<p>This simplifies cleanly to:</p>
<p>\[ (x-3b)^2 + (y+4b)^2 = 75b^2 \]</p>
<p><strong>3. Find the Value of 'b'</strong><br>
Now we have the standard circle equation: \( (x-h)^2 + (y-k)^2 = r^2 \).<br>
Our center is \( (3b, -4b) \).<br>
The problem states the center is \( (12, -16) \).<br>
By setting \( 3b = 12 \), we find that <strong>\( b = 4 \)</strong>.</p>
<p><strong>4. Find the Radius Squared</strong><br>
Substitute \( b=4 \) back into the right side of our equation to find the radius squared (\( r^2 \)):</p>
<p>\[ r^2 = 75(4)^2 = 75(16) = 1200 \]</p>
<p><strong>5. Solve for the Area of the Square</strong><br>
Here is the geometry shortcut: If a square is inscribed in a circle, the <strong>diagonal of the square is equal to the diameter of the circle (\( 2r \))</strong>.</p>
<p>You can use the Pythagorean theorem, or simply remember that the area of a square inscribed in a circle is always <strong>\( 2r^2 \)</strong>.</p>
<p>\[ Area = 2(1200) = \mathbf{2400} \]</p>
Question 2
In the system \( \frac{a}{5}x + \frac{9}{10} = \frac{b}{5}y - \frac{3}{5} \) and \( -24x + 5y = \frac{11}{2} - 9y \), where \(a\) and \(b\) are constants, the system has no solution. The ratio \(a : b\) is equivalent to \(15:k\).
What is the value of \(k\)?
Answer Preview
<h3>Explanation</h3>
<p>To determine the condition for no solution, both linear equations must first be rewritten in slope-intercept form, \( y = mx + c \).</p>
<p>Starting with the first equation, \( \frac{a}{5}x + \frac{9}{10} = \frac{b}{5}y - \frac{3}{5} \), we isolate the \( y \)-term by adding \( \frac{3}{5} \) to both sides and then multiplying by \( \frac{5}{b} \). This simplification yields:</p>
<p>\[ y = \frac{a}{b}x + \frac{15}{2b} \]</p>
<p>This indicates the slope is \( \frac{a}{b} \).</p>
<p>Next, we simplify the second equation, \( -24x + 5y = \frac{11}{2} - 9y \), by grouping the \( y \)-terms on the left side to get \( 14y = 24x + \frac{11}{2} \). Dividing the entire equation by 14 results in:</p>
<p>\[ y = \frac{12}{7}x + \frac{11}{28} \]</p>
<p>This reveals a slope of \( \frac{12}{7} \).</p>
<p>For a system of linear equations to have no solution, the lines must be parallel, meaning their slopes must be equal while their y-intercepts differ. Equating the two slopes gives the proportion \( \frac{a}{b} = \frac{12}{7} \).</p>
<p>Since the problem states that the ratio \( a:b \) is equivalent to \( 15:k \), we set \( \frac{15}{k} \) equal to \( \frac{12}{7} \). Solving for \( k \) via cross-multiplication results in \( 12k = 105 \), which simplifies to:</p>
<p>\[ k = \frac{35}{4} = 8.75 \]</p>
Question 3
\(\sqrt{2x+7} = x+2\)
A
\(-3\)
B
\(-2\)
C
\(1\)
D
\(3\)
<h3>Explanation</h3>
<p>To solve for \( x \), we first square both sides of the equation to eliminate the radical.</p>
<p>\[ (\sqrt{2x+7})^2 = (x+2)^2 \]</p>
<p>\[ 2x+7 = x^2 + 4x + 4 \]</p>
<p>Next, rearrange the equation into a standard quadratic form by subtracting \( 2x \) and \( 7 \) from both sides.</p>
<p>\[ 0 = x^2 + 4x - 2x + 4 - 7 \]</p>
<p>\[ 0 = x^2 + 2x - 3 \]</p>
<p>Now, factor the quadratic equation. We look for two numbers that multiply to \( -3 \) and sum to \( 2 \), which are \( 3 \) and \( -1 \).</p>
<p>\[ 0 = (x+3)(x-1) \]</p>
<p>This yields two potential solutions: \( x = -3 \) and \( x = 1 \). Since we squared both sides of the original equation, it is crucial to check these potential solutions in the original equation to identify any extraneous solutions.</p>
<p>First, check \( x = -3 \):<br>
Substitute \( x = -3 \) into the original equation \( \sqrt{2x+7} = x+2 \).<br>
The left side becomes \( \sqrt{2(-3)+7} = \sqrt{1} = 1 \).<br>
The right side becomes \( -3+2 = -1 \).<br>
Since \( 1 \neq -1 \), \( x = -3 \) is an extraneous solution.</p>
<p>Next, check \( x = 1 \):<br>
Substitute \( x = 1 \) into the original equation.<br>
The left side becomes \( \sqrt{2(1)+7} = \sqrt{9} = 3 \).<br>
The right side becomes \( 1+2 = 3 \).<br>
Since \( 3 = 3 \), \( x = 1 \) is a valid solution.</p>
<p>The only distinct real solution to the given equation is \( x=1 \). Therefore, the sum of all distinct real solutions is 1, meaning that <strong>C</strong> is the correct answer.</p>
Question 4
The graph of \(y = -3f(x+2)\) is shown.
