remainder, or $2. The ratio of Jennifer’s contribution to Kelly’s contribution to Meredith’s
contribution is then 16:6:2, which simplifies to 8:3:1, or (D).
36 . F The general equation of a circle is (x − h)^2 + (y − k)^2 = r^2 , for which (h,k) are the coordinates
for the center of the circle, and r gives the radius length. For the given equation, the h = 0 and
k = −4; (J) and (K) confuse the sign in front of k. Since r^2 = 28, the radius of the circle is
. Choice (G) divides by 2 rather than square rooting 28. Choice (H) gives the value of
r^2 , not r.
37 . B The perimeter of the figure consists of one side of the equilateral triangle and the arc length
of two semicircles. You can immediately eliminate (A), (D), and (E) because the length of
one side of the equilateral triangle is 6 inches. Both semicircles have a diameter of 6 and
given C = dπ = 6π, each semicircle has an arc length of 3π. With the exposed side of the
triangle, the perimeter of the figure should be P = 6 + 3π + 3π = 6 + 6π. If you selected (C),
you may have found the circumferences of two full circles rather than two semicircles.
38 . H Because DEFG is a rhombus, which has equal sides and equal angles, all 8 triangles formed
by drawing the diagonals in the figure are equivalent. HEFM encloses the area of 3 triangles
and the shaded region the area of 5 triangles, thus the ratio is 3:5. Choice (J) is the ratio of
the unshaded area to the total area.
39 . C The midpoint is the average of the endpoints, so the y-coordinate of the midpoint is
= 4. Choice (A) is the x-coordinate of the midpoint. Choices (B) and
(E) only find the sum of the endpoint coordinates. Choice (D) confuses the negative sign and
finds the average of 10 and 2.
40 . F The volume of a cube is s^3 , so this cube is 9^3 = 729 cubic feet. Choice (G) gives the surface
area of the cube. Choice (J) the area of 3 faces of the cube, (H) gives the area of 1 face, and
(J) confuses the side as 3 feet rather than 9.
41 . D Rearrange both equations into y = mx + b form to compare their slopes and y-intercepts. The
first equation, rx + sy = t, becomes y = x + . The second equation, rx + sy = v, becomes
y = x + . Both functions have the slope of , so they cannot intersect at only 1 point,
eliminating (I), and thus (A), (C), and (E). A system of two linear functions can give a single
line graph if the equations have the same slope and y-intercept, which occurs if t = v. Lines
