Scholastic Aptitude Test (SAT)

Scholastic Aptitude Test (SAT) PREP

Practice problemsVersion 3

SAT ( 1 + 1 ) ÷1 = 5 3 2 3 + 5÷1 = 15 2 8•2 = 16 15 1 15

SAT x 2 – 2 x when x = – 2 ? (– 2 ) 2 – 2 (– 2 ) = 4 + 4 = 8

SAT x = 96 60 160 x = 96 • 60 160 x = 6 • 60 10 x = 6 • 6 = 36

SAT 1 > 7 > 1 3 x 4 ( 1 > 7 > 1 ) 12 x 3 x 4 4 x > 84 > 3 x _______________________________________________________________________________________ 4 (22) > 84 > 3 (22) --- 88 > 84 > 66 4 (23) > 84 > 3 (23) --- 92 > 84 > 69 4 (24) > 84 > 3 (24) --- 96 > 84 > 72 4 (25) > 84 > 3 (25) --- 100 > 84 > 75 4 (26) > 84 > 3 (26) --- 104 > 84 > 78 4 (27) > 84 > 3 (27) --- 108 > 84 > 81

SAT 2 x + 5 5 x – 6 – 4 x + 2 3 x + 1 3 x + 1 = x + 1 3 3

SAT A 70⁰ 45⁰ C B < B + < C + < A = 180⁰ < B + 45⁰ + 70⁰ = 180⁰ < B + 115⁰ = 180⁰ < B = 180⁰ – 115⁰ = 65⁰

SAT x2 + x2 + x2 = x2 3 x2 = 3 x2

SAT x2 = 5 x – 4 x2 – 5 x + 4= 0 ( x – 4 ) ( x – 1 ) = 0 x = 4 or 1

SAT 1 + 3 = 4 Ex where odd integers doesn’t work 3 + 5 = 8 Ex where prime doesn’t work 2 + 4 = 6 Even Integers does work

SAT 89 x 4 = 356 90 x 5 = 450 So, the difference between the 2 gives us the total needed to average 90 450 – 356 = 94

SAT Sale price is 25 % off or .75 s Post sale price is increased by 20 % The increase = .75 s x .2 = .15 s Then we add .75 s + .15 s = 90 s

SAT 36 2 x 18 2 x 3 x 6 2 x 3 x 2 x 3

SAT AT = 1hb = 36 2 1 h (9) = 36 2 h (9) = 36 2 h = 36 • 2 9 h = 8

SAT If a  = a – a 4 6  x  = x – x 4 6 If x  = 3  3 = x – x 4 6 12 (3 = x – x ) 4 6 = 3 x – 2 x 36 = x

SAT This is a “TRAP” question: The problem only asks for the total # of coins, NOT the total value of coins, which is what you are lead to think.

SAT • • The slope is 2 and the y-intercept is 2, so the answer is y = 2 x + 2

SAT Prime factors of 6 are 2 x 3, so any # divisible should have at least one2 and one3 as factors Prime factors of 9 are 3 x 3, so any # divisible should have at least two3’s as factors  Any # divisible by both should have at least one 2 and two3’s and the first possibility is 2 x 32or 18 12 and 36 require two2’s in its prime factorization  12 doesn’t have a 2nd prime factor of 3 36 does work in some cases by NOT all, such as the most obvious one, 6 x 9 or 54.

SAT O Q P ? 50⁰ R < POR =180⁰ – 50 ⁰ = 130 ⁰ PO = OR (all radii are =) < RPQ = <PRO (base <‘s of Isosceles are =) and < QPR + <PRO = 50 ⁰  < RPQ =25

SAT 2 64 2 4 E3 = 4 x 4 x 4 = 64 4 2 4 4 4 E3 = 4 x 4 x 4 = 64 Two rectangle solids 4 x 4 x 2, when glued 2 x 4 x 8 Then Top and Bottom = 8 x 4 = 32 or 64 total Then Front and Back = 8 x 2 = 16 or 32 total Then two sides = 4 x 2 = 8 or 16 total Grand total of 6 sides = 64 + 32 + 16 = 112

SAT 6 blue 12 black 15 white 32 Total Black 12 = 3 Total 32 8

SAT This is a TRAP question ! We are NOT given enough information to use a distance formula. So we need to improvise by selecting a distance that is easy to work with. Since we have speeds of 40 and 60 mph, why not use a multiple of both, such as, 120 Distance = rate x time (going trip) 120 = 40 t where t = 3 hours (return trip) 120 = 60 t where t = 2 hours Now 240 = 5 r, where r = 48 miles per hour

SAT c2 = a2 + b2 c2 – a2 = b2 √ c2 – a2= b AT = ½ h b A = 1 a ( √ c2 – a2 ) 2 c2 a2 b2

SAT R Q T S P AT = ½ h b Using altitude QS and side PR= ½ • 9 • 8 = 36 Using altitude RT with side PS = ½ • 7 • PS = 36 7 PS = 36 then, PS = 36 • 2 = 10 2 2 7 7

SAT • River City • 1 2 3 [ 3 ] • x • Bayville • 1 2 3 4 [ 4 ] • x • Eagles Mere • 1 2 3 [ 3 ] • = • Twin Peaks [ 36 ]

Scholastic Aptitude Test (SAT)