1. C Figure A contains 6 whole boxes plus 4 half-boxes, for a total of 8. Figure B contains 8 whole boxes. The answer is choice (C). Alternatively, the area of the right triangle in Figure A is 
bh=(4)(4)=8 and Figure B contains a rectangle plus two units whose total area is bh=(3)(2)=6;6+2=8.
2. A The distance between P and Q is 6
ticks, so the midpoint will be 3 ticks from either endpoint, which is the first tick to
the right of 0, at 
. To find the coordinate of the midpoint, simply average the coordinates of the endpoints: 
. so Quantity A is greater. Alternatively, because the smaller ticks divide each unit into 3 equal parts, they represent lengths of .
3. B The axes split the quadrilateral into four equal 5-12-13 right triangles (note the highly suspicious numbers on the drawing). So the perimeter is 4(13) = 52, and Quantity B is therefore greater.
4. B Subtracting the y-coordinates of the given points gives you the length of leg EF: 4 − (–1) = 5. The area formula for a triangle will give you the length of the other leg, DF: bh=(b)(5);b=6, which equals 15. Subtracting 6 from the x-coordinate of F gives you the x-coordinate of D, and D has the same y-coordinate as F. So the coordinates of D are (–2, –1); the answer is choice (B).
5. A On the x-axis, y = 0; eliminate choices (B), (C), and (D). Then plug in 0 for y in the given equation. Solve for x: 0 = 4x + 20, so x = –5, and the answer is choice (A).
6. C Point C has the same x-coordinate as point A and the same y-coordinate as point B. The coordinates of C are therefore (1, 1), so the quantities are equal.
7. B Draw a rectangular coordinate system with Atlanta as the origin, i.e., the point (0, 0). So Cam’s home is at (2, 3). Going 3 miles east and 9 miles north puts town B at (5, 12). Quantity A is the straight line distance from (0, 0) to (5, 12). Connecting these two points creates the
hypotenuse of a 5-12-13 right triangle, so Quantity A is 13, and the answer is choice (B).
8. B Point (a, b) is in the second quadrant where points have signs of (–, +); thus, a is negative and b is positive. Point (c, d) is in the third quadrant where points have signs of (–, –); thus, c is negative and d
is negative. So, Quantity A is a positive times a negative, which is negative. Quantity B is a negative times a negative, which is positive. Quantity B must be greater.
9. C Plug values into the equation and eliminate graphs that do not include those values. If x = 1, then y = –1; eliminate choices (B) and (D). If x = –1, then y = –1, eliminate choices (A) and (E). Only choice (C) remains.
10. D Plug in points in the appropriate quadrants. If (a, b) = (1, 2) and (c, d) = (–3, –4), then the point in question is (−bd, bc) = [−(2)(–4), (2)(–3)] = (8, –6), which is in quadrant IV; the answer is choice (D).
11. A Use Process of Elimination to solve this one. First, only coordinate pairs with a positive x-value and a negative y-value will fall in the proper quadrant, so eliminate choices (B), (C), and (E). The line that divides the correct quadrant into shaded and unshaded regions has a slope of –1
because it goes through the origin and the point (–10, 10). On this line, the absolute value of the x-coordinate equals the absolute value of the y-coordinate. In the shaded region, then, |x| > |y|, so choice (D) can be eliminated. Only choice (A) remains. Alternatively, realize that this figure is drawn accurately, because of the placement of (10, –10), and plot all 5 points, eliminating all of those that fall outside the
shaded region.
12. B Point X is at approximately (2, –1.5). So Quantity A is about –2 and Quantity B is about –1.5, thus, Quantity B is greater.
13. A Although you have enough information to find the exact values of AO and BO, it’s not necessary to do so to compare the quantities. The slope of the line is 
, which means that the vertical distance, or rise, is
greater than the horizontal distance, or run, by a ratio of 13 to 12 (you’re dealing with distances on a coordinate plane, so disregard the negative sign). Because AO and BO are equal to, respectively, the rise and the run of the same segment of the line, Quantity A is greater.
14. B Note that the line contains three points: (0, 0), (4, 5), and (x, 1.5). The slope between any two of these points is the same. Remember that slope is change in y over change in x. Thus, 
, or 
. Cross-multiply to find 5x = 6.0. Divide by 5 to find x = 1.2. The answer is choice (B).
15. B Just because you don’t know the values of –m and n doesn’t mean you can’t determine which is greater. Using point A and the origin, you can find the slope of segment AB: 
.
