Scholastic Aptitude Test (SAT)

Scholastic Aptitude Test (SAT) PREP

Practice problemsVersion 2

SAT P x⁰ 105⁰ M N Q p3 q2 + p2 q = ? (– 2)3(3)2 + (– 2)2 (3) = ? (– 8) (9) + (4)(3) = ? (– 72) + (12) = – 60

SAT AB = 5 DE = ? A C D E B • • • • • BD = 8 BC = 5 by definition of midpoint, then 8 – 5 = 3 Thus DE = 3 by definition of midpoint

SAT P = 3 N + 1 works for all entries 7 = 3 (2) + 1 13= 3 (4) + 1 19 = 3 (6) + 1 25 = 3 (8) + 1

SAT P x⁰ 105⁰ M N Q < MNP = 180⁰ – 105⁰ = 75⁰ Since < PMN = < MNP , < PMN = 75⁰ Thus, x = 180⁰ – 150⁰ = 30⁰

SAT What is the probability of Red or Black ? Blue 5 Black 6 Red 4 Total 15 10 out of 15 or 2/3

SAT 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

SAT d⁰ c⁰ e⁰ f⁰ L1 b⁰ a⁰ L2 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

SAT .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

SAT 12 C = π d C = 12 π Since figure is a half circle = 6 π, so if d = 6, then r = 3 Thus, A = π r2 or π (3) 2 or 9π

SAT a2 – a = 72 a2 – a – 72 = 0 (a – 9 ) (a + 8 ) = 0 a = – 8, 9 bn = a bn = – 8 or bn = 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

Scholastic Aptitude Test (SAT)