Question 1
In the figure below, AB and GE are parallel. Triangle ACD is isosceles with CA = CD. The measures of angles FDE and GDH are 60° and 65° respectively. What is the measure of angle CAB?
Solution
Since AB ∥ GE, ∠BAD and ∠EDF are corresponding angles and equal. Therefore, m∠BAD = 60°.
Angles ∠CDE and ∠GDH are vertical angles, so m∠CDE = 65°.
In triangle CDA, the angles sum to 180°:
m∠CDA = 180° - (60° + 65°) = 55°.
Since triangle ACD is isosceles with CA = CD, m∠CAD = m∠CDA = 55°.
Therefore, m∠CAB = m∠BAD - m∠CAD = 60° - 55° = 5°.
Question 2
If \(a\), \(b\), and \(y\) are positive real numbers such that none equals 1 and \(b^{2a+6} = y^2\), which of these must be true?
Solution
Since \(a\) and \(b\) are positive:
\(\sqrt{b^{2a+6}} = b^{(1/2)(2a+6)} = b^{a+3}\).
Since \(y\) is positive:
\(\sqrt{y^2} = y\).
Given \(b^{2a+6} = y^2\), take the square root of both sides:
\(y = b^{a+3}\).
Question 3
What is true about the graph of \(f(x) = 4 + |\,|x| - 3\,|\)?
(I) Symmetric with respect to y-axis
(II) Has no x-intercepts
(III) Has a y-intercept at (0,7)
Solution
(I) Compute \(f(-x)\):
\(f(-x) = 4 + |\,|-x| - 3\,| = 4 + |\,|x| - 3\,| = f(x)\).
The function is even, so its graph is symmetric about the y-axis.
(II) Solve \(f(x)=0\):
\(4 + |\,|x| - 3\,| = 0 \Rightarrow |\,|x| - 3\,| = -4\).
An absolute value cannot be negative, so no solution. The graph has no x-intercepts.
(III) The y-intercept is \(f(0) = 4 + |\,|0| - 3\,| = 4 + 3 = 7\).
Thus, the y-intercept is at \((0,7)\).
All three statements are true.
Question 4
A number \(x\) is divided by 0.71 and the result decreased by 0.2. If the fifth root of the final result equals -0.15, what is \(x\) rounded to the nearest thousandth?
Solution
The described operations yield:
\(\sqrt[5]{\frac{x}{0.71} - 0.2} = -0.15\).
Raise both sides to the 5th power:
\(\frac{x}{0.71} - 0.2 = (-0.15)^5\).
Solve for \(x\):
\(x = 0.71\left((-0.15)^5 + 0.2\right)\).
Calculate: \((-0.15)^5 \approx -7.59375 \times 10^{-5} \approx -0.0000759375\).
Then \(x \approx 0.71(0.1999240625) \approx 0.14194608\).
Rounded to the nearest thousandth: \(x \approx 0.142\).
Question 5
Find the constant \(k\) so that the perpendicular bisector of the segment with endpoints \((k,0)\) and \((4,6)\) has slope \(-3\).
Solution
The slope of the segment is \(m_1 = \frac{6-0}{4-k} = \frac{6}{4-k}\).
The perpendicular bisector has slope \(-3\). For perpendicular lines, \(m_1 \times m_2 = -1\). Thus:
\(\frac{6}{4-k} \times (-3) = -1\).
Solve for \(k\):
\(-\frac{18}{4-k} = -1 \Rightarrow 18 = 4 - k \Rightarrow k = -14\).
Question 6
Forty students are each registered for one or more of three courses. 23 students are in one course only and 14 students are in two courses only. How many are registered for all three courses?
Solution
Let \(x\) be the number in all three courses. The total number of students is 40. Those in one course only (23) and two courses only (14) are given. Therefore:
\(23 + 14 + x = 40 \Rightarrow x = 3\).
So, 3 students are registered in all three courses.
Question 7
Which statements are true about the graphs of \(x^2 + y^2 = 9\) and \((x+6)^2 + y^2 = 9\)?
(I) Both graphs are circles.
(II) The two graphs touch at one point.
(III) The two graphs have two distinct intersections.
Solution
Both equations are of the form \((x-h)^2 + (y-k)^2 = r^2\), so they are circles.