Which equation defines the linear function \(f\)?
A
\(f(x) = \frac{2}{3}x - \frac{10}{3}\)
B
\(f(x) = -2x + 6\)
C
\(f(x) = \frac{2}{3}x - 2\)
D
\(f(x) = -\frac{2}{3}x + \frac{10}{3}\)
<p>First, determine the equation of the line shown in the graph. By observing the graph, we can identify two key points: the \( y \)-intercept at \( (0, 6) \) and the \( x \)-intercept at \( (3, 0) \).</p>
<p>To find the slope \( m \), we use the slope formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Substituting our points results in:</p>
<p>\[ m = \frac{0 - 6}{3 - 0} = \frac{-6}{3} = -2 \]</p>
<p>Since the \( y \)-intercept is 6, the equation of the graphed line is \( y = -2x + 6 \).</p>
<p>The problem states that this graph represents the function \( y = -3f(x+2) \). We can now set our linear equation equal to this transformation expression:</p>
<p>\[ -3f(x+2) = -2x + 6 \]</p>
<p>To isolate \( f(x+2) \), divide both sides of the equation by \( -3 \):</p>
<p>\[ f(x+2) = \frac{-2x + 6}{-3} = \frac{2}{3}x - 2 \]</p>
<p>Now we must convert \( f(x+2) \) into \( f(x) \). Let \( u = x + 2 \), which implies that \( x = u - 2 \). Substitute these values into the equation:</p>
<p>\[ f(u) = \frac{2}{3}(u - 2) - 2 \]</p>
<p>Distribute the fraction and simplify the constants:</p>
<p>\[ f(u) = \frac{2}{3}u - \frac{4}{3} - \frac{6}{3} = \frac{2}{3}u - \frac{10}{3} \]</p>
<p>Finally, replacing \( u \) with \( x \) gives the function definition \( f(x) = \frac{2}{3}x - \frac{10}{3} \). This means that <strong>A</strong> is the correct answer.</p>
Question 5
The function \(f\) is defined by \(f(x) = a(-x + b)^2\), where \(a\) and \(b\) are constants. In the \(xy\)-plane, the graph of \(y = f(x)\) passes through \((4,0)\), and \(f(-10) > f(-1)\).
A
\(a < f(b)\)
B
\(a > b\)
C
\(f(2) < f(-2b)\)
D
\(a < b\)
<h3>Explanation</h3>
<p>First, we determine the value of the constant \( b \) using the given point \( (4,0) \). Substituting these coordinates into the function \( f(x) = a(-x + b)^2 \) yields:</p>
<p>\[ 0 = a(-4 + b)^2 \]</p>
<p>Assuming \( f(x) \) is a valid quadratic function, \( a \) cannot be zero. Therefore, the term inside the square must be zero:</p>
<p>\[ -4 + b = 0 \implies b = 4 \]</p>
<p>Next, we determine the sign of \( a \) using the given inequality \( f(-10) > f(-1) \). With \( b=4 \), the function becomes \( f(x) = a(-x + 4)^2 \). Let's evaluate the function at the two specific points:</p>
<p>\[ f(-10) = a(-(-10) + 4)^2 = a(14)^2 = 196a \]</p>
<p>\[ f(-1) = a(-(-1) + 4)^2 = a(5)^2 = 25a \]</p>
<p>Substituting these into the inequality \( f(-10) > f(-1) \):</p>
<p>\[ 196a > 25a \]</p>
<p>Subtracting \( 25a \) from both sides results in \( 171a > 0 \). For this to be true, \( a \) must be positive (\( a > 0 \)).</p>
<p>Finally, we test Answer C, which compares \( f(2) \) and \( f(-2b) \). Since \( b=4 \), we are comparing \( f(2) \) and \( f(-8) \).</p>
<p>\[ f(2) = a(-2 + 4)^2 = a(2)^2 = 4a \]</p>
<p>\[ f(-8) = a(-(-8) + 4)^2 = a(12)^2 = 144a \]</p>
<p>Since we established that \( a > 0 \), it must be true that \( 4a < 144a \). Therefore, \( f(2) < f(-2b) \).</p>
<p>Choice <strong>C</strong> is correct.</p>