Now plug in coordinates for point B that will give you the same slope; the easiest way to pick them would be to simply rise –2 and run 3, bringing you to the point (m, n) = (3, –2). So –m = –3, and n = –2, and Quantity B is greater.
16. B You don’t need to use the distance formula, because the distance between the points (3–
,0) and (5+,0) can be measured horizontally.
Distance is positive, so subtract the smaller x-coordinate from the larger: (5+)–(3–)=2+2; the answer is choice (B).
1. A Plug in a few points that lie below line c, such as (0, –1), (–3, –4), (1, 0). In each case, the x-coordinate is greater than the y-coordinate, so Quantity A is greater. Alternatively, realize that the 45 degree angle and the fact that the line passes
through the origin tells us that the equation of line c is y = x. So the region below the line is the graph of y < x. The coordinates of all points in that region must satisfy the inequality.
2. C Break this one into bite-sized pieces: You need x in order to find the length of AB, so find x first. If you insert the given values into the slope formula, 
, you get 
, so x = 5. Now you need to find the length of AB. Rather than using the distance formula, turn AB into the hypotenuse of a right triangle and find the lengths of the other sides. To make the triangle, add a new vertex at coordinate (5, 2): The length of the horizontal leg is 5 – 2 = 3, and that of the vertical leg is 6 – 2 = 4, yielding the familiar 3-4-5 triangle. The length
of segment AB = 5, so the two quantities are equal.
3. E Draw the points and connect the points to form a right triangle. Subtract the x-coordinate of A from that of C to find the length of AC: 7 − 1 = 6. Subtract the y-coordinate of C from that of B to find the length of BC: 4+6
-4=6. Notice that the ratio of AC to BC is 1 to . Therefore, ABC is a 30-60-90 triangle, and the length of AB, the hypotenuse, will be double the length of the shorter side (6), so AB = 12. The absolute value of the difference will be the positive value,
obtained by subtracting the smaller value (BC) from the larger value (the length of the hypotenuse, AB): AB – BC=12-6, which is choice (E).
4. D Plug in points. Points (3, 4) and (5, 4) both lie inside the circle, so different values give different outcomes.
5. A The slope of MO is 1, so it makes a 45 degree angle with the positive x-axis. Similarly, the slope of NO is –1, so it makes another 45 degree angle with the positive x-axis. The sum of the degree measures of these angles is 90, so
MNO is a right triangle. Therefore, MO and NO are the base and height of triangle MNO. To find the area of the triangle, you need to find the length of MO and NO. Drop a perpendicular from point M to the y-axis, to form an isosceles right triangle whose hypotenuse is MO. Each leg of this triangle has length 1 so, MO=. Similarly, dropping a perpendicular line from the y-axis to point N creates another isosceles right triangle, whose legs have length 3, and whose hypotenuse is NO. Therefore, NO=3. So the area of triangle MNO is bh=()(3)=3, and the answer is choice (A).
6. B To find AB, find the radius of the circle and then subtract OA. If the radius is r, then AB = r − (x − 5) = r − x + 5. Similarly, DC = r − (x + 5) = r − x
− 5. So AB − DC = [r − x + 5] − [r − x − 5] = 10, so the answer is choice (B). If you use POE, you can eliminate choice (A) because you know the answer has to be positive. If you selected choice (E), you selected the radius.
7. A Drawing a horizontal line from point R to the positive y-axis forms a right triangle. The length of the leg that sits on the y-axis is , and the horizontal leg you just drew has length 1. The ratio of the legs is 1 to
, so you have a 30-60-90 right triangle. Therefore, the hypotenuse (OR) has length 2, and the angle between OR and the positive y-axis is 30 degrees. The first quadrant includes 90 degrees total, so rotating OR 120 degrees clockwise puts OS on the positive x-axis, with a length of 2. Therefore, the x-coordinate of OS is 2, which is
slightly larger than , which is approximately 1.7. The answer is choice (A).
8. C, D, and E
First, find the slope of the line, which runs through (–3, 2) and (0, 0). The slope is the change in y over the change in x, or 
. Therefore, the
equation of line AB is y = x. To find the range of possible values for d, plug in the given range of possible values for c to the equation. If c = 2, d = 
, and if c = 10, d = 
. Any value between 
and 
, or –1.33 and –6.67, will work for d. Therefore, the only right answers are choices (C), (D), and (E).
