Solutions to Algebra Placement Test Practice
Solutions with explanations to algebra placement test practice.
-
Solution
Given expression:
\[
\frac{a^3 - 1}{b - 1}
\]
Substitute \(a = -2\) and \(b = -2\):
\[
\frac{(-2)^3 - 1}{(-2) - 1}
\]
Simplify:
\[
\frac{-8 - 1}{-2 - 1} = \frac{-9}{-3} = 3
\]
-
Solution
Given equation:
\[
-2(x + 9) = 20
\]
Expand:
\[
-2x - 18 = 20
\]
\[
-2x = 38
\]
Multiply both sides by \(2\):
\[
-4x = 76
\]
-
Solution
Let \(x\) be the unknown.
20% of \(x\):
\[
0.2x
\]
The fifth of \(x\):
\[
\frac{1}{5}x = 0.2x
\]
"20% of it is added to its fifth":
\[
0.2x + 0.2x
\]
Seven-tenths of \(x\):
\[
\frac{7}{10}x = 0.7x
\]
Statement:
\[
0.2x + 0.2x = 0.7x - 12
\]
Simplify:
\[
0.4x = 0.7x - 12
\]
\[
12 = 0.3x
\]
\[
x = 40
\]
-
Solution
Given formula:
\[
A = 0.5(b + B)h
\]
Divide both sides by \(0.5h\):
\[
\frac{A}{0.5h} = b + B
\]
Solve for \(B\):
\[
B = \frac{A}{0.5h} - b
\]
Since \(\frac{1}{0.5} = 2\):
\[
B = \frac{2A}{h} - b
\]
-
Solution
Let \(x\) be the amount Tom has.
Alex has:
\[
\frac{x}{3}
\]
Linda has twice Alex’s amount:
\[
\frac{2x}{3}
\]
Total:
\[
x + \frac{x}{3} + \frac{2x}{3} = 120
\]
Multiply by 3:
\[
3x + x + 2x = 360
\]
\[
6x = 360 \quad \Rightarrow \quad x = 60
\]
Linda has:
\[
\frac{2}{3} \cdot 60 = 40
\]
-
Solution
Factorize:
\[
6x^2 - 11x - 2
\]
Try:
\[
(6x - 1)(x + 2) \quad \Rightarrow \quad 6x^2 + 11x - 2 \ (\text{incorrect})
\]
Try:
\[
(6x + 1)(x - 2) \quad \Rightarrow \quad 6x^2 - 11x - 2 \ (\text{correct})
\]
-
Solution
Factor:
\[
x^2 - 7x - 8 = (x + 1)(x - 8)
\]
One factor is \(x + 1\).
-
Solution
Simplify:
\[
(2xy^2 - 3x^2y) - (2x^2y^2 - 4x^2y)
\]
Remove parentheses:
\[
2xy^2 - 3x^2y - 2x^2y^2 + 4x^2y
\]
Group like terms:
\[
2xy^2 - 2x^2y^2 + x^2y
\]
-
Solution
Average speed formula:
\[
\text{Speed} = \frac{\text{Distance}}{\text{Time}}
\]
Given speed \(= 70\) mph, total time \(= 5\) hours:
\[
\frac{x + 200}{5} = 70
\]
\[
x + 200 = 350
\]
\[
x = 150
\]
-
Solution
Rewrite in slope-intercept form:
(I) \(y = -\frac{3}{2}x + \frac{3}{2}\)
(II) \(y = -\frac{2}{3}x - \frac{5}{3}\)
(III) \(y = \frac{2}{3}x - \frac{3}{2}\)
(IV) \(y = -3x + \frac{9}{2}\)
Slopes:
(I) \(-\frac{3}{2}\), (II) \(-\frac{2}{3}\), (III) \(\frac{2}{3}\), (IV) \(-3\).
Two lines are perpendicular if \(m \cdot n = -1\).
For (I) and (III):
\[
\left(-\frac{3}{2}\right) \cdot \left(\frac{2}{3}\right) = -1
\]
Hence, (I) and (III) are perpendicular.
-
Solution
Given:
\[
f(x) = (x + 1)^2
\]
Substitute \(x = t + 2\):
\[
f(t + 2) = (t + 3)^2
\]
Expand:
\[
t^2 + 6t + 9
\]
-
Solution
Simplify:
\[
(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y}) - (\sqrt{x} - \sqrt{y})^2
\]
First term:
\[
(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y}) = x - y
\]
Second term:
\[
(\sqrt{x} - \sqrt{y})^2 = x + y - 2\sqrt{xy}
\]
Subtract:
\[
(x - y) - (x + y - 2\sqrt{xy}) = -2y + 2\sqrt{xy}
\]
-
Solution
Given:
\[
\frac{x}{2} - \frac{y}{4} = 7
\]
Multiply by 4:
\[
2x - y = 28
\]
\[
y = 2x - 28
\]
Slope \(= 2\).
-
Solution
Simplify:
\[
\left( \frac{x}{x - 3} + \frac{1}{2} \right) \cdot \frac{2}{x - 1}
\]
Combine inside parentheses:
\[
\frac{2x}{2(x - 3)} + \frac{x - 3}{2(x - 3)} = \frac{3x - 3}{2(x - 3)}
\]
Multiply by \(\frac{2}{x - 1}\):
\[
\frac{3x - 3}{2(x - 3)} \cdot \frac{2}{x - 1} = \frac{3(x - 1)}{x - 3} \cdot \frac{1}{x - 1} = \frac{3}{x - 3}
\]
-
Solution
Point \(A(2,1)\), midpoint \(M(3,2)\), point \(B(a,b)\).
Midpoint formula:
\[
\left( \frac{2 + a}{2}, \frac{1 + b}{2} \right) = (3,2)
\]
From \(x\)-coordinate:
\[
\frac{2 + a}{2} = 3 \quad \Rightarrow \quad a = 4
\]
From \(y\)-coordinate:
\[
\frac{1 + b}{2} = 2 \quad \Rightarrow \quad b = 3
\]
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