3. 1.If p = – 2 and q = 3, then p 3 q 2 + p 2 q = ? (A)– 84 (B)– 60 (C)36 (D)60 (E)84 p 3 q 2 + p 2 q = ? (– 2) 3 (3) 2 + (– 2) 2 (3) = ? (– 8) (9) + (4) (3) = ? (– 72) + (12) = – 60 p 3 q 2 + p 2 q = ? (– 2) 3 (3) 2 + (– 2) 2 (3) = ? (– 8) (9) + (4) (3) = ? (– 72) + (12) = – 60SAT 105⁰ x⁰x⁰ N Q M P

4. 2. D In this figure, B is the midpoint of AC and D is the midpoint of CE. If AB = 5 and BD = 8, what is the length of DE ? (A)8 (B)6 (C)5 (D)4 (E)3 BC = 5 by definition of midpoint, then 8 – 5 = 3 Thus DE = 3 by definition of midpoint BC = 5 by definition of midpoint, then 8 – 5 = 3 Thus DE = 3 by definition of midpointSAT CDA B E AB = 5 BD = 8 DE = ?

5. 3. Which of the following equations describes the relationship of each pair of numbers (N,P) in this table ? (A)P = N + 5 (B)P = 2 N + 3 (C)P = 2 N + 5 (D)P = 3 N + 1 (E)P = 3 N +1 P = 3 N + 1 works for all entries 7 = 3 (2) + 1 13= 3 (4) + 1 19 = 3 (6) + 1 25 = 3 (8) + 1 P = 3 N + 1 works for all entries 7 = 3 (2) + 1 13= 3 (4) + 1 19 = 3 (6) + 1 25 = 3 (8) + 1SAT NP 27 413 619 825

6. 4.In this figure MQ is a straight line. If PM = PN, what is the value of x ? (A)30 (B)45 (C)60 (D)75 (E)90 < MNP = 180⁰ – 105⁰ = 75⁰ Since < PMN = < MNP, < PMN = 75⁰ Thus, x = 180⁰ – 150⁰ = 30⁰ < MNP = 180⁰ – 105⁰ = 75⁰ Since < PMN = < MNP, < PMN = 75⁰ Thus, x = 180⁰ – 150⁰ = 30⁰SAT 105⁰ x⁰x⁰ N Q M P

7. 5. Marty has exactly five blue pens, six black pens, and four red pens in his knapsack. If he pulls out one pen at random from his knapsack, what is the probability that the pen is either red or black ? (A) 11 15 (B) 2323 (C) 1212 (D) 1313 (E) 1515 What is the probability of Red or Black ? Blue 5 Black 6 Red 4 Total 15 10 out of 15 or 2/3 What is the probability of Red or Black ? Blue 5 Black 6 Red 4 Total 15 10 out of 15 or 2/3SAT

8. 6. Two hot dogs and a soda cost $ 3.25 If three hot dogs and a soda cost $ 4.50, what is the cost of two sodas ? (A) $ 0.75 (B) $ 1.25 (C) $ 1.50 (D) $ 2.50 (E) $ 3.00 Since the difference of the 2 costs is 1 hot dog, then the cost for 1 hot dog is $1.25. Thus, 2 hot dogs cost $ 2.50 which leaves $.75 for the soda. And, 2 sodas would be $ 1.50 Since the difference of the 2 costs is 1 hot dog, then the cost for 1 hot dog is $1.25. Thus, 2 hot dogs cost $ 2.50 which leaves $.75 for the soda. And, 2 sodas would be $ 1.50SAT

9. 7. In this figure, if L1 ⁄ ⁄ L2, which of the following must be = to a ? (A)b + c (B)b + e (C)c + d (D)d + e (E)d + f Because f = c + d by exterior < of a ∆ must = the 2 remote interior <‘s and a = f by corresponding <‘s congruent Thus a = c + d by transitive property Because f = c + d by exterior < of a ∆ must = the 2 remote interior <‘s and a = f by corresponding <‘s congruent Thus a = c + d by transitive propertySAT a⁰a⁰ d⁰d⁰ f⁰c⁰c⁰e⁰e⁰ b⁰b⁰ L1 L2

10. 8. A certain phone call costs 75 cents for the first three minutes plus 15 cents for each additional minute. If the call lasted x minutes and x is an integer greater than 3, which of the following expresses the cost of the call, in dollars ? (A)0.75 [3] + 0.15 x (B)0.75 [3] + 0.15 [x + 3 ] (C)0.75 [3] + 0.15 [3 – x ] (D)0.75 + 0.15 [x – 3 ] (E)0.75 + 0.15 x.75 for first 3 minutes +.15 each additional minute Since the call lasted x minutes and x is an integer > 3, then x – 3 represents minutes over 3, such as 4 – 3 = 1 minute over the baseline cost of.75.75 for first 3 minutes +.15 each additional minute Since the call lasted x minutes and x is an integer > 3, then x – 3 represents minutes over 3, such as 4 – 3 = 1 minute over the baseline cost of.75SAT

11. 9. This figure shows a piece of wire in the shape of a semicircle. If the piece of wire is bent to form a circle without any of the wire overlapping, what is the area of the circle ? (A)6 π (B)9 π (C)12 π (D)18 π (E)36 π C = π d C = 12 π Since figure is a half circle = 6 π, so if d = 6, then r = 3 Thus, A = π r 2 or π (3) 2 or 9 π C = π d C = 12 π Since figure is a half circle = 6 π, so if d = 6, then r = 3 Thus, A = π r 2 or π (3) 2 or 9 πSAT 12

12. 10. If a 2 – a = 72, and b and n are integers such that b n = a, which of the following cannot be a value for b ? (A) – 8 (B) – 2 (C) 2 (D) 3 (E) 9 a 2 – a = 72 a 2 – a – 72 = 0 (a – 9 ) (a + 8 ) = 0 a = – 8, 9 a 2 – a = 72 a 2 – a – 72 = 0 (a – 9 ) (a + 8 ) = 0 a = – 8, 9SAT b n = a b n = – 8 or b n = 9 b = n √ – 8 or b = n √ 9 b n = a b n = – 8 or b n = 9 b = n √ – 8 or b = n √ 9 – 8 works as – 8 = 1 √ – 8 – 2 works as – 2 = 3 √ – 8 2 doesn’t work as 2 ≠ 3 √ – 8 or 2 ≠ 3 √ 9 – 8 works as – 8 = 1 √ – 8 – 2 works as – 2 = 3 √ – 8 2 doesn’t work as 2 ≠ 3 √ – 8 or 2 ≠ 3 √ 9