triangle on top, A = = 5 for a total area of 35 + 5 = 40.
48 . J An increased number of products will mean increased expenditures, so the slope of the line
will be positive, and (H) and (K) can be eliminated. Choice (F) can be eliminated because
you don’t have any information about what the expenditure per product is, so you can’t assign
a specific value to it. A line passing between two points can’t have more than one slope, so
(J) is correct.
49 . B A geometric sequence has a constant ratio between terms. To figure out what the ratio in this
sequence is, divide the second term by the first term: 15 ÷ 10 = 1.5. To find the fifth term,
multiply the fourth term by the ratio of 1.5: 33.75 × 1.5 = 50.625.
50 . H The surface area of a cube is SA = 6(s^2 ). Plug in the given surface area, and solve for the
length of a side, s, 54 = 6(s^2 ). Divide both sides by 6, 9 = s^2 , then take the square root of both
sides, s = 3. The volume of a cube is V = s^3 , so V = 3^3 = 27. Choice (F) gives the area of one
face of the cube. Choice (G) multiples the area of one face of the cube by 2 instead of by 3.
Choice (J) squares the area of one face of the cube. Choice (K) uses 9, the area of one face,
instead of 3 for s. Note: If you can’t remember the formula for surface area, draw a figure.
51 . D The area of the shaded region has the same proportion to the area of the entire circle that the
central angle of the shaded region has to 360°. Begin by finding the area of the entire circle:
A = πr^2 = π(5)^2 = 25π. Then set up the proportion, and solve for x: . Cancel the
π on the left side, and then multiply both sides by 360°: = x, or x = 126°. Choice
(B) gives the central angle of the unshaded region.
52 . J The circle formula is (x − h)^2 + (y − k)^2 = r^2 , where (h,k) is the center of the circle, and r is
the radius. Use Process of Elimination. Choices (F) and (G) can be eliminated because they
do not square the radius. Choice (K) can also be eliminated because the x-coordinate of the
center is +2, so the first part of the equation should be (x − 2)^2 , not (x + 2)^2 . Choice (H) can
be eliminated because the y-coordinate is −6, so the second part should be (y + 6)^2 not (y −
6)^2 . Only (J) remains.
53 . E YZ is the side opposite X, and XZ is opposite Y. Set up a proportion according to the law
of sines given in the note, , then solve for XZ. Cross-multiply first, (XZ)