Solve the system:
\((x+6)^2 + y^2 = 9\)
\(x^2 + y^2 = 9\)
Subtract the second from the first:
\((x+6)^2 - x^2 = 0 \Rightarrow x^2+12x+36 - x^2 = 0 \Rightarrow 12x+36=0 \Rightarrow x=-3\).
Substitute \(x=-3\) into \(x^2+y^2=9\):
\(9 + y^2 = 9 \Rightarrow y=0\).
They intersect only at \((-3,0)\), so they touch at one point.
Thus, only (I) and (II) are true.
Question 8
If the perimeter of a regular hexagon is \(6a\), then its area is:
Solution
Each side length is \(a\). A regular hexagon can be divided into 6 equilateral triangles of side \(a\).
The area of one equilateral triangle of side \(a\) is \(\frac{\sqrt{3}}{4}a^2\).
Thus, the hexagon area is \(6 \times \frac{\sqrt{3}}{4}a^2 = \frac{3\sqrt{3}}{2}a^2\).
Question 9
In the figure, AB is a diameter of the circle, and point C is on the circle. The diameter length is 10 units and AC = 5 units. Find the measure of ∠CBA.
Solution
Since AB is a diameter, triangle ABC is a right triangle with ∠ACB = 90°.
Given AC = 5 and AB = 10, we have:
\(\sin(\angle CBA) = \frac{AC}{AB} = \frac{5}{10} = \frac{1}{2}\).
Therefore, \(\angle CBA = \arcsin(1/2) = 30^\circ\).
Question 10
For what positive value of \(k\) does \((x+k)x = -4\) have exactly one solution?
Solution
Rewrite: \(x^2 + kx + 4 = 0\).
A quadratic has exactly one solution when its discriminant is zero:
\(D = k^2 - 4(1)(4) = k^2 - 16 = 0\).
Thus \(k^2 = 16\), and since \(k > 0\), \(k = 4\).
Question 11
The average of the roots of a quadratic is 3 and their difference is 2. Which equation could this be?
Solution
Let the roots be \(A\) and \(B\).
Average: \(\frac{A+B}{2}=3 \Rightarrow A+B=6\).
Difference: \(A-B=2\).
Solving: \(A=4\), \(B=2\).
A quadratic with these roots is \((x-4)(x-2)=0\), or \(x^2 - 6x + 8 = 0\).
Question 12
If points M, B(2,6), and C(4,8) form a right triangle with hypotenuse BC, then points M, B, C lie on a circle of radius:
Solution
By the converse of Thales' theorem, if a right triangle is inscribed in a circle, the hypotenuse is a diameter.
Thus, BC is a diameter. The radius is half of BC:
\(BC = \sqrt{(4-2)^2+(8-6)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}\).
Radius \(R = \frac{BC}{2} = \sqrt{2}\).
Question 13
The volume of a rectangular solid is 1000 m³. If its length, width, and height are each increased by 50%, the new volume is:
Solution
Let original dimensions be \(L, W, H\). Then \(LWH = 1000\).
New dimensions: \(1.5L, 1.5W, 1.5H\).
New volume: \((1.5L)(1.5W)(1.5H) = 1.5^3 LWH = 3.375 \times 1000 = 3375\) m³.
Question 14
In the figure, angle ADE is right, and BC ∥ DE. AC = 5, BC = 4. Find \(\tan(\angle CED)\).
Solution
Since BC ∥ DE, triangles ABC and ADE are similar. Thus, \(\angle CED = \angle ACB\).
In right triangle ABC, by Pythagoras:
\(AB = \sqrt{AC^2 - BC^2} = \sqrt{25 - 16} = 3\).
Therefore, \(\tan(\angle ACB) = \frac{AB}{BC} = \frac{3}{4}\).
Hence, \(\tan(\angle CED) = \frac{3}{4}\).
Question 15
What are the coordinates of the center of a circle entirely in Quadrant II, tangent to lines y=8, y=2, and x=-2?
Solution
The circle is tangent to horizontal lines y=8 and y=2, so its diameter is the vertical distance: \(8-2=6\). Thus, radius \(r=3\), and the y-coordinate of the center is the midpoint: \((8+2)/2 = 5\).
Since the circle is in Quadrant II, its x-coordinate is negative. It is tangent to the vertical line x=-2. For a circle of radius 3 to be tangent to x=-2 and lie to its left, the center's x-coordinate is \(-2 - 3 = -5\).
Center: \((-5, 5)\).
