1 / 27

Scholastic Aptitude Test (SAT)

Scholastic Aptitude Test (SAT). PREP. Practice problems Version 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

quamar-sexton
Télécharger la présentation

Scholastic Aptitude Test (SAT)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Scholastic Aptitude Test (SAT) PREP

  2. Practice problemsVersion 3

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

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

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

  6. 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

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

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

  9. SAT x2 + x2 + x2 = x2 3 x2 = 3 x2

  10. SAT x2 + x2 + x2 = x2 3 x2 = 3 x2

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

  12. 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

  13. 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

  14. 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

  15. SAT 36 2 x 18 2 x 3 x 6 2 x 3 x 2 x 3

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

  17. 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

  18. 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.

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

  20. 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.

  21. 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

  22. 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

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

  24. 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

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

  26. 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

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

More Related