GRE-Math-Questions-to-Know-by-Test-Day-2nd-Edition_pag174

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Answers ‹ 165
74. (B) Make a sketch.
A
B
C
Exterior
angle
Exterior
angle
Exterior
angle
55°
65°
 The exterior angle at B measures ° − ° = °180 55 125 . The exterior angle at C measures 
° − ° = °180 65 115 . The exterior angle at A measures °+ ° = °55 65 120 (the measure of an 
exterior angle of a triangle equals the sum of the measures of the remote interior angles). 
Therefore, for all possible exterior angles, Quantity B is greater.
75. (C) The greatest possible distance between any two points of the box 
is the length of a diagonal across the center of the box, which is given by 
+ + = + + = =4 3 12 16 9 144 169 132 2 2 (Quantity A). Quantity B is 13. The two 
quantities are equal. Select (C).
76. (A) Quantity A, the area of the shaded region, is the area of the larger circle minus the 
area of the smaller circle. Thus, Quantity A equals π π π π− = − ≈R r (6 ) (4 ) 62.82 2 2 2 , where 
R = 1
2
(12) = 6 is the radius of the larger circle and r = 1
2
(8) = 4 is the radius of the smaller 
circle. Quantity A is greater.
77. (B) JKLM is a convex quadrilateral, so the sum of its interior angles 
is − °n( 2)180 = − °(4 2)180 = °360 . Thus, Quantity A is the measure of 
J K Mthe measure of the measure of∠ + ∠ + ∠ = ° − °360 30 = °330 . Quantity B is greater.
78. (D) Let x = the length of one of the two congruent sides of triangle ABC and y = the 
length of the noncongruent side. Then the perimeter, P, of the triangle is P = +x y2 . If AB = 
x and BC = y, then P = ⋅ +2 15 20 = 50. On the other hand, if AB = y and BC = x, then P = 
⋅ +2 20 15 = 55. Hence, you cannot determine a definite value for P. Thus, the relationship 
cannot be determined from the information given. Select (D).
06_McCune_Answer.indd 165 2/21/22 4:45 PM