Question 16
In the figure, BC ∥ DE, AC = x, and CE = 4x. If the area of triangle ABC is 100 cm², what is the area of triangle ADE?
Solution
Triangles ABC and ADE are similar. AC = x, AE = AC + CE = 5x. The similarity ratio (ABC to ADE) is \(\frac{AC}{AE} = \frac{x}{5x} = \frac{1}{5}\).
For similar figures, the ratio of areas is the square of the similarity ratio.
\(\frac{\text{Area}(ABC)}{\text{Area}(ADE)} = \left(\frac{1}{5}\right)^2 = \frac{1}{25}\).
Given Area(ABC) = 100, then Area(ADE) = \(100 \times 25 = 2500\) cm².
Question 17
In right triangle ABC (figure), what is the volume of the cone generated by rotating the triangle about side AB?
Solution
Rotation about AB yields a cone with height \(h = AB = 6\) and radius \(r = AC\).
By Pythagoras in triangle ABC: \(AC = \sqrt{BC^2 - AB^2} = \sqrt{100 - 36} = \sqrt{64} = 8\).
Cone volume: \(V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (8^2)(6) = \frac{1}{3}\pi (64)(6) = 128\pi\).
Question 18
If \(3(x+y)=27\) and \(x,y\) are positive integers, which cannot be the value of \(x/y\)?
A) 1/2 B) 8 C) 2 D) 1/8 E) 1/4
Solution
From \(3(x+y)=27\), we have \(x+y=9\).
Test each ratio:
A) \(x/y=1/2 \Rightarrow y=2x\), then \(x+2x=9 \Rightarrow x=3, y=6\) (positive integers).
B) \(x/y=8 \Rightarrow x=8y\), then \(8y+y=9 \Rightarrow y=1, x=8\).
C) \(x/y=2 \Rightarrow x=2y\), then \(2y+y=9 \Rightarrow y=3, x=6\).
D) \(x/y=1/8 \Rightarrow y=8x\), then \(x+8x=9 \Rightarrow x=1, y=8\).
E) \(x/y=1/4 \Rightarrow y=4x\), then \(x+4x=9 \Rightarrow 5x=9 \Rightarrow x=9/5\) (not an integer).
Thus, E) cannot be the value.
Question 19
Simplify: \(\frac{\cos(x)\sin(2x) - 2\sin x}{\sin^2 x \cos(x)}\).
Solution
Use identity \(\sin(2x) = 2\sin x \cos x\):
Numerator: \(\cos(x) \cdot 2\sin x \cos x - 2\sin x = 2\sin x (\cos^2 x - 1)\).
Since \(\cos^2 x - 1 = -\sin^2 x\), the numerator becomes \(-2\sin^3 x\).
Thus, the expression becomes \(\frac{-2\sin^3 x}{\sin^2 x \cos x} = \frac{-2\sin x}{\cos x} = -2\tan x\).
Question 20
Find the smallest positive integer \(x\) such that \(x^2 + 4\) is divisible by 25.
Solution
We need \(x^2 + 4 = 25k\) for some integer \(k \ge 1\).
Try values:
\(k=1: x^2=21\), no integer.
\(k=2: x^2=46\), no.
\(k=3: x^2=71\), no.
\(k=4: x^2=96\), no.
\(k=5: x^2=121 \Rightarrow x=11\).
The smallest positive \(x\) is 11.
Question 21
Find constant \(a\) such that \(x=1\) is a solution to \(a\sqrt{x+3} - 4a|2x-1| = 4\).
Solution
Substitute \(x=1\):
\(a\sqrt{1+3} - 4a|2(1)-1| = a\sqrt{4} - 4a|1| = 2a - 4a = -2a\).
Set equal to 4: \(-2a = 4 \Rightarrow a = -2\).
Question 22
\(m\) and \(n\) are positive integers with \(m > n\) and \(m^{2n} = 46656\). Find \(m\).
Solution
\(m^{2n} = (m^n)^2 = 46656\).
So \(m^n = \sqrt{46656} = 216\).
Factor 216: \(216 = 6^3\).
Thus \(m^n = 6^3\). Since \(m>n\) and both are positive integers, we can take \(m=6\) and \(n=3\).
Question 23
The root of \(f(x)=2\) is \(x=5\). The solution to \(f(-2x+1)=2\) is:
Solution
Let \(s\) be the solution to \(f(-2x+1)=2\). Then \(f(-2s+1)=2\).