9. C
and F
First, find the equation of the line that defines the shaded region, expressed as y = mx + b, Using the origin and the one given point in the diagram, the slope m equals 
, and the y-intercept b is 0. So y = x
is the boundary line. Looking at the figure, if either x or y are positive, then the point isn’t in the shaded region; you can eliminate choices (A), (B), and (E). For the other choices, plug in the x-values. If the resulting y-value is less than the y-value in the choice, that point lies below the line and outside the shaded region. So you’re looking for points for which y ≥ x. when x ≤ 0. This is true for correct choices (C) and (F).
10. 9 The circle touches (0, 3) and (3, 0), so its center must be at (3, 3) and its radius must be 3. Since the base of triangle passes through the center of the circle, the base of the triangle must be the diameter of length 6. Given the triangle is isosceles and inscribed within the circle, a line from the circle’s
center to the triangle’s corner equals the triangle’s height. This height must be the radius of the circle. The area of a triangle is 0.5 × base × height: 0.5 × 6 × 3 = 9.
11. C, D and E
Given that the angle ranges from 45° to 60°, you need to plug in values for angle a and find a special triangle to solve for y. If a is 45°, the triangle’s sides are x, x, x. It doesn’t matter what the hypotenuse is; x = 4, which means y also is 4. If a is 60°, the triangle’s sides are x, x, 2x. The shortest side of the triangle would be the one on the x-axis. Since x = 4, then y = 4 = 6.92. So the correct answers range from 4 to 6.92. Choices (C), (D), and (E) are all correct.
12. 2 You need to find the length of a leg of a triangle. By finding the lengths of the two other sides, you can use the Pythagorean theorem to find the third side. The hypotenuse of the triangle is the radius of the (quarter-) circle, which, since point A is at (0, 8), is 8. Since point
C is (6, 0), the other leg is 6. From the Pythagorean theorem, 62 + x2 = 82, so x2 = 28, 
, so 
=2.
13. D Use line L to make a triangle, with points at (x, 0), the 45° angle, and the origin. The angle at (x, 0) must be 45° since the sum of the angles of a triangle is 180°. Since line L is tangential to the circle and forms 45° angles
with each axis, a line from the point where line L and circle intersect to the origin will form a right angle with line L. The smaller triangle formed—from (x, 0), to where line L and circle meet, to the origin—will be a 45-45-90 triangle, with two sides equal to the radius of the circle. Find the radius of the circle, and you can find x. The area of the shaded region is equal to the circumference of a circle with radius
between 1 and 2. Circumference = 2πr, so the area of the shaded region is between 2π and 4.5π, which means the area of the circle in the figure is between 8π and 18π. Area = πr2, so the radius of the circle is between 
and 
, which is to say between 2 and 3. 45-45-90 triangles have sides of a-a-a, where, in this case, a is between 2 and 3. So the hypotenuse of the triangle, from the origin to (x, 0), is between 2 and 3, and therefore
between 4 and 6, so x can range from –4 to –6.
14. 4 From the information in the problem you know that 8π = 2πr and r = 4. The area of the circle is equal to πr2, so the area of the circle is 16π. If you draw one radius from the center O of the circle to point
P, and another radius from O to point R, you know that the central angle formed will be 90 degrees because arc PQR equals of the total circle. The triangle formed by inserting these two radii will be an isosceles right triangle with sides of 4, 4, and 4 and an area of 8. The total area taken up by
the triangle and the shaded area together will be equal to 4π, or of the circle. The total area – the unshaded area = the shaded area you want, so the shaded area must equal 4π-8. The value of π can be rounded to approximately 3, so now you have 12 – 8 = 4.
15. A, D, F, and G
Plug in for a. If a = 40, then b = 140, g = 40, and h = 140. Now plug in for c. If c = 100, then d = 80, i = 100, and j = 80. You know that g, j, and
f must add up to 180, so 40 + 80 + f = 180, which means that f = 60, which in turn means that e = 60. Now check your answer choices to see which ones are true. In this case e = f, so keep choice (A). However, i≠d, so you can eliminate choice (B), and choice (C) is also not true and should be eliminated. 100 + 80 = 180, so you cannot eliminate choice (D). 100 + 140 ≠ 180, so choice (E) is
incorrect. 40 + 80 + 60 = 180, so keep choice (F) for now. You have three answers and since this is a must be true question, you should try another set of numbers and test the three choices with the new values to see if the choices are still correct. Choices (A), (D), (F), and (G) are the correct answers.