But we know \(f(5)=2\). So \(-2s+1 = 5 \Rightarrow -2s = 4 \Rightarrow s = -2\).
Question 24
For what value of \(b\) will the system \(2x+5y=3\) and \(-5x+by=14\) have no solution?
Solution
The system has no solution if the lines are parallel (same slope) but different intercepts.
Slope of first line: rewrite as \(y = -\frac{2}{5}x + \frac{3}{5}\), slope \(m_1 = -\frac{2}{5}\).
Slope of second line: \(y = \frac{5}{b}x + \frac{14}{b}\), slope \(m_2 = \frac{5}{b}\).
Set slopes equal: \(-\frac{2}{5} = \frac{5}{b} \Rightarrow -2b = 25 \Rightarrow b = -\frac{25}{2}\).
Check intercepts: first line y-intercept is 3/5, second is \(14/b = 14/(-25/2) = -28/25\), different. So no solution when \(b = -25/2\).
Question 25
Find real numbers \(a,b\) such that \(3a - bi = (2-i)(4+i)\), where \(i=\sqrt{-1}\).
Solution
Compute right side: \((2-i)(4+i) = 8 + 2i - 4i - i^2 = 8 - 2i - (-1) = 9 - 2i\).
So \(3a - bi = 9 - 2i\). Equate real and imaginary parts:
\(3a = 9 \Rightarrow a=3\), and \(-b = -2 \Rightarrow b=2\).
Question 26
A dealer increased an item's price by 20%, then increased the new price by 30%. If original price is \(x\), what is the final price?
Solution
After first increase: \(x + 0.2x = 1.2x\).
After second increase: \(1.2x + 0.3(1.2x) = 1.2x + 0.36x = 1.56x\).
Question 27
Two dice are thrown. What is the probability the sum is greater than 10?
Solution
Total outcomes: 36. Sum > 10 means sum 11 or 12.
Favorable outcomes: (5,6), (6,5), (6,6) → 3 outcomes.
Probability = \(3/36 = 1/12\).
Question 28
If \(3x+2y=5\) and \(5x+4y=9\), then \(4x+3y = ?\)
Solution
Add the two equations:
\((3x+2y) + (5x+4y) = 5+9 \Rightarrow 8x+6y=14\).
Divide by 2: \(4x+3y = 7\).
Question 29
The solution set of \(|2x-4| > x+1\) is which interval?
Solution
Solve inequality. Consider critical points where \(2x-4=0 \Rightarrow x=2\), and where \(2x-4 = \pm(x+1)\).
Case 1: \(x \ge 2\). Then \(|2x-4| = 2x-4\).
Inequality: \(2x-4 > x+1 \Rightarrow x > 5\). So part of solution: \(x>5\).
Case 2: \(x < 2\). Then \(|2x-4| = 4-2x\).
Inequality: \(4-2x > x+1 \Rightarrow 3 > 3x \Rightarrow x < 1\).
Thus, solution: \(x < 1\) or \(x > 5\), i.e., \((-\infty,1) \cup (5,\infty)\).
Question 30
If \(m/n = 2/3\) for positive integers \(m,n\), which pair cannot be \(m,n\)?
A) \(m=12, n=18\) B) \(m=60, n=90\) C) \(m=34, n=51\) D) \(m=7, n=21\) E) \(m=102, n=153\)
Solution
Check each ratio:
A) \(12/18 = 2/3\)
B) \(60/90 = 2/3\)
C) \(34/51 = 2/3\) (since 34=2×17, 51=3×17)
D) \(7/21 = 1/3 \neq 2/3\)
E) \(102/153 = 2/3\) (102=2×51, 153=3×51)
Thus, D) cannot be the pair.
Question 31
If 35 is the median of the data set {21, 7, 45, 33, 62, \(x\)}, then \(x = ?\)
Solution
Order known values: 7, 21, 33, 45, 62. We have 6 numbers total, so the median is the average of the 3rd and 4th when sorted.
For median to be 35, the average of the 3rd and 4th must be 35, so their sum is 70.
Try inserting \(x\) such that when sorted, the two middle numbers average to 35. One working value: \(x=37\).
Sorted set: 7, 21, 33, 37, 45, 62. Median = (33+37)/2 = 35.
Thus, \(x=37\) works. (Note: other values might work, but 37 is a typical answer.)